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M. Bradley • J. Campbell • S. McPetrie
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Contents
Term 1
TOPIC 1:
Unit 1
Unit 2
Unit 3
Unit 4
Unit 5
Unit 6
Unit 7
Term 2
Patterns, sequences and series
Arithmetic sequences
Geometric sequences
The sum of arithmetic series
The sum of geometric series
Sigma notation
Practical applications
Quadratic patterns and combinations of
arithmetic and geometric sequences
4
8
12
15
18
21
25
Revision Test
Formal Assessment: Assignment
TOPIC 2:
Unit 1
Unit 2
29
31
Functions and inverse functions
Functions
Inverse functions
34
37
Revision Test
TOPIC 3:
Unit 1
Unit 2
43
Exponential and logarithmic
functions
Unit 1
Unit 2
Unit 3
Unit 4
Unit 5
Revision of exponential laws and functions 48
Logarithms and logarithmic functions
51
58
Finance, growth and decay
Unit 1
Unit 2
Unit 3
Unit 4
Trigonometry: problem solving
in two and three dimensions
Problems in two dimensions
Problems in three dimensions
60
62
71
79
82
88
Trigonometry: compound and
double angle identities
120
125
Revision Test
TOPIC 7:
Unit 1
Unit 2
130
Polynomials
Factorise third degree polynomials
Factorise and solve cubic polynomials
using the remainder or factor theorems
Revision Test
TOPIC 8:
Unit 1
Unit 2
Unit 4
Unit 5
Unit 6
Unit 7
Unit 8
TOPIC 9:
Unit 1
Unit 2
132
135
140
Differential calculus
Limits
Use limits to define the derivative of a
function f
Differentiation of function from
first principles
Use the specific rules for differentiation
Find the equations of tangents
to functions
The second derivative
Sketch cubic graphs
Optimisation and rate of change
Revision Test
Revision: Grade 11 Finance, growth and
decay
Derivation and use of formulae
for annuities
Annuity application and problem solving
Calculate time periods using logarithms
Analyse investments and loan options
Revision Test
TOPIC 5:
Unit 1
Unit 2
Unit 3
Revision Test
TOPIC 4:
TOPIC 6:
142
146
150
154
158
160
167
176
184
Analytical geometry
Equations of a circle
Equation of a tangent to a circle
Revision Test
Mid-year Exam practice: Paper 1
Mid-year Exam practice: Paper 2
Term 2 summary
188
193
198
202
205
209
Revision: Grade 11 Trigonometry
90
Derive the compound and double angle
identities
92
Prove identities using compound and
double angle identities
101
Solve equations and determine
the general solution
103
Revision Test
Formal Assessment:
Option 1: Investigation: Area of polygons
Option 2: Project: Compound interest
Term 1 summary
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111
114
115
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Term 4
Term 3
TOPIC 10: Euclidean geometry
Unit 1
Unit 2
Unit 3
Unit 4
Unit 5
Unit 6
Revision: Grade 11 Geometry
Similar polygons
Proportionality theorem
Equiangular triangles and similarity
Triangles with proportional sides and
similarity
Pythagoras’ Theorem and similarity
Revision Test
Revision Test
296
Paper 1
Paper 2
231
235
Paper 1
Paper 2
Term summary Examination tips
307
311
316
Answers
319
Glossary
399
Index
401
237
Revision of skewed and symmetric data
Bivariate data: scatter plots, regression
lines and correlation
Structure of final examination
Exam practice A
214
218
221
227
TOPIC 11: Statistics
Unit 1
Unit 2
Unit 1
Unit 2
240
243
253
Unit 3
299
303
Exam practice B
TOPIC 12: Counting principles and
probability
Unit 1
Unit 2
Unit 3
Unit 4
Revision of rules for independent,
mutually exclusive and complementary
events
Use Venn diagrams, tree diagrams and
contingency tables to solve problems
The Fundamental Counting Principle
Applications of the counting principle
to solve probability problems
Revision Test
Preliminary Exam practice: Paper 1
Preliminary Exam practice: Paper 2
Term 3 summary
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256
261
267
275
280
283
286
290
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Term
1
2
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TOPIC 1
Patterns, sequences and series
Unit 1
Unit 2
Unit 3
Unit 4
Unit 5
Unit 6
Unit 7
Arithmetic sequences
4
Geometric sequences
8
The sum of arithmetic series
12
The sum of geometric series
15
Sigma notation
18
Practical applications
21
Quadratic patterns and combinations of arithmetic
and geometric sequences
25
Revision Test
29
FORMAL ASSESSMENT: ASSIGNMENT
31
TOPIC 2
Functions and inverse functions
Unit 1
Functions
Unit 2
Inverse functions
Revision Test
TOPIC 3
Exponential and logarithmic functions
Unit 1
Revision of exponential laws and functions
Unit 2
Logarithms and logarithmic functions
Revision Test
TOPIC 4
Unit 1
Unit 2
Unit 3
Unit 4
Unit 5
Revision Test
Topic 5
Unit 1
Unit 2
Unit 3
Unit 4
34
37
43
48
51
58
Finance, growth and decay
Revision: Grade 11 Finance, growth
and decay
Derive and use formulae for annuities
Annuity application and problem solving
Calculate time periods using logarithms
Analyse investments and loan options
60
62
71
79
82
88
Trigonometry: Compound and
double angle identities
Revision: Grade 11 Trigonometry
Derive the compound and
double angle identities
Prove identities using compound and
double angle identities
Solve equations and determine the
general solution
Revision Test
Formal assessment: Option 1: Investigation: Area of polygons
Formal assessment: Option 2: Project: Compound interest
Term 1 summary
90
92
101
103
109
111
114
115
3
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TOPIC
2
1
Patterns, sequences and series
Unit 1: Arithmetic sequences
In general we define an arithmetic sequence as follows:
a;
KEY WORDS
linear pattern – an
arithmetic sequence with a
common difference between
consecutive terms
arithmetic sequence – a
sequence of numbers with a
common difference between
consecutive terms
common difference – the
constant amount by which
consecutive terms increase (or
decrease)
a + d;
a + 2d;
a + 3d;
a + 4d;
a + 5d; … a + ( n − 1 )d
• a is the value of the first term
• d is the common difference between the terms, d = T2 − T1 = T3 − T2 = Tn − Tn−1
• Tn is the value of the term in position n, so Tn = a + ( n − 1 )d
• n is the position of a term and can only be a positive whole number, also known as
a natural number.
Consider the arithmetic sequence 3; 7; 11; 15; … 99
T1 = 3, T2 = 7, T3 = 11, T4 = 15
d1 = T2 – T1 = 7 – 3 = 4, d2 = T3 – T2 = 11 – 7 = 4 and d3 = T4 – T3 = 15 – 11 = 4.
Since d1 = d2 = d3, we have a common difference of 4.
The first term is given by a = 3 and the common difference is given by d = 4.
We determine the formula for the nth term in the sequence by substituting a = 3 and
d = 4 into Tn = a + (n – 1) d.
This gives us Tn = 3 + (n – 1)(4) = 3 + 4n – 4 = 4n – 1
Check the formula by substituting n = 1 to obtain the value of T1, n = 2 to obtain the
value of T2 and so on.
If n =1, then T1 = 4(1) – 1 = 3
If n = 2, then T2 = 4(2) – 1 = 7
If n = 3, then T3 = 4(3) – 1 = 11
The nth term formula, Tn = 4n – 1, can be used to determine the position of any term
in the sequence if the value of the term is given. To determine which term has a value
of 99, substitute Tn = 99 into Tn = 4n – 1.
99 = 4n – 1 ⇒ 4n = 100 and n = 25, so T25 = 99, which means that the twenty-fifth
term has a value of 99.
WORKED EXAMPLE 1
Consider the arithmetic sequence 2; 6; 10; 14; …
1
What is the common difference?
2
State the values of the next two terms in the sequence.
3
Determine a formula for the nth term of the sequence.
4
Determine the value of the twenty fifth term.
5
Which term has a value of 46?
6
Briefly explain why the value of n in the formula Tn = a + (n – 1)d can only
be a whole number.
7
Is 72 a term in the sequence? Justify your answer.
4
Topic 1 Patterns, sequences and series
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SOLUTIONS
1
2
3
4
5
6
7
d = T2 − T1 = 6 – 2 = 4
The common difference is 4, so the next two terms are 18 and 22.
Substitute a = 2 and d = 4 into the formula Tn = a + (n − 1)d
Tn = 2 + (n − 1)(4) = 4n − 2
Substitute n = 25 into Tn = 4n – 2 to find the value of T25
T25 = 4(25) − 2 = 98
Substitute Tn = 46 into Tn = 4n – 2 to determine the value of n.
4n − 2 = 46 ⇒ 4n = 48 and n = 12
The twelfth term has a value of 46.
n represents the position of a term and it cannot be negative, a fraction or
zero.
If n is a whole number, then 72 is a term in the sequence.
If n is negative, zero or a fraction, then 72 is not a term in the sequence.
Substitute Tn = 72 into the formula into Tn = 4n − 2
4n − 2 = 72 ⇒ 4n = 74 and n = 18,5 so 72 is not a term in the sequence.
WORKED EXAMPLE 2
Consider the arithmetic sequence 3x − 1; 5x − 2; 4x + 3
1
Determine the value of x.
2
If x = 2, determine the values of the first three terms in the sequence and
then determine a formula for the nth term in the sequence.
3
Determine the value of the 15th term.
4
Which term has a value of 302?
5
Is 150 a term in the sequence? Justify your answer fully.
SOLUTIONS
1
2
3
4
5
d = T2 − T1 = T3 − T2
( 5x − 2 ) − ( 3x − 1 ) = ( 4x + 3 ) − ( 5x − 2 )
5x − 2 − 3x + 1 = 4x + 3 − 5x + 2
3x = 6 ⇒ x = 2
T1 = 3( 2 ) − 1 = 5; T2 = 5( 2 ) − 2 = 8 and T3 = 4( 2 ) + 3 = 11
Substitute a = 5 and d = 3 into Tn = a + ( n − 1 )d
⇒ Tn = 5 + ( n − 1 )( 3 ) = 5 + 3n − 3 = 3n + 2
T15 = 3( 15 ) + 2 = 47
Tn = 3n + 2 = 302 ⇒ 3n = 300
Find n if Tn = 302, so
n = 100
substitute Tn = 302.
Tn = 3n + 2 = 150 ⇒ 3n = 148
148
1
n = ____
= 49__
⇒ n ∉ ℕ, so Tn ≠ 150
3
3
d = Tn – Tn – 1
d = T2 – T1
d = T3 – T2
a = T1 = 5
d=8–5=3
Substitute n = 15 to
determine the fifteenth term’s
value, T15
n is the position of term
Tn and must be a natural
number.
Unit 1 Arithmetic sequences
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EXERCISE 1
1
2
3
Consider the sequence 11; 5; −1; −7; … .
1.1
What is the common difference?
1.2
State the values of the next two terms in the sequence.
1.3
Determine a formula for the nth term of the sequence.
1.4
Determine the value of the 20th term.
1.5
Is 43 a term in the sequence? Justify your answer.
Consider the sequence 2; 9; 16; 23; …
2.1
What is the common difference?
2.2
State the values of the next two terms in the sequence.
2.3
Determine a formula for the nth term of the sequence.
2.4
Determine the value of the twelfth term.
2.5
Which term has a value of 135?
2.6
Is 200 a term in the sequence? Justify your answer.
Answer the questions below for sequences A, B, C and D.
A: 2x + 4; − 2x − 3; 3x − 1
B: 4 − 2x; x − 1; 3x − 2
C: 2x − 8; − x − 1; 3x − 4,5
D: 10x; 3x − 2; 6x + 1
3.1
Determine the value of x for which these terms form an arithmetic
sequence.
3.2
Determine a formula for the nth term of the sequence.
3.3
Determine the value of the 19th term.
WORKED EXAMPLE
The sixth term of an arithmetic sequence is −5 and the sum of the second and
eighth terms is −2. Determine the first three terms in the sequence.
SOLUTION
T6= −5; n = 6; Tn = −5 and Tn = a + ( n − 1 )d
−5 = a + ( 6 − 1 )d ⇒ a + 5d = −5
➀
T2 + T8 = −2, T2 = a + d and T8 = a + 7d
( a + d ) + ( a + 7d ) = −2
2a + 8d = −2 ⇒ a = −1 − 4d
➁
Substitute ➁ into ➀:
( −1 − 4d ) + 5d = −5 ⇒ d = −4
Substitute d = −4 into ➁ :
a = −1 − 4( −4 ) = 15
T1 = 15; T2 = 11; T3 = 7
6
Topic 1 Patterns, sequences and series
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EXERCISE 2
1
2
3
4
5
6
7
The fifth term of an arithmetic sequence is 23 and the 12th term is 72.
1.1
Determine the first three terms of the sequence and the nth term.
1.2
What is the value of the tenth term?
1.3
Which term has a value of 268?
The 11th term of an arithmetic sequence is 57 and the sum of the first and fourth
terms is 29.
2.1
Determine the first three terms of the sequence.
2.2
Determine a formula for the nth term.
2.3
Is 100 a term in the sequence? Justify your answer.
The sum of the second and sixth terms of an arithmetic sequence is 4.
The third term is 24 more than the 11th term.
3.1
Determine the first three terms in the sequence.
3.2
Determine a formula for the nth term of the sequence.
3.3
Determine the 17th term in the sequence.
3.4
Is −72 a term in the sequence? Justify your answer.
5
4
Which term in the sequence −2; − __
; − __
; −1; … has a value of 8?
3
3
Consider the sequence 13; 11; 9; 7; 5; … .
5.1
Determine a formula for the nth term of the sequence.
5.2
Show that Tn + T15 − n = 0, for n ∈ ℕ, n < 15
1
1
The second term of an arithmetic sequence 6 __
and the tenth term is 18 __
.
2
2
th
6.1
Determine a formula for the n term of the sequence.
6.2
Is 93 a term in the sequence? Justify your answer.
The letters of the alphabet are arranged in rows as illustrated, starting with A
and ending with Z.
Row 3
Row 2
7.1
7.2
C C C C C C C C C C C
B B B B B B B
A A A
Row 1
Which row will contain the letter K and how many Ks will there be
in that row?
Which letter will appear 79 times and in which row will this letter be?
Unit 1 Arithmetic sequences
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Unit 2: Geometric sequences
KEY WORDS
geometric sequence – a
sequence of numbers with
a common ratio between
consecutive terms
common ratio – the constant
amount by which consecutive
terms are multiplied
In general we define a geometric sequence as follows:
a; ar; ar 2; ar 3; ar 4; ar 5; … ar n − 1
• a is the value of the first term
T3
Tn
T2 ___
= T = ____
• r is the common ratio between the terms, r = ___
T
T
1
2
n−1
• Tn is the value of the term in position n, so Tn = ar n−1
Consider the geometric sequence 5; 10; 20; 40; 80; …
The terms are doubling, so the next term will be 160.
T3
T3
T2
10
20
40
= ___
= 2, r2 = ___
= ___
= 2, r3 = ___
= ___
= 2, so r1 = r2 = r3 = 2 and so the
r1 = ___
T
5
T
T
10
20
1
2
2
common ratio is 2.
The first term is given by a = 5 and the common ratio is given by r = 2.
In order to determine a formula for the nth term, substitute a = 5 and r = 2 into the
formula Tn = ar n−1
The formula for the nth term in this series is given by Tn = 5 × 2n−1
WORKED EXAMPLE 1
Consider the sequence 96; 48; 24; 12; … .
1
State the values of the next three terms.
2
Determine a formula for the nth term of the sequence.
3
Determine the value of the 10th term.
3
4
Which term has a value of ___
?
64
Each term is half the
value of the previous
term.
SOLUTIONS
1
2
3
6; 3; __
2
The first term is 96, so a = 96.
T3 24 1
T2 ___
48 __
1
1
___
=
= and ___ = ___ = __, so this is a geometric sequence with r = __
T1
96
2
T2
48
2
(2)
1
1 n−1
into T = ar n−1 ⇒ T = 96 __
Substitute a = 96 and r = __
n
2
3
( )
1 n−1
Substitute n = 10 into the formula Tn = 96 __
2
3
1 10−1
1 9
⇒ T = 96 __
= 96 __ = ___
10
4
n
2
(2)
(2)
( )
( )
( )
( ) ( )
16
( )
3
3
1 n−1
To determine which term has a value of ___
, substitute Tn = ___
into Tn = 96 __
64
64
2
3
3
1 n−1
1 n−1
___
__
_______
__
∴
= 96
⇒
=
64
2
64 × 96
2
1
1 n−1
1 11
1 n−1
_____
__
__
and so 2 = __
∴ 2 048 = 2
2
| 2 048 = 211
∴ n – 1 = 11 and n = 12
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WORKED EXAMPLE 2
x − 6; 2x − 6; 5x + 3 are the first three terms of a geometric sequence.
1
2
Solve for x, showing all necessary calculations.
If x = −6, state the first three terms of the sequence and the nth term in
the form Tn = …
If x = 9, state the first three terms and the nth term in the form Tn = …
3
SOLUTIONS
T3
T2
___
= ___
1
T1
T2
5x + 3
2x − 6 ______
______
= 2x − 6
x−6
( 5x + 3 )( x − 6 ) = ( 2x − 6 )( 2x − 6 )
In a geometric sequence,
there is always a common
ratio for which the formula is
Tn
r = ____
Tn − 1
5x2 − 27x − 18 = 4x2 − 24x + 36
x2 − 3x − 54 = 0
( x − 9 )( x + 6 ) = 0
x = 9 or x = −6
T1 = −6 − 6 = −12, T2 = 2(−6) – 6 = − 18 and T3 = 5(−6) + 3 = −27
2
T2 ____
3
−18
= −12 = __
a = T1 = −12 and r = ___
T
2
1
3
into Tn = ar n−1
Substitute a = −12 and r = __
2
n
−
1
3
⇒ T = −12 __
(2)
n
3
T1 = 9 − 6 = 3, T2 = 2(9) − 6 = 12 and T3 = 5(9) + 3 = 48
T2 ___
12
= 3 =4
a = T1 = 3 and r = ___
T
1
Substitute a = 3 and r = 4 into Tn = ar n − 1
⇒ Tn = 3(4)n –1
EXERCISE 3
1
Answer the questions below for sequences A, B, C and D.
A
3; 15; 75; 375; … 234 375
1 __
1
__
B
; 1 ; __
; 1; … 128
8 4 2
C
D
2
2; −6; 18; −54; … 13 122
15
15
60; 30; 15; ___;… ____
2
128
1.1
Write down the fifth term in the sequence.
1.2
Determine a formula for the nth term of this sequence.
1.3
Determine the value of the seventh term.
1.4
How many terms are there in the sequence?
Answer the questions below for sequences A, B, C and D.
A: 9x + 8; 2x − 6; x − 8
B: 4x − 3; x; 2x − 5
C: 7x + 1; x + 3; x − 5
D: x − 1; 2x − 14; 3x
2.1
Determine the value(s) of x for which these terms form geometric
sequences.
2.2
Determine a formula for the nth term of each sequence.
2.3
Determine the value of the tenth term of the sequence.
Unit 2 Geometric sequences
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WORKED EXAMPLE 1
The second term in a geometric sequence is −4 and the fifth term is 32.
1
2
3
Determine a formula for the nth term of this sequence, that is, Tn.
Which term has a value of −1 024?
Determine the eighth term in the sequence.
SOLUTIONS
If any term in a
geometric sequence
is divided by any
other term in the
same sequence, the a
values cancel.
1
–1 024 is the value of
Tn in the formula
Tn = 2(–2)n–1
2
T2 = −4 = ar
➀
T5 = 32 = ar 4
➁
ar 4 ___
32
___
➁÷➀⇒ =
ar
−4
r 3 = −8 = ( −2 )3 ⇒ r = −2
Substitute r = −2 into ➀
a( − 2 ) = − 4 ⇒ a = 2
Tn = 2( − 2 )n−1
3
Tn = 2( − 2 )n−1 = −1 024
( −2 )n−1 = −512 = ( −2 )9
n − 1 = 9 ⇒ n = 10
T8 = 2( −2 )7 = −256
WORKED EXAMPLE 2
The first three terms of the sequence 2; x; y; 9 form an arithmetic progression
and the last three terms form a geometric progression. Determine x and y and
state the A.P. and G.P. in each case.
SOLUTION
An arithmetic sequence has a constant first difference:
x − 2 = y − x ⇒ y = 2x − 2
➀
A geometric sequence has a constant ratio:
y
9 __
__
= ⇒ y 2 = 9x
➁
y
x
Substitute ➀ into ➁:
( 2x − 2 )2 = 9x
4x2 − 17x + 4 = 0
( 4x − 1 )( x − 4 ) = 0
1
⇒ x = __
or x = 4
4
3
1
1
__
− 2 = −__
If x = 4 , then y = 2 __
4
2
1
3
__
;
−
A.P. 2; __
4
2
1
3
__
;
−
;
9
G.P. __
4
2
If x = 4, then y = 2( 4 ) − 2 = 6
( )
A.P. 2; 4; 6
G.P. 4; 6; 9
10
Topic 1 Patterns, sequences and series
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EXERCISE 4
1
2
3
4
5
3
The third term of a geometric sequence is 6 and the sixth term is ___
.
32
Determine the first three terms and the general term Tn.
The first term of a geometric sequence is 3 and the sum of the second and third
terms is 60. Determine the first three terms and the general term Tn.
The first three terms of the sequence 6; x; y; 27 form an arithmetic progression
and the last three terms form a geometric progression.
3.1
Determine the values of x and y.
3.2
State the arithmetic and geometric sequences for each of your solutions.
The first three terms of the sequence 6; x; y; 16 form an arithmetic progression
and the last three terms form a geometric progression.
4.1
Determine x and y.
4.2
State the arithmetic and geometric sequences for each of your solutions.
3 5
1 ___
7
Consider the sequence: __
; ; ___; ____
;…
4 16 64 256
5.1
5.2
State the next two terms in the sequence.
Determine the nth term of the sequence.
Unit 2 Geometric sequences
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Unit 3: The sum of arithmetic series
An arithmetic series is formed when the terms in an arithmetic sequence are added
together.
• 3; 7; 11; … 39 is an arithmetic sequence or an arithmetic progression.
• 3 + 7 + 11 + … + 39 is an arithmetic series.
The order of the values in the
first row is reversed in the
second row!
There are 10 columns
because there are 10
terms.
Consider the series 3 + 7 + 11 + 15 + 19 + 23 + 27 + 31 + 35 + 39.
• The first term is 3, so a = 3.
• There is a constant first difference of 4, so d = 4.
• The value of the last term is 39 and we represent this with l = 39.
• We work out the position of the last term using the Tn formula:
Tn = a + ( n − 1 )d, Tn = 39; a = 3 and d = 4
39 = 3 + ( n − 1 )( 4 ) ⇒ 36 = 4( n − 1 )
9 = n − 1 and n = 10
The last term is the tenth term in the sequence, so n = 10.
S10 = 3 + 7 + 11 + 15 + 19 + 23 + 27 + 31 + 35 + 39
S10 = 39 + 35 + 31 + 27 + 23 + 19 + 15 + 11 + 7 + 3
➀
➁
_____________________________________________________________
× S10 = 42 + 42 + 42 + 42 + 42 + 42 + 42 + 42 + 42 + 42 for 10 terms
➀ + ➁: 2___________________________________________________
Adding the rows of values results
in the sum of the values being
doubled, so divide through by 2.
2 × S10 = 10( 42 )
10
⇒ S = ___( 420 ) = 2 100
10
2
The algebraic proof below applies to all arithmetic series and is examinable.
n
Required to prove: S = __[2a + n − 1d]
n
In each column the ds add
up to zero because they are
always equal in value, but
opposite in sign.
The second row of
terms is obtained by
reversing the order of
the terms in first row.
There are n terms, so
there are n columns.
2
Proof:
a is the first term, d is the difference, n is the number of terms and l is the last term,
given by l = Tn = a + ( n − 1 )d
Sn =
Sn =
+ (a + d) + (a + 2d) + … + (l − 2d) + (l − d) +
+ (l − d) + (l − 2d) + … + (a + 2d) + (a + d) +
a
l
➀ + ➁: 2Sn = (a + l) + (a + l)
+ (a + l)
+ … + (a + l)
+ (a + l)
l ➀
a ➁
+ (a + l)
for n terms
n(
a + l)
2Sn = n( a + l ) ⇒ Sn = __
2
n(
a + l)
Substitute l = a + ( n − 1 )d into Sn = __
2
n(
n
__
__
⇒ S = a + l ) = S = [ 2a + ( n − 1 )d ]
n
2
n
2
Before you decide which version of the formula to use, check the information given:
n
• If the value of the last term is given, use the formula sn = __2 ( a + l ).
n
If the last term is not given, use the formula s = __[ 2a + ( n − 1 )d ].
•
n
2
If
the
number
of
terms
is
not
given,
then
use
T
=
a( n − 1 )d to determine the value
•
n
of n.
• If the only information given is the sum formula, remember that:
Sn = T1 + T2 + T3 + … + Tn−1 + Tn and Sn −1 = T1 + T2 + T3 + … + Tn−1,
so Tn = Sn − Sn − 1
12
Topic 1 Patterns, sequences and series
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• In some questions you may need to use both Tn = a( n − 1 )d and
n[
n(
Sn = __
2a + ( n − 1 )d ] or Sn = __
a + l ).
2
2
• Substitute into the appropriate formula(e) and then, if necessary, use simultaneous
equations.
WORKED EXAMPLE 1
Determine the sum of the arithmetic series 4 + 11 + 18 + 25 + … + 368.
SOLUTION
a = 4, d = 7 and l = 368
Tn = a + ( n − 1 )d ⇒ 368 = 4 + ( n − 1 )( 7 )
368 − 4
So n − 1 = _______ and n = 53
7
53 (
n(
a + l ) ⇒ S53 = ___
4 + 368 ) = 9 858
Sn = __
2
2
WORKED EXAMPLE 2
Determine the sum of the first 20 terms of the arithmetic series −15 − 9 − 3 + 3 + …
SOLUTION
a = −15, d = 6 and n = 20
n
S = __[ 2a + ( n − 1 )d ]
n
2
20 [ (
⇒ S20 = ___
2 −15 ) + 19( 6 ) ] = 840
2
∴ 11 + T2 = 26 and T2 = 15
WORKED EXAMPLE 3
Sn = 2n2 + 9n
1
2
3
Determine the first three terms of the sequence.
Determine the 12th term by using the formula Sn = 2n2 + 9n.
Determine the 12th term by determining the formula for the nth term.
SOLUTION
1
S1 = T1 = 2( 1 )2 + 9( 1 ) = 11 ∴ T1 = 11
S2 = T1 + T2 = 2( 2 )2 + 9( 2 ) = 26
∴ 11 + T2 = 26 and T2 = 15
S3 = T1 + T2 + T3 = 2( 3 )2 + 9( 3 ) = 45
∴ 26 + T3 = 45 and T3 = 19
The first three terms are 11, 15, 19.
Unit 3 The sum of arithmetic series
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2
3
S11 = 2( 11 )2 + 9( 11 ) = 341
S12 = 2( 12 )2 + 9( 12 ) = 396
T12 = S12 − S11 = 396 − 341 = 55
a = 11, d = 4 and Tn = a + ( n − 1 )d
Tn = 11 + ( n − 1 )( 4 ) = 4n + 7
T12 = 4( 12 ) + 7 = 55
WORKED EXAMPLE 4
The second term of a sequence is 17 and the sum of the first six terms is 147.
Determine the first three terms and the nth term.
SOLUTION
T2 = 17 and S6 = 147
Tn = a + ( n − 1 )d ⇒ T2 = a + d = 17 and so a = 17 − d
n[
S = __
2a + ( n − 1 )d ]
➀
∴ 3( 2a + 5d ) = 147 and 2a + 5d = 49
➁
n
2
6[
⇒ S6 = __
2a + 5d ] = 147
2
Substitute ➀ into ➁:
⇒ 2( 17 − d ) + 5d = 49
⇒ 34 − 2d + 5d = 49
∴ 3d = 15 and d = 5
Substitute d = 5 into ➀:
⇒ a = 17 − 5 = 12
T1 = 12, T2 = 17 and T3 = 22
Tn = a + ( n − 1 )d = 12 + ( n − 1 )( 5 ) = 5n + 7
EXERCISE 5
1
2
3
4
5
6
14
Determine the sum of the series:
1.1
11 + 8 + 5 + 2 + … − 166.
1.2 1 + 5 + 9 + 13 + … + 113
1.3
−12 − 5 + 2 + 9 + … + 121
Determine the sum of the first 40 terms of the arithmetic series
2.1
2 + 10 + 18 + …
2.2 16 + 10 + 4 − 2 − …
2.3
1 + 2,5 + 4 + …
Determine the first three terms and the nth term of the sequence if
3.1
Sn = 43n − 6n2
3.2 Sn = n + 2n2
31
3 2
___
__
3.3
Sn = 2 n − 2 n
The sixth term of an arithmetic series is 32 and the sum of the first nine terms
is 234. Determine the first three terms and the nth term.
Determine the sum of the series −9 − 5 − 1 + 3 + … + 207.
The fourth term of an arithmetic series is 11 and the sum of the first twelve
terms is 222. Determine the first three terms and the nth term.
Topic 1 Patterns, sequences and series
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Unit 4: The sum of geometric series
REMEMBER
Consider the series 2 + 6 + 18 + 54 + 162.
a = 2, r = 3 and n = 5
The formula for determining the sum of terms in a geometric series requires each term
in the series to be multiplied by the common ratio. Each term is shifted to the right so
that the subtraction of equal values is easier to process.
Below is a numerical illustration of the proof.
T1
S5 = 2 + 6 + 18 + 54 + 162
3 × S5 =
T2
Tn−1
The position of the term in
the sequence is given by n.
6 + 18 + 54 + 162 + 486
S5 − 3 × S5 = 2
The general term of a
geometric sequence is given
by Tn = ar n−1
The first term is represented
by a.
The common ratio is given
T3
Tn
T
by r = __2 = __ = ____
− 486
S5( 1 − 3 ) = 2 − 486 ⇒ S5 = _____
= 242
−2
−484
The algebraic proof below applies to all geometric series and is examinable.
a( 1 − r n ) a( r n − 1 )
Required to prove: S = ________ = ________, for r ≠ 1
n
1−r
r−1
Proof:
Sn = a + ar + ar 2 + … + ar n −2 + ar n − 1
r Sn =
➀ − ➁:
ar + ar 2 + … + ar n − 2 + ar n − 1 + ar n
➀
➁
− ar n
Sn − r Sn = a
a( 1 − r n )
1−r
a( r n − 1 )
Sn( 1 − r ) = a( 1 − r n ) ⇒ Sn = ________
= ________
, r≠1
(
)
r−1
WORKED EXAMPLE 1
Determine the sum of the first seven terms in the geometric series
3 + 15 + 75 + …
SOLUTION
a( 1 − r n )
3( 1 − 57 )
a = 3, r = 5 and n = 7 ⇒ Sn = ________
= ________
= 58 593
1−r
1−5
WORKED EXAMPLE 2
Determine the sum of terms in the series 5 + 10 + 20 + 40 + … + 655 360.
SOLUTION
10
a = 5, r = ___
= 2 and Tn = 5r n−1 = 655 360 ⇒ r n−1 = 131 072 = 217, so n = 18
5
| By trial and error
a( 1 − r n ) 5( 1 − 218 )
S = ________ = _________ = 1 310 715
n
1−r
1−2
Unit 4 The sum of geometric series
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WORKED EXAMPLE 3
The first term of a geometric series is 16 and the sum of the second term and third
term is 12.
Determine the sum of the first ten terms.
SOLUTION
a = 16 and T2 + T3 = 12 ⇒ ar + ar 2 = 12, so 16r + 16r 2 = 12 ⇒ 4r + 4r 2 = 3
3
1
4r 2 + 4r − 3 = 0 ⇒ ( 2r + 3 )( 2r − 1 ) = 0 and so r = − __
or r = __
2
2
( ( 3) )
10
2
32
3
1 − ( − __
2)
1
16( 1 − ( __
2) )
1 023
1
, then S = ___________ = _____
If r = __
3
If r = −__ then S
16 1 − − __
10
2
11 605
= ____________ = −______
10
10
2
REMEMBER
a is the first term.
r is the common ratio.
n is the number of terms or
the position of a term.
Tn = ar n−1 is the formula for
the nth term in a geometric
sequence.
a( 1 − r n )
S = ________ is the formula
n
EXERCISE 6
Round off your answers to two decimal places where necessary.
1
Determine the sum of the first 12 terms of the series 36 + 24 + 16 + … .
2
Determine the sum of the first 10 terms of the series 125 + 100 + 80 + 64 + … .
2
.
3
Determine the sum of the series 54 + 18 + 6 + … + ___
81
4
5
1
.
Determine the sum of the series 3 125 + 625 + 125 + 25+ …+ ____
625
n
243
1
____
__
1−
.
Determine the fifth term of the series if S =
n
1−r
for the sum of the first n
terms in a geometric series.
Tn = Sn − Sn−1
KEY WORDS
infinite series – a series that
has no end and for which it is
impossible to determine the
number of terms
diverge – grow further apart
in value
oscillate – swing back and
forth between large and small
values, and between positive
and negative values
sum to infinity – no limit to
the number of terms being
added to determine the sum
converge – Tn tends to zero
as n tends to infinity
16
32
1
1 − __
2
2
(
(3) )
The sum of an infinite number of terms of a geometric series can be determined
if −1 < r < 1
If r > 1, then r n → ∞ as n → ∞, so a sum to infinity cannot be determined.
• If r = 2, the sequence 2n diverges:
2; 4; 8; 16; 32; 64; 128; 256; 512; 1 024; 2 048; 4 096; 8 192; …
If r < −1, then r n → −∞ when n is odd and n → ∞, but r n → ∞ when n is even and n → ∞.
• If r = −2 the sequence ( −2 )n diverges and oscillates:
−2; 4; −8; 16; −32; 64; −128; 256; −512; …
If −1 < r < 1, r n → 0 as n → ∞, so a sum to infinity can be determined.
; ___; ___; ____; ____; ____; _____; …
• If r = __2 , the sequence ( __2 ) converges __2 ; __4 ; __8 ; ___
16 32 64 128 256 512 1 024
1 n
1
1 1 1
1
1
1
1
1
1
1
• If r = −__2 , the sequence ( −__2 ) oscillates and converges
1
1 n
1 __
1 ___
1 ___
1
−__
; 1 ; −__
; 1 ; −___
; 1 ; −____
;…
2 4
8 16
32 64
128
Topic 1 Patterns, sequences and series
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The proof of the formula for the sum to infinity of a geometric sequence, for
−1<r<1
a
Required to prove S = _____, if −1 < r < 1
∞
1−r
Proof:
a( 1 − 0 ) _____
a
= 1−r
If −1 < r < 1, then r n → 0 as n → ∞, so S∞ = _______
1−r
WORKED EXAMPLE 1
Determine the sum to infinity of the sequence 16 + 8 + 4 + … for an infinite
number of terms.
SOLUTION
a
8
16
1
a = 16 and r = ___
= __ ⇒ S∞ = _____
= _____
= 32
1
16 2
1−r
__
1−2
WORKED EXAMPLE 2
Consider the series 16k + 8k2 + 4k3 + ...
1
For which value(s) of k will the series converge?
2
Calculate the sum of the series to infinity if k = −1,5.
SOLUTIONS
1
2
k
−1 < __
< 1 ⇒ −2 < k < 2
2
−1,5
= −0,75
a = − 24 and r = _____
2
a
−24
_____
___________
⇒ S∞ = 1 − r =
1 − ( − 0,75 )
96
___
=−7
EXERCISE 7
1
2
3
Determine the sum to infinity for each of these geometric series:
1.1
36 + 12 + 4 + …
1.2
−20 − 5 − 1,25 − ...
1.3
40 − 20 + 10 − 5 + ...
1.4
−27 + 18 − 12 + 8 − ...
( x − 2 )2
( x − 2 )3
x − 2; _______ and _______ are the first three terms of a geometric sequence.
4
16
2.1
For which value(s) of x is the sequence convergent?
2.2
Determine the sum to infinity if x = 1.
Consider the geometric series: 45 + 15 + 5 + …
3.1
Does the sequence converge or diverge? Justify your answer.
3.2
Determine the sum to infinity of the series, if it exists.
Unit 4 The sum of geometric series
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Unit 5: Sigma notation
Sigma notation enables us to represent a series in shorthand. For example:
3
∑( 2k − 1 ) = [ 2( 1 ) − 1 ] + [ 2( 2 ) − 1 ] + [ 2( 3 ) − 1 ] = 1 + 3 + 5 = 9
k=1
• k = 1, so substitute 1 into the formula to determine the value of the first term, T1
• Substitute the 3 above the sigma sign to determine the value of the last term.
• There are ( 3 − 1 + 1 ) = 3 terms in the series.
• The formula for each term is given by Tk = 2k − 1
2
∑ 2n = 2 3 + 2 2 + 2 1 + 20 + 21 + 22 = __18 + __14 + __12 + 1 + 2 + 4 = 7__78
−
−
−
n = −3
• Substitute n = −3 to determine the value of the first term, T1.
• There are ( 2 − ( −3 ) + 1 ) = 6 terms in the series.
• The formula for each term is given by Tn = 2n.
REMEMBER
a is the first term for both
an arithmetic series and a
geometric series.
The number at the bottom and top of the sigma sign may need to change in each
series. You are often given these numbers, but you may need to find either the top or
the bottom value if all other values are given. First substitute the bottom value, followed by consecutive integers (always increasing by 1) until you have substituted the
top value.
d is the common difference of
an arithmetic series.
In one of the examples above, the formula is Tn = 2k − 1 and the substitution values
are k = 1, k = 2 and k = 3. So you substitute 1 to determine the first term’s value and
3 to determine the last term’s value.
r is the common ratio of a
geometric series.
For
n is the number of terms or
the position of a term in both
arithmetic and geometric
sequences.
5
∑ ( 4k + 1 ) you substitute −2 to determine the value of the first term
k = −2
and 5 to determine the value of the last term. The total number of terms is:
5 − ( −2 ) + 1 = 8
This is because you substitute in −2; −1; 0; 1; 2; 3; 4; 5, which is a total of 8 numbers.
Tn = a + ( n − 1 )d is the
formula for the nth term in an
arithmetic sequence.
Tn = ar n−1 is the formula for
the nth term in a geometric
sequence.
n
S = __[ 2a + ( n − 1 )d ] and
n
2
n( + )
Sn = __
a l are both sum
2
formulae for arithmetic series.
a( 1 − r n )
S = ________ is the formula
n
1−r
for the sum of the first n
terms in a geometric series.
a
S = _____ , −1 < r < 1 is the
∞
1−r
formula for the sum to infinity
of an infinite geometric series.
WORKED EXAMPLE 1
20
Evaluate
∑ ( 5n − 2 )
n = −4
SOLUTION
20
∑ ( 5n − 2 ) = [ 5( − 4 ) − 2 ] + [ 5( − 3 ) − 2 ] + [ 5( −2 ) − 2 ] + … + [ 5( 20 ) − 2 ]
n = −4
= −22 − 17 − 12 − ... + 98
a = −22, l = 98 and n = 25
n
S = __( a + l )
n
2
25 (
⇒ S25 = ___
−22 + 98 )
2
= 950
Tn = Sn − Sn−1 in both
arithmetic and geometric
series.
18
Topic 1 Patterns, sequences and series
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WORKED EXAMPLE 2
∞
∑3( __12 )n 1
Evaluate
−
n=1
SOLUTION
∞
∑3( __12 )n 1 = 3( __12 )1 1 + 3( __12 )2 1 + 3( __12 )3 1 + … = 3 + __32 + __34…
−
−
−
−
n=1
a
3
1
and S∞ = _____
= _____
=6
a = 3, r = __
1
2
1−r
__
1−2
WORKED EXAMPLE 3
n
If
∑__12( 3r 1 ) = 1 640, determine the value of n.
−
r=1
SOLUTION
n
3
= 1 640
∑__12( 3r 1 ) = __12 + __32 + __92 + … + ____
2
−
n −1
r=1
1
G.P. a = __
, r = 3 and Sn = 1 640
2
a( 1 − r n )
Sn = ________
1−r
1(
__
1 − 3n )
2
= ________
1−3
= 1 640
1(
− __
1 − 3n )
4
= 1 640
1 − 3n = −6 560
3n = 6 561
⇒n=8
Unit 5 Sigma notation
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WORKED EXAMPLE 4
Evaluate each of the series below and express the result in sigma notation:
1
12 + 4 − 4 … −276
2
100 + 40 + 16 + …
3
5 − 10 + 20 − … 5 120
4
1 + 4 + 9 + 16 + 25 + 36 + 49
SOLUTIONS
1
a = 12, d = −8 and Tn = −276
Tn = a + ( n − 1 )d
−276 = 12 + ( n − 1 )( −8 )
−276 = 20 − 8n
8n = 296 ⇒ n = 37
37 (
S37 = ___
12 − 276 ) = −4 884
2
37
∑( 20 − 8n ) = −4 884
n=1
2
40 ___
16 __
2
a = 100 and r = __
| r = ____
=
=2
5
100 40 5
a
100
5
500
S = _____ = ______ × __ = ____ ⇒
∞
3
(
1−r
2
1 − __
5
)
5
3
∞
500
∑100( __25 )n 1 = ____
3
−
n=1
a = 5, r = −2 and Tn = 5 120
Tn = ar n−1 = 5 × ( −2 )n−1 = 5 120 ⇒ ( −2 )n−1 = 1 024 = ( −2 )10
5( 1 − ( −2 )11 )
⇒ n = 11 and S = ___________ = 3 415
11
11
1 − ( −2 )
∑5 × ( −2 )n 1 = 3 415
−
n=1
4
12 + 22 + 32 + … + 72 = 140 ⇒
7
∑n2 = 140
n=1
EXERCISE 8
1
2
3
Given the series 2 + 9 + 16 + …+ 331.
1.1
Determine a formula for the nth term.
1.2
Express the series in sigma notation.
1.3
Calculate the sum of terms in the series.
Consider the series 2 + 18 + 162 + … + 9 565 938.
2.1
Determine a formula for the nth term.
2.2
Determine the number of terms in the series.
2.3
Express the series in sigma notation.
2.4
Calculate the sum of terms in the series.
The first three terms of an infinite series are 36 − 24 + 16 − …
3.1
Determine a formula for the nth term.
3.2
Express the series in sigma notation.
3.3
Determine the sum to infinity.
5
n
4
Find n if
∑
2k−1 = 255.
k=1
20
5
Evaluate
∑r 3
r=2
Topic 1 Patterns, sequences and series
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Unit 6: Practical applications
WORKED EXAMPLE 1
12x
6x
6x
6 2x
6x
3 2x
12x
6x
6x
6 2x
6x
3 2x
12x
The large red square has sides of 12x cm. The midpoints are joined to form the
green square. The midpoints of the green square are joined to form the purple
square. This process is repeated indefinitely. Calculate the sum of the areas of all
the squares that are found this way, including the large red square.
SOLUTION
The sides of the large red square are 12x cm.
By Pythagoras' Theorem:
The sides of the large green square are:
__________
____
__
√ 36x2 + 36x2 = √ 72x2 = 6√ 2 x cm
By Pythagoras' Theorem:
REMEMBER
Pythagoras’ Theorem: In a
right-angled triangle, the
square on the hypotenuse
is equal to the sum of the
squares on the other two
sides.
Remember to convert units
to the same value. All units
should be in m, cm or mm.
The sides of the large purple square are:
_______________
__
__
____
√( 3√2 x )2 + ( 3√2 x )2 = √36x2 = 6x cm
__
__
Sides of squares: 12x; 6√2 x; 6x; 3√2 x
Areas of squares: 144x2; 72x2; 36x2; 18x2...
1
.
This is a geometric series with a = 144x2 and r = __
2
Sum of areas of squares:
144x2
S = _____ = 288x2
∞
1
1 − __
2
1
Note that the sum to infinity can be determined because −1 < __
2 < 1.
Unit 6 Practical applications
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Greenworks, a project started in 2007 by former South African President Thabo
Mbeki, aims to plant one million trees a year in South Africa.
WORKED EXAMPLE 2
At the end of the first year after planting, a tree is one metre tall. In the second
7
year its height increases by 56 cm. In each successive year its height increases by __
8
of the previous year’s increase in height.
1
2
3
4
Determine the increase in the height of the tree in the third year.
Determine the total height of the tree after three years.
Lobelo says that by the end of the fifth year the tree will exceed a height of
3m. Thabo disagrees with him and says that the tree will only exceed a height
of 3m after 6 years. Is either of them correct? Fully justify your answer by
means of calculations.
Calculate the maximum height to which the tree will grow.
SOLUTIONS
1
2
3
7
In the third year the increase in the height of the tree is __
× 56 cm = 49 cm.
8
After three years the total height of the tree = 100 cm + 56 cm + 49 cm
= 205 cm
After five years its height is
(
( ))
7
56 1 − __
4
8
100 + __________ = 285,3906 cm
7
1 − __
8
= 2,85 m
After 6 years, its height is
(
( ))
7
56 1 − __
5
8
100 + __________ = 318,2168 cm
7
1 − __
8
= 3,18 m
Neither of them is correct. It will be less than 3 m by the end of 5 yers and
more than 3 m by the end of 6 years, so it will exceed 3 m during the sixth
year.
4
Maximum height will be 100 + _____
7 = 100 + 448 cm = 5,48 m
__
56
1−8
WORKED EXAMPLE 3
˙=2
Show that 1,9
SOLUTION
1,999999999… = 1 + (0,9 + 0,09 + 0,009 + 0,0009 + 0,00009 + … )
0,9
a
= 1 + _______
| a = 0,9; r = 0,1 and S = _____
1 − 0,1
∞
1−r
=1+1
=2
22
Topic 1 Patterns, sequences and series
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EXERCISE 9
1
2
3
4
5
The large red triangle has a base of 96 cm and a height of 64 cm. The base of the
green triangle is half the base of the red triangle and its height is three quarters
of the height of the red triangle. The base of the purple triangle is half the base
of the green triangle and its height is three quarters of the height of the green
triangle. Determine the sum of areas of all of the triangles that will be created if
this process is repeated indefinitely.
Zukile trained for the Comrades. In the first week of training he ran 5 km per day.
In the second week he ran 7 km per day and in the third week he ran 9 km per day.
He ran on every day except Sundays when he rested.
2.1
In which week did his total distance exceed 100 km for the first time?
2.2
How far did he run in the tenth week?
2.3
Determine the total distance Zukile ran during the first ten weeks of
his training.
Siseko is hoping to buy a car when he turns 18. His grandmother offers to sell
him her car for R131 071. He works on Saturdays only and earns R200 every
Saturday. He saves R1 on the first Saturday. Each Saturday thereafter he doubles
the amount he saved on the previous Saturday. He puts the money into a bottle
on his desk and not into a bank account.
3.1
In which week will his weekly payment exceed his earnings for the first
time, and what is the amount of the final payment he can afford?
3.2
How much will he have saved by the time he is no longer able to afford
the weekly payments?
3.3
If he had been able to afford the payments, how long would it have taken
him to save R131 071?
17
˙7
˙ = ___
Show that 0,1
.
99
A tree grows to a height of 3 metres in the first year. In the second year it grows
4
of the increase in
2,5 metres. In each successive year its height increases by __
5
height of the previous year.
What is the maximum height to which the tree can grow?
Unit 6 Practical applications
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6
REMEMBER
a is the first term for both
an arithmetic series and a
geometric series.
d is the common difference of
an arithmetic series.
r is the common ratio of a
geometric series.
n is the number of terms or
the position of a term in both
arithmetic and geometric
sequences.
A circle with diameter before 6 116 cm is placed inside a rectangle so it touches
the sides of the rectangle in three places. More circles are placed inside the
rectangle. Each circle touches the circle to its left in one place and has its centre
on the line AB, with A and B midpoints of the sides of the rectangle on which
they lie. The area of each new circle is one quarter of the area of the circle to its
left. This process is continued indefinitely.
A
Tn = a + ( n − 1 )d is the
formula for the nth term in an
arithmetic sequence.
6.1
Tn = ar n−1 is the formula for
the nth term in a geometric
sequence.
n
S = __[ 2a + ( n − 1 )d ] and
6.2
B
Determine the dimensions of the smallest rectangle that will allow all the
circles to fit perfectly with no gaps between them on the left and right
sides of the rectangle.
Determine the area of the unshaded region, correct to two decimal places.
n
2
n( + )
a l are both sum
Sn = __
2
formulae for arithmetic series.
a( 1 − r n )
S = ________ is the formula
n
1−r
for the sum of the first n
terms in a geometric series.
a
S = _____ , −1 < r < 1
∞
1−r
is the formula for the sum
to infinity of an infinite
geometric series.
The area of a triangle
1
= __
base × ⊥ height
2
The area of a circle = πr2
7
24
A bouncy rubber ball dropped from any height will lose 25% of its height on
each successive bounce. If dropped from a height of 30 m, the first bounce will
have a height of 22,5 m and the second bounce a height of 16,875 m. Determine
the total vertical distance travelled by the ball from the time it is dropped from
a height of 40 m, until it finally comes to rest. Note that the height from which
the ball is dropped is 40 m, and not 30 m.
Topic 1 Patterns, sequences and series
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Unit 7: Quadratic patterns and combinations
of arithmetic and geometric sequences
An arithmetic sequence has
• a common difference between consecutive terms
• d = Tn – Tn –1
• Tn = a + (n – 1)d
A geometric sequence has
• a common ratio between consecutive terms
Tn
r = ____
•
Tn−1
• Tn = ar n−1
A second differences sequence has
• no constant first difference
• a constant second difference, d2
• Tn = an2 + bn +dc
2
• 2a = d2, so a = __
2
• 3a + b = T2 – T1
• a + b + c = T1
In arithmetic, geometric and second difference sequences
• n ∈ ℕ, so it is a positive whole number.
WORKED EXAMPLE 1
Consider the quadratic sequence: −6; 2 − x; 4; 2x + 5
1
Solve for x.
2
Determine the nth term in the sequence.
3
Which term has a value of 400?
SOLUTIONS
1
−6
2−x
4
2x + 5
2+x
2x + 1
2x − 6
x−1
2x − 6 = x − 1 ⇒ x = 5
The sequence is −6; −3; 4; 15
8−x
2
Tn = an2 + bn + c
T1
T2
T3
T4
T0
T1
T2
T3
T4
−6
−3
4
15
−5
−6
−3
4
15
3
7
4
11
4
3
−1
4
2a = 4 ⇒ a = 2
3a + b = 3 ⇒ b = 3 − 6 = −3
a + b + c = −6
2 − 3 + c = −6 and c = −5
Tn = 2n2 − 3n − 5
7
4
11
4
2a = 4 ⇒ a = 2
T0 = −5 = c
T1 = a + b + c = −6
2 + b − 5 = −6 and b = −3
Tn = 2n2 − 3n − 5
Unit 7 Quadratic patterns and combinations of arithmetic and geometric sequences
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3
2n2 − 3n − 5 = 400 ⇒ 2n2 − 3n − 405 = 0
( 2n + 27 )( n − 15 ) = 0
| Use the quadratic formula if necessary.
27
___
|n∈ℕ
n ≠ − , so n = 15 only
2
The 15th term has a value of 400.
REMEMBER
In an arithmetic sequence
• Sn is the sum of the first n
terms
n
• is the number of terms
• a is the first term, T1
• l is the last term, Tn.
In a geometric sequence
• Sn is the sum of the first n
terms
• n is the number of terms
• a is the first term, T1
• r is the common ratio.
WORKED EXAMPLE 2
Evaluate the following:
1 + 2 + 4 + 5 + 7 + 8 + 10 + 11 + 13 + … + 95 + 97 + 98 + 100
SOLUTION
The sequence is neither arithmetic nor geometric. The multiples of 3 have been
removed. Create a new sequence with the multiple of 3 as part of the sequence.
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + ... + 98 + 99 + 100
100
= ____ ( 1 + 100 )
2
= 5 050
The red numbers which have been inserted are all multiples of 3.
Tn = 3n = 99 ⇒ n = 33
The sum of these multiples is
33 (
3 + 6 + 9 + 12 + ... + 96 + 99 = ___
3 + 99 )
2
= 1 683
1 + 2 + 4 + 5 + 7 + 8 + 10 + 11 + 13 + ... + 95 + 97 + 98 + 100
= 5 050 − 1 683
= 3 367
WORKED EXAMPLE 3
Evaluate the following: 1 + 2 + 4 + 5 + 8 + 10 + 15 + 16 + … + 256 + 260
SOLUTION
Geometric sequence with
a = 1 and r = 2
1 + 2 + 4 + 5 + 8 + 10 + 15 + 16 + … + 256 + 260
( 1 + 2 + 4 + 8 + 16 + … + 256 ) + ( 5 + 10 + 15 + 20 + … + 260 )
Tn = 2n−1 = 256 = 28 ⇒ n = 9
and
Tn = 5n = 260 ⇒ n = 52
1( 1 − 29 ) 52
Sum = ________ + ___( 5 + 260 )
Arithmetic sequence with
a = d = 5 and l = 260
1−2
2
= 511 + 6 890
= 7 401
26
Topic 1 Patterns, sequences and series
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WORKED EXAMPLE 4
The first two terms of a geometric sequence and an arithmetic sequence are
the same. The first term is 4. The sum of the first three terms of the
geometric sequence is one more than the sum of the first three terms of the
arithmetic sequence. Determine two possible values of r, the common ratio
of the geometric sequence.
SOLUTION
Let the first two terms be 4 and x.
Arithmetic sequence:
Common difference: d = x − 4
⇒ T3 = 4 + 2( x − 4 ) = 2x − 4
Geometric sequence:
x
Common ratio: r = __
x 2 __
x2
⇒ T3 = 4 __
= 4
4
(
( )
4
)
x2
4 + x + __ − ( 4 + x + 2x − 4 ) = 1
4
x2
⇒ __ − 2x + 4 − 1 = 0
4
x2 − 8x + 12 = 0
⇒ ( x − 6 )( x − 2 ) = 0
x = 6 or x = 2
Common ratio:
6 3
r = __ = __
4
2
2 __
1
=
or r = __
4 2
Check:
A.P.: 4; 6; 8 and G.P.: 4; 6; 9
or A.P.: 4; 2; 0 and G.P. 4; 2; 1
EXERCISE 10
1
2
3
Consider the sequence 3; −1; −9; −21; −37; … .
1.1
If the pattern continues in this way, write down the next two terms in
the sequence.
1.2
Determine the nth term in the sequence.
1.3
Which term has a value of −541?
20; x; y is a geometric sequence and x; y; 60 is an arithmetic sequence.
Determine the value(s) of x and y.
The first three terms of the sequence −4; p; q; 50 form an arithmetic progression and
the last three terms form a geometric progression. Determine the value(s) of p and q.
Unit 7 Quadratic patterns and combinations of arithmetic and geometric sequences
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4
5
6
7
An arithmetic sequence and a geometric sequence are combined to give
the following series:
3 + 4 + 8 + 9 + 12 + 16 + 20 + 24 + 27 + 28 + … + 729 + 732
4.1
State the first three terms and the last term of the arithmetic series.
4.2
State the first three terms and the last term of the geometric series.
4.3
Determine the sum of the combined series.
An arithmetic sequence and a geometric sequence are combined to give
the following series:
3 + 5 + 6 + 10 + 12 + 15 + 20 + 24 + 25 + 30 + 35 … + 90 + 95 + 96 + 100
5.1
State the first three terms and the last term of the arithmetic series.
5.2
State the first three terms and the last term of the geometric series.
5.3
Determine the sum of the combined series.
Consider the sequence: 2; 7; 16; 29; 46
6.1
If the pattern continues in this way, write down the next two terms
in the sequence.
6.2
Determine the nth term in the sequence.
Consider the triangles sketched below.
2
1
4
2
4
6
8
Triangle 1
Triangle 2
Triangle 3
Triangle 4
7.1
7.2
28
3
Determine a formula for the area of the nth triangle if the pattern
continues in the same way.
What type of sequence is formed by the areas of these triangles?
Topic 1 Patterns, sequences and series
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Revision Test Topic 1
Total marks: 210
1
2
Consider the sequences A, B and C and for each sequence, answer the questions
that follow.
A 1; 6; 11; 16; 21; …
B 3; 6; 12; 24; 48; …
C 0; 5; 12; 21; 32; …
1.1
Is the sequence arithmetic, geometric or quadratic? Fully justify your
answer.
(2)
th
1.2
Determine a formula for the n term of the sequence.
(3)
1.3
Which term has a value of 96?
(4)
1.4
Is 1 536 a term in the sequence? Fully justify your answer.
(2)
th
1.5
Determine the 20 term in the sequence.
(2)
1.6
If the sequence is arithmetic or geometric:
1.6.1
express the sum of the first fifty terms in sigma notation
(2)
1.6.2
determine the sum of the first 50 terms.
(4)
Given the sequence: 32; x; 18
2.1
Determine the value(s) of x if the sequence is arithmetic.
2.2
Determine the value(s) of x if the sequence is geometric.
(3)
(3)
15
3
Determine the value of A if A =
∑5k 6.
−
(4)
k=1
4
The following sequence is a combination of an arithmetic sequence and
a geometric sequence:
5; 5; 15; 10; 25; 20; 35; 40; …
4.1
Write down the next 2 terms.
(4)
4.2
Calculate T42 − T41.
(6)
4.3
Determine the sum of the first 50 terms.
(6)
4.4
Prove that ALL the terms of this infinite sequence will be divisible by 5. (4)
5
A quadratic pattern has a third term equal to 16, a fourth term equal to 35 and
a sixth term equal to 91.
5.1
Determine the second difference of this quadratic pattern.
(4)
5.2
Determine the first term of the pattern.
(4)
5.3
Determine the nth term of the pattern.
(4)
6
The following sequence forms a convergent geometric sequence:
x3
7x; x2; __
; …
7
6.1
Determine the possible values of x.
6.2
If x = −4, calculate S∞.
(3)
(4)
n
7
∑( 3k − 15 ) = 285
k=1
7.1
7.2
8
How many terms are there in the sequence?
The odd numbers are removed from the sequence.
Calculate the sum of the terms of the remaining sequence.
The sequence 3; −2; x; −24; … is a quadratic sequence.
8.1
Calculate x.
8.2
Determine the nth term of the sequence.
(4)
(6)
(4)
(5)
REMEMBER
If x2 = 9, then x = 3 or −3
If x2 = 16, x = ±4
a is the first term for both
an arithmetic series and a
geometric series.
d is the common difference of
an arithmetic series.
r is the common ratio of a
geometric series.
n is the number of terms or
the position of a term in both
arithmetic and geometric
sequences.
n ∈ ℕ as it indicates the
position of a term in the
sequence.
Tn = a + ( n − 1 )d is the
formula for the nth term in an
arithmetic sequence.
Tn = ar n−1 is the formula for
the nth term in a geometric
sequence.
n[
2a + ( n − 1 )d ]
Sn = __
2
n( + )
a l are both
and Sn = __
2
sum formulae for arithmetic
series.
a( 1 − r n )
is the formula
Sn = ________
1−r
for the sum of the first
n terms in a geometric series.
a
if −1 < r < 1
S∞ = _____
1−r
is the formula for the sum
to infinity of an infinite
geometric series.
If a sequence converges, the
ratio lies between −1 and 1.
The area of a triangle
1
= __
base × ⊥ height
2
The area of a circle = πr2
29
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REVISION TEST TOPIC 1 CONTINUED
9
Thabiso plans to participate in an ultracycling race in Kimberley. On the first day
of training she cycles 15 km. Each day she increases her distance by 3 km. She
has 90 days in which to train before the big event!
9.1
On which day will she cycle 120 km?
(3)
9.2
How many kilometres will she cycle in total in her third week
of training?
(5)
9.3
Is it likely that she will be able to keep up this daily rate of increase for the
duration of her training? Substantiate your answer.
(4)
10
Consider the sequence 32; 21; 12; 5; 0; … .
10.1 Write down the next term of the sequence.
10.2 Determine a formula for the nth term of this sequence.
11
a( 1 − r n )
11.1 Prove that: a + ar + ar 2 + … + ar n−1 = ________
for r ≠ 1.
1−r
27
___
+…
11.2 Given the geometric series 15 + 9 +
(5)
11.2.1 Explain why the series converges.
11.2.2 Express the series in sigma notation.
11.2.3 Determine S∞.
11.3 The sum of the first n terms of a sequence is given by Sn = 2(3n − 1)
11.3.1 Determine the sum of the first 12 terms.
11.3.2 Determine the 12th term.
11.3.3 Prove that the nth term of the sequence is 4 × 3n−1.
(3)
(2)
(3)
5
12
13
14
(2)
(4)
(3)
(3)
(4)
3 5 7
1 __
Consider the sequence: __
; ; __; ___
;…
2 4 8 16
12.1 Write down the next two terms of the sequence.
12.2 Determine the nth term of the sequence.
(2)
(4)
Consider the sequence 3; p; 19; q; 43; … which has a second difference of 2.
13.1 Determine the values of p and q.
13.2 Determine the nth term of the sequence.
(6)
(4)
A shrub grew in a greenhouse for one year and reached a height of 108 cm. The
shrub was planted in a garden and in the second year grew 54 cm. During each
1
consecutive year, the height increased by __
of the previous year’s increase.
3
14.1 Calculate the height of the shrub after 5 years.
14.2 Show that the shrub could never reach a height of 2 m.
(5)
(3)
15. The first term of an arithmetic sequence is 1. The first, second and fifth terms of
the arithmetic sequence form a geometric sequence. The common difference is
not zero.
15.1 Find the common difference of the arithmetic sequence.
(4)
15.2 Determine the first three terms of the geometric sequence.
(4)
16
Insert four geometric means between 7 and 224. (Insert the missing terms of the
geometric sequence.)
(5)
17
Insert 3 arithmetic means between 22 and −14. (Insert the missing terms of the
arithmetic sequence.)
(4)
30
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Formal assessment: Assignment
Total marks: 215
1
2
3
4x + 1; 2x + 4 and x + 3 are the first three terms of an arithmetic sequence.
1.1
Determine the value of x.
1.2
Determine a formula for the nth term.
1.3
Which term in the sequence has a value of −123?
1.4
Determine the sum of the first 20 terms.
1.5
Determine the sum of terms eleven to 20.
6x + 12; 2x + 4 and x − 7 are the first three terms of a geometric sequence.
2.1
Solve for x.
2.2
Is this a converging or diverging sequence? Justify your answer.
2.3
If the sequence is convergent, determine the sum to infinity. If it is
divergent, determine the sum of the first twenty terms.
Consider the series: 5 + 8 + 11 + 14 + 17 + … + 302
3.1
Determine a formula for the nth term in the series.
3.2
How many terms are there in the series?
3.3
Express the series in sigma notation.
3.4
Determine the sum of the terms in the series.
3.5
Determine the sum of the second 50 terms in the series.
Evaluate
∑
(5)
(5)
(3)
(3)
(2)
(2)
(3)
(4)
4
∞
4
(3)
(4)
(3)
(3)
(4)
3 × 2 1−r −
r = −3
∑ 3 × 21 r
−
(9)
r = −3
n
5
Find n if
∑( 12 − 5k ) = −2 256
(6)
k=1
6
In the arithmetic series: a + ( a + d ) + ( a + 2d ) + … + ( l − d ) + l, a is the first term,
n[
d the common difference and l the last term. Prove that Sn = __
a + l ].
(5)
2
7
Prove without using a formula, that the sum of the series
9 ( 20
32 + 34 + 36 + … to 20 terms is given by S20 = __
9 − 1 ).
8
(7)
8
The sum to infinity of a convergent series is 243. The sum of the first five terms
is 242. Determine the values of the common ratio and the first term.
(7)
9
A new sequence is formed by adding together the corresponding terms of a
geometric sequence and an arithmetic sequence. The geometric sequence has a
common ratio of 3 and the arithmetic sequence has a common difference of −2.
The first two terms of the new sequence are 4 and 20.
9.1
Calculate the third term of the new sequence.
(7)
9.2
Write down an expression for the nth term in:
9.2.1
the arithmetic sequence
(3)
9.2.2
the geometric sequence
(3)
9.2.3
the new sequence.
(2)
31
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Formal assessment: Assignment continued
10
3 2 __
5
If Sn = __
n − 2 n, determine:
2
10.1 the 11th term of the sequence
10.2 the nth term of the sequence.
10.3 How many terms have a sum of 1 106?
(5)
(5)
(7)
11
The sum of the first two terms of a convergent geometric sequence is 15 and the
sum of all its terms is 20.
11.1 Determine the ratio(s).
(9)
11.2 Determine the first three terms in the series.
(6)
12
Three positive numbers are in the ratio 1:3:8. If 7 is added to the largest number,
the three numbers form a geometric sequence. Find the numbers.
(5)
13
Evaluate:
3
13.1
∑( k + 1 )k
k=0
100
13.2
∑( 9 − 7k )
k=1
∞
13.3
∑5 × 2 1 n
−
n−1
(3 × 4)
14
A bouncy ball is dropped from a height of 25 m. On each bounce the ball reaches
80% of its previous height. Determine the total vertical distance travelled by the
ball from the time it is dropped until it finally comes to rest.
(6)
15
Consider the geometric sequence: 3 − 6x + 12x2 − 24x3 + …
15.1 Determine the values of x for which the sequence will converge.
15.2 Calculate the value of x for which S∞ = 99.
(5)
(5)
The nth term of a sequence is 2n2 − 1 if n is odd, but 20 − 4n if n is even.
16.1 Calculate the sum of the sixth and seventh terms.
16.2 Which term in the sequence is equal to 241?
16.3 Determine the sum of the first 6 terms in the sequence.
(5)
(4)
(7)
17
How many terms are there in an arithmetic sequence if it has a common
difference of 3, a last term of 30 and the sum of its terms is 135?
(7)
18
3 __
9 15 21
Consider the sequence: __
; ; ___; ___
;…
2 4 8 16
16
18.1 Write down the next two terms in the sequence.
18.2 Determine the nth term of the sequence.
(4)
(6)
32
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19
The large red circle has a diameter of 162 cm. The diameter of
8
the green circle is __
of the diameter of the red circle. The
9
8
diameter of the purple circle is __
of the diameter of the green
9
8
circle. If each new circle has a diameter which is __
of the
9
diameter of the previous circle, determine the sum of the areas
of all of the circles obtained in this way if the process continues
indefinitely.
(7)
20
The large red equilateral triangle has sides of 8x units. The
midpoints of the red triangle are joined to form the blue
triangle. The midpoints of the blue triangle are joined to form
the green triangle. The process of joining midpoints to form
new triangles continues indefinitely.
A
I
D
L
J
G
B
F
K
E
H
C
20.1 Determine the lengths of the sides of the first three triangles formed in
this way.
(3)
20.2 State the heights of these triangles.
(3)
20.3 What type of sequence is generated by
the heights?
(2)
20.4 Determine the areas of the red, blue and green triangles.
(3)
20.5 Determine the sum of the areas of all the triangles formed in this way. (4)
__
20.6 Solve for x if the sum of all the areas is given by 19 200√ 3 .
(4)
33
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TOPIC
2
2
Functions and inverse functions
Unit 1: Functions
KEY WORDS
function – a rule or equation
linking x and y so that
for every x value in an
equation there is only one
corresponding y value
domain – the set of x values
of a function
range – the set of y values of
a function
Definition of a function
A function is a relationship or rule between two sets, the domain (x) and the
range (y), where every element of the domain is assigned to one and only one element
of the range.
Functions can be one-to-one or many-to-one relations.
We use function or f(x) notation to represent the relationship between x and y.
y
The straight line: y = mx + c or f(x) = ax + p are examples of
one-to-one functions.
(1;5)
For example:
REMEMBER
You have worked with
functions and function
notation before.
If f(x) = y = 2x + 3
Then f(1) = 2(1) + 3 = 5 and f(−3) = 2(−3) + 3 = −3.
Notice that the vertical dotted line cuts the graph of f(x)
once only, indicating a one-to-one correspondence
between x and y.
x
(−3;−3)
y
The parabola: y = x2 or g(x) = ax2 + bx + c are examples of
many-to-one functions.
For example:
If g(x) = y = 2x2
(−1;2)
2
(1;2)
2
Then g(1) = 2(1) = 2 and g(−1) = 2(−1) = 2.
Notice that the vertical dotted line cuts the graph of g(x)
once only, indicating that g(x) is a function.
x
The horizontal line cuts the graph twice indicating a
many-to-one correspondence between x and y.
The graph of x = −4y2 is not a function because this is a
one-to-many relation.
y
On the graph, you can see that when x = −4, y = ±1.
The vertical line test shows that the graph is cut more
than once which means that the graph is not a function.
If the range of the graph is restricted so that y ≥ 0, the
graph will become a function.
(−4;1)
x
Also, if the range of the graph is restricted so that y ≤ 0,
the graph will become a function.
(−4;−1)
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Topic 2 Functions and inverse functions
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We can represent functions in tables or as sets.
KEY CONCEPT
A one to one relation A many to one relation A one to many relation
1
2
2
4
3
6
Domain x Range y
1
−1
2
−2
3
1
1
2
2
−3
3
Domain x Range y
The vertical line test:
A graph in the Cartesian plane
is a function if any vertical line
drawn cuts the graph once
only.
1
−1
2
−2
−3
Domain x Range y
WORKED EXAMPLES
Given f(x) = −x2 + 4.
1
Determine f(0), f (2) and f (−2) and use their values to draw the graph of f(x) .
2
Use the graph to decide the type of relation that
y
f(x) represents and whether it is a function or not.
3
Write down the domain and range of f(x).
(0;4)
SOLUTIONS
1
2
3
f(0) = 0 + 4 = 4, f(2) = −(2)2,+ 4 = −4 + 4 = 0 and
f(2) = −(−2)2+ 4 = −4 + 4 = 0
The vertical line test shows that f(x) is a function.
The horizontal line test shows that f(x) is a
many-to-one function.
Domain of f : x ∈ ℝ, range of f : y ∈ ℝ, y ≤ 4
(−2;0)
(2;0)
The horizontal line test:
A function is one-to-one if a
horizontal line intersects the
graph once.
A function is many-to-one if a
horizontal line intersects the
graph more than once.
• You use the horizontal
line test to determine if a
function has an inverse that
is also a function.
x
EXERCISE 1
1
The x and y values in the tables represent a relation between x and y. State
whether each relation is:
one-to-one, one-to-many, many-to-one or many-to-many.
Use your answers to say whether each relation in 1.1−1.5 is a function or not.
1.1
1.4
x
y
1
1.2
x
y
3
−2
2
6
3
1.3
x
y
5
2
1
−1
2
2
−1
9
0
1
3
−4
4
12
1
2
3
4
5
15
2
5
4
5
x
y
x
y
4
1
0
1
8
2
1
3
12
3
2
5
16
4
3
3
20
5
4
1
1.5
Unit 1 Functions
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2
Apply the vertical line test to decide whether or not the graphs below
are functions.
2.1
2.2
y
y
x
2.3
x
2.4
y
y
x
3
State how the range of each graph should be restricted so that is becomes a
function.
3.1
3.1
y
5
6
3.1
y
(0;2)
x
(0;2)
4
x
3.1
3.2
3.2y y
(0;2)
x
3.2
y
xx
(2;−1)
3.2
3.3
y y
3.3
3.3y
3.3
(0;2)
(0;2)
x
x x
(2;−1)
(2;−1)
y
(0;2)
x
x
Given f(x) = −3x + 6.
4.1
Calculate f (−1), f (0) and f (2).
4.2
Use the answers in 4.1 to write down three elements of the domain of f
and three elements of the range of f.
4.3
Draw the graph of f(x) for the domain x ∈ ℝ.
4.4
Use the graph to explain why f(x) is a function.
4.5
What kind of function is f(x)?
Given g(x) = −3x2 + 3
5.1
Calculate g(−1), g(0) and g(1).
5.2
Use the answers to 5.1 to write down three elements of the domain of g
and two elements of the range of g.
5.3
Draw the graph of g(x) for the domain x ∈ ℝ.
5.4
Use the graph to explain why g(x) is a function.
5.5
What kind of function is g(x)?
__
Given the relation: y = ± √ x + 4
Copy and complete the table and use it to explain why this relation is not a
function.
x
0
1
4
9
16
25
y
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Topic 2 Functions and inverse functions
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Unit 2: Inverse functions
The inverse of a function
• You obtain the inverse of a function by swapping the x and y values.
• The domain becomes the range and the range becomes the domain.
• A one-to-one function becomes a one-to-one inverse function, but a many-to-one
function becomes a one-to-many inverse relation which is not a function.
• Only one-to-one functions can therefore have inverses which are functions.
• Many-to-one functions require their domains to be restricted for their inverses to
be functions.
• By swapping the x and y values we get (x;y) → (y;x).
This means that the graph is reflected about the line y = x. A graph and its inverse are
therefore always symmetrical about the line y = x.
• You use f −1 to represent the inverse of f(x).
• If the inverse is a function, use the notation f −1(x).
• Be careful not to mistake the −1 in f −1 for an exponent:
1
• f −1(x) does NOT mean the reciprocal ___
.
f(x)
How to find the inverse of a function
1.
2.
3.
4.
Replace f(x) with y.
Swap x and y.
Solve for y in terms of x.
Replace y with f −1(x) if the inverse is a function.
KEY CONCEPT
The horizontal line test
• A function is one-to-one if
a horizontal line intersects
the graphs once.
• A function is many-toone if a horizontal line
intersects the graph more
than once.
• You use the horizontal
line test to determine if a
function has an inverse that
is also a function.
How to plot the inverse function
1.
2.
3.
Sketch the graph of f using key features (intercepts, TPs and asymptotes).
Draw in the line y = x (the line of reflection or symmetry).
Reflect key points on the graph of f about the line y = x to sketch the graph
of ( f −1 ).
WORKED EXAMPLE 1
Given the straight line: y = ax + q
If f(x) = y = 2x + 2, find the inverse f −1.
y f(x)
REMEMBER
• Notation: f (x) for function:
f (x) = …
• Inverse function:
f −1(x) = …
• Interchange x and y and
make y the subject of the
formula.
• f and f −1 are reflections
about the line y = x.
y=x
SOLUTION
Plot f and f −1 on the same set of axes and
show the line of reflection.
x = 2y + 2
2y = x − 2
x−2
∴ y = _____
(0;2)
(−1;0)
(0;−1)
(2;0)
f ˉ1(x)
x
2
1
∴ f −1(x) = __
x
−1
2
f (x) and f −1(x) are both functions.
Unit 2 Inverse functions
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WORKED EXAMPLE 2
Given the parabola: y = ax2
KEY CONCEPT
For f −1(x) to be a function
f(x) has to be a one-to-one
function.
The vertical line test shows
whether or not the graph is a
function.
The horizontal line test on the
function shows whether or
not the inverse is a function.
If f(x) = 2x2 find the inverse f −1.
SOLUTION
y
Plot f and f −1 and the line of reflection
on the same set of axes.
f : y = 2x 2
f(x)
(1;2)
(−1;2)
inverse
To find the inverse: f −1 : x = 2y2
x
y2 = __
2
y=x
(2;1)
x
__
x
f −1 : y = ± √__
2
(2;−1)
Note:
In Figure 1 you can see that the inverse of f(x) is not
a function because the vertical line test shows that
the graph is cut twice; this means that there are two
y values to one x value. The many-to-one function
became a one-to-many relation.
y
inverse
x
REMEMBER
If f (x) = 2x2 and x ≥ 0, the
__
x
.
inverse f −1(x) = y = ± √__
2
However the reciprocal of
1
2x2 is ___
2.
2x
Do not confuse the inverse
function f −1(x) with the
1
.
reciprocal ___
f(x)
Figure 1
In Figure 2 the domain of f(x) is restricted to x ≥ 0 making
f(x) a one-one function and therefore its inverse is also
a function.
y
f(x); x ≥ 0
(1;2)
f 'x
(2;1)
x
Figure 2
In Figure 3 the domain of f(x) is restricted to x ≤ 0
making f(x) a one-to-one function and therefore its
inverse is also a function.
If a horizontal line cuts a function more than once,
the function is a many-to-one function and therefore
its inverse will not be a function unless the domain
is restricted.
f(x); x ≤ 0
y
(−1;2)
x
(2;−1)
f 'x
Figure 3
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Topic 2 Functions and inverse functions
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EXERCISE 2
1
For each function f(x) drawn below:
a)
Copy the graph of the given function and use it to sketch the graph of the
inverse on the same set of axes.
b)
Draw in the line y = x, the line about which f and f −1 are reflections.
c)
Write the equation of the inverse in the form y = …
−1
d)
Write
.
1.1 down the domain and range of f and f 1.2
y
1.1
1.2
y = 3x −3
y = –2x + 6
y
(0;6)
(1;0)
(3;0)
x
(0;−3)
f
1.3
f
1.4
y
1.3
1.4
f
x
(−1;−3)
(−2;5)
(1;−3)
(0;1)
y
y = x² + 1
(2;5)
x
y = –3x²
f
2
x
The graphs of f(x) = −x2 and g(x) = −2x − 3 are sketched below. A and B are the
points of intersection of f and g.
y
g
x
A
f
B
2.1
2.2
2.3
2.4
2.5
Determine the coordinates of A and B.
Determine the equation of g −1 in the form y = ...
Determine the equation of f −1 in the form y = ...
Give a reason why f −1 is not a function and place a restriction on the
domain of f so that f −1 is a function.
1
2
___
Show that: g(x) + 2f __
x + 3 = f(x) − g(−x) − 3
( )
Unit 2 Inverse functions
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3
4
5
6
1
1 2
Given: f(x) = − __
x and g(x) = − __
x
3
3
3.1
Sketch the graphs of f(x) and g(x) and clearly show the points (3;f(3)) and
(3;g(3)) on the graphs.
3.2
Write down the equations of f −1(x) and g −1(x) in the form y = …
3.3
About which line are f(x) and f −1(x) symmetrical?
3.4
Draw the graph of the of f −1(x) on the same set of axes as f(x).
3.5
Explain why f −1(x)is not a function.
3.6
Indicate one way in which the domain of f(x) can be restricted for f −1(x) to
be a function.
1 2
Given: f(x) = __
x and g(x) = 2x + 4
2
4.1
Is f(x) a function? Give a reason for your answer.
4.2
Write down the domain of f (x) for f −1(x) to be a function.
4.3
Write down the equation of g−1(x) in the form y = …
4.4
Write down the range of f (x).
The graph of f(x) = ax2, x ≤ 0 is sketched alongside.
y
f
The point A(−2;5) lies on f.
A(−2;5)
5.1
Determine the value of a.
−1
5.2
Determine the equation of f .
5.3
Write down the domain of f and the range
x
of f −1.
5.4
Copy the graph of f onto a set of axes and
draw the graph f −1 on the same set of axes.
Show the coordinates of two points on the
graph of f −1.
5.5
If the graph of f is reflected in the y-axis and
then reflected in the line y = x, write the equation of the new function in
the form y = …
f(x) is the inverse of g(x) if f ( g(x)) = g( f (x)) = x
Use this statement to test whether f(x) and g(x) are inverses.
6.1
f( x ) = 5x + 6
g(x) = _____
5
x−6
6.2
f(x) = 2x2 − 1
g(x) = ±
6.3
____
x+1
√_____
2
f(x) = _____
x
x+1
1
g(x) = _____
x−1
6.4
f(x) = 3x3 + 1
g(x) =
40
_____
(x − 1)
√______
3
3
Topic 2 Functions and inverse functions
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You will cover these graphs in
greater detail in Topic 3. This
is included for enrichment
purposes only and is not
examinable.
WORKED EXAMPLE 1
This is a parabola of the type: y = a(x − p)2 + q
1
Find the inverse of f(x) = (x − 1)2 − 4.
2
Write down the turning point of f(x) and of f −1(x).
3
Sketch the graphs f and f −1 on the same axes.
4
Decide if the inverse is a function.
If not, state the restrictions on
the domain of f(x) so that the inverse
graph
is a function.
5
Sketch the graph of f −1(x) using
the domain from question 4.
(−4;1)
Then write down the domain and
1
−
range of f (x).
y
x
(−3;0)
SOLUTIONS
1
2
3
4
5
f(x) = (x − 1)2 − 4.
For f −1: x = (y − 1)2 − 4
_____
(y − 1)2 = x + 4 ∴ y = 1 ± √x + 4
The turning point of f(x) is (1;−4)
and of f −1(x) is (−4;1).
See graphs
For f −1 to be a function, the domain of
f(x) must be x ≥ 1 or x ≤ 1.
If the domain of f(x) is restricted to x ≥ 1,
then Figure 1 represents f −1(x).
(0;−3)
(1;−4)
y
y
(−4;1)
(0;3)
x
(0;−1)
(−4;1)
x
The domain of f −1(x) is x ≥ −4; x ∈ ℝ and the
range of f −1(x) is y ≥ 1; y ∈ ℝ.
If the domain of f(x) is restricted to
x ≤ 1, then Figure 2 represents f −1(x).
The domain of f −1(x) is x ≥ −4; x ∈yℝ and the
Figure 1
Figure 2
y
range of f −1(x) is y ≤ 1; y ∈ ℝ.
(−4;1)
(0;3)
x
(0;−1)
(−4;1)
x
Figure 1
Figure 2
Unit 2 Inverse functions
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WORKED EXAMPLE 2
KEY WORDS
a
A hyperbola of the type: y = _____
x−p +q
asymptote – a line which
the function approaches but
never touches
1
1
− 4. Show all asymptotes and intercepts with
Draw the graph of f(x) = ______
(x − 1)
the axes.
2
3
4
5
Use the graph of f to draw the graph of its inverse f −1(x).
Give a reason why f −1(x) is a function.
Write down the domain of f and f −1.
Write the equation of f −1(x) in the form y = …
SOLUTIONS
1
Asymptotes: x = 1 and y = − 4
1
−4=
y-intercept: x = 0 ∴ y = _____
4
3
4
5
y=x
f
5
1
x-intercept: y = 0 ∴ 4 = _____
x−1
Asymptotes: y = 1 and x = −4
5
y-intercept: x = 0 and y = __
x=1
10
0−1
−5
2
y
x = −4
f²
−10
−5
x-intercept: y = 0 and x = −5
f −1(x) is a function because it is a
−5
one-to-one relation. The vertical
f −2
line test shows that the graph is cut
−10
once; this confirms that for every
x value there is only one y value.
Domain of f: x ∈ ℝ, x ≠ 1. Domain of f −1: x ∈ ℝ, x ≠ −4.
1
− 4 and
f(x): y = ______
y=1
5
−;0
4
5
10
x
y = −4
(x − 1)
1
−4
f (x): x = ______
(y − 1)
1
______
∴ (x + 4) =
(y − 1)
1
∴ (y − 1) = ______
(x + 4)
1
∴ f −1(x) = y = ______ + 1
(x + 4)
−1
Questions marked with an
asterisk are for enrichment
purposes only and are not
examinable.
EXERCISE 3
For each given function f(x):
a) Draw the graphs of f(x) and its inverse on the same set of axes. Clearly show all
the intercepts with the axes, asymptotes and lines of symmetry.
b) Write the equation of the inverse of f(x) in the form y = …
1
1
f(x) = ______
+3
(x + 2)
42
2*
f(x) = (x − 3)2 − 1
3
−2
+2
f(x) = ______
(x − 4)
4*
f(x) = −(x + 4)2 + 1
Topic 2 Functions and inverse functions
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Revision Test Topic 2
Total marks: 118
1
Say whether the statements that follow are true or false. Give a reason for
your answer.
1.1
The inverse of f = {(2;3), (4;7)} is equal to {(3;2), (7;4)}.
1.2
f = {(2;3), (4;7), (4;−7), (2;−3)} is a many-to-one function.
1.3
The inverse of the graph in A is a function.
1.4
The graphs in B are reflections in the x-axis, which means they are
also inverses.
1.5
The graph in C is a one-to-many relation.
1.6
1.6.1
If the domain of the graph in D is restricted to y ≥ 1,
then the graph is a function.
1.6.2
The minimum value of the inverse graph in D is (1;2).
A
y
(2)
(2)
Questions marked with an
asterisk are for enrichment
and are not examinable.
(2)
(2)
(2)
y
B
f(x)
x
x
–f(x)
y
C
y
D
(2;1)
x
x
2
For each equation below:
a)
sketch the graphs of f and f −1 on the same set of axes
b)
write the equation of f −1 in the form y = …
c)
state the range of f and the domain of f −1.
1
2.1
y = −2x + 4
2.2
y = __
2x − 2
2.3
9 2
y = __
4x
2.5*
4
y = __
x+4
2.4*
(4)
(1)
(2)
y = 2x 2 − 8
(7 × 5)
43
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REVISION TEST TOPIC 2 CONTINUED
3
Match the functions below with their inverse functions. There is one that does
not match.
3.1
(2)
A
y
y
B
6
6
6
4
4
4
2
2
2
−6
−4
−2
2
4
6
x
−6
−2
2
4
6
x
−6
−4
−2
2
−2
−4
−4
−4
−6
−6
−6
y
−4
−4
−2
D
−6
y
C
6
6
4
4
2
2
4
2
x
6
x
−2
−6
−4
2
−2
−2
−2
−4
−4
−6
−6
4
6
x
3.2
A
6
y
E
−2
4
(2)
y
B
(1;2)
C
y
y
(−2;1)
x
x
x
(−2;−1)
D
E
y
y
(2;1)
x
x
(−1;–2)
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REVISION TEST TOPIC 2 CONTINUED
4
The following inverse functions are given. Find the equation of each function.
1
4.1
q −1(x) = __
(2)
2 x__+ 3
−1
√
4.2
p (x) = x + 1
(2)
_____
4.3
r −1(x) = − √x − 2 − 3
(2)
5
The graphs of f(x) = 2x2 and g(x) = x + 3 are sketched alongside. A and B are the
points of intersection of f and g.
y
f
g
B
A
x
5.1
5.2
5.3
5.4
5.5
5.6
5.7
Determine the coordinates of A and B.
(6)
1
1
−
−
Write down the coordinates of two points of intersection of f and g . (2)
Determine the equation of g −1 in the form y = ...
(2)
Determine the equation of f −1 in the form y = ...
Copy the graphs of f and g and draw the graphs of f −1 and g −1 on the same
set of axes. Clearly label the coordinates of the points of intersection.
(4)
Use your graphs to answer the questions. For which values of x:
5.6.1
are f and g −1 both increasing
(2)
5.6.2
is f(x).g(x) < 0
(2)
5.6.3
must the domain of f be restricted so that f −1 is a function?
(2)
Find the average gradient of f(x) between:
5.7.1
the origin and point B
(2)
5.7.2
the origin and point A.
(2)
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REVISION TEST TOPIC 2 CONTINUED
6
1
The graphs of f(x) = __
x and g(x) = 5x − 4 are
sketched alongside. A and B are the points of intersection of f and g.
y
g
A
f
x
B
6.1
6.2
6.3*
6.4
6.5*
6.6
Determine the coordinates of A and B.
(6)
1
1
−
−
Write down the coordinates of two points of intersection of f and g . (2)
Determine the equation of f −1 in the form y = ... What to you notice
about f and f −1?
(3)
Determine the equation of g −1 in the form y = ...
(2)
Copy the graphs of f and g alongside and draw the graph of g −1 on the
same set of axes. Clearly label the coordinates of the points of intersection. (2)
Use your graphs to answer the following questions. For which values
of x:
6.6.1* f (x) ≥ g −1(x)
(2)
6.6.2
f (x).g(x) ≤ 0?
(2)
46
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REVISION TEST TOPIC 2 CONTINUED
7*
The figure shows the curve of f(x) = (x − 2)2 + 3.
Point P(4;7) lies on the curve of f.
y
P(4;7)
x
The graphs below represent f (−x), f −1(x) and −f(x).
A
B
y
x
P(4;−7)
7.1
7.2
8*
C
y
y
P(7;4)
P(−4;7)
x
x
Match the graph with its function and write down the letter
corresponding to each function.
(3)
State which graph (A, B or C) is not a function and how the range should
be restricted so that it becomes a function.
(2)
Write down the inverses of these functions in the form y = …
x+4
8.1
y = _____
x
8.2
y = (x − 1)2 + 1
8.3
y = (x + 1)3
(3)
(3)
(3)
47
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TOPIC
2
3
Exponential and logarithmic functions
Unit 1: Revision of exponential laws and
functions
Definition of an exponential function
y = a x where a > 0, a ≠ 1 and x ∈ ℝ.
You can sketch the exponential function by using a table or remembering the key
features of the graph.
The graphs of y = 2x and y = 2−x = ( _1 )x are sketched below using a table of values for
2
x and y.
x
y = 2x
−4
1
___
−3
1
__
−2
1
__
−1
1
__
0
1
2
3
4
0
2
4
8
16
16
8
4
2
0
1
__
1
__
1
__
1
___
16
(2)
1
y = 2−x = __
x
8
4
2
2
4
8
16
Key features of exponential functions
All graphs of the form y = ax, a > 0, a ≠ 1 have these features:
• The y intercept is (0;1).
• The x-axis is a horizontal asymptote.
• The domain is x ∈ ℝ and the range is y ∈ ℝ, y > 0.
• If a > 1, the graph increases and if 0 < a < 1, the graph decreases.
• The graph is a one-to-one function.
y = –1 x
2
y = 2x
y
(3;8)
(−3;8)
(−2;4)
1
—
− 4;16
(2;4)
(0;1)
1
4; —
16
x
When you compare y = a x and y = a −x you can see that the two graphs are reflections
in the y-axis.
The graph of y = ax + p has a horizontal asymptote y = p.
The focus of this unit is on y = a x, a > 0, a ≠ 1.
48
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Revision of exponential equations
You solve exponential equations using the property:
If ax = a y then x = y, a > 0, a ≠ 1.
This works only if you can express
both sides of the equation as
powers of the same base.
y
WORKED EXAMPLES
1
1
Solve for x if 27.(3x+1) = ___
x
2
The graph of f(x) = 4x+1 is shown alongside.
2.1
Find the coordinates of P,
the y-intercept of f.
2.2
Find the value of a if (a;8) lies on f.
2.3
Write down the equation g,
the reflection of f in the x-axis
and the coordinates of Q.
f
(a;8)
27
REMEMBER
When working with
exponents and logarithms,
remember that:
20 = 1
P
1
2 1 = 2; 2 −1 = __
2
x
1
2 2 = 4; 2 −2 = __
4
1
2 3 = 8; 2 −3 = __
8
1
2 4 = 16; 2 −4 = ___
16
1
2 5 = 32; 2 −5 = ___
32
Q
3
The graphs of f and g are drawn in the figure alongside.
g is the reflection of f in the y-axis.
1
f(x) = a x, (2;9) and p;___
are points on f.
81
g
3.1
Determine the values of a and p.
3.2
Determine the equation of g in the form
y=…
3.3
Determine the equation of h,
the reflection of f in the x-axis.
(
)
1
p; -
SOLUTIONS
1
81
1
2 6 = 64; 2 −6 = ___
64
g
y
f
REMEMBER
(2;9)
(0;1)
1
27.(3x+1) = ___
x
27
x
| Bring to the same base.
33.3x+1 = (3−3)x
33+x+1 = 3−3x
| Use the laws of exponents.
4 + x = −3x
| Drop the bases.
4x = −4
x = −1
2.1 y-intercept: x = 0 ∴ y = 40+1 = 4 ∴ P(0;4)
2.2 Substitute (a; 8) into y = 4x+1
8 = 4a+1
| Solve for a by writing each side to the same base.
+1)
3
2(x
2 =2
3 = 2x + 2
1
x = __
Definition and laws of
exponents
Definition:
a x = a × a × a × a × a ... to x
factors
Laws:
a x × a y = a x+y
a x ÷ a y = a x−y
(a x) y = a xy
(a × b) x = a x × b x
1
a −x = __
__
ax
x
__
√a x = a n
n
a 0 = 1; a ≠ 0
2
2.3 g: −y = 4x+1
| For a reflection in the x-axis, (x;y) → (x;−y).
∴ g(x):y = −4x+1 and Q = (0;−4)
3.1 Substitute to (2;9) into y = ax: 9 = a2 ∴ a = 3
(
)
1
x
Substitute p;___
81 into y = 3 :
1 ∴ 3p = 3−4 ∴ p = −4
3p = ___
81
( )
1 x
3.2 g(x): y = 3−x ∴ y = __
3
3.3 h(x): y = −3x
Unit 1 Revision of exponential laws and functions
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EXERCISE 1
1
Solve for x in the exponential equations:
1.1
2x(2x−2) = 64
1.2
5x × 252x = 125
1
1.3
3x−1 × 27 = ___
81
2
1.4
12.(2x−2) = 192
In the graphs below, f(x) = b x. For each pair of graphs answer these questions:
a)
Write down the equations of f and g where f and g are reflections in
the x or y-axis.
b)
Write down the domain and range of f and g.
c)
Find the value of a, the x or y-coordinate of the point P on the graph
of f or g.
y
2.1
2.2
f
y
g
f
(2;4)
P(a;25)
–3;
1
125
P (−4;a)
(0;1)
x
(0;1)
x
g
2.3
f
y
−2;−9
4
(0;1)
Exponential functions and
their inverses
You will deal with exponential
functions and their inverses in
Unit 2.
x
P(−2;a)
g
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Topic 3 Exponential and logarithmic functions
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Unit 2: Logarithms and logarithmic functions
Definition of a logarithm
Special logarithm values
loga 1 = 0 because a0 = 1
loga a = 1 because a1 = a
y
loga x = y if and only if x = a where x > 0 and a > 0, a ≠ 1
This means that a logarithm (or a log) is closely related to an exponent:
log2 16 = 4 because 24 = 16
1 = −3 because 5−3 = ____
1
log ____
5 125
125
2
log 100 = 2 because 10 = 100
| Note: 4 = log = exponent and 2 = base
| Note: −3 = log = exponent and 5 = base
| Note: When the base is 10, you do not write it down.
Logarithmic laws
1
For example: log6 2 + log6 18 = log6 36 = 2
2
logc AB = logc A + logc B
A = log A − log B
log __
3
logc Ap = p log c A
For example: log3 81 = log3 34 = 4 log 3 3 = 4
4
c
logb A = ______
log B
3
For example: log4 8 = _____
= ______
= ______
= __
log 4
log 22
2 log 2
c B
c
c
log A
c
For example: log6 12 − log6 2 = log6 6 = 1
log 8
log 23
3 log 2
2
WORKED EXAMPLES
Use the definition of a logarithm to convert logarithms to exponential form and
exponents to logarithmic form.
1
Write these equations in exponential form and solve for x.
1.1
logx 81 = 2
1.2
log5 x = −2
1.3
log2(x2 + 2x) = 3
2
Write the equations in logarithmic form and solve for x, correct to 2 decimal
places where necessary.
1.1
5x = 100
1.2
25 = 10(1,5)x
1.3
2 × 3x−3 = 12
SOLUTIONS
1.1 logx 81 = 2
x2 = 81
x=±9
but x > 0 ∴ x = 9
1.2 log5 x = −2
x = 5 −2
1
x = ___
1.3 log2(x2 + 2x) = 3
x2 + 2x = 23
x2 + 2x − 8 = 0
(x + 4)(x − 2) = 0
x ≠ −4 , x = 2
2.1 5x = 100
x = log5 100
| Use a calculator.
x = 2,86
2.2 25 = 10(1,5)x
1,5x = 2,5
log1,5 2,5 = x
x = 2,26
2.3 2 × 3x−3 = 12
3x−3 = 6
log3 6 = x − 3
x − 3 = 1,63
x = 4,63
25
Note: You will not
be examined on
manipulations involving
the logarithmic laws,
but is useful for
understanding logs
better.
Calculator tips
Some calculators have the
feature logB A (the button
next to x −1).
If your calculator does not
have this feature, remember
log A
that logBA = _____
.
log B
This means that
log 8
log3 8 = _____
= 1,89.
log 3
Note that log x = log10 x.
When the base is 10 you do
not write it down.
(This is similar to not writing
__
__
2
x or
the 2 for √ x = √
the 1 for x = 1x.)
Unit 2 Logarithms and logarithmic functions
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EXERCISE 2
1
Write these logs in exponential form:
2
1.1 log2 32 = 5
1.2 log 1 000 = 3
1.4 log5 125 = 3
1.5 log __1 64 = −6
1 = −4
1.3 log3 ___
81
2
Write in these exponents in logarithmic form:
1
2.1 24 = 16
2.2 3−3 = ___
2.3 a2 = 5
27
2.4 3x = 8
2.5 2y = x
2.6 c 3 = a
Solve for x by applying the logarithmic definition. You do not require a
calculator.
1 =x
3.1 log 25 = 2
3.2 log x = −2
3.3 log ____
3
5 125
x
3.4 logx 16 = 4
3.5 log2 128 = x
Solve for x by applying the logarithmic definition. Give your answer to 2 decimal
places where necessary. You may use a calculator.
4
4.1 2x = 5
4.2 3x + 1 = 25
4.3 15 = 3(1 + 0,2)x
4.4 210 = 520(1 − 0,8)x
4.5 14 × 2x−2 = 50
4.6 log3 x = 0
4.7 logx 64 = 3
4.8 2 log2 x = 4
4.9 log2(x2 − x ) = 1
4.10 log2 8 = x2 − 2x
Exponential functions and their inverses
In Topic 2 you learnt that to find the inverse of a function, you swap x and y and
rewrite y in terms of x.
For all exponential functions:
If f(x) = y = ax, a > 0, a ≠ 1 then for f −1: x = ay and f −1(x) = y = loga x
1 x, 0 < a < 1 then for f −1: x = __
1 y and f −1(x) = y = log x
If f(x) = y = __
1
__
(a)
(a)
a
WORKED EXAMPLES
1
2
Draw the graph of f (x) = 2x and its inverse f −1(x) on the same set of axes.
Write down the inverse of f (x) in the form y = …
SOLUTIONS
1
The table shows the values of x and y for y = 2x.
x
y = 2x
−4
1
___
16
−3
1
__
8
−2
1
__
4
−1
1
__
2
0
1
2
3
4
0
2
4
8
16
The table shows the swap of x and y values into x = 2y.
y
x = 2y
−4
1
___
16
−3
1
__
8
−2
1
__
4
−1
1
__
2
0
1
2
3
4
0
2
4
8
16
The graphs are drawn on the next page.
52
Topic 3 Exponential and logarithmic functions
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2
To find the equation of the inverse of y = 2x:
Swap x and y and use the definition of a
logarithm to write x = 2 y in y form.
∴ y = log2 x
f
y
y=x
(3;8)
(8;3)
1
−4; —
(0;1)
f −1
16
x
(1;0)
1
—
16 ;−4
WORKED EXAMPLES
1
2
3
Given f(x) = 3x. Write down the equations of these graphs in the form y = …
1.1
g, the reflection of f in the y-axis
1.2
h, the reflection of f in the x-axis
1.3
f −1, the inverse of f
1.4
g −1, the inverse of g.
Draw the graphs of f and f −1 on the same set of axes, showing all intercepts
with the axes.
Use the graphs to determine the values of x for which f(x). f −1(x) ≤ 0.
( )
(3)
1 y
∴ g 1: x = ( __
3)
−
Note:
If you use the form y = 3−x for
g then g −1 becomes
x = 3−y and −y = log3 x or
y = −log3 x.
Using the logarithm laws, you
can show that
−log3 x = log__1 x.
3
However, it is better to write:
1 x
1 x
3 −x = __
or a −x = __
(3)
(a)
when working with
exponents, logarithms and
inverses.
SOLUTIONS
1 x
1.1 g: y = 3−x = __
3
1.2 h: y = −3x
1.3 f : y = 3x.
f −1: x = 3y
∴ y = log3 x
1 x
1.4 g: y = 3−x = __
Note:
The graphs are reflections in
the line y = x.
The range of y = 2 x is
y ∈ ℝ, y > 0.
The domain of x = 2 y is
x ∈ ℝ, x > 0.
Both graphs are one-to-one
functions.
Both graphs are increasing
functions
| Use the log definition.
| Use the log definition.
∴ y = log__1 x
3
2
3
The graphs are shown alongside.
f(x).f −1(x) ≤ 0 when 0 < x ≤ 1
Notice that 0 < x ≤ 1 when one graph
is positive and the other is negative.
y
y=3
x
y=x
(2;9)
(9;2) y = log x
(0;1)
3
(1;0)
x
Unit 2 Logarithms and logarithmic functions
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EXERCISE 3
REMEMBER
Increasing function:
As x increases, y also
increases.
1
For each graph in questions 1.1−1.4:
a)
Draw the graph of f (x) using a table or the key features of the exponential
function.
Indicate the coordinates of three points on the graphs of f and
label f clearly.
b)
Use the graph of f (x) to draw f −1(x), the inverse of f (x).
Indicate the coordinates of three points on the graphs of f −1(x) and label
f −1 clearly.
c)
Draw in the line of symmetry about which f (x) and f −1(x) are symmetrical.
d)
Write the equation of f −1(x) in the form y = …
e)
Write down the range of f (x) and the domain of f −1(x) .
f)
State whether f(x) and f −1(x) are increasing or decreasing functions.
Decreasing function:
As x increases, y decreases.
Reflections:
Reflection in the y-axis:
(x;y) → (−x;y) and
f(x) → f (−x)
Reflection in the x-axis:
(x;y) → (x;−y) and f (x) → −f (x)
Reflection in the line y = x:
(x;y) → (y;x) and
f (x) → f −1(x)
f(x) = 4x
1.2
1.3
f(x) = 5x
1.4
(4)
1 x
f(x) = ( __
2)
1
f(x) = __
x
Repeat questions 1a−f for exponential graphs of the form y = ax + p.
Remember the horizontal asymptote for these graphs is y = p.
1 x+3
2.1
f(x) = 2x + 1
2.2
f(x) = 3x − 2
2.3
f(x) = __
2.
Note: You would regard
these questions with the
vertical shift as
non-routine.
1.1
(3)
Logarithmic functions
We define the logarithmic function as y = loga x, a > 0, a ≠ 1, x ∈ ℝ.
In Unit 1 you were introduced to the graph of y = log2 x or x = 2 y
as the inverse of y = 2x.
Key features of logarithmic functions
All graphs of the form y = loga x, a > 0, a ≠ 1,x ∈ ℝ have these features:
• The x intercept is (1;0).
• The y-axis is a vertical asymptote.
• The domain is x ∈ ℝ, x > 0 and the range is y ∈ ℝ.
• If a > 1, the graph increases and if 0 < a < 1, the graph decreases.
When you compare y = loga x and y = log __1 x you can see that the two graphs are
a
reflections in the x-axis.
WORKED EXAMPLE
( )
1 y on
Use a table to plot the graphs of y = log2 x or x = 2 y and y = log __1 x or x = __
2
2
the same set of axes. The graphs are drawn on the next page.
SOLUTION
When using the table method to plot logarithmic graphs, it is easier to rewrite
the logarithmic equation in exponential form and choose the y values first.
y
x = 2y
54
−3
1
__
−2
1
__
−1
1
__
0
1
2
3
4
0
2
4
8
16
16
8
4
2
0
1
__
1
__
1
__
1
___
16
(2)
1
x = __
−4
1
___
y
8
4
2
2
4
8
16
Topic 3 Exponential and logarithmic functions
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y
1
–
;4
16
(8;3)
x
(1;0)
1
–
;– 4
16
y = log 2 x
(8;−3)
1
y = log −x
2
EXERCISE 4
1
Draw the graphs of f(x) and g(x) on the same set of axes. Show clearly the
intercepts with the axes and the coordinates of two points on each curve.
Write down the equation of the line about which f and g are symmetrical, if
there is one.
1.1
f(x) = log4 x and g(x) = log__1 x
1.2
f(x) = log3 x and g(x) = 3x
4
x
x
1.3
f(x) = 3 and g(x) = −3
1.4
f(x) = log2 x and g(x) = −log2 x
x
1.5
f(x) = 5 and g(x) = log __1 x
1.6
5
f(x) = log __1 x and g(x) = log__1 (−x), x < 0
3
3
Logarithmic graphs and their inverses
If the inverse of the exponential function is the logarithmic function, then the inverse
of the logarithmic function is the exponential function.
WORKED EXAMPLES
1
2
If f(x) = y = log3 x, find f −1(x) in the form y = …
If f(x) = y = log __1 x, find f −1(x) in the form y = …
4
SOLUTIONS
1
f : y = log3 x
2
−1
f : x = log3 y | Swap x and y.
∴ y = 3x
| Use log definition to
write in y form.
f : y = log __1 x
4
f −1: x = log __1 y | Swap x and y.
(4)
1 x
∴ y = __
4
| Use log definition to
write in y form.
Unit 2 Logarithms and logarithmic functions
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For all logarithmic functions:
If f (x) = y = loga x, a > 0, a ≠ 1
then f −1(x) = y = ax
y y = ax
y=x
If f (x) = y = log __1 x, a > 1
a
1 x
then f −1(x) = y = ( __
)
1 x
y= —
a
a
y=x
y
(y;x)
(y;x)
y = log a x
(0;1)
(x;y)
(1;0)
x
(0;1)
x
(1;0)
(x;y)
y = log−1 x
a
EXERCISE 5
1
For each of the functions f(x) given in questions 1.1−1.4:
a)
Write down the equation of f −1, the inverse of f in the form y = …
b)
Draw a sketch graph of f, clearly showing two points on the curve of f.
c)
On the same system of axes as 1b, draw a sketch graph of f −1 clearly
showing two points on the curve of f −1.
d)
If g is the reflection of f in the x-axis write the equation of g in the form
y=…
e)
If h is the reflection of f in the y-axis, write the equation of h in the form
y=…
1 x
1.1
f(x) = __
1.2
f(x) = 6x
(5)
2
56
1.3
f(x) = log3 x
1.4
f(x) = log__2 x
3
In questions 2.1−2.4, the function is f(x) = loga x and a point on the graph
is given. Use this information to:
a)
Find the value of a.
b)
Give the equation of f −1 in the form y = …
c)
Draw the graphs of f(x) and f −1 on the same system of axes, clearly
showing all the intercepts with the axes.
d)
Give the equation of the function g(x) if g(x) is symmetrical to f(x) with
respect to the x-axis.
e)
State the domain of g(x).
The point on each graph is:
2.1
(9;−2)
2.2
(8;3)
2.3
(5;1)
2.4
(64;−2)
Topic 3 Exponential and logarithmic functions
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Logarithmic graphs and inequalities
Logarithmic inequalities obey the logarithmic laws, but you can solve inequalities
using graphs.
WORKED EXAMPLE
y
y = log 1_ x
Use the graph alongside to answer the
following:
For which values of x is:
1
log2 x ≤ 3
2
log__1 x ≤ −3
2
S
Q
y =3
2
3
log2 x ≥ −3
4
log__1 x ≥ 3
x
P
2
T
R
y = –3
SOLUTIONS
At Q: log2 x = 3
∴ x = 23 = 8
∴ log2 x ≤ 3 when 0 < x ≤ 8
y = log x
2
At R: log __1 x = −3
2
(2)
1
∴ x = __ −3 = 8
∴ log__1 x ≤ −3 when x ≥ 8
2
At T: log2 x = −3
1
∴ x = 2−3 = __
8
1
∴ log2 x ≥ −3 when x ≥ __
8
At S: log__1 x = 3
2
(2)
1 3 1
∴ x = __ = __
8
1
∴ log__1 x ≥ 3 when 0 < x ≤ __
8
2
EXERCISE 6
Draw the graphs of f(x) and g(x) on the same set of axes and use them to solve the
inequality.
1
f(x) = log3 x and g(x) = 2. Now solve log3 x ≤ 2
2
f(x) = log__1 x and g(x) = −1. Now solve log__1 x ≥ −1
3
f(x) = log5 x and g(x) = 1. Now solve log5 x ≥ 1
4
f(x) = log__1 x and g(x) = x − 1. Now solve log__1 x ≥ x − 1
5
f(x) = log4 x and g(x) = −2x + 2. Now solve log4 x ≤ −2x + 2
4
2
4
2
Unit 2 Logarithms and logarithmic functions
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Revision Test Topic 3
Total marks: 114
1
Solve for x:
1
1.1 log __
=x
(3)
1.2 logx 49 = 2
(3)
1.3 log__1 x = − 2
(3)
1.4 5 x = 150
(3)
2 8
4
1.5 1 500 = 250(1 + 0,6) x
2
3
4
5
6
If log3
28 x
1.6 ___
= 10
7x
(3)
(3)
5 = x, find the value of 3 0 + 3 x + 3 2x.
(4)
( )
1 x
1 2
x and f(x) = __
are drawn alongside.
The graphs of g(x) = __
a
4
4
3.1
The graphs are labelled a and b. Decide which is f and which is g.
(2)
3.2
Write down the coordinates of P, the y-intercept of f.
(1)
3.3
If Q(1;k) is the point of intersection of f and g, determine the value of k. (2)
3.4
For which values of x are the graphs of f and g both decreasing?
(2)
3.5
Write the equation of the inverse of g in the form y = …
(2)
3.6
Explain why the inverse of g is not a function.
(2)
3.7
Write down two ways in which the domain of g could be restricted so
that g −1 is a function.
(2)
3.8
Write down the equation of f −1 in the form y = …
(2)
b
y
P
Q(1;k)
x
y
The graph of f(x) = logk x is drawn alongside.
4.1
If P(16;2) is a point on the of the graph of f, find the value of k.
4.2
Explain why the coordinates of Q are (1;0).
4.3
Use the graph to solve for x if f (x) ≤ 2.
4.4
Write the equation of f −1, the inverse of f in the form y = …
(2)
(1)
(3)
(2)
Given the function p(x) = 2 x + 1.
5.1
Write the equation of the asymptote of p.
5.2
Write the equation of the asymptote of p −1 .
5.3
Write the equation of p −1 in the form y = …
(2)
(2)
(2)
P(16;2)
Q
y = log k x
x
Given the function f(x) = b x, b > 0.
9
6.1
Find the value of b if the point −2;__
lies on f .
(2)
4
6.2
Find the equation of f −1, the inverse of f in the form y = …
(2)
6.3
Give the equation of the line about which f and f −1 are symmetrical. (2)
6.4
Draw the graphs of f and f −1 on the same set of axes. Show the intercepts
with the axes and the coordinates of two other points on each graph. (4)
6.5
Write the equation of h, the reflection of f in the x-axis.
(2)
−1
6.6
Draw the graph of h on the same set of axes as f and f .
(2)
6.7
Write down the range of h.
(1)
6.8
Is h an increasing or decreasing function? Give a reason for your answer. (2)
(
)
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7
8
The graphs of f (x) = 2 x − 1 and g(x) = 2x 2 are drawn.
7.1
Write down the equation of the asymptote of f (x).
7.2
Write down the equation of the asymptote of f −1.
7.3
Write down the equation of g −1 in the form y = …
7.4
For which values of x is f (x). g(x) ≥ 0?
y
g
f
(2)
(2)
(2)
(2)
The graphs of f (x) = b x, b > 0, b ≠ 1 and g(x) = a(x − p) 2 + q are drawn. P and Q
are the points of intersection of the two graphs. Q(1;4) is also the turning point
of g(x) and P is the y-intercept of both graphs.
8.1
Show that the coordinates of P are (0;1).
(1)
8.2
Find the values of a, b, p and q.
(2)
8.3
Find the equation of f (−x) and describe the transformation that took
place from f (x).
(2)
8.4
Describe the transformation from f to h if h(x) = logb x.
(2)
8.5
Write down the range of f and g.
(2)
8.6
For which values of x are the functions f and g both increasing?
(2)
8.7
Explain why the inverse of g(x) is not a function.
(2)
8.8
Write down two ways in which the domain of g could be restricted so
that g −1 is a function.
(2)
x
y
Q(1;4)
f
P
x
g
y
f
B(0;5)
9
10
The graphs of f (x) = 3 x and g(x) = ax + p are drawn. A(1;3) is the points of
intersection of the two graphs. B(0;5) is the y intercept of g.
9.1
Find the values of a and p.
9.2
Write down the range of f (x) + 1.
9.3
Write the equation f −1(x), the inverse of f (x).
9.4
For which values of x is:
9.4.1
f (x).g(x) < 0
9.4.2
f −1(x).g(x) ≤ 0?
A(1;3)
(4)
(2)
(2)
x
g
(2)
(2)
The graphs of f (x) = a x and g(x) = __
x are drawn. Q(1;2) is the point of intersection.
10.1 Find the values of a and k.
(4)
10.2 Write down the equation of f −1(x).
(2)
10.3 Copy the graphs alongside and draw the graph of f −1(x) on the same
set of axes.
(2)
10.4 Draw in any lines of symmetry on your set of axes.
(1)
10.5 For which values of x is:
10.5.1 g(x) ≥ f(x)
(2)
−1
10.5.2 f (x) ≤ g(x)?
(2)
y
f
k
P
Q(1;2)
g
x
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TOPIC
2
4
Finance, growth and decay
Unit 1: Revision: Grade 11 Finance
• Growth takes place with:
simple interest: A = P(1 + in) or
compound interest: A = P(1 + i)n
Decay takes place with:
simple decay (called the straight line method): A = P(1 – in)
compound decay (called the reducing balance method): A = P(1 – i)n
• If the annual interest that is quoted (the nominal interest rate) is compounded
more frequently than once a year, the effective interest rate will be higher than the
nominal interest rate,
and is determined using the formula:
m
i (m)
1 + 1eff = 1 + ___
m
• To determine the amount accumulated after an investment has been growing with
compound interest that is compounded k times per year: divide the quoted interest
rate by k and multiply the number of years by k.
• When more than one transaction occurs, draw a time-line to visualise what has
happened over time. Remember to take all values to any ONE moment in time,
before adding or subtracting values. Use the logic that ‘total of money in = total of
money out’.
• When taking values back in time, you are finding the P value for a known A value,
so the formula becomes A = P(1 + i)–n
(
)
EXERCISE 1
1
2
3
4
60
Determine the rate of depreciation (to one decimal place) for an item to become
a quarter of its original value after 9 years on:
1.1
the straight line method of depreciation
1.2
the reducing balance method.
Determine the length of time that it will take for an investment of R520 000 to
become R860 000 if the simple interest rate is 12% p.a. Give your answer correct
to two decimal places.
Tamryn will need R550 000 to buy a flat in 5 years’ time.
3.1
How much must she deposit now into an account offering 10% p.a.
compounded monthly to have the necessary funds in 5 years’ time?
3.2
What was the effective interest rate that Tamryn received each year (to two
decimal places)?
Shakir has R25 000 to invest over a period of 6 years. He is offered an interest
rate of 13% p.a. simple interest. What rate of interest, compounded annually,
should he be offered for the compound interest option to be better?
Topic 4 Finance, growth and decay
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5
6
7
8
A motorbike costs R95 000. Afika pays a 15% deposit
and will pay the balance over a period of 3 years in
equal monthly payments.
5.1
Find the size of the monthly payments if he
borrows the balance on hire purchase, where
the simple interest rate is 13% p.a.
5.2
How much interest did Afika pay?
Sarah deposits R15 500 into an account. The interest
rate for the first two years is 10% p.a. compounded
quarterly. It then changes to 8,5% p.a., compounded
monthly for the next 3 years, and then to 11% p.a.,
compounded semi-annually thereafter. How much
money will she have in her account after 7 years?
Nomsi deposits R20 000 into an account at an
interest rate of 9% p.a. compounded quarterly.
Three years later she deposits another amount of Rx.
Two years after that the interest rate changes to 10%
p.a., compounded annually. She withdraws R12 000
at the time that the interest rate changes. At the
end of 8 years she has R43 062,27 in her account.
Determine the value of x.
Philemon takes out a loan of R 50 000 to
renovate his house. Interest on the loan is 9% p.a.
compounded monthly for the first two years, and then changes to 9,5% p.a.,
compounded half yearly. He makes a payment of R20 000 one year after taking
out the loan, and another payment of R 25 000 two years later. How much will
he still owe on his loan four years after taking out the loan?
Unit 1 Revision: Grade 11 Finance
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KEY WORDS
nominal interest rate – the
quoted annual interest rate
annuity – a number of regular
payments of a fixed amount
made over a determined time
period
Unit 2: Derivation and use of formulae
for annuities
In Grade 11 you covered these formulae:
• Simple interest growth: A = P( 1 + in )
• Simple interest decay: A = P( 1 − in )
• Compound interest growth: A = P( 1 + i )n
• Compound interest decay: A = P( 1 − i )n
i( m ) m
• Effective annual interest rate: 1 + ieff = ( 1 + ___
m
)
where A = final amount, P = principal or present amount, i = rate of interest
rate
(i = ____
and you write it as a decimal) and n = number of times interest is received.
100
i( m ) is the nominal interest rate when interest is compounded m times per year.
See Revision Test on page 82 for revision questions on Grade 11 work.
Future value annuities
T0 T1 T2 T3
Tn – 2 Tn – 1 Tn
-
Each payment grows
according to the formula
A = P(1 + i)n where n
represents the number of
times compound interest is
received.
Future value annuities refer to regular payments of a fixed amount over time to
save for the future. For example, a retirement annuity is a fixed amount paid into an
account each month to accumulate a lump sum of money or monthly payments on
retirement. The value of each payment grows as compound interest on the payment
accumulates. The final amount available on retirement will be the sum of the future
values of each monthly payment. You can visualise this best on a time line:
-
REMEMBER
x
x
This payment earns interest for (n – 1) time periods
This payment earns interest for (n – 2) time periods
x This payment earns interest for (n – 3) time periods
Payments continue in this manner
x
n–1
x (1 + i ) n – 2
x(1 + i) n – 3
x(1 + i)
2
x
x(1 + i)
x(1 + i)
x This payment does
not earn interest as it
has only just been paid
Total future
value
accumulated
Remember these three important assumptions on which the derivation of the
annuity formula has been based:
• The first payment is made one time period from the present.
• The final payment is made at the time that the total accumulated is calculated
(that is, after n time periods).
The
regularity of compounding the interest is the same time period as the
•
regularity that the payments are made.
Use these assumptions as a ‘three-point check’ before applying the annuity formula in
any question.
To find the total future value accumulated, we must realise that the values to be added
form a geometric sequence (taking the terms from the last future value to the first):
x; x(1 + i); x(1 + i)2; … x(1 + i)n−3; x(1 + i)n−2; x(1 + i)n−1 where r = (1 + i).
62
Topic 4 Finance, growth and decay
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Thus, you can find the sum of these n terms using the formula for the sum of a GP:
a(r n − 1)
REMEMBER
Sn = ________
r − 1 where a = first term = xn and r = 1 + i
(1 + i) − 1
∴ Total value accumulated = x __________
[ 1+i −1]
1+i −1
= x[ __________ ]
i
(
(
[
)
)n
]
( 1 + i )n − 1
∴ F = x __________ where F = Total future value accumulated in an annuity situation
i
x = value of the regular fixed payment
i = rate of interest
n = number of payments made
A geometric sequence is a
number pattern where each
successive term is obtained by
multiplying the previous term
by a constant value.
This value is called the
constant ratio (r).
Consider the implications of having a payment at T0:
This payment would grow to x( 1 + i )n at Tn, which is the next term in the geometric
sequence mentioned earlier. You would calculate the sum of this sequence in the same
way, except that you will have one extra term. Thus, if you are given that the annuity
starts immediately, use the same future value annuity formula, but increase the value
of n by 1.
WORKED EXAMPLES
1
Suppose that Thandeka invests R500 every month for a period of 8 months,
starting one month from now. Her interest is 9% p.a. compounded monthly.
Calculate the value of her investment at the end of 8 months, after she has
made the 8th payment.
SOLUTION
1
You should recognise these important aspects:
• It is an annuity situation since there are regular payments of a fixed
amount.
• The money is saved for the future, so it is a future value annuity.
• All three aspects of the three-point check are as expected, so you can apply
the formula without modifying it. (Payments from T1 to Tn and regular
monthly payments coincide with monthly compounding of interest.)
[
]
0,09
( 1 + i )n − 1
9
F = x __________
where i = ____
per year = 0,09 per year = ____
per month
100
12
i
[
(
)
0,09 8
1 + ____ − 1
= 500 ______________
0,09
____
12
12
]
= R4 106,59
2
Mr Klein opens a savings account for his son’s future education. On opening
the account he deposits R850, and then makes monthly payments of R850 at
the end of each month for a period of 10 years. The interest rate remains at
12% p.a. compounded monthly.
2.1
Calculate how much Mr Klein will have accumulated at the end of
10 years.
2.2
Calculate how much interest was earned.
Note:
This question refers to the
amount of money that was
gained overall due to interest
received on the investment,
and not to the rate of interest.
Unit 2 Derivation and use of formulae for annuities
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SOLUTION
You should recognise these important aspects:
• It is an annuity situation since there are regular payments of a fixed amount.
• The money is saved for the future, so it is a future value annuity.
• When you consider the three-point check, notice that the first payment is made
at T0 instead of T1 which implies one extra payment. Monthly payments over
10 years mean 120 monthly payments, so with the extra payment there will be
121 payments.
Payments made at the end of a month are considered the same as at the start
of the next month, so apart from the initial payment at T0, the rest of the
payments are made as expected.
[
[
]
0,12
( 1 + ____
)
121
−1
( 1 + i )n − 1
12
F = x __________ = 850 _______________
2.1
i
0,12
____
12
]
n = (10 × 12) + 1
= 121
= R198 338,21
2.2 Interest earned = Future value received − payments made
= R 198 338,21 − (R850 × 121) = R95 488,21
3
On her 25th birthday, Sarah decides to accumulate R5 000 000 by her 50th
birthday. She plans to make equal monthly payments into an account
that pays 10% interest p.a. compounded monthly. If Sarah makes her first
payment a month after her 25th birthday and her last payment on her
50th birthday, determine how much she will need to deposit monthly to
accumulate R5 000 000 on her 50th birthday.
SOLUTION
You should recognise these important aspects:
• It is an annuity situation since there are regular payments of a fixed amount.
• The money is saved for the future, so it is a future value annuity.
• All three aspects of the three-point check are as expected, so apply the formula
without modifications. This time you are finding the value of the monthly
payment, x.
[ 1 +1 i+ i− 1 ]
)n
(
F = x ___________
(
)
[
0,1
1 + ___
− 1
(
12 )
______________
5 000 000 = x
300
0,1
___
12
[(
]
)
0,1
0,1 300
___
∴ 5 000 000 × ___
=
x
1
+
− 1
12
12
]
n = (50 − 25) years
= 25 × 12 months
= 300 payments
0,1
5 000 000 × ___
12
∴ ________________ = x
0,1
[ ( 1 + ___
12 )
300
− 1
]
∴ x = R3 768, 37 per month
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Topic 4 Finance, growth and decay
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4
Jeremy decides to save money for ten years. At the end of each month,
starting one month from now, he deposits R1 000 into the fund and
continues to do this for the ten-year period. Interest is 10% p.a. effective.
Calculate the final value of this investment.
KEY WORDS
effective interest – the
actual rate of interest that is
obtained
SOLUTION
You should recognise these important aspects:
• It is an annuity situation since there are regular payments of a fixed amount.
• The money is saved for the future, so it is a future value annuity.
• When you consider the three-point check, notice that the regularity of the
compounding of interest is annual but the payments are monthly. You must
find the equivalent monthly interest before applying the annuity formula.
(
)
i( m ) m
1 + ieff = 1 + ___
m
(
)
i( 12 ) 12
∴ 1 + 0,1 = 1 + ____
12
___
∴ ( 12√ 1,1 − 1 ) × 12 = i( 12 )
∴ i( 12 ) ≈ 9,57 %
[
]
[
0,0957
( 1 + ______
)
120
−1
]
( 1 + i )n − 1
12
= 1 000 _________________
= R199 875,60
F = x __________
0,0957
i
______
12
EXERCISE 2
1
2
3
REMEMBER
Calculate how much will be in a savings account after seven years if monthly
payments of R1 000 are paid into the account, starting in one month’s time.
Interest on the account is calculated at 12% p.a. compounded monthly.
Nusrah opens a saving account and deposits R5 000 every year, starting
immediately.
2.1
How much will she have accumulated after 8 years if interest is
compounded at 8,5% p.a.?
2.2
How much interest will Nusrah have earned after 8 years?
Tim plans to have R3 000 000 in his account when he turns 50 in 20 years’
time.
3.1
Calculate how much Tim must pay into the account each month,
starting in one month’s time if the interest is calculated at 9,5% p.a.
compounded monthly.
3.2
Calculate how much interest Tim earns over the 20-year period.
Interest earned = Future value
received − payments made
Unit 2 Derivation and use of formulae for annuities
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4
Mr Mkhize pays a fixed monthly amount into a savings account so he can buy a
car for his son when he turns 18. His son is now 8 years old.
4.1
The rate of inflation is expected to be 8% over the 10-year time period. If
the car that Mr Mkhize would like to buy for his son currently costs
R180 000, calculate the amount he would pay for the car when his son
turns 18.
4.2
The interest rate of Mr Mkhize’s savings account is 10% p.a. compounded
monthly and his first payment is in one month’s time. Calculate the
monthly amount Mr Mkhize would have to pay into a savings account to
accumulate the amount he expects to pay for the car in 10 years’ time.
4.3
Calculate how much money Mr Mkhize will have in his account after
5 years.
How much will Themba have in his savings account after 5 years if he deposits
R2 500 every three months, starting immediately. Interest is calculated at
12% p.a. compounded quarterly.
Mr January plans to save for his daughter’s education. He estimates that he will
need R85 000 in 8 years time. He will start immediately and make monthly
payments into an account giving an effective interest rate of 9% p.a. Determine
the value of his monthly payment if he is to succeed in his plan.
5
REMEMBER
6
Every 3 months is once a
quarter.
Present value annuities
Present value annuities involve a number of regular payments of a fixed amount
made over time, but unlike future value annuities where the accumulated total is
available in the future, in this case the money is available in the present. For example,
in a loan situation, a client receives a lump sum of money immediately, and the client
repays the loan by regular fixed payments over an agreed period of time.
REMEMBER
When you work back in time
to find the present value of
a payment, divide the future
value of the payment by the
interest growth.
We can derive a formula to determine the total loan amount available in a similar way to
that in which we derived the future value annuity formula. Consider the present value
of each payment that will be made, and find the total present value by considering the
geometric sequence that results. This is described on the time line below:
x(1 + i) – 2
x(1 + i) – 3
x(1 + i)
– ( n – 2)
x(1 + i) – ( n – 1)
x(1 + i) – n
x(1 + i)
x
x
x
Divide by one period of interest to find the
present value of this payment
-
Tn – 2 Tn – 1 Tn
T0 T1 T2 T3
–1
Payments continue
in this manner
Divide by (n – 2) periods of interest to find the present value of this payment
Divide by (n – 1) periods of interest to find the present value of this payment
Divide by n periods of interest to find the present value of this payment
x
x
x
Total present
value
available
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The present value annuity formula is based on the same three assumptions as the
future value annuity formula was based:
• The first payment is made one time period from the present.
• The final payment is made at Tn.
• The regularity of compounding the interest is the same time period as the
regularity of the payments made.
Use these as a three-point check before applying the present value annuity formula in
any question.
To find the total present value available, you should recognise that the values to be
added form a geometric sequence (taking the terms from the first present value to the
last):
x( 1 + i )−1; x( 1 + i )−2; x( 1 + i )−3; ...; x( 1 + i )−(n−2); x( 1 + i )−(n−1); x( 1 + i )−n
where r = ( 1 + i )−1
Thus, you can find the sum of these n terms using the formula for the sum of a GP:
a(r n − 1)
Sn = ________
where a = first term = x( 1 + i )−1 and r = ( 1 + i )−1
r−1
x( 1 + i )−1[ { ( 1 + i )−1 }n − 1 ]
∴ Total present value available = ______________________
( 1 + i )−1 − 1
x( 1 + i )−1[ ( 1 + i )−n − 1 ]
= ____________________
( 1 + i )−1 − 1
x
______
[ ( 1 + i )−n − 1 ]
(1 + i)
(1 + i)
= __________________
× ______
(1 + i)
1
______
−
1
(1 + i)
x[ ( 1 + i )−n − 1 ]
= _____________
1 − (1 + i)
x[ ( 1 + i )−n − 1 ]
= ______________
−i
[
[
)−n
1 − (1 + i
∴ P = x ____________
i
]
1 − ( 1 + i )−n
= x ____________
i
]
P = Total present value available in an annuity situation
x = value of the regular fixed payment that will cover the
value of P and interest growth
i = rate of interest
n = number of payments made
Consider the implications of having a payment at T0:
It is important to remember that the formula derived above for P gives the value of
the loan one time-period before the first payment is made. If payments are started
immediately, consider this as a deposit which is subtracted from the amount owed.
The difference becomes the value of the loan.
Unit 2 Derivation and use of formulae for annuities
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KEY WORDS
amortise a loan – to fully pay
back a loan or debt, including
interest
WORKED EXAMPLES
1
2
3
Siswe takes outs a loan to buy a car and will
Note: The loan is given
pay back the loan over 3 years. He plans to pay
immediately, but the first
R2 000 per month, starting in one month’s time.
payment is made in one
Calculate how much Siswe will borrow now if the
months’ time, which is
interest rate charged is 14% compounded monthly. the assumption on which
the derivation of the
Azrah buys a flat for R760 000. She pays R120 000
formula was based.
as a deposit and secures a loan for the balance.
If she pays back the loan over 20 years and the
interest is calculated at 15% p.a. compounded quarterly, how much will
Azrah pay each month if the monthly payments start in one month’s time?
James takes out a loan of R62 500 at an interest rate of 12% p.a. compounded
quarterly.
3.1
If James makes quarterly payments that start immediately, calculate the
value of each payment to amortise the loan after a full 10 years have
passed (that is, after 41 payments from now).
3.2
How much interest did James pay over the 10-year period?
SOLUTIONS
1
You should recognise these important aspects:
• It is an annuity situation, since there are to be regular payments of a
fixed amount.
• Since the loan money for the car is received now, it is a present value annuity.
• Since all three aspects of the three-point check are as expected, you can
apply the formula without modifications. (Payments are from T1 to Tn
and the regularity of monthly payments coincides with monthly
compounding of interest.)
[
]
[
(
)
0,14 −36
1 − 1 + ____
]
1 − ( 1 + i )−n
12
P = x ____________ = 2 000 ________________ = R58 517,81
i
0,14
____
12
0,14
14
n = 3 years = 36 months and i = ____
per year = ____
per month
100
12
2
68
You should recognise these important aspects:
• After the initial deposit has been deducted from the cost, Azrah pays back
the rest of the loan with regular payments of a fixed amount. It is therefore
an annuity situation.
• Since Azrah receives the money to purchase the flat now, it is a present
value annuity.
• When you consider the three-point check, notice that the payments are
monthly whereas the interest is compounded quarterly. Therefore you
determine the equivalent monthly interest rate before using the annuity
formula. As the first payment (deposit) is made immediately, you deduct
that first payment from the loan.
Topic 4 Finance, growth and decay
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(
)
i( m) m
Since we know 1 + ieff = 1 + ___
for interest compounded m times per year
m
(
)
i( n ) n
or 1 + ieff = 1 + ___
n for interest compounded n times per year
(
)
(
)
i( n ) n
i( m ) m
It follows that 1 + ___
= 1 + ___
m
n
0,15
i
i
∴ ( 1 + ____
= ( 1 + ____
( 1 + ___
) =_________
4 )
4 )
12 )
0,15
∴i
= 12( √ ( 1 + ____
− 1 ) ≈ 14,82%
4 )
1− 1+i
P = x [ ___________
]
i
(
( 12 )
i
∴ 1 + ____
12
12
(4)
( 12 )
4
12
4
12
( 12 )
4
(
)−n
(
Note:
When you convert a nominal
interest rate from one
regularity of compounding
to its equivalent value in
a different regularity of
compounding, use the
formula:
i( m ) m
i( n ) n
1 + ___ = 1 + ___
m
)
(
n
)
After paying the R120 000 deposit, the loan amount
= 760 000 − 120 000 = R640 000
[
(
)
0,1482 −240
1 − 1 + ______
12
640 000 = x _________________
0,1482
______
12
]
n = 20 years
= 20 × 12 months
= 240 monthly payments
[ (
) ]
0,1482
0,1482 −240
∴ 640 000 × ______
= x 1 − 1 + ______
12
12
0,1482
640 000 × ______
12
∴ _________________
=x
0,1482 −240
______
1 − 1 + 12
(
3
)
∴ x = R8 342,46 per month
You should recognise these important aspects:
• James pays back the loan with regular payments of a fixed amount.
It is therefore an annuity situation.
• Since James receives the loan money now, it is a present value annuity.
• When you consider the three-point check, notice that both the payments
and the interest compounding are quarterly as expected. However, the first
payment should be made after one time period (that is, after one quarter,
which is in 3 months’ time as quarter years are every 3 months). Since
James made the first payment immediately, there is one more payment
than expected and you deduct that first payment from the loan.
x[ 1 − ( 1 + i )−n ]
3.1
P = _____________
10 years = 10 × 4 quarters
i
−40
= 40 quarters
0,12
1 − 1 + ____
4
_______________
∴
there
are
41 payments,
62 500 − x = x
0,12
____
the first payment as a deposit.
4
After the deposit is paid, there
are 40 further payments.
(
[
(
)
)
0,12
(
) +1
________________
62 500 = x
1 − 1 + ____
4
0,12
____
4
−40
]
12
i = ____
= 0,12 per quarter
100
0,12
= ____
per quarter
4
x = R2 591,77
3.2
∴ x = R2 591,77 per quarter
Interest paid = Payments made − Loan received
= R2 591,77 × 41 − R62 500
= R43 762,57
Unit 2 Derivation and use of formulae for annuities
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EXERCISE 3
1
2
3
REMEMBER
Interest paid = payments
made − loan received
4
5
6
REMEMBER
(
)
i( m ) m
1 + ieff = 1 + ___
m
7
REMEMBER
i___
( 1 + i___
m ) = (1 + n )
(m)
70
m
(n)
n
Mrs Ndlovu takes out a mortgage bond of R980 000 on a property at a rate of
14% p.a. compounded monthly over 25 years. Calculate the monthly payments
required to settle the bond if Mrs Ndlovu makes the first payment in one
month’s time.
Determine the loan that Kaleb will receive if he plans to pay R1 200 per month
for a period of 10 years to amortise the loan. He makes his first payment in one
month’s time and interest on the loan is 8,5% p.a. compounded monthly.
Mr Taylor takes out a mortgage bond of R1 800 000 to help pay for his house.
The bond is repayable in equal monthly instalments over 20 years at a rate of
12,5% p.a. compounded monthly.
3.1
Calculate Mr Taylor’s monthly payments if he starts his payments in
one months’ time.
3.2
Determine how much interest Mr Taylor will pay altogether.
Laeeqa takes out a loan to buy a car. She pays R2 250 every six months towards
her loan for the next 4 full years, starting immediately (that is, 9 payments).
How much can she borrow if the interest charged is 12% p.a. compounded
half-yearly?
Abraham buys a car for R163 500. He pays a 10% deposit and settles the rest of
his debt with quarterly payments over 5 years, starting in three month’s time.
5.1
Calculate the value of his quarterly payments if interest on the loan is
charged at a rate of 13% p.a. compounded quarterly.
5.2
Calculate how much interest Abraham pays on his loan.
Alison borrows R120 000 to pay for renovations to her house. She pays monthly
instalments over 4 years to amortise the loan, starting in one months’ time.
Interest on the loan is compounded at 12% p.a. effective.
6.1
Determine the interest rate compounded monthly that would be
equivalent to a compound interest rate of 12% p.a.
6.2
Calculate the monthly payment required to settle the loan.
A car dealership sells Mr Matabane a car on these conditions:
He will pay a 15% deposit and monthly payments of R800 for a period of 6 years,
starting in one month’s time. Interest on the loan will be charged at 10% p.a.
compounded half-yearly.
7.1
Show that the equivalent monthly interest rate on the loan is 9,8%
(to 1 decimal place).
7.2
Determine the value of the loan.
7.3
Calculate the original cash price of the car.
7.4
Calculate how much Mr Matabane will pay in total for his car.
7.5
Calculate how much interest Mr Matabane will pay if he agrees to
this plan.
Topic 4 Finance, growth and decay
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Unit 3: Annuity applications and problem
solving
Sinking funds
Companies often purchase expensive equipment they will use for a specified time.
They then sell the old equipment at scrap value and purchase a new replacement
item. To have the money available to purchase new equipment, the company usually
plans ahead and sets up a fund (called a sinking fund). They make regular payments
into the sinking fund to save for the lump sum required to replace the equipment
when it reaches the end of its useful life. The company uses the scrap value of the old
equipment and the money available in the sinking fund to purchase the replacement
item.
Remember these three important aspects of a sinking fund situation:
• Since this sinking fund involves a regular fixed payment providing money for
the future, it is a future value annuity. To determine the amount available in the
account, you can apply the future value annuity formula. (See Remember 1.)
• To determine the scrap value of the old piece of equipment, use the formula for
compound decay. (See Remember 2.)
• To determine the expected cost of the replacement item, use the formula for
compound growth. (See Remember 3.)
KEY WORDS
scrap value – the depreciated
value of an item once it has
reached the end of its useful
life
REMEMBER
[ (1 + i) − 1 ]
n
F = x __________
i
F = P(1 − i)n
F = P(1 + i)n
➀
➁
➂
WORKED EXAMPLE
A company purchases a photocopying machine for R270 000. The company
expects to replace the machine in 5 years’ time. They anticipate the cost of the
machine to escalate at 16% p.a. compound interest. They expect their present
machine to have a scrap value of R100 000 in 5 years’ time when they sell it.
The company sets up a sinking fund to save for a new photocopying machine.
They will use the amount they obtain from the scrap value of the old machine and
the money in the sinking fund after 5 years, to purchase a new machine.
The company will pay a fixed monthly amount into the sinking fund, starting
in one month’s time and will make the final payment at the end of the 5-year
period. The interest earned on the sinking fund is 10% p.a. compounded monthly.
Determine:
1
2
3
4
the cost of a new machine in 5 years’ time
the rate of depreciation of the old machine on the reducing balance method
the amount required in the sinking fund in 5 years’ time
the value of the fixed monthly payments the company must pay into the
sinking fund.
Unit 3 Annuity applications and problem solving
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SOLUTION
1
F = P( 1 + i )n
= 270 000( 1 + 0,16 )5
= R567 092,25
2
F = P( 1 − i )n
100
000 = 270 000( 1 − i )5
______
100 000
5 _______
= 1 − i
√ 270 000
√
______
100 000
= 18,02% p.a.
∴ i = 1 − 5 _______
270 000
3
The amount required = Cost of new machine − scrap value obtained for old
machine
= R567 092,25 − R100 000
= R467 092,25
4
F = x ___________
i
[ (1 + i) − 1 ]
n
[
0,1
( 1 + ___
) −1
60
12
467 092,25 = x ______________
0,1
467 092,25 × ___
12
_______________
= x
0,1
− 1
( 1 + ___
12 )
60
0,1
___
12
]
∴ x = R6 031,89
EXERCISE 4
1
KEY WORDS
book value – the depreciated
value of a vehicle at a point
in time
2
72
A company buys a machine for R370 000. They expect the cost of a new machine
to rise by 14% p.a., while the rate of depreciation is 16% p.a. on the reducingbalance. The life span of the machine is 8 years.
1.1
Calculate the scrap value of the old machine.
1.2
Calculate the cost of a new machine in 8 years’ time.
1.3
Calculate the value of the sinking fund required to purchase a new
machine in 8 years’ time if the proceeds from the sale of the old machine
(at scrap value) will be used as part of the payment of the new machine.
1.4
The company sets up a sinking fund to pay for a new machine. They make
their payments into a savings account that pays 7% p.a. compounded
monthly. Calculate the monthly payments if they start payments one
month after they purchase the present machine and continue paying until
the end of the 8-year period.
A school buys a new bus for R450 000. The governing body decides that they will
replace the bus in 10 years’ time.
2.1
Use the reducing-balance method to calculate the annual depreciation rate
if the expected book value of the bus in 10 years’ time is R125 325,44.
2.2
Determine the rate of inflation if the estimated cost of the replacement
bus in 10 years’ time is R971 516,25.
2.3
Calculate the future value required in a sinking fund to pay for the
replacement bus in 10 years’ time. Assume that the current bus will be
traded in for the replacement.
Topic 4 Finance, growth and decay
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2.4
The school starts payments immediately into a sinking fund and continues
paying until the end of the 10-year period. Calculate how much the
school pays monthly into the sinking fund if they receive 9% p.a. interest
compounded monthly.
Future value annuities that end early
If regular payments of an annuity are stopped before the money is required, the value
obtained from the annuity formula is the total at the time of the last payment. This
total will grow according to the compound growth formula until the time that the
money is taken out of the account.
WORKED EXAMPLES
1
Selwyn pays a monthly amount of R1 500 into an annuity earning an interest
rate of 9% p.a. compounded monthly for 30 years. At the end of 30 years,
he stops his monthly payments but leaves the money in the account for a
further 2 years. Calculate how much he had in his account at the end of
32 years.
SOLUTION
1
After 30 years his annuity would have accumulated to:
[
[
]
0,09
( 1 + ____
)
360
− 1
( 1 + i )n − 1
12
F = x ___________
= 1 500 _______________
0,09
i
____
12
]
n = 30 × 12 = 360
To determine the amount accumulated after a further 2 years of compound
growth, consider the situation on a time line:
2 years of
compound
growth
30 years of monthly payments
T360
1 500
1 500
1 500
Therefore:
F = P( 1 + i )n
(
[(
T384
-
-
T0 T1 T2 T3
)
0,09 360
1500 1 + ____
− 1
]
)(
1 500
)
0,09 24
12
= _____________________
1 + ____
0,09
12
____
12
= R3 285 489,41
F1
F2
F 1 of annuity
becomes P of
compound
growth calculation
Unit 3 Annuity applications and problem solving
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2
Thandi opens a savings account and immediately deposits R1 000 into the
account. One month later and thereafter on a monthly basis, she invests
R1 000 per month for a period of 3 years. One month after her final payment
of R1 000, Thandi increases her monthly payment to R2 500 and continues
these monthly payments for a further 3 years. The interest rate during the
six-year period is 10% p.a. compounded monthly. Calculate the future value
of this investment at the end of the six-year period.
SOLUTION
You should recognise that there are two annuity calculations involved. The first
annuity stops before the total future value is calculated, which means that it will
continue to grow with compound interest for the remaining time period. It is
easiest to understand this on a time-line:
T0
T72
-
T36 T37
R1 000
F 1 1 + 0,1
12
F1
R1 000
R2 500
R2 500
36
F2
Note: Since the payments start at T0 there are 37 payments of R1 000.
[
0,1
( 1 + ___
) −1
37
]
[
0,1
( 1 + ___
) −1
36
12
12
F = 1 000 _____________ and F = 2 500 _____________
1
0,1
___
12
2
[
0,1
( 1 + ___
) −1
37
0,1
___
](
12
)
]
[
0,1
( 1 + ___
) −1
36
0,1 36
12
12
∴ Total future value = 1 000 ______________
1 + ___
+ 2 500 _____________
0,1
0,1
12
___
12
___
12
]
= R162 601,64
EXERCISE 5
1
2
74
Bongani saves for a holiday in 3 year’s time. He starts an annuity at an interest
rate of 9% p.a. compounded monthly, and makes his first monthly payment of
R500 at the end of the month. Determine the future value of the annuity at the
end of 3 years if he pays each month for 30 months.
A hotel owner buys a generator for R 625 000. He expects the generator to last
for 8 years and to depreciate at 11% p.a. on a reducing balance. The hotel owner
expects the cost of a new generator to escalate at 14% p.a. He sets up a sinking
fund with an interest rate of 9,6% p.a. compounded monthly to pay for a
replacement generator in 8 year’s time.
Determine:
2.1
the scrap value of the present generator in 8 years’ time
2.2
the expected cost of the new generator in 8 years’ time
2.3
the future value of the sinking fund in 8 years’ time if the scrap value
of the present generator goes towards the expenses of purchasing a new
generator.
Topic 4 Finance, growth and decay
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2.4
3
4
the value of the monthly payments that must be paid into the sinking
fund if payments start immediately and finish 1 year before a new
generator is bought.
Ntando saves for a new scooter by depositing quarterly payments into an
account that pays an interest rate of 12% p.a. compounded quarterly. If he plans
to save R50 000 in 4 years’ time, calculate the amount that he must deposit each
quarter if he starts his first payment in 3 months and makes his last payment
3 months before purchasing the new scooter.
Jamie opens a savings account and deposits R2 000 immediately. She continues
to pay monthly payments of R2 000 into the account for 5 years. The interest
rate is 10% p.a. compounded monthly for the first 2 years, and then changes to
9% p.a. compounded monthly. Calculate the future value of her investment
after the full 5 years.
Balance outstanding in present value annuities
Hint: Consider two different
annuities because the interest
changed. Do not forget the
extra interest growth on the
first annuity.
In the course of repaying a loan, we could be interested in knowing the balance
outstanding (the actual remaining debt) after a period of time.
There are two methods you can use to determine the balance at any time. The most
logical method is:
Balance outstanding on a loan = money owed − money paid = (loan + interest over
the time period) − (payments made + interest accumulated on those payments)
T 0 T1 T 2 T3
Tk
-
-
Consider a time-line showing a loan, L, to be paid back over n payments of x. You
want to find the balance outstanding after k payments have been paid.
L
Tn
k
x
L(1 + i)
x
k
x
F = x (1 + i) – 1
i
x
The loan debt grows with compound interest. You can determine the total of the
payments made (including the interest growth on those payments) using the future
value annuity formula.
( 1 + i )k − 1
∴ Balance outstanding = L( 1 + i )k − x ___________
L = loan
[
i
]
k = number of payments
made
x = value of payments
The second method is a simpler calculation based on the understanding that the
balance outstanding is the value of the debt at a particular time. This means that the
loan value will be fully paid back by the remainder of the payments left to pay. So,
if there were n payments to amortise the loan originally, after k payments have been
paid, the remainder of the loan (balance outstanding) will be amortised after (n − k)
payments. As the variable P in the present value annuity formula signifies the value of
the loan debt, use the present value annuity formula to find the balance outstanding.
Unit 3 Annuity applications and problem solving
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Consider a time-line showing the same situation as before: a loan, L, must be paid
back over n payments of x. You want to find the balance outstanding after k payments
have been paid:
Tk Tk + 1
Tn
-
T0
x
x x
B.O. at Tk
(amount
still owing)
x
x
x
The remaining debt (balance outstanding) will be covered by the remaining (n − k)
payments of x.
1 − ( 1 + i )−( n − k )
∴ Balance outstanding = x _______________
n = total number of payments to
[
]
i
amortise loan
k = number of payments made
x = value of payments
Note: The second method relies on the fact that all payments will be the same to the
end of the loan agreement. If the payment amount is changed to a different value
from the value calculated at the start of the loan agreement, the final payment will be
different from the previous payments. Avoid this method to determine the balance
outstanding when this occurs as it will be difficult to use.
WORKED EXAMPLE
Nigel takes out a loan of R3 000 000 from the bank to start his own business. The
loan will be amortised after 15 years and the monthly repayments will start one
month after the loan is granted. The interest rate is 9% p.a. compounded monthly.
Determine:
1
the value of the monthly payments (to the nearest Rand)
2
the balance of the loan at the end of 5 years.
SOLUTION
1
[
(
)
0,09 −180
1 − 1 + ____
12
3 000 000 = x ________________
0,09
____
[
12
(
]
)
0,09 −180
1 − 1 + ____
Note: The reason for
the slight difference in
the answers is due to
rounding off the monthly
payment
]
12
∴ x = 3 000 000 ÷ ________________ = R30 428
2
0,09
____
12
Balance outstanding after 60 months
(
[
)
0,09
( 1 + ____
) −1
60
]
0,09 60
12
= 3 000 000 1 + ____
− 30 428 ______________ = R2 402 037,44
12
[
(
)
0,09 −( 180−60 )
1 − 1 + ____
0,09
____
]
12
12
or B.O. = 30 428 ___________________ = R2 402 037,83
76
0,09
____
12
Topic 4 Finance, growth and decay
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EXERCISE 6
1
Martha takes out a loan of R250 000 to finance a car. She repays the loan over
a period of 6 years, starting one month after she bought the car. The interest
charged is 9% p.a. compounded monthly.
1.1
Determine her monthly payments to repay the loan over the required
period.
1.2
If she sells the car after 3 years, calculate how much money she owes on
the loan.
Mr Sebeko retired on 31 December 2005. He has a guaranteed pension of
R10 325 paid monthly for a period of 15 years. If he dies before he has received
15 years of pension, his beneficiaries will receive the balance outstanding.
If Mr Sebeko dies on 31 December 2012, calculate the amount that the
beneficiaries receive if the interest was 8,5% p.a. compounded monthly.
Stefan takes out a five-year loan of R900 000 that he can repay by monthly
payments of R18 248,75 at an interest rate of 8% per annum compounded
monthly. Stefan decides to pay monthly amounts of R20 000, starting in one
month’s time. Calculate his balance outstanding after the 30th payment.
2
3
Delayed start of payments
When the start of payments is delayed in a future value annuity, the number of
payments is affected. For example, if an investor plans to accumulate a certain
amount in 10 years’ time but starts paying monthly payments in 6 months’ time,
instead of paying 120 payments, he will pay 115 payments.
T0 T1
T k – 1 Tk
-
-
It is more complex when loan repayments start late, since the loan grows with
compound interest during the time that payments have not started. Consider the
situation on the time-line:
L
Tn
x
Payments usually start after
one month, so if the first
payment is made after
six months, the first five
payments have been missed.
Note: Payments always start
one time period after the
loan. Therefore if payments
start at Tk, the loan grows
in interest for (k − 1)
time periods.
k–1
L(1 + i)
x
x
REMEMBER
x
WORKED EXAMPLE
Clifford borrows R500 000 to buy furniture for his new flat. The loan is to be
repaid by monthly payments, and amortised at the end of five years from now.
Clifford will make the first payment after 3 months, and the interest charged is
10% p.a. compounded monthly. Calculate the amount of money that he will pay
each month.
SOLUTION
As payments start after 3 months instead of after 1 month, Clifford has missed the
first 2 payments. This means that there will be (5 × 12) − 2 payments, and the
loan will grow by 2 months of compound interest before the repayments start.
Unit 3 Annuity applications and problem solving
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(
)
0,1 2
Therefore the loan becomes R50 000 1 + ___
, so the formula becomes
12
[ 1 − (1 + i) ]
−n
P = x ____________
:
i
(
)
[
(
)
0,1 −58
1 − 1 + ___
0,1 2
12
50 000 1 + ___ = x _______________
12
(
0,1
___
)
[
12
(
]
)
0,1 −58
1 − 1 + ___
]
0,1 2
12
÷ _______________
= R1 108,89
∴ x = 50 000 1 + ___
0,1
12
___
12
EXERCISE 7
1
REMEMBER
2
When payments start late,
multiply the number of years
by the number of payments
per year, and then subtract
the number of missed
payments to determine n.
3
4
REMEMBER
When payments on a loan
start late, the loan increases
with compound interest for
the same number of time
periods as the number of
missed payments.
78
5
How much can be borrowed from a bank if the borrower repays the loan by
means of 48 equal monthly payments of R1 000, starting six months from now,
if interest is 10,5% p.a. compounded monthly?
Jacob saves for a deposit on a house. He estimates that he will need R350 000 in
five years’ time. He plans to make monthly payments into a savings account that
earns 9% interest per annum compounded semi-annually.
2.1
Show that i(12) ≈ 8,8% (to 1 decimal place).
2.2
Calculate his monthly payments if he makes his first payment in
6 months’ time.
2.3
Calculate how much Jacob will have in his account after 3 years from now.
2.4
If Jacob stops his payments after 3 years from now and makes no further
payments, calculate by how much he will be short of the required
R350 000 in 5 years’ time.
A loan of R200 000 will be amortised in 2 years. Monthly payments are made,
starting five months from now, in order to repay the loan. What is the value of
each payment if interest is 9,6% p.a. compounded monthly?
Susan and Pieter both plan trips around Europe in 12 months’ time. They both
save money for their holidays by opening savings accounts. Susan deposits
R500 immediately and continues to do so monthly until the end of the
12 months. Pieter pays monthly amounts of R600 into his savings account, but
starts after 3 months have passed. If both Susan and Pieter have accounts with
interest of 10% p.a. compounded monthly, who will have saved the most at the
end of 12 months, and by how much?
The bank granted Thebogo a loan of R3 500 000 to start a business. The bank
required him to amortise his loan 12 years after the loan was granted, and
agreed that he could start his monthly payments after 12 months had passed.
Interest on the loan was 9% p.a. compounded monthly. Determine the value of
Thebogo’s monthly payments.
Topic 4 Finance, growth and decay
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Unit 4: Calculate time periods using
logarithms
In each formula involving compound interest, the time period, n, is an exponent.
Therefore, if we need to calculate n, the calculation will often use logarithms (logs).
WORKED EXAMPLE 1
How long will it take an amount of money to halve in value when it is depreciated
at 12% p.a. on the reducing-balance?
SOLUTION
Let P = x then F = 0,5x ∴ 0,5x = x( 1 − 0,12 )n
∴ 0,5 = ( 0,88 )n
∴ n = log0,88 0,5 ≈ 5,42 years
1
years to halve in value.
So it will take just short of 5__
REMEMBER
If a = bc then logb a = c
2
WORKED EXAMPLE 2
Gail saves money to buy a new computer. Calculate how long it will take her to
accumulate R10 000 if she deposits R300 each month into a savings account that
gives interest at 10,5% p.a. compounded monthly. Her payments will commence
in one month’s time.
SOLUTION
Note: It is important to recognise that the time period for the value of n will be the
same as that of the interest compounding. So in this case, the answer that we
obtain for n will be the number of months that must pass. If we want an answer in
years, divide by 12.
[
]
[
0,105
( 1 + _____
) −1
n
( 1 + i )n − 1
12
F = x __________ ∴ 10 000 = 300 ______________
(
i
)
(
(
)
)
0,105 n
0,105
10 000
∴ ______
× _____
+ 1 = 1 + _____
300
12
12
31
807 n
∴ ___
= ____
24
800
0,105
_____
12
]
( )
31
___
807
∴ n = log( ____
) 24 ≈ 29,38 months (or 2,45 years)
800
Unit 4 Calculate time periods using logarithms
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WORKED EXAMPLE 3
Rafael buys a car that costs R280 000. He pays a 10% deposit and takes a loan at an
interest rate of 11% p.a. compounded monthly for the balance. He agrees to pay
R8 000 at the end of each month, starting in one month’s time.
1
Calculate the value of Rafael’s loan.
2
How many payments of R8 000 will he have to make?
3
If Raphael clears his debt with one final payment one month after his last
payment of R8 000, calculate the value of his final payment.
SOLUTIONS
1
Loan = 0,9 × 280 000 = R252 000
2
0,11
1 − 1 + ____
1 − ( 1 + i )−n
12
___________
______________
P=x
∴ 252 000 = 8 000
0,11
i
____
[
[
]
(
(
)
−n
12
]
(
)
)
0,11
569
569
____
∴ ____
= ( 1 + ____
∴ log(
) ( 800 ) = − n
800
12 )
−n
0,11
0,11
252 000
∴ _______
× ____
− 1 = − 1 + ____
8 000
12
12
−n
1 211
_____
1 200
∴ − n ≈ − 37,34 ∴ n ≈ 37,34 months (or 3,11 years)
So he will make 37 monthly payments of R8 000.
T37 T38
-
T0 T1
-
3
Loan
8 000
P1
8 000
P2
y
If the payments of R8 000 had fully covered the loan, then the loan value
would equal the value of P1. If that were the case, then:
[
(
)
0,11 −37
1 − 1 + ____
12
Loan = P = 8 000 _______________
1
0,11
____
12
]
But an extra payment, y is required to fully cover the loan. Whenever you
want to create an equation with regard to money transactions, ensure that
the values being equated are at the same moment in time. Therefore, add the
0,11 −38
present value of the final payment, that is, y 1 + ____
[
(
)
] (
(
12
)
0,11 −37
1 − 1 + ____
0,11 −38
12
_______________
252 000 = 8 000
+ y 1 + ____
0,11
12
____
12
(
[
(
)
0,11 −37
1 − 1 + ____
12
∴ y = 252 000 − 8 000 _______________
0,11
____
12
)
])
0,11
= R2 734,55
( 1 + ____
12 )
38
Note: We found in worked example 3, question 2 that 37,34 payments were
required. The value of this last payment is more than 0,34 × 8 000 because there
was interest during the last month before the final payment is made.
80
Topic 4 Finance, growth and decay
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WORKED EXAMPLE 4
Greg borrows R1 000 000 from a bank. He plans to pay R9 000 per month on the
loan. Interest is 12% per annum compounded monthly. Decide whether the bank
will grant him the loan. Explain your answer.
SOLUTION
If we try to determine how long it would take to repay the loan under these
conditions we discover:
[ (
)
0,12 −n
1 − 1 + ____
12
](
0,12
)
(
1 000 000 = 9 000 ______________
∴ _________
× ____
− 1 = − 1 + ____
0,12
9 000
12
12
1
∴ −__ = ( 1,01 )−n ∴ log
9
____
1,01
12
1 000 000
)
0,12 −n
REMEMBER
loga b is only defined for
a > 0; a ≠ 1 and b > 0
( −__19 ) = −n
( )
1
But log1,01 −__
is not valid, so there is no solution to this equation.
9
This means that it will be impossible for the loan to be paid back under these
conditions, and the bank would not grant the loan. We could have come to the
same conclusion by noticing that the interest growth on the loan each month is
more than the monthly payment:
0,12
1 000 000 1 + ____ = R1 010 000 ∴ interest = R10 000
(
12
)
As the monthly payment Greg suggests is R9 000, the loan grows each month by
more than the monthly payment, so the debt increases and will never be repaid.
EXERCISE 8
1
2
3
Vusi invests R1 000 at an interest rate of 10% p.a. compound interest. Calculate
how many years it will take for this investment to be worth R1 610,51.
A new car costs R375 000. Calculate how long it will take to reach a scrap value of
R176 344,70 if the rate of depreciation on the reducing balance method is 9% p.a.
Sydney borrows R300 000 at 10% p.a. compounded monthly. He pays back
R1 000 per month for the first year, starting in one month’s time, and R5 000
per month for the remaining time until he has repaid the loan in full (including
interest).
Determine:
3.1
the balance outstanding at the end of the first year
3.2
how many payments of R5 000 Sydney must pay
3.3
the value of his final payment
3.4
whether the bank would have allowed him to continue paying R1 000 on
an on-going basis.
Unit 4 Calculate time periods using logarithms
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Unit 5: Analyse investments and loan options
When you make a decision about which investment option or loan option to choose,
you must recognise that there are a number of factors that should be taken into
account.
Investments
Compound interest results in exponential growth whereas simple interest results in
linear growth. It will almost always be advantageous to choose a compound interest
investment, unless the simple interest offered is much higher than the compound
interest and the investment is short term. Also, because of the affect of exponential
growth, the more frequently the interest is compounded, the better for the investor.
WORKED EXAMPLE
Alice, Ben and Fatima each win R500 000 in a competition. Alice invests her
money in an investment offering 20% p.a. simple interest. Ben invests his money
at 10% p.a. compound interest and Fatima invests her money at 9,8% p.a.
compounded daily. Determine how much money each person has after 15 years.
Then decide who made the best choice.
SOLUTION
Alice: F = 500 000( 1 + 0,2 × 15 ) = R2 000 000
Ben: F = 500 000( 1 + 0,1 )15 = R2 088 624,09
(
)
0,098 5 475
Fatima: F = 500 000 1 + _____
365
= R2 174 188,55
Fatima has the most money after 20 years, although she had the lowest interest
rate. However, the difference is not great, so the choice will depend on the length
of the investment.
KEY WORD
retirement annuity – a future
value annuity where regular
payments are invested into an
account to save for retirement
82
Retirement annuities
When you save for retirement, a retirement annuity is a good option. The longer an
investor pays into such an annuity without withdrawing from the investment, the
better. Some people prefer to invest their money in the stock exchange; here you can
buy partial ownership in a company and benefit from the profits of that company.
A stock exchange investment can be risky since it relies on the performance of the
company in which you have chosen to invest. A less risky way of investing in the
stock exchange is to buy units in a unit trust. A unit trust is the pooled money of
many investors that is invested in the financial markets through a single collective
investment scheme. An investor can choose to buy units as a once-off investment or
can put aside monthly amounts into a unit trust. Unlike an annuity, however, there
is no guarantee on the rate of return you will have on your investment. If the market
hits a low or crashes, then your investment will suffer; but if you can leave your
money in the trust until the market has stabilised, you can sometimes do well in this
kind of investment. Generally, the longer your money can stay invested, the more
likely you are to have a good investment return.
Topic 4 Finance, growth and decay
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WORKED EXAMPLE
Kamil and Ismail are twins. On their 30th birthday they decide to save for their
retirement. Kamil invests R1 000 each month into a retirement annuity that gives
him 10% p.a. compounded quarterly. Kamil’s first payment starts one month after
his birthday. Ismail invests R1 000 each month into a unit trust.
1
They both consider retiring on their 60th birthdays, and investigate the
amount their investments have yielded. Since the world money market is at
a low at that time, Ismail’s investment is worth only R1 500 000. Determine
the value of Kamil’s investment on his 60th birthday after he has made his
monthly payment for that month. Compare Kamil’s investment with Ismail’s
investment at that stage.
2
The twins both decide to work for a further 5 years. On his 65th birthday
Ismail receives a payment of R3 500 000 from his unit trust investment.
Determine whether he will be better off than Kamil, who has continued
paying into his annuity for a further 5 years.
SOLUTION
1
[
0,092
( 1 + _____
)
360
−1
12
F = 1 000 _______________
0,092
_____
12
]
= R1 908 835,09
∴ R408 835,09 more than Ismail
2
Note: We need interest compounded monthly, so:
0,1
i
= ( 1 + ___
( 1 + ____
4 )
12 )
( 12 )
4
12
∴ i( 12 ) ≈ 9,92%
[
0,092
( 1 + _____
)
420
−1
12
F = 1 000 _______________
0,092
_____
12
]
= R3 094 257,30
∴ R405 742,70 less than Ismail
Pyramid schemes
A pyramid scheme is a risky way of trying to make money quickly. There are various
forms of pyramid schemes, and some have variations to hide their true nature. These
schemes promise money to people who choose to participate by enrolling other
people into the scheme, rather than supplying real investment themselves. The
schemes are not sustainable, and are illegal in many countries, including South Africa.
The idea behind this money-making formula is that a person starts a ‘business’ by
recruiting people to make a payment and to then recruit other people. This continues
until there is a pyramid of people. Each new recruit makes a payment to the person at
the top of the list, and continues the pyramid by recruiting further people. There is
seldom any sale of real products involved, or services. Often these scams are made to
look more creditable by including false testimonials. Only the original ‘investor’ and a
Unit 5 Analyse investments and loan options
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few people at the top levels of the pyramid make any money; those further down the
pyramid often lose money as it becomes increasingly more difficult to find new
recruits. The basic problem is that no wealth is created, but that existing wealth is
moved around. Every R1,00 that one person gains through such a scheme is R1,00
that someone else has lost.
Levels
1-6
2- 36
3- 216
4-1 296
5- 7 776
6- 46 656
7- 279 936
8- 1 679 616
9- 10 077 696
6
10- 60 466 17
6
5
0
7
11- 362 79
36
3
2
8
7
12
2 176
016
13- 13 060 694
The diagram alongside shows a typical pyramid scheme
and how it is impossible to sustain.
One example of a pyramid scheme is the ‘aeroplane
game’. In this scheme there are four levels. The
person in the top level is the ‘captain’, the second
Number of participants
level participants are the ‘co-pilots’, the third level are
the ‘crew’ and the bottom level are the ‘passengers’.
The eight ‘passengers’ each pay a sum of money to
the ‘captain’ to join the scheme. The captain leaves
the scheme after receiving his payments from each
‘passenger’, and everyone moves up a level. As there are
now two ‘captains’ the groups splits into two groups;
and each group must find eight more passengers. If the scheme collapses due to not
finding more ‘passengers’, all participants who have not become ‘captains’ lose their
money.
“Captain”
(paid by new passengers)
“Co-pilot”
“Crew”
“Passengers”
(pay the “captain”
when they join the
scheme)
WORKED EXAMPLE
Mr X starts an ‘aeroplane game’. He sets up the first three levels using false names,
and sends an e-mail to enlist recruits. Each new recruit will become a ‘passenger’
by paying R1 000 to the captain (Mr X).
1
2
84
How much money will Mr X make by the time that the first real investors
become ‘captains’, assuming that he receives all the money paid to the false
names that he made up?
How many further new recruits will he need for the first round of ‘passengers’
to receive R8 000 from their ‘passengers’ once they have become ‘captains’?
Topic 4 Finance, growth and decay
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SOLUTIONS
1
Consider the pyramid scheme shown
in Figure 1 where A, B, C, D, E and F
are the false names made up by Mr
X, and G to N are the first round of
real investors. Each of these people
pay R1 000 to Mr X, so he will get R8
000 from the first round of investors.
Figure 2 shows the next stage, where
A and B have become ‘pilots’. But as
these false names are Mr X, he will
receive another R16 000.
A
D
H
A
B
C
G
D
H
I
E
J
K
F
L
M
N
B
C
G
Mr X
E
I
J
K
F
L
M
N
Figure 2
Figure 3 shows how Mr X will receive another R32 000 when the false names
C, D, E and F become ‘pilots’.
He will therefore receive R56 000 in total without pay in any money himself.
C
G
D
H
I
E
J
K
F
L
M
N
Figure 3
2
G to N are the first paying recruits. They must each recruit two new members
as shown in Figure 2 (there are 16 recruits in total). Figure 3 shows how
32 further recruits would be necessary for G to N to become ‘co-pilots’.
Each new recruit would need to recruit a further two recruits (thus 64 more
recruits) for G to N to become ‘pilots’. So there would need to be 112 recruits
in total.
Loan options
In a loan situation, instead of benefitting from the interest, the borrower pays the
interest. It is always important as a borrower to choose an option where the interest
paid is minimised. These points should be taken into account:
• In a simple interest loan agreement, interest is calculated on the original debt,
whereas in compound interest the interest is calculated on the reducing balance,
which benefits the borrower.
• The more frequently the interest is compounded, the more the borrower will
benefit from the balancing reducing.
REMEMBER
A hire purchase agreement
is a loan agreement made
with a company, that allows
the buyer to take the item
purchased on loan, often
after paying a deposit. The
buyer pays the remainder
of the cash price in monthly
instalments over time. Simple
interest is calculated and
added to the cash price, and
the total debt owed is divided
between the number of
payments.
Unit 5 Analyse investments and loan options
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• It is always better to amortise a loan in as short a time as possible. The longer you
take to pay back a loan, the more interest you will pay in total.
Usually loan agreements apply compound interest to the loan, except for hire
purchase (which you studied in Grade 10), in which simple interest rates are often
very high, resulting in large amounts being paid in interest.
WORKED EXAMPLE
Len takes out a loan of R300 000 at an interest rate of 11% p.a. compounded
monthly. He must amortise the loan within 5 years with equal monthly payments
starting in one month’s time. Len calculates his monthly payments to be
R6 522,73 if he pays the loan over 5 years.
1
Determine how much interest Len would pay under this agreement.
2
Len decides to pay R7 000 per month. Determine how many payments of
R7 000 he must pay.
3
Determine the value of Len’s final payment.
4
Calculate how much Len will save in terms of interest paid by paying
R7 000 instead of R6 522,73.
SOLUTION
1
2
Interest = (R6 522,73 × 60) − 300 000
= R91 363,80
[
(
)
0,11 −n
1 − 1 + ____
12
300 000 = 7 000 ______________
(
)
17
1 211 −n
∴ ___
= _____
28
1 200
0,11
____
12
]
( )
17
___
∴ −n = log( _____
1 211
) 28
1 200
∴ n = 54,68 ∴ 54 payments of R7 000
3
[
(
)
0,11 −54
1 − 1 + ____
] (
)
0,11 −55
12
300 000 = 7 000 _______________ + y 1 + ____
(
0,11
____
12
[
(
)
0,11 −54
1 − 1 + ____
12
∴ y = 300 000 − 7 000 _______________
0,11
____
12
12
]) (
)
0,11 55
1 + ____
12
= R4 798,51
4
86
Interest paid = [(R7 000 × 54) + R4 798,51] − R300 000
= R82 798,51
Therefore by paying R7 000 instead of R6 522,73 Len saved:
R91 363,80 − R82 798,51 = R8 565,29
Topic 4 Finance, growth and decay
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EXERCISE 9
1
2
Keith takes out a retirement annuity on his 35th birthday for R1 500 per month.
He makes his first payment immediately on his 35th birthday. Interest on the
annuity is 10,2% p.a. compounded quarterly. On Keith’s 35th birthday, his
friend, Barry, invests R80 000 as a once-off payment into a unit trust for his
own retirement.
1.1
Determine the monthly interest rate that is equivalent to 10,2% p.a.
compounded quarterly.
1.2
Determine how much Keith will have saved by his 50th birthday just after
he has made his payment.
1.3
On his 50th birthday Keith increases his payments to R2 500 from the
following month. Calculate how much he will be paid out from the
annuity when he retires on his 65th birthday.
1.4
At the end of the 15-year period Barry averages 14% p.a. compound
interest on his investment. Determine how much Barry receives at the end
of 30 years. Compare his final value with Keith’s final value.
Nina bought an apartment for R1 400 000. She was offered these options:
Option 1: Pay a 15% deposit and have a mortgage bond over a 20-year period for
the balance at an interest rate of 9,5% p.a. compounded monthly.
Option 2: Take out a mortgage bond for the full amount at an interest rate
of 9,5% compounded monthly over a 15 year period.
In both options payments will commence in one month’s time.
2.1
Determine Nina’s monthly payment for each option.
2.2
Determine the interest Nina pays in each option.
2.3
Which option is better for Nina?
Unit 5 Analyse investments and loan options
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Revision Test Topic 4
Total marks: 115
1
2
3
Calculate how long it would take for an investment to triple in 5 years if the
interest rate is:
1.1
10% p.a. simple interest
1.2
10% p.a. compound interest.
(3)
(4)
A loan is amortised a full 8 years after a loan is granted by equal monthly
instalments of R1 230,50. Payments begin immediately and interest is
charged at 10,5% p.a. compounded monthly.
2.1
Determine the value of the loan.
2.2
Calculate how much interest was paid in total.
(6)
(3)
Mpho invests R12 000 now. Determine the annual rate of interest,
compounded quarterly, that yields the same amount in 5 years’ time as
investing R12 000 at 15% simple interest per year.
(6)
4
Savanna wins R100 000 in a competition and invests it in a savings account.
She plans to withdraw money in four years’ time to travel overseas. She also
plans to have R150 000 available for house alterations in 5 years’ time. She
knows that she is receiving another payment of R15 000 three years from
now, which she plans to deposit into the same account. The interest rate at
the start of the investment is 12% p.a. compounded semi-annually. If the
interest rate changes after 2 years to 8,5% compounded quarterly, determine
the amount Savanna will be able to withdraw for her overseas holiday.
(10)
5
Mr Dumani pays R550 000 for a minibus taxi in June 2013. He plans to
replace it in June 2020. He estimates that the cost of a new minibus will
escalate at a of rate 9% compound interest each year.
5.1
Calculate what Mr Dumani expects to pay to replace his minibus
in 2020.
(3)
5.2
In 2020 he hopes to sell his present minibus for its depreciated value
at that time, and to use the money to partly finance his new minibus.
He estimates the rate of depreciation as 14% p.a. on the reducing
balance. Calculate the book value of his present minibus when he
sells it.
(3)
5.3
Mr Dumani sets up a sinking fund to finance his new minibus in
2020. Determine the amount that he needs to save in the sinking fund. (2)
5.4
He makes equal monthly payments into the sinking fund on the first
day of each month, starting in one month’s time (1 July 2013).
He makes the last payment into the account on 1 June 2020. The
bank pays interest at a rate of 8% p.a. compounded monthly.
Determine the value of the monthly payment that Mr Dumani must
make to save what he needs by 1 June 2020.
(6)
6
A loan of R50 000 is amortised over a period of 5 years. Payments are made
monthly starting six months after the loan is granted. The interest rate is
10,5% p.a. compounded monthly.
6.1
Calculate the monthly repayments.
6.2
Determine the balance outstanding after 2 years.
(6)
(6)
88
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7
Desmond starts an investment by immediately depositing R750 into an
account. He pays R750 at the beginning of each month thereafter. The
interest is calculated at 8% pa compounded monthly.
7.1
Calculate the balance Desmond will have in his account 5 years later,
straight after he makes his final payment.
(6)
7.2
If Desmond stops his payments but leaves the money in the account,
calculate how much he will have in his account after a further 2 years. (3)
8
Mrs Seleke plans to buy a flat. She requires a mortgage bond of
R125 000. The interest rate on the bond is 9% p.a. compounded monthly.
Mrs Seleke plans to repay the loan with equal monthly payments of R900
with payments starting one month after the loan is granted.
8.1
Decide whether the bank will allow Mrs Seleke to take out a bond
under these conditions.
8.2
If Mrs Seleke pays R2 000 per month until the bond is cleared,
calculate the number of payments required to amortilise the loan.
8.3
Calculate Mrs Seleke’s final payment.
8.4
Determine how much interest Mrs Seleke pays.
9
10
11
Johan owns a small business and buys a car for R225 500. The expected
replacement cost of the car in 6 years’ time is R338 414,70 and the expected
book value of the car after 6 years is R101 203,34.
9.1
Calculate the estimated rate of depreciation using the reducingbalance method of depreciation.
9.2
Johan invests in a sinking fund with interest that is calculated at
9,8% pa compounded monthly. Determine the equal amounts Johan
must invest into the sinking fund, starting in three months, to have
sufficient funds in the bank to purchase the car in 6 years time.
Assume that Johan will trade in his current car at the same time and
use the money towards the cost of the new one.
(4)
(6)
(6)
(5)
(4)
(7)
Ken has R80 000. He plans to accumulate R145 000 to pay for his university
fees. If he has 4 years in which to accumulate the money, calculate the
nominal interest rate, compounded monthly, that he must be offered.
(6)
Bushrah and Philip plan to renovate their house. They take out a loan of
R55 000 immediately and a second loan of R78 000 two years later. They
plan to repay the loan in three equal instalments. The first payment must
be made after 4 years, the second instalment must follow 3 years later and
the final payment at the end of 10 years. The interest is 10,5% p.a. for
the first 3 years, 10% p.a. compounded quarterly for the next 6 years and
9% compounded half-yearly for the last year. Calculate the value of each
instalment.
(10)
89
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2
5
Trigonometry: Compound and double
angle identities
Unit 1: Revision: Grade 11 Trigonometry
Definitions and signs of ratios in all four quadrants
y
y
y
sin θ = __
y
2
r
x
cos θ = __r
y
tan θ = __
x
r
θ
O
1
r
y
y
r
–x
x
x
x
–y
θ is the acute angle between
the arm and the x-axis
r
2
1
2
S
All
Sin
y
x
x
Tan
–y
r
3
4
T
Cos
3
4
3
y
1
180° – θ
y
Reduction formulae:
θ
x
Quadrant 2
sin (180° − θ) = sin θ
cos (180° − θ) = −cos θ
tan (180° − θ) = −tan θ
x
–y
180° + θ
360° – θ
4
Quadrant 3
sin (180° + θ) = −sin θ
cos (180° + θ) = −cos θ
tan (180° + θ) = tan θ
Quadrant 4
sin (360° − θ) = sin (− θ) = −sin θ
cos (360° − θ) = cos (− θ) = cos θ
tan (360° − θ) = tan (− θ) = −tan θ
Special angles in all four quadrants (signs change
according to CAST diagram)
30°
150°
210°
330°
45°
135°
225°
315°
60°
120°
240°
300°
0°
180°
sin
1
__
2
1__
___
√3
___
0
1
√3
___
2
1__
___
1
__
1
0
1__
___
1
__
√3
0
undefined
–1 -
90
90° 180° 270° 360°
y = sin x
30°
3
y
1-
0-
x
90° 180° 270° 360°
–1 -
y = cos x
1
45°
45°
1
1
3
90°
180°
45°
2
1
-
30°
543210–1 –2 –3 –4 –5 -
-
x
45°
2
1
y
-
-
-
-
-
1-
-
√3
y
0-
2
√2
-
__
2
60°
2
270°
360°
2
-
tan
__
√2
60°
90°
270°
-
Q1
Q2
Q3
Q4
cos
-
x
TOPIC
360°
x
y = tan x
Topic 5 Trigonometry: Compound and double angle identities
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1
y
Co-ratios
90° + θ
θ
θ
c
cos(90° − θ) = __
a = sin θ
b
sin(90° − θ) = __
a = cos θ
cos(90° + θ) = −sin θ
sin(90° + θ) = cos θ
a
b
θ
x
90° – θ
Identities
c
cos θ
sin θ
1
_____
_____
tan θ = _____
cos θ or tan θ = sin θ
sin2 θ + cos2 θ = 1 or sin2 θ = 1 − cos2 θ or cos2 θ = 1 − sin2 θ
EXERCISE 1
1.1 Use the diagram alongside to find the length OP.
1.2 Now write down the value of:
1.2.1 sin θ
1.2.2 61 cos2 θ + 1
1.2.3 tan(180° − θ)
2
If 7 sin θ + 4 = 0 and cos θ > 0 find, without a calculator, the value of
2.1
1 − 49 cos2 θ
2.2
cos θ. tan θ
5
tan A.tan B.sin A
.
3
If cos A = __ and A + B = 90°, use a sketch to find the value of _______________
4
5
Prove the following identities:
4.1
cos2 x + sin x. cos x. tan x = 1
4.2
1 − tan2 θ
_________
= 2 cos2 θ − 1
5.2
7
P(–6;5)
x
O
1 + tan2 θ
Simplify each of the following.
5.1
6
cos B
6
y
2 sin(90° − x) − cos(360° − x)
_________________________
2 cos(90° + x) − sin(180° + x)
tan(180° − x) tan 45°.cos(−x)
_________________________
sin(180° − x)
If tan 38° = k, write down the following in terms of k (Hint: Draw a diagram.)
6.1
tan 142°
6.2
cos (−38°)
6.3
sin 52°
Simplify without the use of a calculator.
sin 150°. sin 55° sin 135°. tan 240°
_______________________________
7.1
cos 330°. cos 145°. tan 150°. cos 225°
7.2
8
sin 25°
1__
___
cos 210° + 2 tan 135° cos2 135° − ________
√3
cos 115°
Solve the following equations.
8.1
sin(θ + 20°) = 0,577 and 0° ≤ x ≤ 360°
__
8.2
2 sin 2θ + √ 3 = 0 and −180° ≤ x ≤ 180°
8.3
3 cos θ + 3 = 4sin2 θ. Find the General Solution.
8.4
tan(θ + 40°) = tan 52° and −360° ≤ θ ≤ 360°
8.5
sin(θ − 20°) = cos 3θ. Find the general solution.
Unit 1 Revision: Grade 11 Trigonometry
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Unit 2: Derive the compound and double angle
identities
In Grade 12 you are introduced to new identities called the compound angle or addition
formulae.
Compound angle identities
cos(α − β) = cos α cos β + sin α sin β
cos(α + β) = cos α cos β − sin α sin β
sin(α − β) = sin α cos β − cos α sin β
sin(α + β) = sin α cos β + cos α sin β
Proof of identity for cos(α − β).
The formulae for cos(α + β), sin(α − β) and
sin(α + β) are derived from cos(α − β).
y
Proof: cos(α − β) = cos α cos β + sin α sin β
In the diagram A and B are points on a circle
of radius 1 unit.
^ X = β.
^ X = α and BO
AO
^ B = α − β.
Therefore AO
A
1
Follow these steps:
B
α−β
1. Use the cosine rule in △AOB to find an
expression for the length AB.
α
1
2. Express the coordinates of A and B in
β
x
terms of sines and cosines of α and β.
O
X
3. Find another expression for the length
AB, using the coordinate geometry
distance formula.
4. Equate the two expressions for AB and you have proved the trigonometric identity.
1.
2.
^B = α − β
Given that OA = OB = 1, and AO
Using the cosine rule:
AB2 = 12 + 12 − 2(1)(1)cos(α − β)
AB2 = 2 − 2 cos(α − β)
The coordinates of A in terms of sines and cosines of α are:
xA
yA
__
= cos α and __ = sin α ∴ A = (cos α;sin α)
1
1
The coordinates of B in terms of sines and cosines of β are:
xB
yB
__
= cos β and __ = sin β ∴ B = (cos β;sin β)
1
92
1
Topic 5 Trigonometry: Compound and double angle identities
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3.
4.
Using the distance formula:
AB2 = (cos α − cos β)2 + (sin α − sin β)2
AB2 = cos2 α − 2 cos α cos β + cos2 β + sin2 α − 2 sin α sin β + sin2 β
AB2 = (cos2 α + sin2 α) + (cos2 β + sin2 β) − 2(cos α cos β + sin α sin β)
AB2 = 1 + 1 − 2(cos α cos β + sin α sin β) −2 − 2(cos α cos β + sin α sin β)
= 2 − 2(cos α cos β + sin α sin β)
Making the two expressions for AB2 equal:
2 − 2 cos(α − β) = 2 − 2(cos α cos β + sin α sin β)
∴ cos(α − β) = cos α cos β + sin α sin β
You have now proved the first of the compound angle formulae.
WORKED EXAMPLE
Determine cos(90° − β) using the formula for cos(α − β):
SOLUTION
Let α = 90°:
cos(90° − β) = cos 90° cos β + sin 90° sin β
cos(90° − β) = (0) cos β + (1) sin β
cos(90° − β) = sin β
To derive the other formulae use the formula for cos(α − β)
Derivation of cos(α + β) = cos α cos β − sin α sin β:
cos(α + β) = cos(α − (−β)
cos(α + β) = cos α cos(−β) + sin α sin(−β)
but cos(−β) = cos β and sin(−β) = −sin β
∴ cos(α + β) = cos α cos β − sin α sin β
Derivation of sin(α + β) = sin α cos β + cos α sin β:
sin(α + β) = cos(90° − (α + β))
sin(α + β) = cos((90° − α) − β)
sin(α + β) = cos(90° − α)cos β + sin(90° − α) sin β
but cos(90° − α) = sin α and sin(90° − α) = cos α
∴ sin(α + β) = sin α cos β + cos α sin β
REMEMBER
Co-ratios:
sin(90° − θ) = cos θ
cos(90° − θ) = sin θ
sin(90° + θ) = cos θ
cos(90° + θ) = −sin θ
Numerical example:
sin 70° = cos 20°
cos 30° = sin 60°
Derivation of sin(α − β) = sin α cos β − cos α sin β:
sin(α − β) = sin(α + (−β))
sin(α − β) = sin α cos(−β) + cos α sin(−β)
but cos(−β) = cos β and sin(−β) = −sin β
∴ sin(α − β) = sin α cos β − cos α sin β
Unit 2 Derive the compound and double angle identities
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WORKED EXAMPLES
Compound angle formulae are used to simplify the following examples.
Simplify the expressions:
1
sin 20° cos 40° + cos 20° sin 40°
2
cos 50° cos 140° − sin 130° sin 220°
3
cos 3A cos A + sin 3A sin A
In these worked examples you will ‘expand’ the compound angle formula.
4
Show that: sin(A + B).sin(A − B) = sin2 A − sin2 B
5
Find the value of sin 75° without using a calculator.
SOLUTIONS
1
REMEMBER
2
60°
2
cos 50°
2 cos 140° − sin 130° sin 220°
= cos 50°(−cos 40°)
1 − sin 50°(−sin 40°)
= −cos 50° cos 40° + sin 50° sin 40°
= −(cos 50° cos 40° − sin 50° sin 40°)
45°
= −cos(50° + 40°) = −cos(90°) = 0
3
cos 3A cos A + sin 3A sin A
= cos(3A − A)
= cos 2A
4
sin( A + B ).sin(A − B)
= (sin A cos B + cos A sin B)(sin A cos B − cos A sin B) | Difference of two
squares
2
2
2
2
2
= sin A cos B − cos A sin B
| cos x = 1 − sin2 x
2
2
2
2
= sin A(1 − sin B) − (1 − sin A)sin B
= sin2 A − sin2 A sin2 B − sin2 B + sin2 A sin2 B
= sin2 A − sin2 B
5
sin 75° = sin(30° + 45°)
30°
3
45°
2
1
1
45°
2
1
0°
sin 20°cos 40° + cos 20° sin 40°
= sin(20° + 40°)
= sin
__ 60°
√3
= ___
45°
1
Special angle triangles
1
| The sum of two special angles
+ cos 30° sin 45°
sin 75° = sin 30° cos 45°
__
√ 3 ___
1 ___
1__
1__
__
___
=
+
( ) ( 2 )( 2 )
2 √2
__
√3
1 __ ____
__
= ____
+
2√2
2√2
__
__
√2
1+√
3
__ × ___
__
= _______
√2
2√2
__
__
√2 + √6
________
=
94
√
4
Topic 5 Trigonometry: Compound and double angle identities
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EXERCISE 2
1
2
3
Use the compound angle formulae to rewrite:
1.1
cos(θ + 2β)
1.2
cos(3x − y)
1.3
sin(3θ + 2β)
1.4
sin(x − 4y)
Rewrite the expressions that follow as the sine or cosine of a single angle and
evaluate where possible without a calculator.
2.1
sin 68° cos 22° + cos 68° sin 22°
2.2
cos 28° cos 2° − sin 28° sin 2°
2.3
cos 108° cos 188° + sin 252° sin 172°
2.4
sin 70° cos 10° − cos 80° cos 70°
2.5
sin 4x sin 3x + cos 4x cos 3x
2.6
sin(90° − x) cos x + cos(90° + x) sin x
2.7
sin 80° cos 50° − cos 80° sin 50°
2.8
cos 45° sin 75° + sin 135° sin 15°
Use the compound angle formulae to prove that:
3.1
cos(90° + θ ) = −sin θ
3.2
sin(360° − θ) = −sin θ
3.3
cos(A + B) + cos(A − B) = 2 cos A cos B
3.4
sin(A + B) + sin(A − B) = 2 sin A cos B
3.5
cos(A + B)cos(A − B) = cos2 A − sin2 B
3.6
sin(A + B) sin(A − B) = cos2 B − cos2 A
3.7
cos(A − 30°) = sin(60°
+ A)
__
√ 3 cos x − sin x
______________
3.8
cos(30° + x) =
2
4
5
3.9
sin(x + 30°) − sin(x − 30°) = cos x
2__
3.10 cos(A − 45°) − sin(A − 45°) = ___
cos A
√
2
__
__
Find the value of the ratios below in terms of √2 and √ 3 using the sum or
difference of two special angles.
4.1
cos 15°
[Hint: 15° = 60° − 45°]
4.2
sin 75°
4.3
cos 105°
4.4
sin 195°
4.5
sin 165°
4.6
tan 15°
Use the given ratios to draw diagrams in the Cartesian plane. First find the
individual ratios and then calculate the ratios of the compound angles. Do not
solve for angles with a calculator.
4
12
5.1
If sin α = __
, α is acute and cos β = − ___
, 0° < β < 180° evaluate cos(β − α).
5
13
5.2
If 13 cos P = 5, P ∈ [ 0°;270° ] and 15 tan Q − 8 = 0 for sin Q < 0, evaluate:
5.2.1
17 tan P.sin Q
5.2.2 sin(P + Q)
5.2.3
cos(P − Q)
3
5
5.3
If tan A = __
, A > 90° and tan B = ___
, B < 90°, evaluate:
4
12
5.3.1
cos(A + B)
5.3.2 sin(A + B)
5.3.3
tan(A + B)
3
5.4
If sin x = __
5 and x > 90° determine the value of:
5.4.1
sin 2x
5.4.2 cos 2x
5.4.3
tan 2x
[Hint: let 2x = x + x]
Unit 2 Derive the compound and double angle identities
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REMEMBER
Identities:
sin2 θ + cos2 θ = 1
cos2 θ = 1 − sin2 θ
sin2 θ = 1 − cos2 θ
Double angle identities
sin 2θ = 2 sin θ cos θ
AND
cos 2θ = cos2 θ − sin2 θ
OR
1 − 2 sin2 θ
OR
2 cos2 θ − 1
These identities are derived from the compound angle identities by letting:
2θ = θ + θ
OR
2x = x + x.
1)
Derivation of sin 2x = 2 sin x cos x
sin 2x = sin(x + x)
sin 2x = sin x cos x + cos x sin x
sin 2x = 2 sin x cos x
2)
Derivation of cos 2x = cos2 x − sin2 x
cos 2x = cos(x + x)
cos 2x = cos x cos x − sin x sin x
cos 2x = cos2 x − sin2 x
3)
Derivation of cos 2x = 1− 2 sin2 x
cos 2x = cos(x + x)
cos 2x = cos x cos x − sin x sin x
cos 2x = cos2 x − sin2 x
cos 2x = (1 − sin2)x − sin2 x
cos 2x = 1 − 2 sin2 x
4)
Derivation of cos 2x = 2 cos2 x − 1
cos 2x = cos(x + x) − 2 cos2 x − 1
cos 2x = cos x cos x − sin x sin x
cos 2x = cos2 x − sin2x
cos 2x = cos2 x − (1 − cos2 x)
cos 2x = 2 cos2 x − 1
WORKED EXAMPLES
1
Write sin 10A in terms of ratios with angles of 5A.
2
Write cos 8A in terms of sin 4A.
3
Evaluate: cos2 15° − sin2 15°
A
Write 2 cos2 __
− 1 in terms of cos A.
4
6
12
Find cos 2A if sin A = − ___
and 0° < A < 270°.
13
1
__
If sin 40° = determine cos 80° and sin 280° in terms of a.
7
Prove that cos 3x = 4 cos3 x − 3 cos x
5
96
(2)
a
Topic 5 Trigonometry: Compound and double angle identities
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SOLUTIONS
1
sin 10A = sin 2(5A)
= 2 sin 5A cos 5A
2
cos 8A = cos 2(4A)
y
= 1 − 2 sin2 4A
3
cos2 15° − sin2 15° = cos 2(15°)
__
√3
= cos 30° = ___
2
( )
( )
–5
4
A
A
2 cos2 __
− 1 = cos 2 __
= cos A
2
2
5
12 2
cos 2A = 1 − 2 sin2 A = 1 − 2 −___
13
−119
144
= 1 − 2 ____ = _____
( 169 )
(
)
–12
A
x
13
169
or
cos 2A = cos2 A − sin2 A
−5 2
−12 2
= ___ − ____
( 13 ) ( 13 )
25
144 −119
= ____
− ____ = _____
169
169 169
6
Question 5
cos 80° = cos 2(40°)
2
= 1 − 2 sin2 40° = 1 − __2
50°
a
sin 280° = −sin 80° = −sin 2(40°)
= −2s in 40°
cos 40°
_____
2
√
1 a −1
= −2( __ )_______
a_____a
√
−2 a2 − 1
= __________
a2
a
1
40°
a² – 1
Question 6
7
LHS = cos 3x = cos(2x + x)
= cos 2x cos x − sin 2x sin x
= (2 cos2 x − 1) cos x − 2 sin x cos x sin x
= 2 cos3 x − cos x − 2 cos x sin2 x
= 2 cos3 x − cos x − 2 cos x(1 − cos2 x)
= 2 cos3 x − cos x − 2 cos x + 2 cos3x
= 4 cos3 x − 3 cos x
Unit 2 Derive the compound and double angle identities
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EXERCISE 3
1
Use the double angle identities to write each ratio or expression in terms of the
angle and ratio in the bracket.
1.1
sin 4A (2A)
1.2
cos 12A (6A)
1.3
cos 6A (3A)
1.4
cos 8A (cos 4A)
1.5
sin 4A (sin A)
1.6
cos 10A (sin 5A)
1
1
B
__
__
1.7
2 sin 2 x.cos 2 x sin(x)
1.8
2 cos2 __
− 1 (B)
2
( )
2
1.9
4 sin 3A.cos 3A (6A)
1.10 sin x.cos x (2x)
Write each expression as a single angle and evaluate if possible.
2.1
2 sin 30°.cos 30°
2.2
cos2 45° − sin2 45°
2
2.3
1 − 2 sin 30°
2.4
2 sin 22,5°.cos 22,5°
2.5
2 sin 150°.sin 300°
2.6
2 cos2 15° − 1
2.7
3
6
4 sin 75°.cos 75°
(cos x + sin x)2 − 1
3.4
2 sin x. cos x
___________
2 cos2 x − 1
3.6
cos 2x + sin2 x
3.5
sin 2x. tan x + sin2 x
1
__
2
2
3.7
(1 − tan x)(1 − sin x)
3.8
(1 − cos 2x)
2
4
If sin 20° = a, find these expressions in terms of a. [Hint: Draw a diagram.]
4.1
sin 160°
4.2
cos 70°
4.3
tan 200°
4.4
sin 40° [Hint: Use double angles.]
4.5
4 sin 10°.cos 10°
4.6
1 − 2 sin2 35°
4.7
sin 80° [Hint: Use a special angle.]
4.8
cos 50°
Use compound and double angles to prove that:
5.1
sin 3x = 3 sin x − 4 sin3 x
5.2
cos 3x = cos x (1 − 4 sin2 x)
5.3
cos 4x = 8 cos4 x − 8 cos2 x +1
5.4
1 − cos 4x = 2 sin2 2x
5.5
sin 4x = cos x(4 sin x − 8 sin3 )
5.6
sin 4x = 4 sin x cos3 x − 4 cos x sin3 x
Prove that:
sin 3x ______
cos 3x
______
6.1
− cos x = 2
sin x
6.2
98
2.8
4
Simplify each expression as a single trigonometric ratio in x or 2x.
3.2
cos4 x − sin4 x
3.1
(cos2 x − sin2 x)(cos2 x + sin2 x)
3.3
5
4 − 8 sin2 15
___________
(cos x + sin x)(cos x − sin x)
cos 3x ______
sin 3x _______________________
______
+
=
sin x
cos x
sin x cos x
Topic 5 Trigonometry: Compound and double angle identities
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Compound and double angle identities for
tan(α ± β) and tan 2α
tan α + tan β
Note: This is for enrichment
only and is not examinable
Derivation of tan(α + β) = _____________
1 − tan α tan β
sin(α + β)
tan(α + β) = __________
cos(α + β)
1
___________
sin α cos β + cos α sin β ___________
cos α cos β
______________________
×
=
cos α cos β − sin α sin β ___________
1
cos α cos β
sin α cos β
cos α sin β
___________
+ ___________
cos
α
cos
β
cos α cos β
= ________________________
cos
α
cos
β
sin α sin β
___________ − ___________
cos α cos β
cos α cos β
tan α + tan β
= _____________
1 − tan α tan β
tan α − tan β
Derivation of tan(α − β) = _____________
1 + tan α tan β
sin(α − β)
tan(α − β) = __________
cos(α − β)
1
___________
sin α cos β − cos α sin β ___________
cos α cos β
______________________
=
×
cos α cos β + sin α sin β ___________
1
cos α cos β
sin α cos β
cos α sin β
___________
− __________
cos α cos β
cos α cos β
________________________
=
cos
α
cos
β
sin
α sin β
___________ + ___________
cos α cos β
cos α cos β
tan α − tan β
= ______________
1 + tan α tan β
2 tan α
1 − tan α
Derivation of tan 2α = _________
2
tan 2α = tan(α + α)
tan α + tan α
= ______________
1 − tan α tan α
2 tan α
= __________
1 − tan2 α
Unit 2 Derive the compound and double angle identities
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WORKED EXAMPLES
1
2
Use the formula for tan(α − β) to find the value of tan 165°.
If cos x = k, find tan 2x in terms of k.
SOLUTIONS
1
The alternative solutions are
for enrichment only and are
not examinable.
Alternative solution:
tan 165° = −tan 15°
sin 15°
= −_______
cos 15°
sin(45° − 30°)
= −_____________
cos(45° − 30°)
sin 45° cos 30° − cos 45° sin 30°
= − ____________________________
cos 45° cos 30° + sin 45° sin 30°
(
(
__
√3
1__ ___
1__ __
___
.
− ___
.1
)
__
tan 165° = −tan 15°
= −tan(45° − 30°)
(
)
tan 45° − tan 30°
= − ________________
1 + tan 45° tan 30°
1__
__
1 − ___
√3
√3
___
__________
=−
× __
1__
√3
1 + 1 × ___
√
(
√2 2
√2 2
2√__
2
__
= − _____________
× ____
√
√
3
1__ ___
1__ __
___
.
+ ___
.1
2
√2
√2 2
2 2
__
__
(√3 − 1)
(√ 3 − 1)
__
__
= = −________
× ________
√3 + 1
(√ 3 − 1)
__
√
(3 − 2 3 + 1)
_____________
=−
2
√3 + 1
__
= −2 + √ 3
tan 2A formula:
2√1 −
k2 .k
______
= ______________
2_______
k
2k√1 − k
k −1+k
2
= ___________
2
2
______
2k√1 − k2
2k − 1
= __________
2
(√3 − 1)
= −____________
3−1
__
______
__
__
(3 − 2√3 + 1)
= − 2 + √3
sin 2x
tan 2x = _______
cos 2x
sin 2x
= _______
cos 2x
2 sin x. cos x
= _____________
cos2 x − sin2 x
k2 − (√1 − k2 )2
)
(√3 − 1)
−(√ 3 − 1) ________
__
= _________
× __
3−1
______
3
__
)
Alternative solution using
_____
cos x = k ∴ sin x = √ 1 − k2
tan 2x = _________
2
2 tan x
1 − tan x
_____
√ 1 − k2
1
1 – k²
k2
= __________ × __
( 1 −k k )
2
1 − ______
2
_____
2k√ 1 − k2
= _________
k2 − 1 + k2
_____
k2
x
k
2k√ 1 − k2
= _________
2k2 − 1
EXERCISE 4 (EXTENSION WORK)
1
1 + tan β
Use the formula for tan(α + β) to prove that: tan(45° + β) = ________
1 − tan β
__
5
Use the formula for tan(α + β) to prove that tan 75° = 2 + √ 3
If tanx = p, find tan2x in terms of p.
2
Prove that tan(45° + x) + tan(45° − x) = ______
cos 2x
a
If tan x = __ and x is an acute angle, find sin 2x and tan 2x in terms of a and b.
6
6.1
2
3
4
b
Using the expansions for sin(A + B) and cos(A + B), prove the identity
sin(A + B) _____________
tan A + tan B
_________
=
1 − tan A tan B
sin(A + B)
If tan(A + B) = __________
,
cos(A + B)
cos(A + B)
6.2
prove that in any △ABC, tan A.tan B.tan C = tan A + tan B + tan C.
100
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Unit 3: Prove identities using compound and
double angle identities
You can use compound and double angle identities to prove more complex identities.
To prove identities follow these steps:
• Ensure all the ratios are in terms of sin x and cos x: write tan x in terms of sin x
and cos x.
• Change all double angles to single angles so that you are working with ratios
of one angle.
• Use the square identities and compound angle identities where necessary.
• See the LHS or RHS as an algebraic expression and use algebraic manipulations to
simplify both sides as far as possible.
• If there are fractions, find the LCD and add.
• If there are fractions/fractions, simplify as you would in algebra by multiplying
LCD
by ____
.
LCD
Factorise
where possible and simplify.
•
• End your proof with LHS = RHS.
Note: Identities involving compound angles do not necessarily follow the steps above.
WORKED EXAMPLES
1
REMEMBER
Prove that 1 − cos 2x = tan x sin 2x.
Proof:
RHS = tan x sin 2x
sin x
= _____ × 2 sin x.cos x
LHS = 1 − cos 2x
= 1 − (1 − 2 sin2 x)
= 2 sin2 x
cos x
= 2 sin2 x
∴ LHS = RHS
2
cos 2x
cos x − sin x
Prove that _________
= ___________
cos x + sin x
1 + sin 2x
REMEMBER
Proof:
cos 2x
cos2 x − sin2 x
= __________________________
LHS = _________
2
2
1 + sin 2x
(sin x + cos x) + 2 sin x cos x
(cos x − sin x)(cos x + sin x)
= _______________________
(cos x + sin x)(cos x + sin x)
| Difference of two squares
| Perfect square trinomial
(cos x − sin x)
= ____________ = RHS
(cos x + sin x)
3
Double angle identities:
sin 2θ = 2 sin θ cos θ
cos 2θ = cos2 θ − sin2 θ
cos 2θ = 1 − 2 sin2 θ
cos 2θ = 2 cos2 θ − 1
Other identities:
sin θ
tan θ = _____
cos θ
sin2 θ + cos2 θ = 1
cos2 θ = 1 − sin2 θ
sin2 θ = 1 − cos2 θ
1 − cos 2x
Prove that tan2x = _________
1 + cos 2x
Proof:
1 − cos 2x
RHS = _________
1 + cos 2x
1 − (1 − 2 sin2 x)
= ______________
1 + 2 cos2 x − 1
2 sin2 x
= _______
2 cos2 x
| Use the double angle identities that cancel the 1.
= tan2x
= LHS
Unit 3 Prove identities using compound and double angle identities
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EXERCISE 5
1
Prove the identities by using mostly the double angle identities.
1.1
cos 2x = cos4 x − sin4 x
1.2
1 − sin 2x
___________
= sin x − cos x
1.3
sin x + sin 2x
_______________
= tan x
1.4
1 + cos x + cos 2x
(cos2 x − sin2 x)2
______________
= cos 2x
cos4 x − sin4 x
1.5
1 − cos 2x
_________
= tan x
1.6
sin 2x − cos x
cos x
_______________
= _____
1.7
1 − cos 2x + sin 2x
________________
= tan x
1.8
sin x − cos 2x ________
sin x + 1
____________
=
1.9
1.10
2
sin x − cos x
sin 2x
sin x
1 − cos 2x − sin x
1 + cos 2x + sin 2x
cos x
sin 2x − cos x
2
(cos x + sin x) − 1
_______________________
= tan 2x
(cos x + sin x)(cos x − sin x)
2
1 − cos x − sin(−2x)
__________________
= sin x
sin x + 2cos x
Prove the identities using compound angles and double angle identities.
2.1
sin(x + y)
_________
= tan x + tan y
2.2
cos 2x _____
sin 2x ______
1
______
−
=
2.3
1
sin(45° + x)sin(45° − x) = __
− sin2 x
2
cos x cos y
sin x
cos x
cos x
2.4
1
cos(60° + x)cos(60° − x) = __
− sin2 x
4
2.5
2.6
2 cos 6x cos 4x − cos 10x + 2 sin2 x = 1
2 sin 5x.cos 4x − sin 9x = sin x
2.7
tan x − tan y = _________
cos x cos y
2.8
sin 3x − sin x
____________
= cos 2x
2.9
cos 2x + 2 sin 2x + 2 = (3 cos x + sin x)(cos x + sin x)
sin(x − y)
2 sin x
cos x 2sin(x − y)
sin x _____
2.10 _____
− cos y = _________
sin y
sin 2y
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Unit 4: Solve equations and determine
the general solution
In Grade 11 you learnt how to solve trigonometric equations by finding the general
solution and using it to find specific solutions in a given interval.
In Grade 12 double and compound angles appear in trigonometric equations. You can
solve these by substituting in the double angle and compound angle formulae into
the equations.
Always find the general solution first and solve for specific solutions in a given
interval only when required.
Solve equations that end with one ratio
WORKED EXAMPLE 1
y
Solve for x if 3 cos(x − 25°) + 2 = 0
and −360° ≤ x ≤ 360°.
SOLUTION
3 cos(x − 25°) + 2 = 0
2
cos(x − 25°) = −__
3
KEY WORDS
x – 25°
2
1
x – 25°
48,19°
x
48,19°
– ( x – 25°)
3
4
solution – a value of the
angle which satisfies a given
trigonometric equation
specific solutions –
solutions that satisfy a given
trigonometric equation in a
restricted interval, such as
–360° ≤ x ≤ 360°
general solution – a formula
that lists all possible solutions
to a trigonometric equation;
it takes into account the
period of the trigonometric
functions, so the angle can be
positive or negative
Method 1: Use the positive ratio, acute angle and quadrant method.
2
Calculator angle = 48,19°
| Find the acute angle by keying in shift cos−1 __
.
3
Quadrant 2 (positive angle):
(x − 25°) = 180° − 48,19° + n.360°
x = 131,81° + 25° + n.360°
x = 156,81° + n.360°, n ∈ ℤ
Quadrant 3 (positive angle):
(x − 25°) = 180° + 48,19° + n.360°
∴ x = 228,19° + 25° + n.360°
x = 253,19° + n.360°, n ∈ ℤ
( )
| Determine the quadrants in which cos is
negative (2 and 3).
| The solution will be (x − 25°) = 180° −
calculator angle (Quadrant 2).
| Do not leave out the general solution (n.360°).
| Solve for the unknown x.
Note: This is a Grade 11 type
question.
| The solution will be (x − 25°) = 180° +
calculator angle (Quadrant 3).
| Solve for the unknown x.
For the negative angles:
x = 156,81° + (−1).360° = −203,19°
x = 253,19° + (−1).360° = −106,81°
Unit 4 Solve equations and determine the general solution
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REMEMBER
If you choose to use the
method where you key in
the given ratio (positive or
negative), the ‘calculator’
angle could be acute,
negative or obtuse.
For sin x = a the general
solution is:
x = sin−1(a) + n.360°, n ∈ ℤ
or
x = 180° − sin−1(a) + n.360°,
n∈ℤ
For cos x = b the general
solution is:
x = cos−1(b) + n.360°, n ∈ ℤ
or
x = −cos−1(b) + n.360°, n ∈ ℤ
For tan x = c the general
solution is:
x = tan−1(c) + n.180°, n ∈ ℤ
Method 2: Key in the given ratio (+ or −) and use ‘2 options’ method.
Calculator angle = 131,81°
1) (x − 25°) = 131,81° + n.360°
| For cos: Option 1 (x − 25°) = calculator angle
x = 156,81° + n.360°
| Solve for the unknown x
2) (x − 25°) = −131,81° + n.360° | For cos: Option 2 (x − 25°) = −calculator
angle
x = −106,81° + n.360°
| Solve for the unknown x.
WORKED EXAMPLE 2
This example uses double angles.
__
Solve for x if 4 sin xcos x = √ 3 and −180° ≤ x ≤ 360°
SOLUTION
__
4 sin x cos x = √3
__
2(2 sin x cos x) = √ 3
__
2(sin 2x) =__√ 3
√3
sin 2x = ___
| Recognise 2 sin x cos x = sin 2x.
| Isolate the ratio.
Calculator angle = 60°
1) 2x = 60° + n.360°
x = 30° + n.180°, n ∈ ℤ
2) 2x = 180° − 60° + n.360°
120°
x = ____
+ n.180°
2
| Recognise the special angle ratio.
| Option 1: 2x = calculator angle + n.360°
| Solve for x. Note n.360° ÷ 2 = n.180°
| Option 2: 2x = 180° − calculator angle + n.360°.
| Solve for x. Note n.360° ÷ 2 = n.180°.
2
x = 60° + n.180°, n ∈ ℤ
Now −180° ≤ x ≤ 360° :
x = 30° or 30° + 180° = 150° or x = 60° or x = 60° + 180° = 240°
x = 30° − 180° = −150° or x = 60° − 180° = −120°
WORKED EXAMPLE 3
This example uses compound angles.
Give the general solution to the equation: sin 3xcos x − cos 3xsin x = 0,4
SOLUTION
sin 3x cos x − cos 3x sin x = 0,4 | Recognise this as sin(A − B) where A = 3x and B = x
sin(3x − x) = 0,4
sin 2x = 0 ,4
| To solve sin A = 0,4 press shift sin−1(0,4)
1) 2x = 23,58° + n.360
| Remember that two options are required (quadrants
1 and 2).
x = 11,79° + n.180°
2)
104
2x = 180° − 23,58° + n.360
2x = 156,42 + n.360°
x = 78,21 + n.180°, n ∈ ℤ
Topic 5 Trigonometry: Compound and double angle identities
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WORKED EXAMPLE 4
REMEMBER
Give the general solution to the equation.
cos 2x
sin 2x
_______
− _______ = 0
cos 50°
cos 40°
SOLUTION
cos 2x cos 40° − sin 2x cos 50° = 0 | Multiply by the LCD = cos 50° cos 40°.
cos 2x cos 40° − sin 2x sin 40° = 0 | Recognise: co-ratio cos 50° = sin 40°
cos(2x + 40°) = 0
| Recognise: compound angle formula cos(A + B)
where A = 2x, B = 40°
If sin x = 0
x = 0° + n.180°
If sin x = 1
x = 90° + n.360°
If sin x = −1
x = −90° + n.360°
y
1-
y = sin x
–1 -
EXERCISE 6
-
-
-
0-
x = 25° + n.90°
-
2x = 50° + n.180°
-
2x + 40° = 90° + n.180°
x
90° 180° 270° 360°
y = sin x
If cos x = 0
x = 90° + n.180°
If cos x = 1
x = 0° + n.360°
If cos x = −1
x = 180° + n.360°
Determine the general solution for each equation.
1
sin x cos 20° − cos x sin 20° = 0,38
2
cos x cos 25° − sin x sin 25° = 0,65
3
sin x cos 60° + cos x sin 60° = 0,66__
√3
4
cos 2x cos 40° + sin 2x sin 40° = ___
2
5
2sin x cos x = −0,42
6
sin x cos x = 0,25
7
cos2 x − sin2 x = 0,66
8
2 cos2 x − 1 = −0,75
9
sin2 x − cos2x = 0,67
cos x
sin 4x
_____
10 sin x + ______
=0
cos 4x
y
1-
Solve equations with double angles and more
than one ratio
–1 -
-
-
-
-
0-
-
y = cos x
x
90° 180° 270° 360°
y = cos x
Follow these steps to solve equations with double angles and more than one ratio.
1. Convert all ratios to sine and cosine so you have only one ratio if possible.
sin x
2
2
2
2
Use the identities tan x = _____
cos x , cos x = 1 − sin x, sin x = 1 − cos x
2.
Convert all angles to the same size angles if possible.
Use the identities sin 2x = 2 sin x cos x, cos 2x = 1 − 2 sin2 x or cos 2x = 2 cos2 x − 1
where you have double and single angles in one equation.
3.
Simplify and factorise to change into simpler equations.
4.
Find the general solution and solve for the given interval.
Unit 4 Solve equations and determine the general solution
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REMEMBER
Double angle identities:
sin 2θ = 2 sin θ cos θ
cos 2θ = cos2 θ − sin2 θ
cos 2θ = 1 − 2sin2 θ
cos 2θ = 2cos2 θ − 1
WORKED EXAMPLE 1
Determine the general solution of sin 2x + cos x = 0.
SOLUTION
sin 2x + sin x = 0
| Change double angles to single angles:
2 sin x cos x + sin x = 0
| Use sin 2x = 2 sin x cos x
sin x(2 cos x + 1) = 0
| Factorise.
1
sin x = 0 or cos x = −__
2
| Solve for cos x and sin x.
x = 0° + n.180° or x = ± 120° + n.360°, n ∈ ℤ
| Solve for x − remember the two
options.
WORKED EXAMPLE 2
Solve for x if 3sin x = 1 + cos 2x and −180°≤ x ≤ 360°.
SOLUTION
| Change double angles to single angles.
3 sin x = 1 + cos 2x
3 sin x
Note: Remember the
two options:
x = A or x = 180° − A
where A = special or
calculator angle.
| Use cos 2x = 1 − 2 sin2 x because sin x is in the
equation.
= 1 + 1 − 2 sin2 x
2 sin2x + 3 sin x − 2 = 0
| Factorise the trinomial (2k2 + 3k − 2)
where sin x = k.
(2 sin x − 1)(sin x + 2) = 0
1
or sin x = −2
sin x = __
| Solve for sin x.
1
: x = 30° + n.360°
For sin x = __
or
| Solve for x (general solution).
2
2
x = 150° + n.360°
For sin x = −2: No solution because −1 ≤ sin x ≤ 1
For −360° ≤ x ≤ 360°: x = 30°; −330°; 150° or −210°
WORKED EXAMPLE 3
Find the general solution of the equation 2cos x = sin(x + 30°).
SOLUTION
| Expand sin(x + 30°) using
sin(A + B) = sin A cos B + cos A sin B.
2 cos x = sin(x + 30°)
2 cos x = sin x cos__ 30° + cos x sin 30° | Use special angle values for cos 30°and sin 30°.
√3
1
+ cos x __
| Collect like terms.
2 cos x = sin__ x ___
2
2
√3
3
2
__
___
_____
cos x =
sin x
| Multiply both sides by _______
( )
2
3__ _____
sin x
___
=
( )
2
√ 3cos x
cos x
__
3__
∴ tan x = √ 3
tan x = ___
√3
√3
x = 60° + n.180°, n ∈ ℤ
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WORKED EXAMPLE 4
Solve 3 sin2 x − sin 2x − cos2 x = 0 and 0° ≤ x ≤ 360°.
SOLUTION
3 sin2 x − sin 2x − cos2 x = 0
3 sin
2 x − 2 sin x cos x − cos2 x = 0
| Use double angle sin 2x = 2sin x cos x
| Recognise the quadratic equation
3s2 − 2sc − c2 = 0.
(3 sin x + cos x)(sin x − cos x) = 0
| Solve by factorising.
3 sin x = −cos x or sin x = cos x
cos x
sin x
−cos x
3 sin x ______
______
=
or _____ = _____
sin x
| Divide by cos x and use identity _____
cos x = tan x
cos x
cos x
cos x
1
__
tan x = − 3 or tan x = 1
cos x
| Now solve for x.
x = −18,4° + n.180° or x = 45° + n.180°, n ∈ ℤ
For 0° ≤ x ≤ 360°: x = 161,6° or 45° or 57° or 225° or 341,5°
EXERCISE 7
Solve for x, giving the general solution first and specific solutions if an interval
is given.
1
2
3
4
5
6
7
8
9
10
11
12
cos 2x + cos x = 0
sin2 x + sin 2x = 0, and −360° ≤ x ≤ 360°
3 cos 2x + cos x + 2 = 0
2 sin 2x − 2 sin x = 6 cos2 x − 3 cos x, and x ∈ [−360°;360°]
cos 2x + cos x + 1 = 0
cos 2x + sin x − 1 = 0
11 cos2 x − 4 cos 2x = 6 + cos x
cos 2x − 4 sin x + 5 = 0
sin 2x = 2cos2 x
cos 2x + 3 cos x − 1 = 0 and −180° ≤ x ≤ 180°
sin 2x − sin x = 1 − 2 cos x
1
2 cos 2x + __
sin 2x = sin2 x and 0° ≤ x ≤ 360°
2
13
14
4 sin x sin 2x + sin 2x − cos x = 0
sin 2x + cos 2x − 1 = 0 and −180° ≤ x ≤ 180°
Solve equations using compound angles and
no calculator
In Grade 11 you solved equations of the type:
sin A = sin B, cos A = cos B or tan A = tan B
To find the general solution for these equations, you used this ‘method’:
1. If sin A = sin B then A = B + n.360° or A = 180° − B + n.360°
2. If cos A = cos B then A = B + n.360° or A = −B + n.360°
3. If tan A = tan B then A = B + n.180°
4. If sin A = cos B then sin A = sin(90° − B)
and A = (90° − B)+ n.360° or A = 180° − (90° − B) + n.360°
Unit 4 Solve equations and determine the general solution
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WORKED EXAMPLE 1
REMEMBER
Remember the ‘rule’ for
sin A = sin B
A = B + n.360°
or A = 180° − B + n.360°
Solve for x if sin(3x − 20°) = sin(x + 10°) and −360° ≤ x ≤ 360°
SOLUTION
Quadrant 1
3x − 20° = x + 10° + n.360°
2x = 30° + n.360°
x = 15° + n.180°, n ∈ ℤ
For −360° ≤ x ≤ 360° use integral values
for n
x = 15°; 195°; −165°; −345°
Find the general
solution first by
considering the two
options where sin is
positive.
Quadrant 2
3x − 20° = 180° − (x + 10°) + n.360°
4x = 190° + n.360°
x = 47,5° + n.90°, n ∈ ℤ
For −360° ≤ x ≤ 360° use integral
values for n
x = 47,5°; 137,5°; 227,5°;
317,5°; −42,5°; −132,5°; −222,5°; −312,5°
WORKED EXAMPLE 2
Find the general solution of cos 2x cos 30° − sin 2x sin 30° = sin x.
SOLUTION
cos 2x cos 30° − sin 2x sin 30° = sin x | Recognise the compound angle cos(A + B).
cos(2x + 30°) = sin x
| Use the co-ratio sin x = cos(90° − x).
cos(2x + 30°) = cos(90° − x)
Consider the two options where cos is positive (Quadrants 1 and 4).
REMEMBER
Quadrant 1
2x + 30° = 90° − x + n.360°
3x = 60° + n.360°
x = 20° + n.120°, n ∈ ℤ
Remember the rule for
cosA = cosB
A = B + n. 360°
or A = −B + n.360°
Quadrant 4
2x + 30° = −(90° − x) + n.360°
2x + 30° = −90° + x + n.360°
x = −120° + n.360°, n ∈ ℤ
For Quadrant 4 you may also use
2x + 30° = 360° − (90° − x) + n.360°
x = 240° + n.360° (which is the same as −120°)
EXERCISE 8
Solve for x, giving the general solution first and specific solutions if an interval
is given.
1
sin x cos 25° + cos x sin 25° = sin 2x and −180° ≤ x ≤ 180°
2
cos x cos 30° + sin x sin 30° = sin 2x
3
cos x cos 330° + sin x cos 120° = cos 2x and 0° ≤ x ≤ 360°
4
cos(45° + x)cos(45° − x) + sin(45° + x)sin(45° − x) = cos(45° + x)
5
sin(50° + x)cos 20° + cos(50° + x) sin 200° = cos(10° + x) and −180° ≤ x ≤ 180°
6
cos x sin 63° + sin x sin 27° = 2 sin x cos x
cos 2x
sin 2x
_______
7
+ _______
= −1
sin 45°
cos 45°
8
108
cos 40°
sin 40° _______
_______
−
=2
sin x
cos x
Topic 5 Trigonometry: Compound and double angle identities
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Revision Test Topic 5
Total marks: 233
1
This test includes
Grade 11 work.
If tan x = p and x is acute, match the expressions
2p
1+p
1 − p2
1+p
2
_____
A: _______
; B: ______2 and C: ______2
2
√1 + p
to the answers for the ratios below in terms of p:
1.1
1.2
1.3
2
3
4
5
cos 2x
sin 2x
cos(90° − 2x)
___________
(3)
(3)
(4)
cos(90° − x)
If cos 32° = m, find these ratios in terms of m:
2.1
sin 148°
2.2
cos(−58°)
2.3
sin 244°
2.4
cos 296°
2.5
cos 26° [Hint: 26° = 58° − 32°]
(2)
(2)
(3)
(3)
(3)
4 ^
12 ^
If sinA = ___
, A is obtuse and tan B = __
, B > 90° determine these ratios without
3
13
calculating values for the angles:
3.1
sin(A − B)
(4)
3.2
cos(A − B)
(4)
3.3
tan(A − B)
(5)
Use the compound angle formulae to prove that:
1
__
[
]
4.1
2 cos(A + B) + cos(A − B) = cos A cos B
4.2
cos(90° +A) = −sin A
__
1 + √3
4.3
sin(30° + x) + cos(30° − x) = _______
(sin x + cos x)
2
2
4.4
sin(A + B) sin(A − B) = sin A − sin2 B
4.5
cos(45° − x) − sin(45° + x) = 0
__
4.6
2 sin(30° − x) = cos x − √3 sin x
(3)
(2)
(4)
(5)
(3)
(3)
Convert these expressions to one trigonometric ratio and evaluate if possible.
5.1
sin 105° cos 45° + cos 105° sin 45°
(3)
5.2
sin 25° cos 55° + cos 205° cos 35°
(3)
5.3
sin 150° cos 30° + cos 150° sin 330°
(4)
5.4
cos2 15°− cos2 75°
(3)
5.5
1 − 2 sin2 225°
(3)
cos 10° cos 340° − sin 190°. sin 200°
______________________________
5.6
(5)
sin 10° sin 70° + sin 100° sin 20°
6
Show that:
6.1
6.2
6.3
6.4
__
tan 120°. sin 390° cos 156°.cos(−135°) ___
√3
________________________________
=
sin 315°. sin 66°
sin 105° cos 75°. tan 135°
1
______________________
= −__
2
sin(−330°)
2
sin 80° − sin 40° = sin 20° [Hint: 80° = 60° + 20° and 40° = 60° − 20°]
3
sin2 20° + sin2 40° + sin2 80 = __
2
(5)
(4)
(4)
(6)
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TOPIC 5: REVISION CONTINUED
7
8
Prove that:
1
7.1
sin(45° + x).sin(45° − x) = __
cos 2x and hence
2
determine the value of sin 75°. sin 15°
7.2
sin 105° + cos 105° = cos 45°.
(5)
(5)
Prove these identities:
8.1
1 − sin 2x = (sin x − cos x)2
(2)
2 sin x cos x
____________
= tan 2x
cos2 x − sin2 x
sin x + sin 2x
1 + 2 cos x
_______________
= _________
1 + 2 sin x
1 + sin x − cos 2x
(2)
8.4
cos2 3x − cos 6x = sin2 3x
(2)
8.5
sin 2x tan x
1
1
________
+ ________ = ___________
1 + cos x 1 − cos x
sin4x
(6)
8.6
cos 2x − cos x _____
1
1
____________
=
− _____
(5)
8.7
cos4 x + sin2 x cos2 x
__________________
= 1 + sin x
1 − sin x
(4)
8.8
1
(1 − cos 2x) 1 + _____
=2
2
(4)
8.2
8.3
sin 2x + sin x
(
tan x
tan x
(5)
sin x
)
8.9
cos 2x + 2 sin 2x + 2 = (3 cos x + sin x)(cos x + sin x)
(3)
8.10 For which values of x ∈ [0°; 360°] are the identities in 8.2, 8.3 and 8.6
undefined?
(5 × 3)
9
10
11
12
Solve the equations in the given intervals:
3 sin 2x + 1
__________
9.1
= 1 and −180° < x < 180°
2
9.2
6 − 10 cos x = 3 sin2 x’ and x ∈ [ −360°;360° ]
9.3
sin2 x − cos2 x = 1 and x ∈ [0°;360°]
9.4
cos2 x + cos 2x = 2 and x ∈ [−180°;180°]
9.5
3 sin 2x − 3 sin x = 4 cos2 x − 2 cos x’ and x ∈ [ − 360°;360° ]
(8)
(8)
(5)
(6)
(9)
Find the general solutions to these equations.
10.1 sin(80° − x) = cos(3x − 76°)
10.2 sin(80° − x) cos 2x + cos(80° − x) sin 2x = 0,874
10.3 1 − 2 sin2 x = 2 sin x cos x
10.4 2 cos2 x + cos 2x + sin 2x = 0
sin 30° _______
cos 30°
10.5 _______
+ cos x = 2
sin x
(7)
(5)
(5)
(7)
(7)
Prove that _______________
= tan x, and then solve for _______________
= −1
sin 2x − cos x
sin 2x − cos x
1 − sin x − cos 2x
1 − sin x − cos 2x
(7)
cos 2x
sin 2x
Given the equation _______
+ _______
= 2, show that the equation can be
sin 60°
sin 30°
__
√3
. Then find the general solution for x.
written as sin(2x + 60°) = ___
2
(8)
110
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Option 1: Investigation: Find the maximum area
of polygons with a fixed perimeter
Time: 2 hours
Total marks: 70 or 75
The problem: To investigate the polygon that has the largest area with a
fixed perimeter
Instructions
1) Use a pencil and ruler to draw the polygons, which you do not
have to construct accurately.
2) Use simple geometry and trigonometry to calculate the areas of
the polygons.
3) Try to keep the answers in surd form as long as possible and round
off to two decimal places at the end.
4) All calculations and diagrams must accompany your investigation.
5) When comparing areas it may be to your advantage to set your
work out in a table.
6) You may use Geogbra software in this investigation to assist you
with drawing the polygons. You may also use an Excel spreadsheet to do
the area calculations. But we recommend that you use pen and paper.
7) Your investigation will be marked according to the given rubric (70 marks)
or with a mark scheme (75 marks).
Information to assist you in your investigation:
A
A polygon is a many-sided closed geometric figure with straight lines.
c
A regular polygon has all sides and interior angles equal.
b
The sum of the interior angles of a polygon with n sides is (n − 2)180°.
The sum of the exterior angles of any polygon is 360°.
1
1
base × height; A = __
ab sin C
Formulae for area of triangle: A = __
2
B
a
C
2
Use Heron’s formula when you know the lengths of all three sides of a triangle.
Heron’s formula states that:
_________________
Area △ABC = √ p(p − a)(p − b)(p − c)
where a, b and c are the lengths of the sides of the triangle and p is half the perimeter
a+b+c
of the triangle: p = ________
.
2
a
c
b
Trigonometric formulae for triangles: _____
= _____
= _____
; a2 = b2 + c2 − 2bc.cos A
sin B
sin A
sin C
Option 1: Investigation: Find the maximum area of polygons with a fixed perimeter
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Investigation continued
Mark allocation
0−3
4−7
8−10
Choice and
length of sides
Most lengths of sides
are incorrect
At least half of
the choices and
lengths of sides
are correct
All choices and
lengths of sides
are correct
Many diagrams
incorrect and do not
accurately represent
the polygons
At least half of
the diagrams
are correct, but
poorly drawn,
and construction
lines are not
shown
Diagrams
Area
Calculations
Inaccurate
calculation
Or answers only and
no steps are shown
All conclusions
correct, but
conjectures are
not worded
appropriately
All correct
Neatness and
presentation
Satisfactory layout
and presentation,
Layout, logic and
presentation are poor but arguments are
average
10
All construction
lines are shown
All calculations
are accurate
and all steps are
shown
Conclusions
and Conjectures
70
10
Diagrams
are neat and
accurate
Most calculations
are correct, but
not all working
out is shown
Some conclusions
are correct but no
conjectures are
correct
Total
(10 × 3) = 30
Conjectures
are well
communicated
and accurate
10
General
presentation
and layout are
excellent
10
Task 1
1
2
3
How many different triangles can you make with a perimeter of 12 units?
Use integer values for the lengths of the sides or use 12 matchsticks to
investigate how to make up your triangles.
Find the area of each triangle and decide which has the greatest area.
(3)
(6)
(9)
Task 2
1
2
3
112
How many different rectangles can you make with a perimeter of 12 units?
(3)
Use integer values for the lengths of the sides or use 12 matchsticks to investigate
how to make up your rectangles.
(3)
Find the area of each rectangle and decide which has the greatest area.
(7)
Term 1 Assessment
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Task 3
1
Use your results from Tasks 1 and 2 to make a conjecture about the type
of polygon that, with the same number of sides, will give a maximum area.
(3)
Task 4
1
2
3
4
Investigate the areas of regular polygons with a fixed perimeter of 12 units.
Calculate the length of the side of each polygon. This does not have to be
an integer value; use this information to find the area.
(Hint: divide the polygons into triangles and use these to find the area of each
polygon.)
Investigate polygons with 3, 4, 6, 8 and 12 sides.
Lastly, calculate the area of a circle with a perimeter or circumference of
12 units.
(5)
(5)
(22)
(4)
Task 5
1
2
If you knew that two regular polygons had the same perimeter, describe
in words how you would predict which one had the greater area.
Use your results from Task 4 to make a conjecture about the type of polygon
or figure that yields a maximum area, given a fixed perimeter.
(2)
(3)
Option 1: Investigation: Find the maximum area of polygons with a fixed perimeter
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Option 2: Project: Financial maths
Time: 2 hours
Total marks: 100
Note:
•
•
1
2
3
4
5
6
114
This project must be your own work, but you
may ask any of the maths teachers for guidance.
Your teacher will give you a copy of the rubric.
The mark allocation is clearly specified in the
rubric. Make sure you know how your project
will be assessed. Return your fully signed rubric
with your project.
Create a scenario about yourself to explain your current
financial circumstances.
Investigate two ways of providing financing for one of
the options listed below.
•
the purchase of a car (new or second hand)
•
the purchase of a house
•
a loan to finance your tertiary studies
•
a loan for extensive overseas travels
Having made your choice, investigate and supply all necessary
information:
•
price; make of car; year, insurance costs, deposit, and so on
•
price of house; address; plot number, size; deposit, and so on
•
academic costs anticipated for 3 years of study
•
return airfare; destination, accommodation
costs, and so on.
Show all your calculations. Support your
conclusions by your calculations. Remember not
to round off during calculations as this causes
inaccuracies in answers.
Your project should be 2−3 pages long.
Your project must contain:
•
an interesting introduction explaining your
scenario
•
a body (your actual work, including two
different financial options)
•
a conclusion which weighs up your options
and justifies your final decision.
Term 1 Assessment
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Term 1 summary
Topic 1
Patterns and sequences
Topic 2
Functions and inverse functions
Note: It is important to know the derivations of
formulae for the examinations.
• Arithmetic sequences: a sequence of numbers with
a common difference between consecutive terms (a
linear pattern)
• Tn = a + ( n − 1 )d; d = Tn+1 − Tn
• where a = first term; d = common difference;
n = number of terms;
Tn = value of nth term.
• Geometric sequences: a sequence of numbers with
a common ratio between consecutive terms (an
exponential pattern)
A function is a relationship or rule between two
sets, the domain (x) and the range (y), where every
element of the domain is assigned to one and only one
element of the range.
• Functions can be one-to-one (for example, straight
lines) or many-to-one relations (for example,
parabolas).
• You can use the vertical line test (which must cross
the graph once) to determine whether a graph is a
function, but the definition of a function must be
given as the reason for it being a function or not.
The inverse of a function is the graph obtained by
reflecting the function about the line y = x.
• This means swapping the x and y values:
(x;y) → (y;x)
• f −1 is used to represent the inverse of f (x).
• The inverse of a many-to-one function will not be a
function, unless its domain is restricted.
Tn+1
• Tn = ar n−1 ; r = ____
T
n
• where a = first term; r = constant ratio;
n = number of terms;
Tn = value of nth term.
• Arithmetic series: sum of an arithmetic sequence
n
__
• Sn = 2 ( a + l ) if last term (l) is know OR
n
S = __[ 2a + ( n − 1 )d ]
n
2
• Geometric series: sum of a geometric sequence
a( 1 − r n ) ________
a( r n − 1 )
________
=
,r≠1
• Sn = (
)
r−1
1−r
n
• If − 1 < r < 1, then r → 0 as n → ∞, and a sum to
a
infinity can be determined: S∞ = = ____
1−r
• Sigma notation: The answer always represents the
n
sum:
∑ Tk = Sn
k=1
Term 1 summary
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Term 1 summary continued
Topic 3
Exponential and logarithmic
functions
Exponential graphs have the form y = ax, a > 0, a ≠ 1
• The y-intercept is (0;1).
• The x-axis is a horizontal asymptote.
• The domain is x ∈ ℝ and the range is y ∈ ℝ, y > 0.
• If a > 1, the graph increases and if 0 < a < 1, the
graph decreases.
• The graph is a one-to-one function.
The graph of y = ax + p has a horizontal asymptote
y = p.
loga x = y if and only if x = ay where x > 0 and a > 0,
a≠1
The logarithmic function is defined as y = loga x, a > 0,
a ≠ 1, x ∈ ℝ
• The x-intercept is (1; 0).
• The y-axis is a vertical asymptote.
• The domain is x ∈ ℝ, x > 0 and the range is y ∈ ℝ.
• If a > 1, the graph increases and if 0 < a < 1, the
graph decreases.
• The graph is a one-to-one function
• When you compare y = loga x and y = log__1 x you can
a
see that the two graphs are reflections in the x-axis.
• If f(x) = y = ax, a > 0, a ≠ 1 then for f −1: x = ay and
f −1(x) = y = loga x
( )
( )
1 x
1 y
__
__
• If f(x) = y = a , 0 < a < 1 then for f −1: x = a and
f −1(x) = y = log__1 x
a
Topic 4
Finance, growth and decay
• In Grade 11 you learnt how to use the simple and
compound interest formulae and to solve for any
variables in those formulae, except for finding n,
the time period. Knowledge of logarithms in
Grade 12 enables you to apply logs to solve for n
in any of the finance formulae.
• Effective interest is extended so that you can
compare interest rates that are compounded at
different frequencies.
(m) m
i___
) for effective annual interest
• 1 + i = (1 +
m
eff
•
i___
( 1 + i___
m ) = ( 1 + n ) for comparing interest
(m)
m
(n)
rates at different compounding frequencies.
• An annuity is a number of regular payments of
a fixed amount made over a determined time
period.
x[ ( 1 + i )n − 1 ]
____________
• Future value annuities: F =
i
• Present value annuities: P =
x[ 1 − ( 1 + i ) ]
_____________
−n
i
where x = the regular payment,
i = interest rate (which must have the
same regularity of compounding as
the payments are made),
n = number of payments made (which
should start one time period after the
process starts).
If
payments
start
immediately, add 1 to the value
•
of n; if payments start late, subtract the number
of missed payments from what should have been
the value of n. For loans, add interest to the loan if
payments start late.
• Sinking funds are an application of future value
annuities.
• Balance outstanding on loans after k payments
have been made:
( 1 + i )k − 1
B.O. = L( 1 + i )k − x __________
where
i
L = loan amount
or
1 − ( 1 + i )−( n−k )
B.O. = x _____________
i
[
[
116
n
]
]
Term 1 summary continued
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Term 1 summary continued
Topic 5
Trigonometry: compound and
double angles
Note: You must know the derivations of formulae for
examination purposes.
• Compound angle formulae:
cos(α − β) = cos α cos β + sin α sin β
cos(α + β) = cos α cos β − sin α sin β
sin(α − β) = sin α cos β − cos α sin β
sin(α + β) = sin α cos β + cos α sin β
tan α + tan β
tan α − tan β
tan(α + β) = ____________tan(α − β) = ____________
1 − tan α tan β
• Double angle formulae:
1 + tan α tan β
sin 2θ = 2 sinθ cos θ cos 2θ = cos2 θ − sin2 θ
or cos 2θ = 1 − 2 sin2 θ or cos 2θ = 2 cos2 θ − 1
2 tan α
tan 2α = ________
2
1 − tan α
• Other identities
sin θ
tan θ = _____
cos θ
sin2 θ + cos2 θ = 1
cos2 θ = 1 − sin2 θ
sin2 θ = 1 − cos2 θ
• When solving a general solution:
For cosA = k → A = ± B + n. 360° where B is any
acceptable solution for the sign of k and B = cos−1 k.
For tan A = k → A = B + n. 180° where B is any
acceptable solution for the sign of k and B = tan−1 k.
For sinA = k → A = B + n. 360° or 180° − B + n.
360° where B is any acceptable solution for the sign
of k and B = sin−1 k.
Term 1 summary continued
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117
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Term
2
118
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TOPIC 6
Trigonometry: Problem solving in
two and three dimensions
Unit 1 Problems in two dimensions
Unit 2 Problems in three dimensions
Revision Test
TOPIC 7
120
125
130
Polynomials
Unit 1
Unit 2
Factorise third degree polynomials
132
Factorise and solve cubic polynomials
using the remainder or factor theorems 135
Revision Test
140
TOPIC 8
Differential calculus
Unit 1
Unit 2
Limits
142
Use limits to define the derivative of
a function f
146
Unit 3 Differentiation of functions from
first principles
150
Unit 4 Use the specific rules for differentiation 154
Unit 5 Find the equations of tangents
to functions
158
Unit 6 The second derivative
162
Unit 7 Sketch cubic graphs
167
Unit 8 Optimisation and rate of change
176
Revision Test
184
TOPIC 9
Analytical geometry
Unit 1 Equation of a circle
Unit 2 Equation of a tangent to a circle
Revision Test
Mid-year Exam practice: Paper 1
Mid-year Exam practice: Paper 2
Term 2 summary
188
193
198
202
205
209
119
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TOPIC
6
2
Trigonometry: Problem solving in
two and three dimensions
REMEMBER
angle of elevation
horizontal
angle of depression
Unit 1: Problems in two dimensions
In this topic you will solve problems in two and three dimensions using the sine,
cosine and area rules you learnt in Grade 11.
You will apply your knowledge of trigonometry, including trigonometric identities
and compound angle identities.
The trigonometric definitions and triangle rules
horizontal
angle of
depression
The basic definitions: Use when you have a 90°
triangle:
opposite
sin A = __________
B
hypotenuse
angle of
elevation
horizontal
The angles of elevation and
depression are the angles
between the line of sight and
the horizontal.
hypotenuse
adjacent
cos A = __________
hypotenuse
opposite
tan A = ________
adjacent
A
sin B
a
sin C
120
b
C
A
b
c
sin A
sin B
sin C
or _____ = _____ = _____
Steps for solving
two-dimensional problems:
1. Use geometry to fill in as
many angles as you can.
2. Start with the triangle
which has a given length.
3. Find a side which links the
two triangles (keep your
answer in the calculator
if necessary).
4. Use this side to solve sides
or angles in the other
triangle.
5. Repeat the process until
you have found what is
required.
6. Use double and compound
angle identities to prove or
arrive at the final answer.
adjacent
The sine rule: Use when you have side, side, angle (SSA) or
angle, side, angle (ASA):
a
c
b
_____
= _____ = _____
sin A
REMEMBER
opposite
c
The cosine rule: Use when you have SAS or SSS.
a2 = b2 + c2 − 2bc cos A
or b2 = a2 + c2 − 2ac cos B
or c2 = a2 + b2 − 2ab cos C
B
a
C
b2 + c2 − a2
cos A = __________
2bc
a2 + c2 − b2
or cos B = __________
2ac
a2 + b2 − c2
or cos C = __________
2ab
The area rule: Use when you have SAS
1
1
1
Area △ABC = __
ab sin C = __
ac sin B = __
bc sin A
2
2
2
Topic 6 Trigonometry: Problem solving in two and three dimensions
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WORKED EXAMPLE
P
In △PQS , R is the midpoint of QS.
^R = θ
QR = RP = RS = d and PS = a. PQ
Prove that: a = 2d sin θ
SOLUTION
^R = θ
QP
^
PRS = 2θ
^ S = P^SR = 90° − θ
RP
| Isosceles △
| Exterior angle of △
| Angles in isosceles △
a
Q
S
R
d
Using the sine rule in △PRS:
a
d
______
= __________
sin 2θ sin(90° − θ)
d sin 2θ
a = __________
sin(90° − θ)
d × 2 sin θ cos θ
a = ______________
cos θ
| Use co-ratio and double angle identities.
a = 2d sin θ
P
You could also have solved this question by
^ S = 90° − θ + θ = 90°.
observing that QP
△QPS is therefore a right-angled triangle
and you can use the basic definitions to
find a.
a
PS ___
___
=
= sin θ
90° –
2
Q
a
90° –
R
d
S
QS 2d
∴ a = 2d sin θ, which is simpler!
EXERCISE 1
E
You will first do a few basic examples to revise
Grade 11 work. Then you will use the double angle or
compound angle identities to find the final answer
for problems.
A
1
2
ABCD is a building with a tower, DE, erected
on top. The angle of elevation from A to E is
21° and the angle of elevation from B to E
is 48°. If the height of the building is 30 m,
calculate the height of the tower.
ABDC is a field in the shape of a trapezium.
^ C = 120°.
BD = DC = x and BD
__
2.1
Show that BC = √3 x.
2.2
If BC = 300 m, find the value of x.
^ if
2.3
Now calculate the value of A
AB = 200 m.
2.4
Then calculate the area of the field.
21°
D
30 m
48°
C
B
A
B
C
x
120°
x
D
Unit 1 Problems in two dimensions
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3
^ R = SR
^ Q = θ and QR = p.
In △QRS, SQ
^
In △QSP, SQP = θ
Q^SP = 90° and SP = k.
p sin θ
Prove that k = _______
R
p
2 cos2 θ
Q
S
k
P
4
^ P = θ, P^SQ = 60° and PS = a.
In △QPS, SQ
^ R = θ, Q^SR = 90° and SR = k.
In △QSR, SQ
__
√3
1
tan θ
Prove that k = ___a + __
2
P
a
2
60°
Q
S
k
R
5
CD is a vertical tower. From A, the angle of
elevation of D is α and from B the angle of
elevation of D is β. The distance between
B and C, the foot of the tower, is 5 m. The
distance between A and B is x.
5.1
5.2
6
Find BD in terms of α and β.
tan β
Now show that x = 5 _____ − 1
( tan α
)
D
A
x
^B = θ
^ C = AC
In△ABC, AB = AC = d and AB
^ D = β,BC = x and DC = k
^ D = θ,BC
In △BCD, CB
6.1
6.2
C
5m
B
A
Use the cosine rule to show x = 2d cos θ
d sin 2θ
Now show that k = ________
sin(θ + β)
d
d
B
x
C
k
D
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WORKED EXAMPLE
P
In △PQR, PT ⊥ QR. Prove that:
1
2
p = q cos R + r cos Q
where p = QR, q = PR and r = QP
r
q
q − r cos P
cos R
______
= _________
r − q cos P
cos Q
^ ≠ 90°
| On condition that Q
Q
p
R
T
SOLUTIONS
1
^ +R
^ = 180° − (Q
^ ).
In △PQR, P
p
r
________________
= _____
| Use the sine rule.
sin R
sin(180° − (Q + R))
rsin(Q + R)
__________
p=
sin R
r(sin Q cos R + cos Q sin R)
_______________________
=
sin R
r sin Q cos R ___________
r cos Q sin R
___________
=
+
sin R
sin R
r sin Q
______
= tan R + r cos Q
cos x _____
1
| _____
= tan
x
sin x
PT
In △PTR, tan R = ___
.
TR
| Trigonometric definitions
PT
In △PQT, sin Q = ___
r .
| Trigonometric definitions
| Use reduction formulae.
| Use the compound angle formula.
| Write the fraction as separate terms.
PT
)
r( ___
p = _____
PT + r cos Q = TR + r cos Q | Substitute from above.
___
r
TR
TR
but in △PTR, ___
q = cos R
∴ TR = q cos R
∴ p = r cos Q + q cos R
2
In △PQR, draw RT ⊥ PQ and QS ⊥ PR.
QT
SR
cos R = ___ and cos Q = ___
p
p
q − PS
r − TP
∴ cos R = ______ and cos Q = ______
p
p
P
PS
TP
but cos P = ___ and cos P = ___
r
q
∴ PS = r cos P and TP = q cos P
q − r cos P
r − q cos P
∴ cos R = _________
and cos Q = _________
p
p
r
T
S q
q − r cos P
_________
p
cos R
______
= ________
cos Q
r − q cos P
_________
p
Q
p
R
q − r cos P
= _________
r − q cos P
Unit 1 Problems in two dimensions
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EXERCISE 2
All questions in this exercise refer to the diagram below.
Use the sine and cosine rules to prove what is required. You may need to use
compound and double angle formulae at the end.
A
b
c
B
1
2
3
4
5
6
7
8
124
a
C
If in △ABC, AC = CB, find a2 in terms of c and cos C using the cosine rule.
If in △ABC, AB = AC, find a in terms of c and cos C using the sine rule.
b2
If in △ABC, AB = BC, prove that cos B = 1 − ___
2a2
^ prove that area △ABC = b2 sin B.cos B
^=C
If in △ABC, B
4.Area △ABC ^
≠ 90°
If △ABC is scalene prove that tan B = ____________ B
a2 + c2 − b2
b2 + c2 − a2
2 sin B __________
If △ABC is scalene prove that ______
=
ac
tan A
2ab sin B
________
^
^
If in △ABC, B = C prove that tan 2B = 2
a − 2b2
a sin C
^ ≠ 90°).
If △ABC is scalene prove that tan A = __________
(on condition that A
b − a cos C
9
If △ABC is scalene prove that:
a + b + c = (b + c) cos A + (c + a) cos B + (a + b) cos C
10
sin C
If △ABC is scalene prove that a = b(cos C + _____
)
tan B
(Hint: Use A = (180° − (B + C))
Topic 6 Trigonometry: Problem solving in two and three dimensions
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Unit 2: Problems in three dimensions
Three-dimensional problems can be complex because you are working in two planes,
vertical and horizontal. Right angles do not always look like right angles.
The diagrams below represent two typical three-dimensional diagrams where AB is a
vertical height and B, C and D are in the same horizontal plane. The horizontal plane
^ C = 90°.
^ D = 90° and AB
is shaded. AB
A
A
B
D
B
C
B
C
D
WORKED EXAMPLE
A
Chela and Desmond stand some distance away
from a building AB that has a height of 12 m.
The foot of the building, B, Chela and Desmond
are in the same horizontal plane. The angle of
elevation from D to the top of the building, A,
^ D = 55°.
is 40° and from C to A is 45°. CB
12 m
Calculate the distance between Desmond and
Chela.
SOLUTION
^ D = 90° and AB
^ C = 90°
AB
12
___
= tan 45°
40°
B
55°
D
45°
C
BC
12
∴ BC = _______
= 12
tan 45°
12
___
= tan 40°
BD
12
∴ BD = _______
= 14,30
tan 40°
DC2 = BC2 + BD2 − 2BC.BD cos 55°
DC2 = 122 + (14,30)2 − 2(12)(14,30) cos 55°
DC2 = 151,6386
DC = 12,31 m
Three-dimensional problems can involve prisms and pyramids which you worked
with in Grades 10 and 11. Remember that right angles will not always look like
right angles.
Unit 2 Problems in three dimensions
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WORKED EXAMPLE
The figure shows a cube with sides 20 mm.
1.1 Calculate the angle that plane ABRS
makes with plane ABCD.
1.2 Calculate the angle diagonal BS makes
with plane ABCD.
20 mm
S
R
Q
P
20 mm
C
D
SOLUTION
20 mm
A
B
This solution does not use the sine or
cosine rule.
^ C is the angle that plane ABRS makes with plane ABCD.
1.1 RB
^ B = 90° ∴ RB
^ C = 45°
In △RCB, RC = BC = 20 and RC
| Isosceles △
^
1.2 SBD is the angle that diagonal BS makes with plane ABCD.
^ B = 90°
In △ADB,________
AD = AB = 20 and DA
__
2
2
√
√
∴ BD = 20 + 20 = 20 2
| Pythagoras’ Theorem
__
^ B = 90°
In △SDB, SD = 20, BD = 20√2 and SD
20 __ ___
1
^ D = ______
^ D = 35,26°
∴ tan SB
= __ ∴ SB
√
√
20 2
2
EXERCISE 3
1
2
The figure represents the top of a sloping
school desk. ABCD represents the
horizontal plane with dimensions 60 cm
by 40 cm. ABFE represents the sloping desk
A
top. Calculate the angle between the
horizontal plane and AF, the diagonal of
^ C = 15°.
the sloping desk top if FB
F
D
Calculate the area of △ABC.
Calculate the length of BC.
Calculate the height CF of the prism if
^ F = 40°.
CB
C
15°
60 cm
A
40 cm
B
C
70°
The figure represents a triangular prism with
^ C = 70°.
AB = AC = 10 cm and BA
2.1
2.2
2.3
3
E
B
D
40°
F
Xandi is standing at point X. She observes
two vertical poles AB and CD. The angle of
A
elevation from X to C is 20° and from X to A is
25°. The distance between Xandi and the foot of
pole AB is 15 m. Pole CD is 5 m high.
3.1
Calculate the distance from X to A.
3.2
Calculate the distance from X to C.
B
3.3
Calculate the distance, AC, between
^ C = 46°.
the two poles if AX
E
C
5m
15 m
25°
46°
20°
D
X
126
Topic 6 Trigonometry: Problem solving in two and three dimensions
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4
In the diagram Amy and Boetie stand 200 m
apart at A and B which are in the same
horizontal plane as D. They both observe that
the angle of elevation of a balloon C from A
C
and B is 48°. Calculate the height, DC, of the
^ B = 64°.
balloon above the ground if CA
D
A
5
A
In the diagram AB is a vertical tower. C and D are
points in the same horizontal plane as B with
BC = 220 m and CD = 350 m. The angle of elevation
^ D = 36°.
from D to A, the top of the tower is 25°. BC
5.1
Calculate AB, the height of the tower.
5.2
Calculate the area of △BCD.
B
220 m
C
6
B
200 m
In the diagram B, C and D lie in the same horizontal
plane. AB is a vertical pole. BD = BC = 96 m.
The angle of elevation from C to A is 20° and
^ D = 47°.
CA
6.1
Why is AC = AD?
6.2
Calculate AC.
6.3
Calculate CD.
6.4
Calculate AB, the height of the tower.
25°
36°
D
350 m
A
47°
96 m
B
96 m
D
20°
C
7
A
In the diagram B, C and D lie in the same
horizontal plane. AB is a vertical pole. The angle
^ D = 54°,
of elevation from C to A is 41°, BC
^ C = 48° and BD = 36 m.
BD
7.1
7.2
Calculate the area of △BCD.
Calculate AB, the height of the tower.
C
41°
54°
BB
36 m
48°
D
Unit 2 Problems in three dimensions
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WORKED EXAMPLE
In the figure, B, C and D lie in the same horizontal plane
AB is a vertical pole. The angle of elevation from D to A
^ D = 30° and BD
^ C = θ.
is 60°. BC = x, CB
Prove that:
x √__ _____
1
3 + tan
1
BD = __
θ
2
1 2 √__ _____
x
3+ 1
2
Area △BDC = __
(
3
8
__
√3
x
__
_____
AB = 2 3 + tan θ
(
)
(
)
tan θ
)
A
60°
B
D
30°
x
SOLUTIONS
1
C
^ D = 180° − (θ + 30°)
BC
x
BD
_________________
= _____
sin θ
sin(180° − (θ + 30°))
x
sin(θ
+
30°)
BD = ____________
| Use compound angle formula.
sin θ
x(sin θ cos 30° + cos θ sin 30°)
= _________________________
| Divide by sin θ.
sin θ
x sin 30°
= x cos 30° + ________
| Use special angle values.
tan θ
__
√
3
x
= ___
x + ______
2
2 tan θ
x √__ _____
1
= __
3 + tan
θ
2
(
2
)
1
x.BD. sin 30°
Area △BCD = __
2
1
1
Area △BCD = __
x.BD.__
2
2
(
__
x __
x
1
Area △BCD = __
. √3 + _____
tan θ
4 2
(
x2 √__ _____
1
Area △BCD = __
3 + tan
θ
8
AB
___
= tan 60°
3
)
BD
AB = BD tan 60°
__ x
__
1
AB = √ 3 .__ √ 3 + _____
)
(
2
(
tan θ
__
√3
_____
x
AB = __
3 + tan θ
2
D
EXERCISE 4
This exercise involves proving sides and areas in
three-dimensional problems equal to algebraic
expressions. You may be required to use the
compound and double angle identities.
1
2
h
C
Two boats, A and B, on a river are equidistant
from the foot of tower CD, on the bank of the
river. The boats and the foot of the tower are in
A
the same horizontal plane. From each boat the
angle of elevation of the top of the tower is θ.
^ B = β.
CD is h m high and AC
Show that
the distance between the two boats is
__________
h√ 2(1 − cos β)
_____________
AB =
B
D
tan θ
In the diagram, A, B and C are points in the same
horizontal plane. CD is a vertical tower. The
angle of elevation from A to D is x and
^ A = y. AD = BD and AC = k.
DB
2k cos y
Prove that AB = _______
cos x
128
)
)
C
k
x
y
B
A
Topic 6 Trigonometry: Problem solving in two and three dimensions
9780636143319_plt_mat_g12_lb_eng_zaf.indb 128
A
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3
In the diagram B, C and D lie in the same horizontal plane and AB is vertical to
the plane.
^ B = θ and AD
^ C = 2θ
AD = 2p and CD = p. AC
Show that:
3.1
Area △ADC = 2p2 sin θ cos θ
3.2
3.3
4
A
2p
__________
AC = p√ 9 − 8 cos2 θ
__________
AB = p sin θ√ 9 − 8 cos2 θ
B
AB is a vertical wall. B, C and D lie in the same horizontal plane. The angle of
^ D = y and BC = 2p.
^ C = 120°, CB
elevation of A from D is x. BD
4.1
Find DC in terms of p and sin y.
4p2
__ sin y.sin(y + 120°).
4.2
Show that area △BDC = ___
4.3
p
√3
1__
sin y).
Prove that AB = 2p tan x(cos y − ___
C
Question 3
√3
F
D
2
A
A
B
x
y
120°
E
D
B
D
y
2p
k
C
Question 4
5
6
x
C
Question 5
^ B = x and
In the diagram B, C and D lie in the horizontal plane with BC = k. DC
^
CBD = y. FAB and ED are vertical to the plane. The angle of elevation from
E to point F is θ.
^ C in terms of x and y.
5.1
Write BD
k sin x
5.2
Show that EF = __________________________
cos θ(sin x cos y + cos x sin y)
D
h
5.3
Calculate the value of EF if x =48°, y =38°, θ = 28° and k = 30 m.
In the figure alongside A, B and C lie in the same horizontal plane. The vertical
pole DC is h metres high and the angle of elevation from B to D is θ.
sin(α + β)
^ B = β show that AC = h__________
^ B = α, AC
6.1
If CA
A
C
sin α tan θ
6.2
6.3
7
If α = θ and β = 30° use 6.1 to show that:
__
h
1
√ 3 + _____
AC = ______
tan α
2 tan α
(
)
B
2h cos2 α
If α = θ = β use 6.1 to show that AC = ________
sin α
A rectangular block of wood has vertices P, Q, R,
S, T, U, V and W. The dimensions of this block
____
of wood are √ 200 cm × 12 cm × 5 cm. A cut is
made through the vertices Q, T and V, revealing a
triangular plane QTV.
^ V.
7.1
Determine the magnitude (size) of TQ
7.2
Determine the area of the triangular plane
QTV.
W
T
V
5
P
200
Q
12
R
Unit 2 Problems in three dimensions
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Revision Test Topic 6
Total marks: 191
1
V
The figure represents a pyramid with a horizontal square base,
ABCD, with sides 8 cm. The vertex V is vertically
above the midpoint, M, of the base. VM = 10 cm.
Calculate the angle that the slanting edge of the
pyramid makes with the horizontal (angle VAM).
(4)
A
2
3.
3.1
3.2
3.3
3.4
c sin B
If △ABC is scalene prove that AC = ______
sin C
If △ABC is scalene prove that area
c2 sin B.sin(B + C)
△ABC = ________________
8 cm
(2)
F
B
If △ABC is scalene prove that
(c − a + b)(c + a − b)
1 − cos C = _________________
2ab
30°
50 cm
D
c
B
C
a
(5)
In the diagram B, C and D lie in the same horizontal
plane and AB is vertical to the plane. AD = 4a and
^ B = θ and AD
^ C = 2θ.
CD = a. AC
5.2
__________
BC = a cos θ√9 + 16 sin2 θ
A
50 cm
b
(4)
(4)
B
C
A
27°
B, C and D lie in the same horizontal plane with BD = DC = 36 m. AB is a
vertical pole anchored at C and D. The angle of elevation from C to A is 22°.
^ C = 44° calculate:
If BD
4.1
the length of BC
(2)
4.2
the height of AB
(3)
^ D = 70°
4.3
AD if AC
(5)
4.4
area △BDC.
(3)
A
Show that:
__________
5.1
AC = a√9 + 16 sin2 θ
E
A
(2)
^ prove that
^=C
If in △ABC, B
BC = 2b cos B
by using the cosine rule
c2 = a2 + b2 − 2ab cos C.
5
8 cm
M
The figure represents a triangular prism. It is placed on one of its faces, ABCD, a
square with sides 50 cm. Face AEFD is a rectangle which is inclined at 30° to the
horizontal plane ABCD. Calculate the size of the angle that line DE makes with
plane ABCD.
(6)
2 sin C
4
C
D
B
36 m
22°
C
4a
B
D
44°
a
(2)
F
D
2
E
A
a
C
6
The diagram shows a cube with sides a metres in length.
__
√ 3 a2
Show that the area of the shaded triangle ACE is equal to _____
.
2
D
(5)
a
B
C
130
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7
B, C and D are points in the same horizontal plane. AB is a vertical tower with
^ D = β. BD = BC = x.
the angle of elevation from C to A equal to α and AC
(2)
7.1
Why is AC = AD?
(2)
7.2
Write AC in terms of x and α.
2x cos β
7.3
Show that: CD = _______
(4)
A
cos α
7.4
8
9
If x = 50 m, CD = 40 m and β =75°, determine
the height of the tower, correct to two decimal places.
(5)
BD is a telephone pole 3 m high. It is supported by
two cables, AD and CD which are both anchored
2,5 m away from B, the base of the pole. Calculate the
^ C if:
size of AD
^
8.1
ABC = 90°
(5)
^ C = 120°.
8.2
AB
(6)
In the diagram below, two kites hover directly over a
level field. The two kite flyers are 2,5 km apart. The
angle of elevation from A to kite C is 21° and from B
to kite D is 31°.
D
x
B
D
x
C
3m
A
2,5 m
B
2,5 m
C
D
C
76°
21°
9.1
9.2
10
72°
31°
2,5 km
A
B
Calculate the distance between the two kites.
Determine which kite is higher from the ground and by how
many metres.
^ B = 90°, AB
^ C = θ, BD
^ C = 2θ, DB
^ C = 90° − θ and DC = x.
In the diagram AC
Show that:
10.1 BC = 2x sin θ
2
2x sin θ
10.2 AC = ________
cos θ
A
(5)
(6)
B
C
90° –
2
(3)
D
(4)
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TOPIC
7
2
Polynomials
Unit 1: Factorise third degree polynomials
A polynomial is an algebraic expression in which the powers of the variable
(let’s use x) must be integers greater than or equal to 0, and the coefficients of
the variable must be constants which are real numbers.
REMEMBER
• The degree of a polynomial
is the degree of the term
with the highest power.
• The constant term has no
variable so its value does
not change. You could
think of the degree of the
constant term as being
0 because 5x° = 5 because
x° = 1.
• Positive integers are the
numbers 1, 2, 3, 4 …
Examples:
3x2 − 4x + 2 is a polynomial of degree 2 and has 3 terms.
3x__4 − 7x3 + 2x2 − 11 is a polynomial of degree 4 and has 4 terms.
√ 3 x3
_____
− x is a polynomial of degree 3 and has 2 terms.
2
Division by a variable will not give you a polynomial.
The following expressions are not polynomials:
3x + y + 3xy−2 + y
___
There is a power that is not a positive integer, (−2).
y2__
1
1
__
−__
1
1
2
___
2
2
__ + x + 2x
√x +
and −__
.
There are powers that are not positive integers, __
√x
2x + 2x(x − 1)−1
_____
x−1
2
There is a power that is not a positive integer, (x − 1)−1.
The standard form of a polynomial shows the terms of all
possible degree combinations.
2
y
For a linear or 1st degree polynomial the standard form is ax + b.
For a quadratic or 2nd degree polynomial the standard form is
x
ax2 + bx + c.
y
y = ax + b
x
y = ax² + bx + c
For a cubic or 3rd degree polynomial the standard form is
ax3 + bx2 + cx + d.
y
x
y = ax³ + bx² + cx + d
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At school you will seldom deal with polynomials of degree higher than 3.
However, the standard form for a polynomial of nth degree is:
axn + bxn−1 + cxn−2 + dxn−3 + …. sx + t
A polynomial function is the rule for determining y given x.
For every value of x there is one value for y.
y = ax + b is a linear function or straight line graph.
y = ax2 + bx + c is a quadratic function or parabola.
y = ax3 + bx2 + cx + d is a cubic function.
Function notation or f(x) notation
You indicate polynomials by P(x) or R(y) or Q(a) where the symbol in the bracket is
the variable of the polynomial. For example, Q(a) = a3 − 2a + 2.
For polynomial functions y = …, it is convenient to give each function a name if you
are working with more than one function. You can do this using function notation
and replacing ‘y’ with f(x), g(x) or h(x).
f(2) means that all the x values are replaced by 2 and f(a + 1) means x is replaced by
(a + 1).
WORKED EXAMPLE 1
If f(x) = x2 − 2x + 1, find the value of f(1) and f(−3).
SOLUTION
f(1) = 12 − 2(1) + 1 = 0
f(−3) = (−3)2 − 2(−3) + 1
= 9 + 6 + 1 = 16
WORKED EXAMPLE 2
If g(x) = −x2 − 3x − 1, find g(a − 3).
SOLUTION
g(a − 3) = −(a − 3)2 − 3(a − 3) − 1
= −(a2 − 6a + 9) − 3a + 9 − 1
= −a2 + 6a − 9 − 3a + 9 − 1
= −a2 + 3a − 1
Unit 1 Factorise third degree polynomials
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EXERCISE 1
REMEMBER
Sum of cubes: x3 + y3
= (x + y)(x2 − xy + y2)
Difference of cubes: x3 − y3
= (x − y)(x2 + xy + y2)
Determine the value of each function numerically or in terms of the variable.
1
f(x) = 4x − 5, find f(−2)
2
f(x) = x2 + x − 3, find f(4)
3
g(a) = 3a − 6, find g(2a)
4
g(a) = 2a2 − 3a, find g(a + 1)
5
h(x) = −x3 − 2x2 + 8x, find h(2)
6
h(x) = 2x3 + 6x2 − 2x, find h(−2p)
7
f(x) = 4x2 − 5x, find f(x + h) − f(x)
8
f(x) = x3 − 8x, find f(a + p) − f(a)
1 − x2 , find p(−3)
9
p(x) = ______
1−x
10
1 + x , find p(1 − a)
p(x) = _____
1−x
In Topic 7 you focus on factorising and solving 3rd degree polynomial equations.
In Topic 8 you learn how to plot cubic functions using calculus to find the turning
points of the graphs.
Factorising cubic polynomials by using common factors, grouping or sum/difference
of cubes.
You are familiar with these methods of factorising:
| Take out the common factor.
1
2x3 − x2 − 3x
= x(2x2 − x − 3)
| Factorise the trinomial.
= x(2x − 3)(x + 1)
2
2x3 − x2 − 2x + 1
= x2(2x − 1) − (2x − 1)
= (2x − 1)(x2 − 1)
= (2x − 1)(x − 1)(x + 1)
| Group by taking out the common factor.
| Take out the common bracket.
| Factorise the 2nd bracket as the difference of squares.
3
2x3 − 16
= 2(x3 − 8)
= 2(x − 2)(x2 + 2x + 4)
| Take out the common factor.
| Factorise the difference of cubes.
EXERCISE 2
Factorise the polynomials using common factors, grouping or sum/difference
of cubes.
1
x3 + x2 − 4x − 4
2
x3 + 2x2 − 8x
3
3
8x − 64
4
x3 − 125
5
9x4 + 72x
6
x3 − 3x2 − 4x + 12
3
2
7
x − 2x − 49x + 98
8
−4x3 + 36x2 − 80x
3
2
9
x + x − 16x − 16
10 −x3 + 2x2 + x − 2
11 3x3 − x2 + 3x − 1
12 5x3 + 5x2 − 20x − 20
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Unit 2: Factorise and solve cubic polynomials
using the remainder or factor theorem
You can express any polynomial function f(x) as the product of two other polynomials
and a remainder: P(x) = D(x)Q(x) + R
REMEMBER
19
___
= 9 remainder 1 or
2
19 = 2 × 9 + 1
polynomial = divisor ×
quotient + remainder
A cubic polynomial can be the product of a linear and a quadratic function:
(x + 1)(x2 − 5x + 6) = x3 − 5x2 + x2 − 5x + 6x + 6 = x3 − 4x2 + x + 6
In reverse you can write the polynomial P(x) as:
P(x) = x3 − 4x2 + x + 6 = (x + 1)(x2 − 5x + 6)
P(x) = (divisor)(quotient) + 0
Note: x + 1 is a factor of x3 − 4x2 + x + 6 because there is no remainder.
P(−1) = (−1)3 − 4(−1)2 + (−1) + 6 = −1 − 4 − 1 + 6 = 0
A cubic polynomial can be the product of a linear and a quadratic function plus
another term or constant:
(x − 2)(x2 − 2x + 6) + 3 = x3 − 2x2 − 2x2 + 4x + 6x − 12 + 3 = x3 − 4x2 + 10x − 9
In reverse you can write the polynomial P(x) as:
P(x) = x3 − 4x2 + 10x − 9 = (x − 2)(x2 − 2x + 6) + 3
P(x) = (divisor)(quotient) + remainder
Note: x − 2 is not a factor of x3 − 4x2 + 10x − 9 because there is a remainder of 3.
P(2) = (2)3 − 4(2)2 + 10(2) − 9 = 8 − 16 + 20 − 9 = 3
The Remainder Theorem states:
b
If a polynomial f (x) is divided by ax + b, then the remainder will be f(− __
a ).
Or more simply, when you divide a polynomial f(x) by (x − a) the remainder is f(a).
The Factor Theorem states:
If a polynomial is divided by ax + b, and the remainder = 0, then ax + b is a factor of
the polynomial.
b ) = 0, then ax + b is a factor of f(x) or if f (a) = 0 then x − a is a factor of f(x).
or if f(−__
a
Note: You are not required to prove the remainder or factor theorems, but you apply
them to factorise cubic polynomials.
WORKED EXAMPLES
1
If f(x) = x3 + 3x2 − 6x − 8
1.1
Show that x − 2 is a factor of f(x).
1.2
Find the remainder when f (x) is divided by x − 1.
SOLUTIONS
1.1 f(x) = x3 + 3x2 − 6x − 8
f(2) = (2)3 + 3(2)2 − 6(2) − 8
= 8 + 12 − 12 − 8 = 0
∴ (x − 2) is a factor.
1.2 f(x) = x3 + 3x2 − 6x − 8
For (x − 1):
R = f(1) = (1)3 + 3(1)2 − 6(1) − 8
= 1 + 3 − 6 − 8 = −10
Unit 2 Factorise and solve cubic polynomials using the remainder or factor theorem
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2
If f(x) = 2x3 + x2 − ax + 5 find a if:
2.1
The remainder is 2 when f(x) is divided by x + 1.
2.2
2x − 1 is a factor of f(x).
SOLUTION
2.1
f(x) = 2x3 + x2 − ax + 5
For (x + 1):R = f(−1) = 2
f(−1) = 2(−1)3 + (−1)2 − a(−1) + 5 = 2
∴−2+1+a+5=2
∴ a = −2
2.2
1) = 0
For (2x − 1): R = f(__
2
1
1 3
1 2
1
f (__
) = 2(__
) + (__
) − a(__
)+5=0
2
2
2
2
−11
1 + __
1 − __
1 a + 5 = 0 ∴ −__
1 a = ____
∴ __
2 ∴ a = 11
4 4 2
2
EXERCISE 3
1
2
3
4
5
6
7
If f(x) = 2x3 + 9x2 + 3x − 4:
1.1
show that x + 4 is a factor of f(x)
1.2
find the remainder when f(x) is divided by x − 2.
Given f(x) = x3 + kx2 + 3x − 5 find k if:
2.1
the remainder is 4 when f(x) is divided by x + 3
2.2
x −1 is a factor of f(x).
Given f( x ) = −x3 + 2x2 − 3x + 3 and g( x ) = 2x − 1:
3.1
Find the remainder when f(x) is divided by g(x).
3.2
If f( x ) = −x3 + 2x2 − 3x + p find p so that g(x) becomes a factor of f(x).
If f( x ) = 2x3 − x2 − 13x − 6:
4.1
Find f(3) and explain the meaning of your answer.
4.2
Find f(–1) and explain the meaning of your answer.
Given f( x ) = 2x4 − x3 − 11x2 + kx + 12, find the value of k if the remainder is 60
when f(x) is divided by x − 3.
What must be added to 2x3 − 5x2 − 12x − 4 so that it is exactly divisible by
2x + 1?
Determine the values of a and b if x + 1 and x − 2 are both factors of
x3 + ax2 + x + b?
When you factorise cubic (3rd degree) polynomials using division by inspection,
follow these steps:
a) Test for linear factors using the factor theorem. Possible factors are factors of
the constant term 6: x ± 1; x ± 2; x ± 3; x ± 6
b) Use division by inspection to find the quadratic factor ax2 + bx + c.
c) Factorise the quadratic factor if possible.
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WORKED EXAMPLES
1
KEY CONCEPT
Factorise: x3 − 2x2 − 5x + 6
2
Factorise: 2x3 + x2 − 13x + 6
2x3 − 5x2 − x + 6
_______________
= Q(x)
SOLUTIONS
1
a)
b)
x−2
Using the factor theorem:
f(x) = x3 − 2x2 − 5x + 6
f(1) = (1)3 − 2(1)2 − 5(1) + 6 = 0
∴ (x − 1) is a factor
Using division by inspection:
x3 − 2x2 − 5x + 6 = (x − 1)(x2 + bx − 6)
−x2
bx2
−x2 + bx2 = −2x2 ∴ b = −1
c)
2
a)
b)
Division by inspection
x3 − 2x2 − 5x + 6 = (x − 1)(x2 − x − 6)
Factorising the 2nd bracket:
x3 − 2x2 − 5x + 6 = (x − 1)(x − 3)(x + 2)
Using the factor theorem:
f(x) = 2x3 + x2 − 13x + 6
f(1) = 2(1)3 + (1)2 − 13(1) + 6 ≠ 0
f(2) = 2(2)3 + (2)2 − 13(2) + 6 = 0
∴ (x − 2) is a factor.
Using division by inspection:
2x3 + x2 − 13x + 6 = (x − 2)(2x2 + bx − 3)
or
2x3 − 5x2 − x + 6 = (x − 2)Q(x)
where Q(x) is a 2nd degree
polynomial
2x3 − 5x2 − x + 6
= (x −2)(ax2 + bx + c)
a = 2 (x × 2x2 = 2x3) and
c = −3
(−2 × −3 = +6)
−3x
−2bx
2x3 − 5x2 − x + 6 = (x − 2)(2x2 + bx − 3)
−4x2
bx2
− 5x2 = −4x2 + bx2 ∴ b = −1
Check: − x = −3x − 2bx ∴ b = −1
2x3 − 5x2 − x + 6
= (x − 2)(2x2 − x − 3)
2x3 − 5x2 − x + 6
= (x − 2)(2x − 3)(x + 1)
−4x2
bx2
2
2
2
−4x + bx = x ∴ b = 5
2x3 + x2 − 13x + 6 = (x − 2)(2x2 + 5x − 3)
c)
Factorising the 2nd bracket:
x3 − 2x2 − 5x + 6 = (x − 2)(x + 3)(2x − 1)
EXERCISE 4
1
Use division by inspection to factorise these cubic polynomials where one factor
is given. Not all these polynomial have three linear factors.
1.1
x3 − x2 − 10x − 8 and x + 1 is a factor
1.2
x3 + 3x2 − 3x − 10 and x + 2 is a factor
1.3
−x3 − 3x2 + 10x + 24 and x + 2 is a factor
1.4
2x3 − 7x2 + 9 and x − 3 is a factor
1.5
x3 − 13x + 12 and x − 1 is a factor
1.6
−x3 − 3x2 + x + 3 and x + 3 is a factor
1.7
2x3 + 2x2 − 18x − 18 and x + 3 is a factor
1.8
4x3 − 7x − 3 and 2x − 3 is a factor
Unit 2 Factorise and solve cubic polynomials using the remainder or factor theorem
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2
Factorise the given cubic polynomials.
2.1
2x3 + x2 − 13x + 6
2.3
2x3 + 6x2 −2x − 6
2.5
4x3 − 7x2 − 7x + 10
2.7
2.9
REMEMBER
The quadratic
formula is:
_______
−b ± √ b2 − 4ac
______________
x=
2.2
2.4
2.6
x3 − 7x2 + 36
4x3 − 23x2 + 14x + 5
x3 − 3x2 + 3x − 1
2x3+3x2 − 4x − 1
3 x2 + 6x − 2
−x3 + __
2
2.8
−3x3 + x2 + 3x − 1
2.10 −6x3 − 7x2 + 1
Solving third degree polynomial equations
2a
This is used to find the roots
of ax2 + bx + c = 0
In Topic 8 you learn to plot the graph of a cubic (3rd degree) function. You will find
the x intercepts of the graphs. This means you must be able to solve equations of the
type ax3 + bx2 + cx + d = 0.
You have seen that a cubic polynomial has three factors.
A cubic equation will therefore have three solutions or a graph will have three roots.
A cubic equation will always have at least one root.
The other two roots may be real and rational or real and irrational.
A cubic equation may have one real root if the solutions to the quadratic factor are
non-real.
In the graphs below:
KEY WORDS
Graph A has three real roots, where two are equal.
For ax2 + bx + c = 0:
real roots − occur when
b2 − 4ac ≥ 0 and can be
rational or irrational; for
example x2 − 3x + 1 = 0__has
3 ± √5
irrational roots x = _______
2
rational roots − occur when
the equation factorises
or b2 − 4ac is a perfect
square; for example (x − 2)
3
(2x + 3) = 0 ∴ x = 2 or −__
Graph B has three real, unequal roots.
Graph C has three real roots that are all resulting in one x-intercept.
Graph D has one real root and two non-real roots.
A
B
C
y
y
y
x
2
or for the___
same equation
1 ± √ 49
________
x=
4
D
y
x
x
x
non-real roots − occur when
b2 − 4ac < 0; for example
___
x = 3 ± √ −4
WORKED EXAMPLES
Solve for x and in each case say what type of roots the polynomial has.
1
x3 − 2x2 − 4x + 8 = 0
138
2
2x3 − 5x2 − x + 6 = 0
3
x3 + x2 − 11x − 3 = 0
4
x3 + 2x2 − 3x − 10 = 0
Topic 7 Polynomials
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SOLUTIONS
1
2
3
x2(x − 2) − 4(x − 2) = 0
| Factorise by grouping.
2
(x − 2)(x − 4) = 0
(x − 2)(x − 2)(x + 2) = 0
x = 2 or x = 2 or x = −2
There are 3 real rational roots, two of which are equal.
f(2) = 16 − 20 − 2 + 6 = 0
∴ (x − 2) is a factor
2x3 − 5x2 − x + 6 = 0
(x − 2)(2x2 − x − 3) = 0
| Division by inspection
(x − 2)(2x − 3)(x + 1) = 0
3 or x = −1
x = 2 or x = __
2
There are 3 real, rational roots.
f(1) = 1 + 1 − 11 − 3 ≠ 0
f(3) = (3)3 + (3)2 − 11(3) − 3
27 + 9 − 33 − 3 = 0
∴ (x − 3) is a factor.
x3 + x2 − 11x − 3 = 0
(x − 3)(x2 + 4x + 1) = 0
| Use division by inspection.
Use the quadratic formula:
__________
− 4 ± √ 42 − 4(1)(1)
_________________
x = 3 or x =
2(1)
__
x = 3 or x = −2 ± √3
4
REMEMBER
If x − a is a factor of f(x), then
f(a) = 0 and x = a is a root of
the equation f(x) = 0.
There are 3 real roots of which 1 is rational and 2 are irrational.
f(2) = (2)3 + 2(2)2 − 3(2) − 10
8 + 8 − 6 − 10 = 0
∴ (x − 2) is a factor
x3 + 2x2 − 3x − 10 = 0
(x − 2)(x2 + 4x + 5) = 0
| Use division by inspection.
Use the quadratic formula:
__________
−4 ± √ 42 − 4(1)(5)
x = 2 or x = _________________
2(1)
___
x = 2 or x = −2 ± √ −1
___
There is 1 real, rational root, x = 2, and x = −2 ± √ −1 are non-real or
imaginary roots.
EXERCISE 5
Solve for x and in each case say what type of roots the polynomial has.
1
(x + 1)(x − 2)(2x − 3) = 0
2
−x3 + 4x2 = 0
3
x3 − 2x2 + x = 0
4
2x3 − x2 − 15x + 18 = 0
5
6
−x3 + 6x2 − 5x − 6 = 0
7
2x3 + 9x2 − 17x + 6 = 0
6
x2 + 4x + 1 = __
8
x3 + 4x2 − 2x − 3 = 0
9
−x3 + 2x2 + 9x − 18 = 0
10
x3 − 2x2 − 5x + 6 = 0
11
x3 − 6x2 + 9x − 2 = 0
12
−x3 + 5x2 = 6x + 12
x
Unit 2 Factorise and solve cubic polynomials using the remainder or factor theorem
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Revision Test Topic 7
Total marks: 194
1
Determine the value of each function numerically or in terms of the variable.
1.1
f(x) = −6x − 15, find f(−3).
(2)
1.2
f(x) = −x2 + 2x − 1, find f(2).
(2)
3x − 6
1.3
g(x) = ______
(2)
x , find g(3a).
1.4
p(x) = 5x2 − x, find p(t − 1).
(2)
1.5
h(x) = x3 + 3x2 + 2x, find h(−1).
(2)
1.6
g(x) = 2x3 − 5x2 − x + 5, find g(−a).
(2)
1
__
3
2
(2)
1.7
h(x) = 2x + 9x − 17x + 6, find h .
1.8
1.9
(2)
3
.
h(x) = 2x3 − x2 − 15x + 18, find h( __
2)
f(x + h) − f(x)
f(x) = −3x2 + x, find ____________
.
h
= 2x3, find f(a + 1) − f(a).
1.10 f(x)
1
1.11 p(x) = __
x , find p(a + h) − p(h).
2+x
_____
1.12 p(x) = 2 − x , find p(2 + b).
(2)
(4)
(3)
(3)
(3)
2
Factorise the polynomials using common factors, grouping or sum/difference of
cubes.
2.1. 4x3 + 4x2 − 9x − 9
(3)
2.2
2x3 + 10x2 − 28x
(2)
2.3. 3x3 − 81
(3)
2.4. x3 + 216
(2)
5
2
2.5. 5x + 40x
(3)
2.6. x3 − 4x + 3x2 − 12
(2)
2.7. x3 − 3x2 − 8x + 12
(3)
4
3
2
2.8. −5x + 45x − 100x
(3)
2.9. x3 + x2 − 16x − 16
(3)
2.10. −x3 + 2x2 + 3x − 6
(2)
3
2
2.11. 5x − x − 20x + 4
(3)
2.12. −3x4 + 3x3 + 81x − 81
(3)
3
Use the remainder and factor theorems to answer the questions that follow.
3.1
If f(x) = 2x3 + x2 − 13x + 6:
3.1.1
show that 2x − 1 is a factor of f(x)
3.1.2
find the remainder when f(x) is divided by x + 1.
3.2. Given f(x) = x3 + 2x2 + kx − 6 find k if:
3.2.1
the remainder is 6 when f(x) is divided by x + 2
3.2.2
x − 3 is a factor of f(x)
3.3. Given f(x) = 4x3 − 7x2 − 7x + 30 and g( x ) = 4x + 5.
3.3.1
What is the remainder when f(x) is divided by g(x).
3.3.2
If f(x) = 4x3 − 7x2 − 7x + k, find k so that g(x) becomes a factor
of f(x).
(3)
(2)
(4)
(4)
(2)
(4)
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3.4.
3.5.
4
5
Find the value of a in each of the following:
3.4.1
x3 + 2x2 + ax + 4 leaves a remainder of 4 when divided by x − 2.
(4)
3.4.2
x3 − 2x2 + ax − 4 leaves a remainder of 0 when divided by x − 1.
(4)
3
2
3.4.3
x + ax − 3x + 4 leaves a remainder of 14 when divided by x + 2. (4)
3.4.4
x + 2 is a factor of 2x4 + x3 + ax − 12.
(4)
Solve for a and b in each of the following:
3.5.1
(x − 2) and (x + 1) are factors of x3 + ax2 + bx + 8
(6)
3.5.2
(x − 1) and (x +1) are factors of ax3 + bx2 + 5x − 5
(6)
Use the factor theorem and division by inspection to factorise
the cubic polynomials.
4.1
x3 − 7x2 + 14x − 8
4.2
x3 − 2x2 − 5x + 6
4.3
x3 + 3x2 − 10x − 24
4.4
6x3 − 11x2 −3x + 2
4.5
2x3 − 7x2 + 9
4.6
2x3 − 3x2 − 3x + 2
5.1
5.2.
5.3
Solve for x where one root is given.
5.1.1
x3 − x2 − 10x − 8 = 0; x = −1 is a root.
5.1.2
2x3 + x2 − 5x + 2 = 0; x = −2 is a root.
1
5.1.3
4x3 − 7x + 3 = 0; x = __
2 is a root.
3
2
5.1.4
2x − 5x − 21x + 36 = 0; x = −3 is a root.
Solve for x where one root is rational. Leave the answers
to the other roots in surd form.
5.2.1
x3 − 4x2 − x + 10 = 0
5.2.2
x3 + 3x2 − 6x + 2 = 0
5.2.3
2x3 + 3x2 − 6x − 8 = 0
1
5.2.4
2x3 − 3x2 − 13x + 7 = 0
(x = __
2 is a solution)
Solve for x where one root is rational and the other roots may
be real or non-real. Give answers correct to two decimal places
where appropriate.
5.3.1
x3 − 2x2 − 6x + 9 = 0
5.3.2
2x3 − 5x2 + 7x − 4 = 0
5.3.3
x3 − 27 = 0
(4)
(4)
(4)
(4)
(4)
(4)
(5)
(5)
(5)
(5)
(6)
(6)
(6)
(6)
(6)
(6)
(6)
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TOPIC
8
2
Differential calculus
Unit 1: Limits
REMEMBER
Division by 0 is undefined
4
is undefined
• __
0
is undefined if x = −3
• _____
x+3
The limit of a function f ( x )
is the value it approaches as
x gets closer and closer to a
particular value.
x−2
KEY WORDS
identical – exactly the same
asymptote – a straight line
that a graph gets very close
to, but never touches
REMEMBER
a
f ( x ) = _____
x − p + q is a hyperbola
• vertical asymptote: x = p
• horizontal asymptote: x = q
When a function is undefined for a particular value x, it is important to know how
the function behaves near the value of x for which it is undefined.
• A function has a limit if it has almost identical values on either side of
the undefined value.
• A function does not have a limit if it has different values on either side of
the undefined value.
y
f(x) = x² – 4
4x – 8
Consider the function sketched alongside:
2
x −4
f ( x ) = ______
'
(2;1)
4x − 8
• f ( 2 ) is undefined because 4x − 8 = 0
when x = 2.
• x → 2 means that x is almost 2, but not
exactly 2, and that the values chosen for x
can approach 2 from both sides.
f
• ( x ) has a limit when x → 2 if the result is
almost the same on both sides of 2.
x
x
1,9
1,99
1,999
1,9999
2
2,0001
2,001
2,01
2,1
f(x)
0,975
0,9975
0,99975
0,999975
Undefined
1,000025
1,00025
1,0025
1,025
As x gets closer to 2:
• from the negative side (values less than 2), the value of f ( x ) gets closer to 1
• from the positive side (values greater than 2), the value of f ( x ) gets closer to 1.
lim f ( x ) = 1 in spite of the fact that f ( 2 ) is undefined.
x→2
It is important to note that you can simplify f ( x ):
(x − 2)(x + 2)
x2 − 4
x+2
f ( x ) = ______ = ____________ = _____, x ≠ 2
•
•
4
4(x − 2)
x
+2
2
+
2
_____
_____
lim f ( x ) = lim 4 = 4 = 1
4x − 8
x→2
x→2
Not all expressions have limits:
3
3
does not exist and the graph of y = _____
has a vertical asymptote
• lim _____
x−2
x−2
x→2
given by x = 2
= _____ = 5 because when x = 3 the denominator is 1, not 0
• lim _____
x−2 3−2
3
x→3
5
does not exist and the graph y = _____
has a vertical asymptote
• lim _____
x+3
x+3
5
5
x→−3
•
142
given by x = –3.
5
5
lim _____ = ___ = −5 because when x = −4, the denominator is −1, not 0.
x→−4
x+3
−1
Topic 8 Differential calculus
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You often need to factorise fractions before you can cancel:
x2 − 2x − 3 cannot be simplified in this form.
__________
• x2 − 9
leads to simplification:
• Factorisation
2
x − 2x − 3
__________
x2 − 9
(x − 3)(x + 1)
= ____________
(x − 3)(x + 3)
x+1
_____
= x+3
y
4
Now consider the function: g ( x ) = − _____
x−1
• g ( 1 ) is undefined because x − 1 = 0 when
x=1
• x → 1 means that x is almost 1, but not
quite 1.
• g ( x ) has a limit when x → 1 only if the values
of f ( x ) are almost identical for x values which
are close to 1, but not equal to 1.
x=1
4
g(x) = – x 4– 1
2
y=0
x
–2
–4
x
0,9
0,99
0,999
0,9999
1
1,0001
1,001
1,01
1,1
g(x)
40
400
4 000
40 000
undefined
– 40 000
– 4 000
– 400
– 40
As x gets closer to 1:
• from the negative side (values less than 1), the value of f ( x ) gets bigger
• from the positive side (values greater than 1), the value of f ( x ) gets smaller
(
)
4
The results show that g does not have a limit when x = 1, so lim − _____
does
x−1
x→1
not exist.
EXERCISE 1
1
For each of the functions A–J, determine the limit, if it exists, by setting up a
table like the one below. Select the appropriate x-values from the table.
x → −2
−2,1
−2,01
−2,001
−2,0001
−2
−1,9999
−1,999
−1,99
−1,9
0,001
0,01
0,1
x→0
−0,1
−0,01
−0,001
−0,0001
0
0,0001
x→2
1,9
1,99
1,999
1,9999
2
2,0001
2,001
2,01
2,1
x→3
2,9
2,99
2,999
2,9999
3
3,0001
3,001
3,01
3,1
x→5
4,9
4,99
4,999
4,9999
5
5,0001
5,001
5,01
5,1
A
x2 − 2x
B
x2 + 2x − 8
lim __________
D
lim _______
2
lim _____
s+2
F
s3 + 8
s −4
s→−2
x→0
C
x−2
x→2
E
3
lim __________
5x − 15
x→3
s→−2
2
x2 − 5x + 6
lim _______
x
x2 − 7x
x
− 3x
x→0
lim _____
2
G
lim ___________
2
2s2 − 3s − 5
2s − 5s
s→3
H
lim _________
2
I
lim ___________
2
k2 − 7k + 12
2
k − 4k − 6
k→3
J
lim _______
x−2
5k3 − 25k2
k − 25
k→5
x2 − 2x
REMEMBER
Factorise fractions to simplify
them. ALWAYS take out the
highest common factor first.
If there are two terms,
consider the options of
trinomials, difference of
squares, sum or difference of
cubes.
If there are three terms see if
you have a trinomial which
factorises.
If there are four terms, try
grouping them into pairs.
x→2
Simplify each function in Question 1 and then determine the limit, if possible.
Unit 1 Limits
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In the sketch f ( x ) = −x2 + 2x + 3
3 − ( −5 )
y = −4x + 11 passes through the points (2;3) and (4;–5), with m = ________
= −4
2−4
3
3 − 0 ___
=
= −3
y = −3x + 9 passes through the points (2;3) and (3;0), with m = _____
2 − 3 −1
3 − 4 ___
−1
y = −x + 5 passes through the points (2;3) and (1;4), with m = _____
= 1 = −1
2−1
None of these lines is a tangent at (2;3) because in each one passes through
two points on f.
y
y = – 3x + 9
11
y = – 4x + 11
9
y = –x + 5
5
KEY WORDS
(1;4)
A (2;3)
3
tangent – a line which
touches a curve at the point
of contact
gradient – slope of a line
–2
–1
1
3
2
5
5
4
x
(4;–5)
f(x) = –x² + 2x + 3
To be a tangent, the line may only pass through one point on f.
The gradient of the tangent at (2;3) must lie between the gradients m = –1
and m = –3.
In the table below, we calculate the gradient between the point ( x; f ( x ) ) and the point
3 − f(x)
(2;3) using the average gradient formula: m = _______ where f(x) = –x2 + 2x + 3.
2−x
REMEMBER
y2 − y1
m = ______
x2 − x1 gives the gradient
of a straight line.
The average gradient of a
curve is also given by
y2 − y1
.
m = ______
x2 − x1
The equation of a straight line
is given by
y − y1 = m ( x − x1 ) and by
y = mx + c.
Vertical lines are given by
x = k.
Horizontal lines are given by
y = k.
144
x
1,9
1,99
1,999
1,9999
2
2,0001
2,001
2,01
2,1
f(x)
3,19
3,0199
3,001999
3,00019999
3
2,99979999
2,997999
2,9799
2,79
m
−1,9
−1,99
−1,999
−1,9999
undefined
−2,0001
−2,001
−2,01
−2,1
3 − 3,00019999
If x = 1,9999 ⇒ m = ______________
= −1,9999
2 − 1,9999
3 − 2,99979999
If x = 2,0001 ⇒ m = ______________
= −2,0001
2 − 2,0001
We cannot use the average gradient to determine the gradient at x = 2, but
3 − f(x)
lim _______
≈ −2
2−x
x→2−
3 − f(x)
3 − f(x)
≈ −2 ⇒ lim_______
= −2
lim _______
2−x
2−x
x→2 +
x→2
and so the gradient is m = –2.
Consider the function: f ( x ) = −x2 + 2x + 3
Topic 8 Differential calculus
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3−y
• A ( 2;3 ) and B ( x;y ) ⇒ mAB = _____
2−x
• A ( 2;f ( 2 ) ) and B ( 2 + h;f ( 2 + h ) ) ⇒ mAB
f ( 2 + h ) − f ( 2 ) ____________
f(2 + h) − f(2)
= ____________
=
(2 + h) − 2
h
h represents the change in x from A to B
f ( 2 + h ) is the y value when x = 2 + h.
f ( x ) = −x2 + 2x + 3
f ( 2 ) = − ( 2 )2 + 2 ( 2 ) + 3 = 3
f ( 2 + h ) = − ( 2 + h )2 + 2 ( 2 + h ) + 3
= −4 − 4h − h2 + 4 + 2h + 3
= −h2 − 2h + 3
f(2 + h) − f(2)
2+h −2
( −h2 − 2h + 3 ) − 3
________________
=
2+h−2
mAB = ____________
(
)
−h2 − 2h
= ________
h
(
h
−
h − 2)
= _________
h
y
A(2;f (2))
–2 –1
1
2 3
f(x) = –x² + 2x + 3
h
4
x
B(2+h;f (2+h)
= −h − 2
The gradient of AB will equal –h – 2 for all real values of x:
• if h = 1: f ( 2 + h ) = f ( 2 + 1 ) = f ( 3 )
= − ( 3 )2 + 2 ( 3 ) + 3 = 0 and B ( 3;0 )
0−3
⇒ mAB = _____
= −3 and mAB
3−2
= −h − 2 = −1 − 2 = −3
if
h
=
5:
f(2 + h) = f(2 + 5) = f(7)
•
= − ( 7 )2 + 2 ( 7 ) + 3 = −32 and B ( 7;−32 )
−32 − 3 ____
−35
⇒ mAB = _______
= 5
7−2
= −7 and mAB − h − 2 = −5 − 2 = −7
If the line is a tangent to the graph at point A, then B should lie on A.
• A and B cannot be the same point because the gradient would be undefined.
• h must be limited so that h → 0 and mtangent at A = lim ( −h − 2 ) = 0 − 2 = −2
h→0
EXERCISE 2
f ( x ) = x2 − x − 6 and g ( x ) = −x2 + 2x
A ( −2; f ( −2 ) ), B ( −2 + h; f ( − 2 + h ) ), C ( 3; g ( 3 ) ) and D ( 3 + h; g ( 3 + h ) )
1
Determine the gradients of AB and CD in terms of h.
2
State mAB if h = –1.
3
State mCD if h = –2.
4
Determine the value(s) of h for which mAB = mCD.
5
State mAB if h = 0.
REMEMBER
A parabola has three
equations:
The general formula:
y = ax2 + bx + c
The x-intercept formula:
y = a ( x − x1 ) ( x − x2 )
The turning point formula:
y = a ( x − p )2 + q
If a < 0, the graph has a
maximum.
If a > 0, the graph has a
minimum.
To determine the x-intercept
of any graph:
Let y = 0 and solve for x.
To determine the y-intercept
of any graph:
Let x = 0 and solve for y.
The symmetry line of a
parabola passes through its
turning point. You can find
this line in two ways:
x1 + x2
b
or x = − ___
x = ______
2
2a
Unit 1 Limits
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Unit 2: Use limits to define the derivative
of a function f
The gradient between two points on a curve
is referred to as an average gradient.
If A and B lie on f, then the average
gradient between
• A ( x1;y1 ) and B ( x2;y2 ) is given by
y2 − y1
mAB = ______
x2 − x 1
y
f
B(x + h;f(x + h))
f(x + h) – f(x)
A(x;f(x))
h
x
• A ( x;f ( x ) ) and B ( x + h;f ( x + h ) ) is given
f(x + h) − f(x)
by mAB = ____________
h
If s ( t ) = −4,9t 2 where s is the distance in metres and t is the time in seconds, then:
• t is the independent variable
• s ( t )is the dependent variable.
KEY WORDS
independent variable – a
variable that does not depend
on other values
dependent variable – a
variable that is affected by
other values
If f ( x ) = x2 + 2x − 3, then:
• x is the independent variable
• f ( x )is the dependent variable.
If y = x3 − 8, then:
• x is the independent variable
• y is the dependent variable.
The derivative of a function measures the rate at which the dependent variable
changes as the independent variable changes.
• The derivative is the gradient of the tangent to the function at a point on the curve
f(x + h) − f(x)
is the gradient of a tangent to the curve f at x
• f ’ ( x ) = lim ____________
h
h→0
• h represents the increase in x from A to B
• if h → 0 then A and B will effectively be the same point
f(x + h) − f(x)
• f ’ ( x ) is called the derivative of f and is given by f ’ ( x ) = lim ____________
h
f(x + h) − f(x)
h→0
is the gradient of the tangent to f at x
• f ’ ( x ) = lim ____________
h
h→0
f(a + h) − f(a)
is the gradient of the tangent to f at a
• f ’ ( a ) = lim ____________
h
h→0
f(2 + h) − f(2)
is the gradient of the tangent to f at 2.
• f ’ ( 2 ) = lim _____________
h
h→0
146
Topic 8 Differential calculus
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WORKED EXAMPLE
1.
Consider the function: f ( x ) = 2x2 − 5x − 7
1.1
Sketch f, indicating the intercepts with the axes and the coordinates
of the turning point.
1.2
Consider the points: P ( −2;f ( −2 ) ) and Q ( 3;f ( 3 ) )
1.2.1
If h = 4 plot P and Q on f and draw the line passing
through them.
1.2.2
Determine the gradient of PQ.
1.3
Consider the points: A ( f ( 1 ) ) and B ( 1 + h;f ( 1 + h ) )
1.3.1
Determine the average gradient between A and B in terms of h.
1.3.2
Determine the average gradient between A and B if h = 4.
1.3.3
If h = 4, plot A and B on f and draw the line passing through
them.
1.3.4
Use your graph to determine the gradient of AB.
1.4
Consider the points: D ( x;f ( x ) ) and E ( x + h;f ( x + h ) )
1.4.1
Determine the average gradient between D and E in terms of x
and h.
f(x + h) − f(x)
1.4.2
Determine: f ’ ( x ) = lim ____________
h
h→0
Determine f ’ ( 1,5 ) and explain your answer.
1.4.3
y
SOLUTIONS
1.1 Calculations
x-intercepts: ( −1;0 ) and ( 3,5;0 )
2x2 − 5x − 7 = 0
( 2x − 7 ) ( x + 1 ) = 0
x = 3,5 or x = −1
y-intercepts: ( 0;−7 )
Turning point: ( 1,25; −10,125 )
3,5 + ( −1 )
f(x) = 2x² –5x – 7
B(5;18)
P(–2;11)
3,5
–1
x
Q(3;–4)
–7
A(1;–10)
2,25
x = _________
= ____
= 1,25
2
2
(1,25;–10,125)
y = f ( 1,25 ) = 2 ( 1,25 )2 − 5 ( 1,25 ) − 7 = −10,125
1.2 1.2.1 f ( −2 ) = 2 ( −2) )2 − 5 ( −2 ) − 7 = 11 ⇒ P ( −2; 11 )
f ( 3 ) = 2 ( 3 )2 − 5 ( 3 ) − 7 = −4 ⇒ Q ( 3;−4 )
11 − ( − 4 )
15
1.2.2 m = _________ = ___ = −3
PQ
−2 − 3
−5
1.3 1.3.1 f ( 1 ) = 2 ( 1 )2 − 5 ( 1 ) − 7 = −10 ⇒ A ( 1; −10 )
f ( 1 + h ) = 2 ( 1 + h )2 − 5 ( 1 + h ) − 7
= 2 ( 1 + 2h + h2 ) − 5 − 5h − 7
= 2 + 4h + 2h2 − 5h − 12
= 2h2 − h − 10 ⇒ B ( 1 + h; 2h2 − h − 10 )
( 2h2 − h − 10 ) − ( −10 )
1+h −1
2h2 − h
h ( 2h − 1 )
= _______ = _________ = 2h − 1
mAB = ____________________
(
)
h
h
1.3.2 If mAB = 2h − 1 =2 ( 4 ) − 1 = 7
1.3.3 See the graph.
18 − ( −10 ) ___
28
1.3.4 A ( 1;−10 ) and B ( 5;18 ) ⇒ mAB = __________
= 4 =7
5−1
Unit 2 Use limits to define the derivative of a function f
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1.4 1.4.1 f ( x ) = 2x2 − 5x − 7
⇒ D ( x; 2x2 − 5x − 7 )
f ( x + h ) = 2 ( x + h )2 − 5 ( x + h ) − 7
= 2 ( x2 + xh + xh + h2 ) − 5x − 5h − 7
= 2x2 + 4xh + 2h2 − 5x − 5h − 7
⇒ E ( x + h; 2x2 + 4xh + 2h2 − 5x − 5h − 7 )
f(x + h) − f(x)
mDE = ____________
h
( 2x2 + 4xh + 2h2 − 5x − 5h − 7 ) − ( 2x2 − 5x − 7 )
= _________________________________________
h
2
4xh
+
2
h
−
5h
= _____________
h
h ( 4x + 2h − 5 )
_____________
=
= 4x + 2h − 5
h
1.4.2 f ’ ( x ) = lim ( 4x + 2h − 5 ) = 4x + 2 ( 0 ) − 5 = 4x − 5
h→0
1.4.3 f ’ ( 1,25 ) = 4 ( 1,25 ) − 5 = 0
This is the gradient at the turning point (1,25;–10,125) and it is not an average
gradient.
The gradient is zero, so the line (tangent) passing through the turning point is
horizontal.
EXERCISE 3
1
148
Consider g ( x ) = x2 − 5x − 6.
1.1
Sketch g, indicating the intercepts with the axes and the coordinates of the
turning point.
1.2
Determine g(−1) and g(2) and then plot P(−1; g(−1)) and Q(2; g(2)) on g.
1.3
Determine the average gradient of g between P(−1; g(−1)) and Q(2; g(2)).
1.4
Consider the points: P(−1; g(−1)) and B(−1 + h; g(–1 + h))
1.4.1
Determine the average gradient between P and B in terms of h.
1.4.2
Determine the average gradient between P and B if h = 1.
1.4.3
Plot B on g and draw the line passing through P and B.
1.4.4
Determine the equation of the line passing through P and B.
1.5
Consider the points: D ( x; g ( x ) ) and E ( x + h; g ( x + h ) )
1.5.1
Determine the average gradient between D and E in terms of x and h.
1.5.2
Determine the average gradient between D and E if x = 1 and h = 5.
1.5.3
Determine the average gradient between D and E if xD = 1 and
xE = 6.
1.5.4
Discuss your results to 1.5.2 and 1.5.3.
1.5.5
Plot D and E on g and draw the line which passes through them.
1.5.6
Determine the equation of the line passing through D and E.
1.5.7
Determine g ’(1).
1.5.8
Briefly explain what g ’ ( 1 ) represents.
1.5.9
Determine the equation of the tangent to g at x = 1.
1.5.10 Determine g ’(2,5).
1.5.11 Briefly explain what g ’(2,5) represents.
1.5.12 Determine the equation of the tangent to g at x = 2,5.
Topic 8 Differential calculus
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2
3
4
Consider f ( x ) = x2 − 4x − 5.
2.1
Sketch f, indicating the intercepts with the axes and the coordinates
of the turning point.
2.2
Determine the average gradient of f between x = −2 and x = 3.
2.3
Plot A ( −2; f ( −2 ) ) and P ( 3;f ( 3 ) ) on f and determine the equation
of the line passing through A and P.
2.4
Consider the points: A ( −2; f ( −2 ) ) and B ( −2 + h; f ( −2 + h ) )
2.4.1
Determine the average gradient between A and B in terms of h.
2.4.2
Determine the average gradient between A and B if h = 4.
2.5
Consider the points: D ( x;f ( x ) ) and E ( x + h;f ( x + h ) )
2.5.1
Determine the average gradient between D and E in terms of x and h.
2.5.2
Determine the average gradient between D and E if x = 2 and h = 1.
2.5.3
Determine the equation of the line passing through D and E
if x = 2 and h = 1.
2.5.4
Determine: f ’ ( 2 )
2.5.5
Briefly explain what f ’ ( 2 ) represents.
2.5.6
Determine the equation of the line which is a tangent to f at x = 2.
Consider f ( x ) = x3.
3.1
Determine the average gradient of f between x = −1 and x = 2.
3.2
Determine the equation of the line passing through the points with x = –1
and x = 2.
3.3
Consider the points A ( −1; f ( −1 ) ) and B ( −1 + h; f ( −1 + h ) ).
3.3.1
Determine the average gradient between A and B in terms of h.
3.3.2
Determine the average gradient between A and B if h = 4.
3.3.3
Determine f ’ ( −1 ).
3.3.4
Determine the equation of the tangent at x = –1.
3.4
Consider the points D ( x; f ( x ) ) and E ( x + h; f ( x + h ) ).
3.4.1
Determine the average gradient between D and E in terms of x and h.
3.4.2
Determine the average gradient between D and E if x = –2 and h = 1.
3.4.3
Determine f ’ ( −2 ).
3.4.4
Briefly explain what f ’ ( −2 ) represents.
3.4.5
Determine the equation of the tangent to f at x = –2.
P(–3;4) is one of the stationary points of
y
f ( x ) = ( x + 1 )2 ( x + 4 ).
4.1
State two values of x for which f ( x ) = 0.
P(–3;4)
4.2
Determine f ( 0 ).
4.3
Use your sketch to determine:
x
R
4.3.1
f ’ ( −3 )
f(x) = (x + 1)²(x + 4)
4.3.2
f ’ ( −1 )
4.4
4.5
4.6
4.7
4.8
4.9
4.10
4.11
State the equation of the tangent to f at P.
Determine A ( 2;f ( 2 ) ) and B ( −1;f ( −1 ) )
Determine the gradient of AB.
Determine the coordinates of C ( 2 + h;f ( 2 + h ) ).
Determine the gradient of AC.
Determine f ’ ( 2 ).
What does f ’ ( 2 ) represent?
Determine the equation of the tangent at x = 2.
Unit 2 Use limits to define the derivative of a function f
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REMEMBER
The derivative of a function
measures the rate at which
the dependent variable
changes as the independent
variable changes:
• the derivative is the
gradient of the tangent to
the function at a point on
the curve at x
f(x + h) − f(x)
• f ’ ( x ) = lim ____________
h
h→0
is the gradient of a tangent
to the curve f at x
• h represents the increase in
x from A to B
• if h → 0 then A and B will
effectively be the same
point
• f ’ ( x ) is called the derivative
of f and is given by
f(x + h) − f(x)
f ’ ( x ) = lim ____________
Unit 3: Differentiation of functions from first
principles
f( x + h) − f( x)
The definition of the derivative of any function f is given by f ’ ( x ) = lim ____________
.
h
h→0
Use the definition if you are asked to:
• determine the derivative from first principles
• use the definition to determine the derivative.
You cannot determine the derivative unless h cancels out, since division by 0 is undefined.
Basic algebraic simplification is important in determining the derivative.
You often need to simplify the expressions such as these below and you should
practise them!
• ( x + h )2 = ( x + h ) ( x + h ) = x2 + 2xh + h2
• ( x + h )3 = ( x + h ) ( x + h )2 = ( x + h ) ( x2 + 2xh + h2 ) = x3 + 3xh2 + 3xh2 + h3
(x + h)
h
1
1 − __
_____
= x − ________ = − ________
• x+h
h
h→0
f(x + h) − f(x)
• f ’ ( x ) = lim ____________
h
h→0
is the gradient of the
tangent to f at x.
f(a + h) − f(a)
x
x(x + h)
x(x + h)
WORKED EXAMPLE 1
Consider the function f ( x ) = 2x2 − 3x − 5.
1
Determine f ’ ( x )from first principles.
2
Determine f ’ ( 3 ).
3
What is represented by f ’ ( 3 )?
4
Determine the equation of the tangent to f at x = 3.
5
Determine the average gradient between x = –2 and x = 3.
6
Determine the equation to the line passing through x = –2 and x = 3.
• f ’ ( a ) = lim ____________
h
SOLUTIONS
is the gradient of the
tangent to f at a.
When you determine a
derivative from first principles
or using the definition, use
the formula
1
h→0
f ( x + h ) = 2 ( x + h )2 − 3 ( x + h ) − 5
= 2 ( x2 + 2xh + h2 ) − 3x − 3h − 5
= 2x2 + 4xh + 2h2 − 3x − 3h − 5
f(x + h) − f(x)
f ’ ( x ) = lim ____________
h→0
h
2x2 + 4xh + 2h2 − 3x − 3h − 5 − ( 2x2 − 3x − 5 )
= lim _______________________________________
h
h→0
4xh + 2h2 − 3h
_____________
= lim
h
f(x + h) − f(x)
.
f ’ ( x ) = lim ____________
h
h→0
h→0
You may not use the rules of
differentiation.
h ( 4x + 2h − 3 )
= lim _____________
h
h→0
= lim ( 4x + 2h − 3 )
h→0
KEY WORD
derivative – the gradient of a
tangent to the function at a
point on the curve
= 4x + 2 ( 0 ) − 3 = 4x − 3
2
3
4
1
f ’(3) = 4(3) − 3 = 9
f ’ ( 3 ) = 9 gives the gradient of the tangent to f at x = –3.
f ( 3 ) = 2 ( 3 )2 − 3 ( 3 ) − 5 = 4, so (3;4) lies on f.
Substitute m = 9 and (3;4) into y − y1 = m ( x − x1 )
y − 4 = 9 ( x − 3 ) ⇒ y = 9x − 23
f ( 3 ) − f ( −2 ) 4 − 9
Average gradient = ___________ = _____ = −5 = −1
1
Substitute m = − 1 and (3;4)
3 − ( −2 )
5
y − 4 = = −4 − 1 ( x − 3 ) ⇒ y − 4 = −x + 3 and so x + y = 7
150
Topic 8 Differential calculus
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WORKED EXAMPLE 2
Consider the function g ( x ) = −x3.
1
Find the derivative of g ( x ) at the point where x = −2, using the definition
of the derivative.
2
What is represented by g ’ ( −2 ).
3
Determine the equation of the tangent to g at x = −2?
4
Determine the average gradient between x = −2 and x = 2.
5
Determine the equation of the line passing through x = −2 and x = 2.
SOLUTIONS
1
g ( x + h ) = − ( x + h )3
= − ( x3 + 3x2h + 3x2 + h3 )
= −x3 − 3x2h − 3xh2 − h3
g(x + h) − g(x)
g ’ ( x ) = lim _____________
h
h→0
( −x3 − 3x2h − 3xh2 − h3 ) − ( x3 )
= lim __________________________
h
h→0
−3x2h − 3xh2 − h3
= lim ________________
h
h→0
h ( −3x2 − 3xh − h2 )
= lim ________________
h
h→0
= lim ( −3x2 − 3xh − h2 )
h→0
2
3
4
5
= −3x2 − 3x ( 0 ) − ( 0 )2
= −3x2
g ’ ( −2 ) = −3 ( −2) )2 = −12
g ’ ( −2 ) = −12 tell us that the gradient of the tangent to g at x = –2
is m = –12.
Substitute m = –12 and (–2;8) into y − y1 = m ( x − x1 )
y − 8 = −12 ( x + 2 ) ⇒ y = −12x − 16
g ( 2 ) − g ( −2 ) ______
−8 − 8 ____
−16
Average gradient =___________
=
= 4 = −4
4
2 − ( −2 )
Substitute m = – 4 and (–2;8) into y − y1 = m ( x − x1 )
y − 8 = −4 ( x + 2 ) ⇒ y = −4x
WORKED EXAMPLE 3
Consider the function g ( x ) = 5.
1
Determine g ’ ( x )from first principles.
2
Determine g ’ ( −7 ).
3
What is represented by g ’ ( −7 )?
4
State the equation of the tangent to g at x = –7.
Unit 3 Differentiation of functions from first principles
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SOLUTIONS
1
g ( x ) = 5 and g ( x + h ) = 5
g(x + h) − g(x)
5−5
0
g ’ ( x ) = lim _____________
= lim _____ = lim __ = 0
h
h
h
h→0
2
3
4
h→0
h→0
g ’(7) = 0
The gradient of the tangent to g is given by m = g ’ ( 7 ) = 0
The tangent is a horizontal line given by y = 5.
Take note that the tangent to g(x) = 5 is also y = 5.
WORKED EXAMPLE 4
2
Consider the function f ( x ) = __
x.
1
Determine f ’ ( x )from first principles.
2
Determine f ’ ( 4 ).
3
What does f ’ ( 4 ) represent?
4
Determine the equation of the tangent to f at x = 4.
5
Determine the average gradient between x = –1 and x = 4.
6
Determine the equation of the line which passes through x = –1 and x = 4.
SOLUTIONS
1
2
2
_____
f ( x ) = __
x and f ( x + h ) = x + h
f(x + h) − f(x)
f ’ ( x ) = lim ____________
h
h→0
(
1 2
2
= lim __ _____ − __
h→0
x
h x+h
)
(
)
1 − 2h
= lim __ ( _______ )
h x x+h
1 2x − 2 ( x + h )
= lim __ ____________
h
x(x + h)
h→0
(
h→0
)
−2
= lim ________
x(x + h)
h→0
−2
= _______
x(x + 0)
2
= −__2
x
2
3
4
2
2
1
___
__
f ’ ( 4 ) = −__
2 = − 16 = − 8
4
1
.
f ’ ( 4 ) gives us the gradient of the tangent to f at x = 4 as m = −__
8
( )
1
1
and 4;__
into y − y1 = m ( x − x1 ).
Substitute m = −__
2
8
1
1(
__
__
y − = − x − 4 ) ⇒ 8y − 4 = −x + 4 so x + 8y = 8
2
5
6
1
__
− ( − 2)
( __1 + 2 ) × 2
f ( 4 ) − f ( − 1 ) _________
2
2
1 + 4 __
1
Average gradient =___________
=
= __________
= _____
=2
(
)
5
5×2
10
4− −1
( )
1
1
Substitute m = __
and 4;__
into y − y1 = m ( x − x1 )
2
2
1 __
1
= ( x − 4 ) ⇒ 2y − 1 = x − 4 and so x − 2y = 3
y − __
2
152
8
2
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EXERCISE 4
Use first principles to find the derivative of f ( x ) = x2 − x − 12 at the point
where x = 2.
1
Using the definition, find the derivative of f ( x ) = −__
x and then determine
(
)
f ’ −2 .
Use first principles to find the derivative of g ( x ) = 5x3 and then determine g ’ ( 4 ).
3
Use the definition to find the derivative of f ( x ) = __
x and then determine f ’ ( −5 ).
Use first principles to find the derivative of g ( x ) = −5x2 + 2x − 3 at the point
where x = –1.
Use the definition to find the derivative of g ( x ) = −x2 − 5.
1
2
3
4
5
6
You use the derivative to determine the gradient of a point on a curve, whereas you
use the average gradient to determine the gradient between two points on a curve.
You cannot differentiate all functions.
f( x + h) − f( x)
.
Consider the definition of the derivative of a function given by f ’ ( x ) = lim_____________
h
0
__
h→0
(
)
(
)
If
f
x
+
h
−
f
x
=
0
then
lim
=
0
.
•
h
h→0
• If f ( x + h ) − f ( x ) ≠ 0, then you can only determine the derivative if h is a common
factor in the numerator.
• You can determine f ’ ( a ) by determining f ’ ( x ) first and then substituting a into the
result.
EXERCISE 5
1
Consider the function f ( x ) = x2 − 5x.
1.1
Determine the average gradient between x = –3 and x = – 3 + h.
1.2
f ( −3 + h ) − f ( −3 )
Determine the lim _______________
h
h→0
2
3
4
5
6
7
8
9
10
11
12
13
14
1.3
Use first principles to determine f ’ ( −3 ).
1.4
Compare your answers to 1.2 and 1.3 and explain the results.
1.5
Determine the equation of the tangent to f at x = –3.
7
Consider the function g ( x ) = −__
x.
2.1
Use the definition to find the derivative of g ( x ).
2.2
Determine g ’ ( −7 ).
2.3
What is represented by g ’ ( −7 )?
2.4
Determine the equation of the tangent to g at x = –7.
Use first principles to find the derivative of f ( x ) = −5x3 at the point where x = –1.
Use first principles to find the derivative of g ( x ) = 2x − 5 at the point x = 4.
Using the definition, find the derivative of f ( x ) = −3 and then determine f ’ ( −5 ).
Use first principles to find the derivative of g ( x ) = −3x2.
Use first principles to determine f ’ ( x ) if f ( x ) = −4x3.
Use the definition to differentiate g ( x ) = 1 − 9x2.
Use first principles to differentiate f ( x ) if f ( x ) = 2.
Use first principles to determine g ’ ( x ) if g ( x ) = −6x2 − 4.
9
Use first principles to differentiate f ( x ) if f ( x ) = __
x.
3
Use the definition to differentiate g ( x ) = −2x .
If f ( x ) = 2 − 5x − 3x2, use first principles to determine f ’ ( 2 ).
Use first principles to find the derivative of g ( x ) = −2x2 + x − 4.
Unit 3 Differentiation of functions from first principles
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Unit 4: Use the specific rules for
differentiation
The formula for the differentiation of the function f ( x ) = axn is given by
f ’ ( x ) = anxn−1 for n ∈ ℝ.
There are many rules for differentiation, but you are required to use only one.
You can only use f ’ ( x ) = anxn−1 to differentiate functions if you can express each
term in the function in the form axn. It is therefore often necessary to write terms in
the correct format before you can apply the formula. You can denote derivatives in
different ways, so make sure you know how to recognise and work with each.
Original expression
Simplified expression
Derivative
f ( x ) = ( 2x − 3 ) ( 4x + 1 )
f ( x ) = 8x2 − 10x − 3
f ’ ( x ) = 16x − 10
3
y = __
x
y = 3x−1
dy
3
___
= −3x−2 = − __
Dx[ x 2 ]
5 __32
__
x
__
dx
5
__
Dx[ √x5 ]
[ x − 5x − 6 ]
2
[ x−6 x+1 ]
2
(
Dx __________
x−6
)(
)
Dx ____________
= Dx[ x + 1 ]
x−6
1
__
__
√x = x2
1
__
__
3
1
__
__
√x
x2
___
= __ = x 2 −2 = x− 2
x2
x2
x2
__
=1
x2
4 − x __
4 x
_____
= − __ = 4x− 1 − 1
x
1
__
x
3
__
4+3
____
x
7
___
x 3 .x 4 = x 12 = x 12
1
ds
__
= −10t + 3
s = −5t 2 + 3t
REMEMBER
x2
dt
WORKED EXAMPLES
Determine, using the rules of differentiation:
1
f ’ ( x ) if f ( x ) = 7x2 − 5x + 4h
dy
___
if y = 3x2 − 9x − 2
2
dx
3
y = ( x − 3 ) ( 2x − 1 )
4
Dx __________
2
5
6
[ x x− +x −3x12 ]
x − 2x + x
d ____________
___
)
dx (
x
2
3
2
2
__
√
s = ( t 2 − 3 ) ( t 2 + 2 ).
1
__
3
__
SOLUTIONS
1
2
f ’ ( x ) = 14x − 5
dy
___
= 6x − 9
dx
3
y = ( x − 3 ) ( 2x − 1 )
= 2x2 − 7x + 3
dy
___
= 4x − 7
4
4
Dx ____________
= Dx ( 1 − 4x−1 ) = 4x−2 = __2
(
)
5
3
__
x3 − 2x2 + √ x
d ____________
d
3 __5
3
___
= ___ ( x − 2 + x− 2 ) = 1 − __x− 2 = 1 − ___
dx
6
dx
)(
(
(
)
__
( __
x2
) __
x
) dx
2
s = (t 2 − 3 ( t 2 + 2 ) = t 2 + 2t 2 − 3t 2 − 6
1
__
ds
9 __1
9 __1
1
__
= 2t + t − 2 − __ t 2 = 2t + __ − __ t 2
1
dt
154
[ x −x 4x +x3+ 3 ]
3
2
1
__
3
__
1
__
2
t2
5
__
2x 2
Topic 8 Differential calculus
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EXERCISE 6
Write all the answers in this exercise with positive exponents. Use the rules of
differentiation to determine:
[x − x + 5]
1
dy
2
___
if y = 3x2 − __
x
2
Dx _________
x
3
Dx ____________
2x − 1
[ 2x − 11x + 5 ]
4
dy
4
___
if y = ___
__
d (
___
5x − 2√ x )2
6
d ( 2
__
3t − 4t )
8
dy
___
if y = ( 5x − 2 )2
dx
dx
2
5
dx
+ 12
]
[ 6x 3−x 17x
− 4x
2
Dx _____________
2
7
3
dx
REMEMBER
1
( 2x3 )−1 = ___
3
3x2
2
2x−3 = __
3
x
dt
(
dx
__
__
Dx[ ( 2x − 3 )3 ]
10
√ x5 − √ x
d _________
___
11
Dx[ ( x − 2 )(2x + 3 ]
12
dy
___
if y = 5x ( x − 3 )
13
dy
3
x2
___
if y = ___ − __
14
2
1
__
__
dy
___
if y = x 3 − x 2
15
dy
3
5
___
if y = √x2 − √x3
16
( t − 3 )3
ds
__
_
if s = _______
dx
2x
3
__
__
dx
17
__
dy
1
___
if xy = √3 − 4x3 + __
19
f ’ ( x ) if f ( x ) = ( √x − 3 ) ( 2√x + 3 )
21
23
x
x0 = 1
d ( )
___
x =1
dx
)
9
3
Only drop the Dx notation
in the step in which you
determine the derivative.
If k is a constant, then
d ( )
___
k = 0.
dx
dx
dx
dt
[
__
3
√
t
__
√ x5 − 3x2 + 2√ x
√x
]
18
__
Dx _______________
3
20
1 3
f ’(x) if f ( x ) = 2x5 − 3x4 + __
x −4
3
g ’ ( x ) if g ( x ) = ___________
x+1
22
d [(
___
2x − 3 ) ( 4x2 + 6x + 9 ) ]
1
f ’ ( t ) if f ( t ) = t 3 − 3t 2 − __
t
24
2
f ’ ( x ) if f ( x ) = 2x 2 − 5x 2 + __
1
__
x
dx
__
__
3x2 − 4x − 7
__
dx
3
__
1
__
x2
25
3__
f ’ ( x ) if f(x) = ( √x − 2x ) 2x − ___
√x
)
26
dy
__
___
if y = ( 3x2 − √ x )2
dx
27
Dx[ [ ( x2 − 9 ) ( x2 − 3x + 9 ) ( x − 3 )−1 ]
28
f ’ ( x ) if f ( x ) = ( x 2 − 2 ) ( x 2 + 3 )
(
2x
1
__
1
__
Important rules
d ( )
d
d
___
[ f x ± g ( x ) ] = ___[ f ( x ) ] ± ___ g ( x )
1
dx
dx [
dx
]
WORKED EXAMPLE 1
If f ( x ) = x3 + 2x and g ( x ) = −2x2 − 5, show that
d ( )
d
d
___
[ f x + g ( x ) ] = ___[ f ( x ) ] + ___ g ( x )
dx
dx
dx [
]
SOLUTION
d ( )
d ( )
LHS = ___
f x ] + ___
g x ] = ( 3x2 + 2 ) + ( −4x ) = 3x2 − 4x + 2
dx [
dx [
d ( )
d[ 3
f x + g ( x ) ] = ___
x − 2x2 + 2x − 5 ] = 3x2 − 4x + 2
RHS = ___
dx [
dx
d
d
d
⇒ ___[ f ( x ) + g ( x ) ] = ___[ f ( x ) ] + ___[ g ( x ) ]
dx
dx
dx
Unit 4 Use the specific rules for differentiation
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d ( )
d
d
___
[ f x − g ( x ) ] = ___[ f ( x ) ] − ___ g ( x )
2
dx
dx [
dx
]
WORKED EXAMPLE 2
d ( )
d ( )
d ( )
If f ( x ) = 5x3 − 2 and g ( x ) = 2x2 + x, show that ___
f x − g ( x ) ] = ___
f x ] − ___
g x ]
dx [
dx [
dx [
SOLUTION
d ( )
d ( )
RHS = ___
f x ] − ___
g x ] = ( 15x2 ) − ( 4x + 1 )
dx [
dx [
= 15x2 − 4x − 1
d [( 3
d ( )
LHS = ___
f x − g ( x ) ] = ___
5x − 2 ) − ( 2x2 + x ) ]
dx [
dx
= 15x2 − 4x − 1
d
d
d
⇒ ___[ f ( x ) − g ( x ) ] = ___[ f ( x ) ] − ___[ g ( x ) ]
dx
dx
dx
d ( )
d
___
k f ( x ) ] = k___
f x ] where k is a constant.
dx [
dx [
d
d
___
[ k f ( x ) ] = k___[ f ( x ) ]
3
dx
dx
WORKED EXAMPLE 3
d
d ( )
If f ( x ) = x3 + 2x, show that ___
k f ( x ) ] = k___
f x ]
dx [
dx [
SOLUTION
d
d ( 3
LHS = ___
kf ( x ) ] = ___
kx + 2kx ) = 3kx2 + 2k
dx [
dx
RHS = kf ’ ( x ) = k ( 3x2 + 2 ) = 3kx2 + 2k
d
d
⇒ ___[ k f ( x ) ] = k___[ f ( x ) ]
dx
dx
d ( )
d
d
___
[ f x × g ( x ) ] ≠ ___[ f ( x ) ] × ___ g ( x )
4
dx
dx [
dx
]
WORKED EXAMPLE 4
If f ( x ) = 2x − x3 and g ( x ) = 2x2 − 3x, show that
d ( )
d
d
___
[ f x × g ( x ) ] ≠ ___[ f ( x ) ] × ___ g ( x )
dx
dx
dx [
]
SOLUTION
d [(
d[
LHS = ___
2x − x3 ) ( 2x2 − 3x ) ] = ___
− 2x 5 + 3x 4 + 4x 3 − 6x 2 ]
dx
dx
= −10x4 + 12x3 + 12x2 − 12x
d ( )
d ( )
RHS = ___
f x ] × ___
g x ]
dx [
dx [
= ( 2 − 3x2 ) ( 4x − 3 ) = −12x3 + 9x2 + 8x − 6
d
d
d
⇒ ___[ f ( x ) × g ( x ) ] ≠ ___[ f ( x ) ] × ___[ g ( x ) ]
dx
dx
dx
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EXERCISE 7
1
2
REMEMBER
2
3
2
Consider the functions f ( x ) = 3x − 3 and g ( x ) = 2x + 2x .
1.1
Determine f ( x ).g ( x ).
1.2
d ( )
d ( )
d ( ) ( )
Show that ___
f x ] × ___
g x ] ≠ ___
f x .g x ].
dx [
dx [
dx [
1.3
f(x)
Determine ____
g ( x ).
1.4
f(x)
d ( )
d ____
d ( )
f x ] ÷ ___
g x ] ≠ ___
Show that ___
dx [
dx [
dx g ( x )
1.5
d ( )
d ( )
d ( )
f x ] + ___
g x ] = ___
f x + g ( x ). ]
Show that ___
dx [
dx [
dx [
Rule of differentiation:
f ( x ) = axn ⇒ f ’ ( x ) = naxn−1
Definition:
f(x + h) − f(x)
f ’ ( x ) = lim ____________
h
h→0
[ ]
Remember that you can
factorise both a sum and
difference of cubes.
a3 + b3
= ( a + b ) ( a2 − ab + b2 )
a3 − b3
= ( a − b ) ( a2 + ab + b2 )
Consider the functions f ( x ) = x2 − 4x and g ( x ) = 2x − 8.
2.1
Fully simplify each expression:
2.2
2.3
2.1.1
f(x) + g(x)
2.1.2
f(x) − g(x)
2.1.3
10f ( x )
2.1.4
f(x) × g(x)
2.1.5
f(x) ÷ g(x)
Use first principles to determine the derivative of each expression in 2.1.
Use the rules of differentiation and your results from 2.1 and 2.2 to
determine which of the statements below are true. Show all necessary
working.
2.3.1
d ( )
d
d
___
[ f x + g ( x ) ] = ___[ f ( x ) ] + ___ g ( x )
]
2.3.2
d ( )
d
d
___
[ f x − g ( x ) ] = ___[ f ( x ) ] − ___ g ( x )
]
2.3.3
d
d
___
[ k f ( x ) ] = k___[ f ( x ) ]
2.3.4
d ( )
d
d
___
[ f x × g ( x ) ] = ___[ f ( x ) ] × ___ g ( x )
2.3.5
___[ f ( x ) ]
f(x)
d ____
dx
___
= _______
dx
dx
dx
dx
dx
dx [
dx [
dx
dx
dx
dx [
]
d
3
4
dx [ g x ] ___ g x
(
d
dx [
)
(
)]
Consider the functions f ( x ) = 2x3 − 16 and g ( x ) = x2 + 2x + 4.
f(x)
3.1
Simplify ____
g ( x ).
3.2
f(x)
( )
If p ( x ) = ____
g ( x ) , determine p ’ 3 .
3.3
Simplify f ( x )g ( x ).
3.4
If h ( x ) = f ( x )g ( x ), determine h ’ ( −1 ).
3.5
Determine Dx[ p ( x ) + h ( x ) ] in two different ways.
x+3
x2 − 4
and g ( x ) = _______
f ( x ) = ______
2
x+3
x − 2x
4.1
Determine f ( x ) × g ( x ).
4.2
If h ( x ) = f ( x )g ( x ), determine h ’ ( x ).
Unit 4 Use the specific rules for differentiation
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Find the equations of tangents to
functions
Unit 5:
Average gradient is the gradient between two points on a curve and is given by
y2 − y1
f(x + h) − f(x)
____________
m = ______
.
x − x or m =
h
2
1
The derivative gives the gradient of a point on a curve and is therefore the gradient of
the tangent to the curve at the specified point.
f( x + h) − f( x)
The derivative is given by f ’ ( x ) = lim ____________
. You can determine the derivative by:
h
h→0
• using the definition, which means you are using first principles
• using the rule f ( x ) ⇒ f ’ ( x ) = anxn−1.
If you are asked to find the derivative but you are not told what method to use, then
use the rules.
Use the definition or first principles method only if you are instructed to do so.
WORKED EXAMPLE
Consider the function f ( x ) = −x2 + 2x + 8.
1
Determine f ’ ( x ).
2
Determine the equation of the tangent to f at the point where:
2.1
3
4
5
6
x = −3
2.2
x = −2
2.3
x=0
2.4
x=1
2.5
x=2
2.6
x=4
Which tangents in Question 2 pass through the turning point of f ? Justify
your answer.
Sketch f ( x ) = −x2 + 2x + 8 and the 6 tangents in 1.1 on the same set of axes.
Show all the intercepts with the axes and the coordinates of any turning
point(s). Label each graph.
For which value(s) of x is f:
5.1
an increasing function
5.2
a decreasing function
5.3
stationary.
Briefly explain how you could use the gradients of the tangents in 1.2 to help
you answer 1.5.
SOLUTIONS
1
f ’ ( x ) = −2x + 2
2.1 m = f ’ ( −3 ) = −2 ( −3 ) + 2 = 8
y = f ( −3 ) = − ( −3) )2 + 2 ( −3 ) + 8 = −7
Substitute m = 8 and ( −3;−7 ) into y − y1 = m ( x − x1 )
y − ( −7 ) = 8 ( x − ( −3 ) ) ⇒ y + 7 = 8x + 24 and so y = 8x + 17
2.2 m = f ’ ( −2 ) = −2 ( −2 ) + 2 = 6
y = f ( −2 ) = − ( −2) )2 + 2 ( −2 ) + 8 = 0
Substitute m = 6 and ( −2;0 ) into y − y1 = m ( x − x1 )
y − 0 = 6 ( x − ( −2 ) ) ⇒ y = 6x + 12
2.3 m = f ’ ( 0 ) = −2 ( 0 ) + 2 = 2 and y = f ( 0 ) = 8
Substitute m = 2 and ( 0;8 ) into y = mx + c ⇒ y = 2x + 8
158
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2.4 m = f ’ ( 1 ) = −2 ( 1 ) + 2 = 0 and y = f ( 1 ) = − ( 1 )2 + 2 ( 1 ) + 8 = 9
m = 0 so y = 9
2.5 m = f ’ ( 2 ) = −2 ( 2 ) + 2 = −2 and y = f ( 2 ) = − ( 2 )2 + 2 ( 2 ) + 8 = 8
Substitute m = −2 and ( 2;8 ) into y − y1 = m ( x − x1 )
y − 8 = −2 ( x − 2 ) ⇒ y = −2x + 12
2.6 m = f ’ ( 4 ) = −2 ( 4 ) + 2 = −6 and y = f ( 4 ) = − ( 4 )2 + 2 ( 4 ) + 8 = 0
Substitute m = −6 and ( 4;0 ) into y − y1 = m ( x − x1 )
y − 0 = −6 ( x − 4 ) ⇒ y = −6x + 24
3
The tangent y = 9 passes through the turning point because it passes through
(1;9).
4
y y = 8x + 17
y = –6x + 24
y = 2x + 12
24
y = 2x + 8
y = –2x + 12
17
16
12
y=9
9
8
–6
–4
–2
(–3;–7)
(1;9)
(2;8)
2
4
6
x
–8
y = –x² + 2x + 8
5.1
5.2
5.3
6
x ∈ ( −∞;1 )
x ∈ ( 1;∞ )
x=1
Gradients of tangents are:
• 0 at the stationary points, so y = 9 indicates that (1;9) is a stationary point
• positive where the function increases and the gradients of the tangents
passing through x-values –3, –2 and 0 are all positive
• negative where the function decreases and the gradients of the tangents
passing through the x-values 2 and 4 are negative.
Unit 5 Find the equations of tangents to functions
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EXERCISE 8
1
2
Consider the function f ( x ) = 2x3 + x2 − 4x + 6.
1.1
Determine the equation of the tangent to the
function at x = 2.
1.2
Is f increasing, decreasing or stationary at x = 2?
Justify your answer.
1.3
Determine the equation of the tangent to the
function at x = –1.
1.4
Is f increasing, decreasing or stationary at x = –1?
Justify your answer.
4
− 2.
Consider the function f ( x ) = __
x
2.1
3
y
x
f(x) = 2x³ + x² – 4x + 6
y
Find f ’ ( x ), stating your answer with positive
f(x) = 4
– –2
exponents.
x
2.2
Determine the gradient of the tangent to f at
the point x = 2.
x
2.3
Is f an increasing or decreasing function at x = 2?
y = –2
2.4
Determine the equation of the tangent of f at
x = 2.
2.5
Determine the gradient of the tangent to f at the
x=0
point x = –1.
2.6
Is f an increasing or decreasing function at x = −1?
2.7
Determine the equation of the tangent to f at x = −1.
2.8
Will f ever be an increasing function? Fully justify your answer.
2
−3
Consider the functions: f ( x ) = x2 − 3x − 4, g ( x ) = −x2 − x + 6 and k ( x ) = __
3.1
3.2
Determine the derivatives of f, g and k.
Determine the equations of the tangents to each function at:
3.2.1
x = –2
3.2.2
x=1
3.2.3
x
x=5
The derivative gives the gradient of a tangent to a curve at a specified point.
• If you know the value of x, you can work out the gradient of the tangent.
• If you know the gradient of the tangent, you can work out the value of x at the
contact point.
At
• stationary points and turning points, the tangent is horizontal and has a
gradient = 0.
WORKED EXAMPLE
Consider the function:
f ( x ) = x2 − 2x − 15 and g ( x ) = x3 − 7x
1
Determine the value of x and the
equation of the tangent if:
1.1
f ’(x) = 2
1.2
f ’(x) = 0
1.3
g ’ ( x ) = −4
2
Solve for x if f ’ ( x ) = g ’ ( x ).
y g(x) = x³ – 7x
x
f(x) = x² – 2x – 15
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SOLUTIONS
1.1
f ’ ( x ) = 2x − 2 = 2 ⇒ 2x = 4 and x = 2
y = f ( 2 ) = ( 2 )2 − 2 ( 2 ) − 15 = −15 ⇒ ( 2; − 15 )
Substitute m = 2 and ( 2;−15 ) into y − y1 = m ( x − x1 )
y − ( −15 ) = 2 ( x − 2 ) ⇒ y + 15 = 2x − 4 and y = 2x − 19
1.2
f ’ ( x ) = 2x − 2 = 0 ⇒ x = 1, y = f ( 1 ) = ( 1 )2 − 2 ( 1 ) − 15 = −16
⇒ ( 1;−16 ) and y = −16
1.3
2
g ’ ( x ) = 3x2 − 7 = −4 ⇒ 3x2 = 3, x2 = 1 and x = ± 1
If x = 1, y = g ( 1 ) = ( 1 )3 − 7 ( 1 ) = − 6 ⇒ ( 1;− 6 )
Substitute m = −4 and ( 1;6 ) into y − y1 = m ( x − x1 )
y + 6 = −4 ( x − 1 ), so y = −4x − 2
If x = −1, y = g ( −1 ) = ( −1 )3 − 7 ( −1 ) = 6 ⇒ ( −1;6 )
Substitute m = − 4 and ( 1;6 ) into y − y1 = m ( x − x1 )
y − 6 = −4 ( x + 1 ), so y = −4x + 2
2x − 2 = 3x2 − 7 ⇒ 3x2 − 2x − 5 = 0 and ( 3x − 5 ) ( x + 1 ) = 0
5
⇒ x = __
or x = −1
3
EXERCISE 9
1
2
3
4
2
3
2
f ( x ) = __
x and g ( x ) = x + x − 8x
1.1
Determine f ’ ( x ) from first principles.
1.2
For which value(s) of x is f ’ ( x ) = −8?
1.3
Determine the equation of the tangent to f at x = –1.
1.4
Determine the equation(s) of the tangent(s) to g if g ’ ( x ) = 0.
1.5
Determine the equation(s) of the tangent(s) to g if g ’ ( x ) = −3.
1.6
Determine the equation of the tangent to g at x = 2.
1.7
Determine the equation of the tangent which is parallel to the tangent to g
at x = –3.
1.8
Determine the average gradient of f between x = –2 and x = 1.
1.9
Determine the average gradient of g between x = –1 and x = 2.
f ( x ) = 2x2 + x − 6 and g ( x ) = −3x2 − 10x − 8
2.1
For which value of x will the tangents to f and g be parallel to each other?
2.2
Determine the equation of each of the parallel tangents.
2.3
Determine the equation of the tangent to g if x = 1.
2.4
Determine the equation of the tangent to f at its turning point.
2.5
Determine the equation of the tangent to f if x = –3.
2.6
Determine the equation of the tangent to f if the gradient is –3.
1
Determine the equation of the tangent to y = ( 2x − 1 )2 ( x + 3 ) where x = __
.
2
4−x
Determine the equation(s) of the tangent(s) to y = _____
:
x
dy
4.1
where ___ = −1
f(x) = 2x² + x – 6 y
x
g(x) = – 3x² – 10x – 8
dx
5
4.2
where x = 4.
x3 − 8
Determine the equation of the tangent to y = ______
:
x−2
5.1
where x = 5
dy
5.2
where ___ = 3.
dx
Unit 5 Find the equations of tangents to functions
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Unit 6: The second derivative
It is possible to determine the stationary points of a graph and to discuss its concavity
without sketching it.
Use the first derivative test to identify and classify stationary points.
• Determine the value(s) of x for which f ’ ( x ) = 0.
• Determine the sign of f ’ ( x ) for x < a, but close to a and for x > a, but close to a.
• If f ’ ( x ) has the same sign for x < a and for x > a, then ( a;f ( a ) ) is a point
of inflection.
• If f ’ ( x ) < 0 for x < a, but f ’ ( x ) > 0 for x > a then ( a;f ( a ) ) is a local minimum
and f is concave up.
• If f ’ ( x ) > 0 for x < a, but f ’ ( x ) < 0 for x > a then ( a;f ( a ) ) is a local maximum
and f is concave down.
f ’(x) = 0
f ’(x) > 0
f ’(x) < 0
f ’(x) > 0
f ’(x) = 0
minimum
f ’(x) = 0
f ’(x) < 0
maximum
f ’(x) > 0
f ’(x) > 0
point of inflection
Use the second derivative to determine the point(s) of inflection and concavity of a
function.
To determine point(s) of inflection:
• determine the value(s) of x for which f ” ( x ) = 0
• if f ”(a) = 0, determine the signs of f ” ( x ) for x < a but close to a and for x > a but
close to a
• if f ’ ( x ) changes sign, then ( a;f ( a ) ) is a point of inflection
To determine concavity, solve for x if f ’ ( x ) = 0:
• if f ’ ( a ) = 0, determine the sign of f ” ( a )
• if f ” ( a ) > 0, then f is concave up on that interval and ( a;f ( a ) ) is a local minimum
• if f ” ( a ) < 0, then f is concave down on that interval and ( a;f ( a ) ) is a local
maximum
• if f ” ( a ) = 0, test the sign of f ” ( x ) for x < a but close to a and for x > a but close to a.
local maximum
f ’’(x) < 0
f ’(x) = 0
f ’’(x) > 0
f ’’(x) < 0
f ’’(x) > 0
f ’’(a) = 0
f ’’(x) < 0
f ’(x) = 0
f ’’(x) > 0
local minimum
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Consider the graphs f ( x ) = x3 + 3x2 − 9x − 27, f ’ ( x ) = 3x2 + 6x − 9 and f ” ( x ) = 6x + 6.
y
f ’(x) = 3x² + 6x – 9
local maximum
f(x) = x³ + 3x² – 9x – 27
6
(–3;0)
–5
–4
–3
–2
1
–1
2
3
4
5
x
–9
(–1;–16)
point of inflection –18
–27
f “(x) = 6x + 6
(1;–32)
local minimum
You find the stationary point(s) by solving f ’ ( x ) = 0.
• f ’ ( x ) = 3 ( x2 + 2x − 3 ) = 0 ⇒ 3 ( x + 3 ) ( x − 1 ) = 0, so x = −3 or x = 1
• f ( −3 ) = 0 and f ( 1 ) = −32, so there are stationary points are at (−3;0) and (1;−32).
First derivative test at stationary points (−3;0) and (1;−32):
• f ’ ( x ) > 0 if x < –3 but f ’ ( x ) < 0 x > −3, so (−3;0) is a local maximum
• f ’ ( x ) < 0 if x < 1, but f ’ ( x ) > 0 if x > 1, so (1;−32) is a local minimum.
Second derivative test for concavity:
• f ” ( −3 ) = 6 ( −3 ) + 6 = −12 < 0, so f ( x ) is concave down at x = −3 and (−3;0) is a
local maximum
• f ” ( 1 ) = 6 ( 1 ) + 6 = 12 > 0, so f ( x ) is concave up at x = 1 and (1;−32) is a local
minimum.
REMEMBER
The derivative gives:
• the gradient of a tangent
to a curve at a specified
point
• the rate of change of the
function at the specified
point.
You find the point(s) of inflection by solving f ” ( x ) = 0 x > –3 x > –3:
• f ” ( x ) = 6x + 6 = 0 ⇒ 6x = −6 and x = −1, so f ” ( −1 ) = 0
• f ” ( x ) < 0 for x ∈ ( − ∞;−1 ), but f ” ( x ) > 0 for all x ∈ ( −1;∞ )
⇒ change in concavity at x = −1
• f ( −1 ) = −16 is a point of inflection because there is a change of concavity
at x = −1.
When you have sketched a graph you can see:
• where the stationary points are
• where the local minimum can be found (where the graph is concave up)
• where the local maximum can be found (where the graph is concave down)
• where the point of inflection (change of concavity) occurs.
Unit 6 The second derivative
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WORKED EXAMPLES
1
2
3
Consider the function f ( x ) = x3 − 6x2 − 36x + 216 = ( x − 6 )2 ( x + 6 ).
1.1
Determine the coordinates of the stationary points of f.
1.2
Use the second derivative test to determine the concavity of f at each
stationary point and hence state whether the stationary point is a local
minimum or a local maximum.
1.3
Determine the coordinates of the point of inflection.
1.4
Sketch f, showing the intercepts with the axes, the stationary points
and the point of inflection.
Classify the stationary point(s) of g ( x ) = x3.
Classify stationary points and discuss the concavity of f ( x ) = x3 − 3x2.
SOLUTIONS
1.1 f ’ ( x ) = 3x2 − 12x − 36 = 0 ⇒ x2 − 4x − 12 = 0
( x − 6 ) ( x + 2 ) = 0 and so x = 6 or x = −2
y = f ( −2 ) = ( −2 )3 − 6 ( −2 )2 − 36 ( −2 ) + 216 = 256 ⇒ ( −2;256 )
y = f ( 6 ) = ( 6 )3 − 6 ( 6 )2 − 36 ( 6 ) + 216 = 0 ⇒ ( 6;0 )
The stationary points are (−2;256) and (6;0).
1.2 f ”(x) = 6x − 12
f ” ( −2 ) = 6 ( −2 ) − 12 = −24 < 0
f is concave down at (–2;256) and this point is a local maximum.
f ” ( 6 ) = 6 ( 6 ) − 12 = 24 > 0
f is concave up at (6;0) and this point is a local minimum.
1.3 f ” ( x ) = 6x − 12 = 0, so x = 2 ⇒ f ” ( 1,9 ) = 6 ( 1,9 ) − 12 = −0,6 and
f ” ( 2,1 ) = 6 ( 2,1 ) − 12 = 0,6
There is a change in concavity at x = 2.
f ( 2 ) = ( 2 )3 − 6 ( 2 )2 − 36 ( 2 ) + 216 = 128 so the point of inflection
is (2;128).
1.4
y
(–2;256)
–6
2
3
164
216
(2;128)
(6;0)
x
g ’ ( x ) = 3x2 = 0, so there is a stationary point at (0;0).
g ” ( x ) = 6x ⇒ g ” ( 0 ) = 6 ( 0 ) = 0, so (0;0) is neither a local maximum nor a local
minimum.
g ” ( −1 ) = −1 < 0, but g ” ( 1 ) = 1 > 0, so (0;0) is a point of inflection.
f ’ ( x ) = 3x2 − 6x = 0 ⇒ 3x ( x − 2 ) = 0, so x = 0 or x = 2
f ” ( x ) = 6x − 6
f ” ( 0 ) = −6 < 0, so (0;0) is a local maximum and f is concave down at (0;0)
f ” ( 2 ) = 6 ( 2 ) − 6 = 6 > 0, so (2;−4) is a local minimum and f is concave up at
(2;−4).
f ” ( x ) = 6x − 6 = 0 ⇒ x = 1 and f is concave down if x < 1 and concave up if
x>1
There is a point of inflection at (1;−2) as there is a change of concavity at
this point.
Topic 8 Differential calculus
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EXERCISE 10
1
Consider the function f ( x ) = x3 + 3x2 − 45x + 9.
1.1
Determine the coordinates of the stationary points of f.
1.2
Use the second derivative test to determine the concavity of f at each
stationary point and then state whether the stationary point is a local
minimum or a local maximum.
1.3
Determine the coordinates of the point of inflection.
2
Consider the function g ( x ) = −4x3 + 15x2 + 72x + 16.
2.1
Determine the coordinates of the stationary points of g.
2.2
Use the second derivative test to determine the concavity of g at each
stationary point. Then state whether the stationary point is a local
minimum or a local maximum.
2.3
Determine the coordinates of the point of inflection.
3
Consider the function f ( x ) = −x3 + 3x.
3.1
Determine the coordinates of the stationary points of f.
3.2
Use the second derivative test to determine the concavity of f at each
stationary point. Then state whether the stationary point is a local
minimum or a local maximum.
3.3
Determine the coordinates of the point(s) of inflection.
4
1 3
Consider the function g ( x ) = __
x − x2 − 35x.
3
4.1
Determine the coordinates of the stationary points of g.
4.2
Use the second derivative test to determine the concavity of g at each
stationary point. Then state whether the stationary point is a local
minimum or a local maximum.
4.3
Determine the coordinates of the point(s) of inflection.
5.
2 3 __
Consider the function f ( x ) = − __
x +1
x2 + 10x + 2.
2
3
5.1
Determine the coordinates of the stationary points of f.
5.2
Use the second derivative test to determine the concavity of f at each
stationary point. Then state whether the stationary point is a local
minimum or a local maximum.
5.3
Determine the coordinates of the point of inflection.
6.
Consider the function: g ( x ) = −x3 − 3x2.
6.1
Determine the coordinates of the stationary points of g.
6.2
Use the first derivative test to classify the stationary points.
6.3
Use the second derivative test to determine the concavity of g at each
stationary point. Then state whether the stationary point is a local
minimum or a local maximum.
6.4
Determine the coordinates of the point of inflection.
6.5
Factorise g ( x ) = −x3 − 3x2 and then solve for x if g ( x ) = 0.
6.6
Sketch g ’, showing the intercepts with the axes and the coordinates
of the turning point.
6.7
Sketch g on the same set of axes as g ’, showing the intercepts with
the axes and the coordinates of the stationary points.
6.8
State the values of x for which g ’’ ( x ) > 0 and the values of x for
which g ’ ( x ) < 0.
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6.9
Briefly comment on the relationship between the stationary points of g
and the roots of g ’.
6.10 Sketch g ”on the same set of axes and comment on the relationship
between g ( x ) and g ” ( x ).
7.
166
Consider the function f ( x ) = x3 + 2x2 − 7x + 4.
7.1
Determine x-coordinates of the stationary points of f.
7.2
Use the first derivative test to classify the stationary points, showing all
necessary calculations.
7.3
Use the second derivative test to classify the stationary points, showing all
necessary calculations.
7.4
Determine the point of inflection.
7.5
Discuss the concavity of the function.
7.6
Draw f ’, showing the intercepts with the axes and the coordinates of the
turning point.
7.7
If f ( 1 ) = 0 and f ( −4 ) = 0, sketch f on the same set of axes as f ’, showing the
intercepts with the axes, the coordinates of the stationary points and the
point of inflection.
7.8
Sketch f ” on the same system of axes as f and f ’.
7.9
Briefly discuss the relationship between the graphs f, f ’ and f ”.
Topic 8 Differential calculus
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Unit 7: Sketch cubic graphs
Cubic graphs:
• have one y-intercept
• have at least one x-intercept, but at most three x-intercepts.
• may have stationary points which are local maximum and/or local minimum point(s)
• may have a point of inflection which is a stationary point, or which is a point of
inflection.
WORKED EXAMPLE 1
Consider the function f ( x ) = x3 − 5x2 − 8x + 12.
1
Determine and classify the stationary point(s).
2
Determine the point(s) of inflection.
3
Determine the roots of f ( x ) = 0.
4
Sketch f, showing the intercept(s) with the axes and the coordinates of
stationary point(s) and point(s) of inflection.
5
What is the maximum value of f ( x ) for x ∈ [ −3;10 ]?
6
Use your graph to determine lim f ( x ).
x→−∞
SOLUTIONS
1
f ’ ( x ) = 3x2 − 10x − 8 and f ” ( x ) = 6x − 10
2
At the stationary points f ’ ( x ) = ( 3x + 2 ) ( x − 4 ) = 0, so x = −__
or x = 4
3
3
2
400
2
2
2
2
2 ____
__
__
__
__
__
f − = −
−5 −
− 8 − + 12 = ⇒ − ;
≈ ( − 0,67;14,81 )
( 3 ) ( 3)
( 3)
( 3 27 )
( 3)
f ( 4 ) = ( 4 )3 − 5 ( 4 )2 − 8 ( 4 ) + 12 = 36 ⇒ ( 4; −36 )
(
)
( 3 27 )
400
2 ____
Stationary points are − __
;
and ( 4;−36 )
3 27
2
2 400
= −14 < 0, so −__;____ is a local maximum and f is concave down
f ” −__
( 3)
400
2 ____
;
at ( −__
3 27 )
2
3
f ” ( 4 ) = 14 > 0, so ( 4;−36 ) is a local minimum and f is concave up at ( 4;−36 )
5
f ” ( x ) = 6x − 10 = 0 ⇒ x = __
3
2
From question 1 we know that f ” − __
< 0 and f ” ( 4 ) > 0, so there is a change
3
( )
5
5 ____
286
in concavity at x = __
and ( __
;− 27 ) is a point of inflection.
3
3
f ( −2 ) = ( −2) )3 − 5 ( −2) )2 − 8 ( −2 ) + 12 = 0 ⇒ x + 2 is a factor of f ( x )
f ( x ) = ( x + 2 ) ( x2 − 7x + 6 ) = ( x + 2 ) ( x − 1 ) ( x − 6 ) = 0
Roots are −2, 1 or 6
4
y
(–0,67;14,81)
–2
y = x³ – 5x² – 8x + 12
12
6
1
(1,67;–10,59)
x
(4;–36)
5
6
f ( 10 ) = ( 10 )3 − 5 ( 10 )2 − 8 ( 10 ) + 12 = 412, so the maximum value is 412.
lim f ( x ) = − ∞
x→−∞
Unit 7 Sketch cubic graphs
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WORKED EXAMPLE 2
Consider the function g ( x ) = x3 − 9x2 + 27x + 37.
1
Determine and classify the stationary point(s) of g.
2
Discuss the concavity of g.
3
Show that x = −1 is the only real root of g(x) = 0.
4
Sketch g, showing the intercept(s) with the axes and the coordinates of
stationary point(s) and point(s) of inflection.
5
Use your graph to determine lim g ( x ).
x→∞
6
7
Sketch g ’ and g ’’ on the same set of axes, showing the intercept(s) with the
axes and the coordinates of the turning point.
Comment on the relationship between g, g ’ and g ’’ by comparing the graphs.
SOLUTIONS
g ’ ( x ) = 3x2 − 18x + 27 and g ’’ ( x ) = 6x − 18
g ’ ( x ) = 3x2 − 18x + 27 = 0 ⇒ 3 ( x2 − 6x + 9 ) = 0 and ( x − 3 ) ( x − 3 ) = 0
g ’ ( 3 ) = 0 and g ( 3 ) = 64 ⇒ ( 3;64 ) is a stationary point
g ’’ ( 3 ) = 6 ( 3 ) − 18 = 0
⇒ we cannot classify the stationary point using the second derivative test
g ’’ ( 2,9 ) = −0,6 < 0 and g ’’ ( 3,1 ) = 0,6 > 0
There is a change of concavity at x = 3
⇒ ( 3;64 )is a point of inflection
2
g ’ ( 2,9 ) = 63,999 > 0 and g ’ ( 3,1 ) = 64,001 > 0, so the first derivative tests tells
you that there is neither a maximum nor a minimum at (3;64). g increases
from left to right and the concavity of g changes at x = 3 from concave down
to concave up.
3
g ( −1 ) = ( −1) )3 − 9 ( −1) )2 + 27 ( −1 ) + 37 = −1 − 9 − 27 + 37 = 0
x = −1 is a root and so x + 1 is a factor of g ( x )
( x + 1 ) ( x2 − 10x + 37 ) = 0
x2 − 10x + 37 has no factors and b2 − 4ac = ( −10 )2 − 4 ( 1 ) ( 37 ) =
−48 < 0 ⇒ roots are non-real
x = –1 is the only real root of g(x).
4 and 6
g’(x) = 3x² – 18x + 27 y
g(x) = x³ – 9x² + 27x + 37
1
(3;64)
g“(x) = 6x – 18
37
27
–1
5
6
7
168
–18
3
x
As x → ∞ , g ( x ) → ∞
From question 1 we know that g ’ ( x ) = 3x2 − 18x + 27 =3 ( x − 3 )2
Roots are (3;0), y-intercept is (0;27) and TP is (3;0).
g ’ ( x ) ≥ 0 for all values of x. Although x = 3 is the x coordinate of a stationary
point on g, there is neither a local minimum nor a local maximum when
x = 3. g ” ( x ) changes sign at x = 3, so there will be a point of inflection on
g when x = 3.
Topic 8 Differential calculus
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WORKED EXAMPLE 3
Consider the function f ( x ) = −x3 + x2 − x − 3.
1
Determine and classify the stationary point(s) of f.
2
Determine the point(s) of inflection.
3
Discuss the concavity of f.
4
Determine the roots of f(x) = 0.
5
Sketch f, showing the intercept(s) with the axes and the coordinates of the
stationary point(s) and point(s) of inflection.
6
Sketch f ’ and f ” on the same set of axes as f, showing the intercept(s) with
the axes and the coordinates of the turning point(s).
7
Discuss the relationship between f, f ’ and f ” by referring to the graphs.
SOLUTIONS
2
f ’ ( x ) = −3x2 + 2x − 1 = 0 ⇒ 3x2 − 2x + 1 = 0
b2 − 4ac = ( −2) )2 − 4 ( 3 ) ( 1 ) = −8 < 0 ⇒ no real roots
f has no stationary points
88
1
1 ___
f ” ( x ) = −6x + 2 = 0 ⇒ x = __
and __
;− 27 is a point of inflection
3
3
3
f ” ( 0 ) = 2 > 0 and f ” ( 1 ) = −4 < 0, so f changes from concave up to concave
1
(
(
1
7
down at __
;−3___
3
27
)
)
f ( −1 ) = − ( −1) )3 + ( −1) )2 − ( −1 ) − 3 = 0 ⇒ −1 is a root and x − 1 is a factor.
x3 − x2 + x + 3 = 0 ⇒ ( x + 1 ) ( x2 − 2x + 3 ) = 0
x2 − 2x + 3 = 0 ⇒ b2 − 4ac = ( −2) )2 − 4 ( 1 ) ( 3 ) = −8 < 0 ⇒ no real roots
x = −1 is the only root
5 and 6
y
f “(x) = – 6x + 2
4
2
1
–
3
(13–;– 23–
–1
–3
(
f(x) = –x³ + x² – x – 3
x
88
(13–;– —
27
(
f ’(x) = –3x² + 2x – 1
7
f ’ ( x ) < 0 for all x ∈ ℝ, so f has no stationary points.
1
f ” __
=0
(3)
1
f ” ( x ) > 0 if x < __
3
1
f ” ( x ) < 0 if x > __
3
1
1
The change of sign of f ” at x = __
indicates a point of inflection on f at x = __
.
3
3
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EXERCISE 11
Sketch each of the graphs below, showing the intercepts with the axes, the
coordinates of any stationary points and the coordinates of the point of inflection.
You may need to use the quadratic formula in questions 15–20.
1 3 __
1
f ( x ) = ( x − 2 )2 ( x + 3 )
2
f ( x ) = __
x −1
x2 − 12x + 18
3
2
3
2
3
3
f ( x ) = −x − 3x + 4
4
f ( x ) = −4x − 13x2 − 10x
3
2
5
f ( x ) = −x − 2x + 7x − 4
6
f ( x ) = (2x + 1) ( 4x2 − 44x + 169 )
62
4
1 3
7
f ( x ) = __x3 − 2x2 + x + ___
8
f ( x ) = −__
x + x2 − x
9
11
13
15
17
19
3
3
f ( x ) = −4x3 + 23x2 − 40x + 21
f ( x ) = 2x3 − 3x2 + 8x − 7
f ( x ) = ( x − 2 )3
1(
f ( x ) = __
x − 3 ) ( 4x2 + 15x − 45 )
6
f ( x ) = 2x3 − 2x2 − 5x − 1
1 3
f ( x ) = __
x + x2 − 24x + 54
3
10
12
14
16
18
20
3
f ( x ) = −x3 + 8
f ( x ) = x3 + 9x2 + 24x + 16
f ( x ) = x3 − 4x
f ( x ) = 4x3 + 37x2 + 110x + 104
1(
f ( x ) = − __
x + 2 ) ( 4x2 + 13x − 50 )
6
3
f ( x ) = 5x + 21x2 − 9x − 62
EXERCISE 12
Consider the functions below and answer the questions that follow for each function.
1
Determine and classify the stationary point(s).
2
Determine the point of inflection.
3
Discuss, with reasons, the concavity of f.
4
Determine the roots of f ( x ) = 0.
If you can find only one rational root:
• check whether the other roots are irrational or non-real, showing all necessary
calculations
• use the quadratic formula to determine the roots of the quadratic factor if
they are irrational.
5
Sketch f, showing the intercept(s) with the axes and the coordinates of a
stationary point(s) and the point(s) of inflection.
6
Sketch f ’ and f ” on the same set of axes as f.
7
Briefly discuss the relationship between f, f ’ and f ” by referring to the graphs.
8
Determine the minimum and maximum values of f ( x ) for x ∈ [ −5;5 ].
9
Use your graph to determine lim f ( x ) and lim f ( x ).
x→−∞
170
( x ) = x3 − 6x2 + 9x − 4
x→∞
A
f
C
D
E
1(
2x + 1 ) ( 4x2 − 44x + 169 )
f ( x ) = __
8
1(
x − 5 ) ( 2x − 1 ) ( x + 7 )
f ( x ) = −__
f ( x ) = x3 − 9x2 + 24x − 20
23
4 3 ___
f ( x ) = −__
x + 11 x2 + 3x − ___
F
f ( x ) = x 3 − 3x 2 − x + 3
G
f ( x ) = x3 − 2x2 − 4x + 8
H
f ( x ) = −x3 − 5x2 − 9x − 9
I
f ( x ) = −x3 + 14x2 − 49x + 36
J
f ( x ) = x3 − 11x2 + 24x
K
f ( x ) = −x3 − 3x2 − 3x − 2
L
f ( x ) = 4x3 − 11x2 + 6x
M
f ( x ) = ( x + 2 ) ( − x2 − x − 7 )
N
f ( x ) = ( x − 1 ) ( x2 + 5 )
6
B
3
2
6
Topic 8 Differential calculus
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You may need to determine the equation of a cubic graph, with or without using
a sketch.
If the x-intercepts are given, then use the formula y = a ( x − x1 ) ( x − x2 ) ( x − x3 )
WORKED EXAMPLE 1
f ( x ) = ax3 + bx2 + cx + d passes through these points: (−3;0), (−1;0), (0;−6) and
(4;0). Determine the values of a, b, c and d.
SOLUTION
−3, −1 and 4 are all x-intercepts, so substitute:
y = a ( x − x1 ) ( x − x2 ) ( x − x3 )
y = a(x + 3)(x + 1)(x − 4)
y
f
Now substitute ( 0;−6 )
1
−6 = a ( 3 ) ( 1 ) ( −4 ) = −12a ⇒ a = __
2
1(
__
2
y = x + 3 ) ( x − 3x − 4 )
–3 –1
2
1( 3
x − 13x − 12 )
= __
2
4
x
–6
13
1 3 ___
= __
x − 2 x−6
2
1
13
a = __
, b = 0, c = −___
, d = −6
2
2
WORKED EXAMPLE 2
Consider the function: f ( x ) = ax3 + bx2 + cx + d
f(1) = f(4) = 0
f(3) = 8
f ’(1) = f ’(3) = 0
1
Draw a sketch graph of f and indicate the x-intercepts and
2
the turning point(s).
Determine the values of a, b, c and d.
SOLUTION
1
2
y
(3;8)
8
1
4
( 1;0 ) is a turning point
⇒ two equal roots at x = 1
1; 1 and 4 are all x-intercepts, so
substitute:
x
f ( x ) = a ( x − x1 ) ( x − x2 ) ( x − x3 )
f(x) = a(x − 1)(x − 1)(x − 4)
f(3) = 8 ⇒ 8
= a(3 − 1)(3 − 1)(3 − 4)
8 = −4a ⇒ a = −2
f ( x ) = −2 ( x2 − 2x + 1 ) ( x − 4 )
= −2 ( x3 − 6x2 + 9x − 4 )
= −2x3 + 12x2 − 18x + 8
a = −2, b = 12, c = −18 and d = 8
Unit 7 Sketch cubic graphs
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WORKED EXAMPLE 3
f is a cubic function.
Consider the sketch of the derivative of f, given by
f ’ ( x ) = ax2 + bx + c.
y
f ’(x)
2
1
2
3
4
x
4
Write down the x-coordinates of the stationary points of f.
Classify each stationary point and justify
your answer.
Determine the x-coordinate of the point of inflection of f.
Then draw a sketch graph of f.
SOLUTIONS
1
2
3
4
x = 2 and x = 4
f ’’ ( 2 ) < 0 so the graph is concave down at x = 2 and the stationary point
is a local maximum.
f ’’ ( 4 ) > 0 so the graph is concave up at x = 4 and the stationary point is
a local minimum.
Point of inflection when f ’’ ( x ) = 0 ⇒ x = 3.
y
f
2
172
4
x
Topic 8 Differential calculus
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WORKED EXAMPLE 4
f ( x ) = x3 + px2 + qx + r has an x-intercept at (2;0) and a point of inflection at
(0;−8).
1
Determine p, q and r, showing all necessary calculations.
2
Sketch f ( x ), f ’ ( x ) and f ’’ ( x ) on the same system of axes, showing all intercepts
with the axes as well as the coordinates of the stationary point(s) and point(s)
of inflection.
SOLUTIONS
1
r = −8
f ( 2 ) = 8 + 4p + 2q − 8 = 0
⇒ q = −2p
f ’ ( x ) = 3x2 + 2px + q and
f ” ( x ) = 6x + 2p
⇒ f ’’ ( 0 ) = 6 ( 0 ) + 2p = 0 ⇒ p = 0
q = 2(0) = 0
p = 0, q = 0 and r = −8
2
f ’(x) = 3x²
y
(2;12) •
f “(x) = 6x
2
(0;0)
x
(0;–8)
f(x) = x³ – 8
WORKED EXAMPLE 5
Consider the function g ( x ) = x3 + ax2 + bx + c which passes through the origin
and has a tangent given by the equation y = 24x − 54 at x = 3.
1
Determine the values of a, b and c, showing all necessary calculations.
2
If a = 0, b = −3 and c = 0:
2.1
Determine the equation of the tangent to g which is parallel to
the tangent at x = 3.
2.2
Determine and classify the stationary points of g, showing all
necessary calculations.
2.3
Determine the point(s) of inflection of g. Justify your answer(s).
2.4
Sketch g, showing the intercepts with the axes and the coordinates
of the stationary point(s) and point(s) of inflection. Show all
necessary calculations.
SOLUTIONS
1
g(0) = c = 0 and at x = 3, y = 24(3) – 54 = 18, so x = 3
⇒ (3;18) is a point on g ( x ) = x3 + ax2 + bx + c.
Substitute (3;18) into g ( x ) = x3 + ax2 + bx + c
⇒ g ( 3 ) = 27 + 9a + 3b = 18.
9a + 3b = −9 ⇒ 3a + b = −3 and so b = −3a − 3
➀
g ’ ( x ) = 3x2 + 2ax + b and we know that g ’ ( 3 ) = 24
∴ 3 ( 3 )2 + 2a ( 3 ) + b = 24 ⇒ 6a + b = −3
➁
Substitute ➀ into ➁ ⇒ 6a + ( − 3a − 3 ) = −3 and a = 0
Substitute a = 0 into ➀ ⇒ b = −3 ( 0 ) − 3 = −3
a = 0, b = –3 and c = 0
Unit 7 Sketch cubic graphs
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2.1
2.2
2.3
Parallel tangents have equal gradients with
the gradient of the tangent as 24.
m = g ’ ( x ) = 3x2 − 3 = 24 ⇒ x2 = 9 and x = ± 3
g ( −3 ) = ( −3) )3 − 3 ( −3 ) = −18
The tangent passes through ( −3;−18 )
y − ( −18 ) = 24 ( x − ( − 3 ) ) ⇒ y = 24x + 54
At the stationary points g ’ ( x ) = 0
3x2 − 3 = 0 ⇒ x2 = 1 and x = ± 1
g ( −1 ) = ( −1) )3 − 3 ( −1 ) = 2
g ” ( −1 ) = 6 ( −1 ) = −6 < 0
g is concave down ⇒ (−1;2) is a local maximum.
g ( 1 ) = ( 1 )3 − 3 ( 1 ) = −2 and g ” ( 1 ) = 6 ( 1 ) = 6 > 0
g is concave up ⇒ (1;−2) is a local minimum.
g ” ( x ) = 6x = 0 if x = 0 and we know from question 2.2 that g is concave
up at x = 1 and concave down at x = −1, so there is a change of
concavity at (0;0) and therefore (0;0) is a point of inflection.
2.4
y
g(x) = x³ – 3x
(–1;2)
– 3
x
(0;0)
3
(1;–2)
EXERCISE 13
y
1
f ( x ) = ax3 + bx2 + cx + d passes through the following points:
(–2;0), (2;0), (6;0) and (3;15). Determine the values of a, b, c and d.
2
f ( x ) = ax3 + bx2 + cx + d passes through the following points:
(–5;0), (1;0), (4;0) and (2;14). Determine the values of a, b, c and d.
3
Consider the function f ( x ) = ax3 + bx2 + cx + d and determine the values
of a, b, c and d if f ( 1 ) = f ( −2 ) = f ’ ( 0 ) = 0 and f ( 0 ) = −4.
4
Consider the function f ( x ) = ax3 + bx2 + cx + d and determine the values
of a, b, c and d if f ( 2 ) = 0, f ’ ( 1 ) = 0, f ” ( −2 ) = −17
and f ” ( 2 ) = 7.
5
f is a cubic graph with f ( 3 ) = f ( −3 ) = f ’ ( 3 ).
f ’ ( x ) is sketched alongside
5.1
State the x-coordinates of the stationary points of f and classify the
stationary points.
5.2
State the x-coordinate of the point of inflection and justify your answer.
5.3
Discuss the concavity of f and explain your answers fully.
5.4
Draw a sketch of f.
f ’(x)
–1
174
3
x
Topic 8 Differential calculus
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6
f ( x ) = −x3 + ax2 + bx + c has a point of inflection at (0;−27) and f ( −3 ) = 0
6.1
Determine a, b and c, showing all necessary calculations.
6.2
If a = b = 0 and c = −27, sketch f ( x ), f ’ ( x ) and f ’’ ( x ) on the same system
of axes, showing all intercepts with the axes as well as the coordinates
of the stationary point(s) and point(s) of inflection.
7
Consider the function g ( x ) = ax3 − x2 + bx + c.
g(2) = 0 and y = 8x + 16 is a tangent at x = –2.
7.1
Determine the values of a, b and c, showing all necessary calculations.
1
7.2
If a = __
, b = −2 and c = 4:
2
7.2.1
Determine the equation of the tangent to g which is parallel
to the tangent at x = −2.
7.2.2
Determine and classify the stationary points of g, showing all
necessary calculations.
7.2.3
Determine the point(s) of inflection of g. Justify your answer(s).
7.2.4
Sketch g ( x ), g ’ ( x ) and g ’’ ( x ) on the same system of axes, showing
the intercepts with the axes and the coordinates of the stationary
point(s) and point(s) of inflection. Show all necessary calculations.
8
y
2
x
g(x) = –x³ – 2x² + 11x + 12
Given g ( x ) = −x3 − 2x2 + 11x + 12.
8.1
Determine the equation of the tangent to g at x = 2.
8.2
Determine the coordinates of the point where the tangent intersects g(x) a
second time.
9
23 2 ___
19
Consider the graphs f ( x ) = x3 − 3x2 + 4 and g ( x ) = ___
x − 3 x.
9
9.1
Find the point where the graphs share a common tangent.
9.2
Find the equation of the common tangent at this point.
9.3
Sketch f, g and their common tangent on the same set of axes.
10
Consider the graphs f ( x ) = −x3 + 3x2 − 2 and g ( x ) = 3x2 − 3x.
10.1 Find the point where the graphs share a common tangent.
10.2 Find the equation of the common tangent at this point.
10.3 Sketch f, g and their common tangent on the same set of axes.
Unit 7 Sketch cubic graphs
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Unit 8: Optimisation and rate of change
KEY WORDS
optimisation – the process
you follow to solve practical
problems using calculus
The derivative of an expression gives the following:
• the formula for the gradient of a tangent at any point on the curve
• the x-values of the maximum and minimum, or local maxima and local minima
• the instantaneous rate of change for a particular value.
If the distance fallen by a stone which has been dropped from a bridge is given by
s ( t ) = 4,9t 2 where t is the time in seconds and s (t ) is the distance in metres:
• average speed is determined by dividing total distance travelled by total time taken,
s(t2) − s(t1)
so the average speed between t 1 and t 2 is given by average speed = __________
t −t
2
1
• velocity is the rate of change of speed, determined by the derivative of distance
with respect to time, so the velocity of the stone at t 3 is given by s’ ( t 3 )
• acceleration is the rate of change of velocity determined by the derivative of velocity
with respect to time, so the acceleration of the stone at t 4 is given by s’’ ( t 4 ).
WORKED EXAMPLE 1
The distance fallen by a stone which has been dropped
from a hot air balloon is given by s ( t ) = 4,9t 2, where t
is the time in seconds and s ( t ) is the distance in metres.
1
Determine the average speed at which the stone
fell between the 2nd and 7th seconds.
2
Determine the velocity of the stone:
2.1
after 2 seconds
2.2
after 7 seconds.
3
Determine the acceleration of the stone after:
3.1
2 seconds
3.2
7 seconds.
4
If the stone hit the ground after 15 seconds,
determine:
4.1
the height of the hot air balloon when the stone was dropped
4.2
the speed at which the stone hit the ground
4.3
the acceleration of the stone as it hit the ground.
SOLUTIONS
1
2
REMEMBER
You may have to convert
a word problem into a
mathematical function so
that you can determine the
minimum or maximum by
means of differentiation.
• maximise volume, area,
profit
• minimise area, volume,
cost, distance, time
176
3
4,9 ( 7 )2 − 4,9 ( 2 )2
s ( 7 ) − s ( 2 ) _______________
average speed = _________
=
= 44,1 m/s
5
7−2
velocity of stone = s’ ( t ) = 9,8t m/s
2.1
s’ ( 2 ) = 9,8 ( 2 ) = 19,6 m/s
2.2
s’ ( 7 ) = 9,8 ( 7 ) = 68,6 m/s
acceleration of stone = s” ( t ) = 9,8 m/s2
3.1
4
9,8 m
_____
s2
3.2
9,8 m/s2
s ( t ) = 4,9t 2 m/s, s’ ( t ) = 9,8t m/s and s” ( t ) = 9,8 m/s2
4.1
distance: s ( t ) = s ( 15 ) = 4,9 ( 15 )2 = 1 102,5 m
4.2
velocity: s’ ( t ) = 9,8 ( 15 ) = 147 m/s
4.3
acceleration: s” ( t ) = 9,8 m/s2
Topic 8 Differential calculus
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The rectangular prisms below have not been drawn to scale.
4
10
30
2
10
10
15
12
5
The volume of each right prism is 600 units3, but their surface areas differ in spite of
their volumes being equal.
TSA Green prism = 500 units2
TSA Red prism = 760 units2
TSA Blue prism = 460 units2
Rather than determine the smallest area by trial and error, we use calculus.
WORKED EXAMPLE 2
A company selling washing powder wants to produce boxes which hold
exactly 3 litres of washing powder. Each box must be twice as long as it is wide and
be in the shape of a right prism.
1
Determine the dimensions of the box in terms of x.
2
Determine the surface area of the box in terms of x.
3
Determine, correct to two decimal places, the value of x for which the surface
area will be a minimum. Then state the dimensions of the box.
4
What is the minimum surface area of the box, correct to two decimal places?
SOLUTIONS
1
h
x
2x
1 ℓ = 1 000 cm3 ⇒ 3 ℓ = 3 000 cm3
V = area of base × perpendicular height = ( 2x ) ( x ) ( h ) = 2x2h
3 000 _____
1 500
⇒ 2x2h = 3 000 and so h = _____
2 =
2
2x
x
The dimensions are x, 2x and _____
2
2
3
1 500
x
1
500
1
500
9 000
TSA = 2 2x2 + x _____
+ 2x _____
= 4x2 + _____
x
x2
x2
(
(
)
(
))
A = 4x2 + 9 000x−1 ⇒ A ’ = 8x − 9 000x−2 = 0 at miminum
_____
3
8x3 − 9 000 = 0 ⇒ x3 = _____
=1 125 and x = √ 1 125 = 10,4 cm
8
9 000
13,87
20,8
10,4
The dimensions are 10,4 cm × 20,8 cm × 13,87 cm
9 000
4
The minimum surface area is A = 4 ( 10,4 )2 + _____
= 1 298,02 cm2
10,4
Unit 8 Optimisation and rate of change
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WORKED EXAMPLE 3
A flat sheet of cardboard has a length of 32 cm and a breadth of 20 cm.
Four identical squares, each with sides of x cm, are cut out of the four corners.
The sides are folded up to form a box in the form of a right prism.
1
Determine the value(s) of x for which the volume will be a maximum.
2
State the dimensions of the box.
3
What is the maximum volume?
SOLUTION
1
x
32 – 2x
x
x
x
20 – 2x
x
32 – 2x
20 – 2x
x
x
x
x
The length of the box is (32 – 2x) cm, the breadth is (20 – 2x) cm
and the height is x cm.
V = area of base × perpendicular height
= ( 32 − 2x ) ( 20 − 2x )x
= 640x − 104x2 + 4x3
V ’ = 640 − 208x + 12x2 = 0
⇒ 3x2 − 52x + 160 = 0
( x − 4 ) ( 3x − 40 ) = 0
40 (
x = 4 or x = ___
x < 10 because 20 − 2x > 0 )
3
x = 4 only
178
2
The dimensions are 24 cm × 12 cm × 4 cm
3
Maximum volume:
V = 640 ( 4 ) − 104 ( 4 )2 + 4 ( 4 )3
= 1 152 cm3
or
V = 24 × 12 × 4
= 1 152 cm3
Topic 8 Differential calculus
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WORKED EXAMPLE 4
A cylindrical right prism has a radius of r cm and height of h cm.
The dimensions are limited in such a way that that the sum of
the circumference of its base and its height is always equal to
21 cm.
1
2
3
4
r
h
Show that the volume of the cylinder is given by
V = 21πr2 − 2π2r3.
Determine r in terms of π if the volume is a maximum.
Determine the maximum volume.
If the volume is 109,18 determine the dimensions for which the total surface
area of the cylinder would be a minimum.
SOLUTIONS
1
2
3
2πr + h = 21
⇒ h = 21 − 2πr
V = πr2h
= πr2 ( 21 − 2πr )
= 21πr2 − 2π2r3
V ’ = 42πr − 6π2r2 = 0
⇒ 2πr ( 21 − 3πr ) = 0
21 __
7
3πr = 21 and r = ___
=
3π π
(r ≠ 0)
V = 21πr2 − 2π2r3
7 2
7 3
= 21π ( __ ) − 2π2 ( __ )
π
π
1 029 686 343
= _____ − ____ = ____
π
π
π
3
4
= 109,18 cm
A = 2πr2 + 2πrh, so we need to find a value for h in terms of r.
V = πr2h = 109,18
109,18
⇒ h = ______
πr2
A
REMEMBER
The area of a circle is A = πr2
The circumference of a circle
is C = 2πr
The volume of a right cylinder
is V = πr2h
The total surface area of a
cylinder is A = 2πr2 + 2πrh
( πr )
109,18
= 2πr2 + 2πr ______
2
218,36
= 2πr2 + ______
r
= 2πr2 + 218,36r−1
218,36
⇒ A ’ = 4πr − 218,36r−2 = 4πr − ______
= 0 at minimum
r2
=0
4πr3 − 218,36
______
218,36
3 ______
⇒r=
≈ 2,59 cm
√ 4π
109,18
π ( 2,59 )
h = ________2 = 5,18 cm
Unit 8 Optimisation and rate of change
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WORKED EXAMPLE 5
y
3
Q
T
3
–3
R
x
g
S(p;–1)
–1
f
f ( x ) = ax2 + bx + c and g ( x ) = k.
QRST is rectangle with T and Q on f and R and
S (p;–1) on g. f cuts the x-axis at –3 and 3 and the y-axis at 3.
1
Determine the values of a, b, c and k.
2
State the coordinates of Q, R and T in terms of p.
2 3
3
Show that the area of QRST is given by A = 8p − __
p .
3
4
5
Determine the value of p for which the area of QRST is a maximum.
Determine the equation of the tangent to f at x = 3.
SOLUTIONS
1
f ( x ) = a ( x − x1 ) ( x − x2 )
= a(x − 3)(x + 3)
Substitute ( 0;3 )
1
⇒ 3 = −9a, so a = −__
3
1
1 2
__
f ( x ) = − ( x2 − 9 ) = −__
x +3
3
k = −1 ⇒ y = −1
2
(
1 2
p +3
Q −p;− __
3
R ( −p;−1 )
1 2
T p; − __
p +3
(
3
3
)
)
1 2
p +4
PQ = 2p and PT = −__
3
Area PQRT
1
= 2p − __p2 + 4
(
3
2 3
__
= − p + 8p
3
180
3
)
4
2 3
p + 8p
A = −__
3
A ’ = −2p2 + 8 = 0
p2 = 4 ⇒ p = 2
(p > 0)
5
2
x ⇒ m = f ’ ( 3 ) = −2
f ’ ( x ) = −__
3
Substitute ( 3;0 ) into y − y1 = m ( x − x1 ):
y = −2 ( x − 3 ) = −2x + 6
Topic 8 Differential calculus
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WORKED EXAMPLE 6
f ( x ) = a ( x − p )2 + q and g ( x ) = kx2 + bx + c
f has a turning point at (–1;32), g passes through
(–8;0) and (0;–24)and f and g both pass through
(3;0).
PQ is parallel to the y-axis, with P on f and Q on g.
1
Determine the values of a, p, q, k, b and c.
2
Determine an expression for the length
of PQ in terms of x.
3
Determine the value of x at which PQ
is a maximum.
4
What is the maximum length of PQ?
(–1;32)
y
g
P
–8
x
3
–24
Q
f
SOLUTIONS
1
3
f ( x ) = a ( x + 1 )2 + 32
Substitute ( 3;0 )
a ( 4 )2 + 32 = 0 and a = −2
f ( x ) = −2 ( x + 1 )2 + 32
⇒ a = −2, p = −1 and q = 32
PQ = yP − yQ
= [ −2 ( x + 1 )2 + 32 ] − [ x2 + 5x − 24 ]
= −2x2 − 4x − 2 + 32 − x2 − 5x + 24
= −3x2 − 9x + 54
3
PQ ’ = −6x − 9 = 0 ⇒ x = −__
4
3 2
3
PQ = −3 − __
− 9 − __
+ 54
2
2
2
( )
( )
g(x) = k(x + 8)(x − 3)
Substitute ( 0;−24 )
−24 = a ( −24 ) ⇒ a = 1
g ( x ) = x2 + 5x − 24
⇒ k = 1, b = 5 and c = −24
2
27 27
= −___ + ___
+ 54
2
4
= 60,75
EXERCISE 14
Unless otherwise stated, give all answers correct to two decimal
r
places. State answers to all calculations involving π in terms of π
as well as correct to two decimal places. Do not round off your
answers until the end.
1
A right cylinder has a volume of 2 ℓ.
1.1
Determine the dimensions of the cylinder for which the
total surface area will be a minimum.
1.2
Calculate the minimum surface area of the cylinder.
2
A right cylinder has a total surface area of 858π cm2.
2.1
Determine the dimensions of the cylinder for which the volume will be
a maximum.
2.2
Calculate the maximum volume of the cylinder.
h
Unit 8 Optimisation and rate of change
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3
4
5
6
7
182
The box below is a right prism with a volume of 5 000 cm3.
The length is four times the breadth.
3.1
Determine the total surface area of the box in
h
x
terms of x.
4x
3.2
Determine the dimensions of the box for which
the total surface area will be a minimum.
3.3
Calculate the minimum surface area.
A storage container in the form of a right prism has a total surface area of 34 m2.
4.1
If the length is five times the breadth, determine the volume of the
container in terms of x if the breadth is x metres.
4.2
Determine the dimensions of the container for which the volume will be a
maximum.
4.3
Determine the maximum volume of the container.
59
A ball is thrown into the air and its height is given by h ( t ) = −5t 2 + ___
t + 3,
2
where t is the time in seconds and h(t ) is the height in metres.
5.1
Determine the height of the ball above the ground after 2 seconds.
5.2
Determine the average speed of the ball during the first two seconds.
5.3
What was the velocity of the ball after 2 seconds?
5.4
What was the acceleration of the ball after 2 seconds?
5.5
At what time did the ball reach its maximum height and what was
the maximum height?
5.6
When did the ball hit the ground?
5.7
What was the speed of the ball at the moment that it hit the ground?
f ( x ) = x3 + ax2 + bx has a stationary point at (2;32).
Determine the values of a and b.
In the diagram alongside, f ( x ) = ax2 + bx + c
y
and g ( x ) = k.
f
TS || QR || y-axis, RS || QT || x-axis and
Q
T
TS intersects the x-axis at p.
2
g
7.1
Determine the values of a, b, c and k.
x
7.2
State the coordinates of T, Q, R and S in
p 4
–4
terms of p.
7.3
What type of quadrilateral is QRTS?
R
S
Briefly justify your answer.
7.4
Show that the area of QRST is given by
A = 20p − p3 units2.
–8
7.5
Determine the value of p for which
the area of QRST will be a maximum.
7.6
State the maximum area of QRST.
7.7
Determine the equation of the tangent to f at x = 2.
Topic 8 Differential calculus
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8
In the diagram alongside f ( x ) = −8 − x3.
H and B are the x- and y-intercepts of f.
E is a point on f such that ED ⊥ OH and
EC ⊥ OB. D(x;0) is a point on the x-axis
between O and H.
State the coordinates of H and B and then
state the lengths of OH and OB.
State the length of DE in terms of x.
Show that the area of the quadrilateral HEBO
is equal to 8 − 4x + x3.
Determine the value of x for which the area
of quadrilateral HEBO will be a maximum.
State the maximum area of HEBO.
8.1
8.1
8.2
8.3
8.4
8.5
9
B
A
10
x
C
y
f(x) = – x³ – 8
(x;0)
H
D
E
O
x
C
B
21 cm
The radii of the smallest circles is x cm and the ratio of the areas of the circles
is 1:4:16.
One of the dimensions of the large rectangle is 21 cm.
9.1
State the radii of circles A and B in terms of x.
9.2
State the dimensions of the large rectangle in terms of x.
9.3
Show that the area of the shaded region is A ( x ) = 294x − 27πx2.
9.4
Determine the value of x for which the shaded region is a maximum.
9.5
Determine the total area of the circles when the shaded region is
a maximum.
A company plans to manufacture buoys for a boat race. Each
scone
buoy is made up of a hemisphere, a right cylinder and a cone.
hcone
The slant height of the cone forms an angle of 60° with the base,
60˚
the radius is r and the total surface area of the buoy is
5 400π cm2.
H
10.1 Express the slant height, s and the height, h of the cone in
r
terms of r.
10.2 Show that H, the height of the cylinder, is given by
2 700
H = _____
r − 2r.
10.3 Show that the volume of the buoy is given by
__
πr3 (
V = 2 700πr − ___
4 − √3 ).
3
10.4 Determine the value of r for which the volume will be a maximum.
10.5 Determine the maximum volume in litres.
REMEMBER
2 3
Vhemisphere = __
πr
3
Vcylinder = πr2h
1 2
πr h
Vcone = __
3
TSAhemisphere = 3πr 2
TSAcylinder = 2πr 2 + 2πrh
TSAcone = πr 2 + πrs
Unit 8 Optimisation and rate of change
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Revision Test Topic 8
Total marks: 200
1
x2 − 3x − 4
Consider the function: f( x ) = __________
x−4
1.1
For which value of x is f( x ) undefined?
1.2
Complete the table below:
3,9
x
3,99
3,999
3,9999
4
(2)
4,0001
4,001
4,01
4,1
(9)
f( x )
1.3
2
Is there a limit at x = 4? Fully justify your answer.
Determine the limit of each of the following:
x2 − 9
2.1
lim ______
(3)
2.2
x+3
x→−3
3
(3)
2+
−
x
2x 8
lim __________
2
+ x − 12
x→−4 x
(4)
Consider the function f( x ) = 3x2 − 2x − 1.
3.1
Sketch f, indicating the intercepts with the axes and the coordinate
of the turning point.
(5)
3.2
Consider the points P(–3;f(−3)) and Q(1;f(1)).
3.2.1
Plot P and Q on f and draw the line passing through them.
(1)
3.2.2
Determine the average gradient of f between P and Q.
(2)
3.3
Consider the points: A(−1;f (−1 ) ) and B(−1 + h;f (−1 + h ) )
3.3.1
Determine the average gradient between A and B in terms of h. (3)
3.3.2
Determine the average gradient between A and B if h = 5.
(2)
3.3.3
Plot A and B on f and draw the line passing through them.
(2)
3.3.4
Use your graph to determine the gradient of AB.
(3)
3.4
Consider the points D( x;f( x ) ) and E( x + h;f ( x + h ) ).
3.4.1
Determine the average gradient between D and E
in terms of x and h.
(2)
f(x + h) − f(x)
____________
3.4.2
Determine f ’ ( x ) = lim
(3)
h
h→0
Determine f ’( 2 ) and explain your answer.
(3)
Consider the function f( x ) = −x2 + 3x.
4.1
Determine the average gradient between x = −2 and x = −2 + h.
(3)
3.4.3
4
4.2
f ( −2 + h ) − f ( −2 )
Determine lim _______________
h
h→0
(4)
5
4 Then determine g ’( −2 ).
Use the definition to find the derivative of g( x ) = __
x
(6)
6
Use first principles to find the derivative of f( x ) = 2x3 at the point x = –3.
(6)
7
Use first principles to find the derivative of g( x ) = −3x + 2 at the point x = 4. (3)
8
Use the definition to find the derivative of f( x ) = 5 and then determine f ’( 2 ). (3)
2 and g( x ) = x3 + x2 − 21x
f( x ) = − __
9
9.1
9.2
9.3
9.4
9.5
9.6
9.7
x
Determine f ’( x ) from first principles.
For which value(s) of x is f ’( x ) = 2?
1.
Determine the equation of the tangent to f at x = __
2
Determine the equation(s) of the tangent(s) to g if g ’( x ) = 0.
Determine the equation(s) of the tangent(s) for which g ’( x ) = 12.
Determine the equation of the tangent to g at x = –2.
Determine the equation of the tangent which is parallel to the tangent
to g at x = 1.
(5)
(3)
(4)
(4)
(4)
(4)
(5)
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10
Use the rules of differentiation to determine:
dy
7x2 − 5x + 4 (4)
2 − 2√__
10.2 ___ if y = 2x2 − __
x
10.1 f ′( x ) if f( x ) = ___________
x
x
[ −x − 5−x ]
3
2
x
3x
10x
10.3 Dx ____________
3
2
dx
− √x
x3 − x2__
d ___________
10.4 ___
3
(6)
dx
√x
d ( 2x − 3√__
(3) 10.6 ___
x )2
10.7 Dx[ ( 3x − 1 )3 ]
(4)
√ x3 − √ x
d _________
__
10.8 ___
√
(3)
t _2
ds if s = _______
10.10 __
3
10.9
dx
5x
1
2
__
__
dy
___
if y = x 5 − x 3
dx
d [ ( 3x + 2 )( 9x2 − 6x + 4 ) ] (4)
10.11 ___
dx
dx
dt
(
5
__
__
3
x
( +
)
√t
(5)
)
2
10.5. ___ if y = − ___
3
dy
dx
11
(
__
(5)
(6)
(6)
)3
(4)
(
)
__
1__ (4)
10.12 f ’( x ) if f(x) = ( √x + 1 ) x − ___
√
x
1 x3 − 2x2 − 32x + 33__
2.
Consider the function g( x ) = __
3
3
11.1 Determine the coordinates of the stationary points of g.
11.2 Use the second derivative test to determine the concavity of g at
each stationary point. Then state whether the stationary point is
a local minimum or a local maximum.
11.3 Determine the coordinates of the point of inflection.
(6)
(6)
(4)
12. Consider the function: f( x ) = −x3 − x2 + 16x + 16.
12.1 Determine and classify the stationary point(s).
12.2 Determine the point of inflection.
12.3 Determine the roots of f( x ) = 0.
12.4 Sketch f, showing the intercept(s) with the axes and the coordinates
of stationary point(s) and point(s) of inflection.
12.5 What is the maximum value of f( x ) for x ∈ [ − 8;8 ].
12.6 Use your graph to determine lim f( x )
(4)
(4)
(2)
f( x ) = ax3 + bx2 + cx + d passes through the following points:
(–4;0), (–2;0), (0;12) and (3;0). Determine the values of a, b, c, and d.
(7)
Consider the function g( x ) = −x3 + ax2 + bx + c and determine the values
of a, b and c if g( 2 ) = 6, g ’( 2 ) = 5 and g ’( −1 ) = −10.
(6)
Consider the function f( x ) = ax3 + bx2 + cx + d.
If f ( 1 ) = f ’( 1 ) = 0, f ( 0 ) = 4 and f ’( 0 ) = −7:
15.1 Determine the equation of the tangent to f at x = 1.
15.2 Determine the equation of the tangent to f at x = 0.
15.3 Determine the values of a, b, c and d.
(2)
(3)
(6)
(8)
(4)
(4)
x→−∞
13
14
15
16
f ′( x ) is sketched alongside.
y
16.1 State the x-coordinates of the stationary points
of f and classify the stationary points.
(6)
–2
16.2 State the x-coordinate of the point of inflection
and justify your answer.
(4)
–4
16.3 Discuss the concavity of f and explain your
P
answers fully.
(6)
16.4 Draw a sketch of f.
16.5 What is the gradient of the tangent to f at x = 0?
16.6 State another value of x at which the tangent to f will be parallel
to the tangent at x = 0.
f ’(x)
4
x
(4)
(3)
(4)
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REVISION TEST TOPIC 8 CONTINUED
17
A cylindrical capsule with hemispherical ends has a radius of r and
a height h.
17.1 State the volume and surface area of
the capsule in terms of π, r and h.
(6)
17.2 If the outer surface area of the capsule
h
is 1,56π m2.
r
0,78
17.2.1 Show that h = ____
(3)
r − 2r
17.2.2 Show that the volume of the
2 πr 3.
capsule is V = 0,78πr − __
3
17.2.3 Determine the maximum volume of the capsule.
9 π:
17.3 If the volume of the capsule is V = ___
16
9 − __
4r
17.3.1 Show that h = ____
16r2
3
9π
4 πr 2 + ___
17.3.2 Show that the total surface area TSA = __
2
3
17.4
8r
(4)
(4)
(4)
(4)
17.3.3 Determine the minimum surface area of the capsule.
(4)
Determine the value of h in 17.2 and 17.3 and briefly explain your results. (8)
18
A ball is thrown into the air. Its height above the ground after t seconds is
63 t + 1 metres.
h(t ) = −t 2 + ___
8
18.1 When will the ball be 12,75 m above the ground?
(3)
18.2 How high will the ball be after one second?
(2)
18.3 When will the ball be at a height of 16,5 m above the ground?
(3)
−1
18.4 When will the instantaneous speed of the ball be −12,125 ms ?
(4)
18.5 After how many seconds will the ball reach its maximum height?
(3)
18.6 Determine the maximum height which the ball can reach.
(4)
18.7 Determine the speed of the ball during the third second.
(3)
18.8 Is the speed of the ball during the third second an average speed or
an instantaneous speed?
(??)
18.9 Determine the height of the ball above the ground after three seconds. (2)
18.10 Is the speed of the ball after the third second an average speed or
an instantaneous speed?
(1)
18.11 Determine the speed of the ball at that after three seconds.
(3)
18.12 Determine the acceleration of the ball after three seconds.
(4)
18.13 How long will it take for the ball to hit the ground?
(2)
18.14 Determine the speed at which the ball will be travelling when it hits
the ground.
(3)
19
Thandi throws a ball into the air from the branch of a tree which he has
climbed. The height of the ball above the ground after t seconds is
h( t ) = −4t 2 + 19,2t + 4 metres.
19.1 Determine the height of the ball above the ground after two seconds.
19.2 Determine the average speed of the ball during the first two seconds.
19.3 When will the ball be at a height of 26,4 m above the ground?
19.4 How long will it take for the ball reach its maximum height?
19.5 Determine the maximum height reached by the ball.
19.6 After how many seconds will the ball hit the ground?
19.7 At what speed was the ball travelling at the moment of impact?
19.8 What was the velocity of the ball after 2,4 seconds? Briefly explain
your answer.
(3)
(3)
(3)
(3)
(2)
(3)
(3)
(4)
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REVISION TEST TOPIC 8 CONTINUED
20
Consider the functions f( x ) = −( x − 5 )( x + 3 ) and g( x ) = ( x + 2 )( x − 7 )
sketched below.
AB || PQ || y-axis, with A and P on f and B and Q on g.
y g(x) = (x+2)(x – 7)
P
A
x
QB
f(x) = – (x – 5)(x + 3)
20.1
20.2
20.3
20.4
20.5
20.6
Determine all possible coordinates of A and B if AB = 25 units.
(6)
Determine the value of x for which PQ will be a maximum length.
(4)
What is the maximum length of PQ?
(3)
For which value(s) of x will the tangents to f and g be parallel to each other? (3)
Determine the equations of the parallel tangents.
(4)
For which values of k will y = k have three or more points of intersection
with f and/or g?
(4)
20.7 For which values of x is f ’( x ).f( x ) > 0?
(3)
20.8 For which value(s) of x is g ’( x ) − g( x ) a maximum?
(4)
21
The temperature of a liquid during an experiment is given by
T( t ) = −t 2( t − 12 ) + 10, where t is the time in seconds and T is the temperature
measured in degrees centigrade.
21.1 What is the temperature of the liquid at the start of the experiment?
(2)
21.2 Determine the rate of change of the temperature of the liquid during
the first three seconds.
(3)
21.3 Determine the rate of change of the temperature of the liquid after
three seconds.
(3)
21.4 After how many seconds did the liquid reach a maximum temperature? (3)
21.5 What was the maximum temperature reached?
(2)
22
( −1;17 ) is a stationary point of f( x ) = x3 + ax2 + bx. Determine the values
of a and b.
23
The straight line cuts the axes at (0;12) and (4;0). BADO is a rectangle with A
on the line, B and D on the x- and y-axes
y
respectively and O at the origin.
12
23.1 Determine the coordinates of A
E
for which the rectangle will have a
maximum area.
(6)
A
D
23.2 Then determine the sum of the
F
minimum areas of the triangles △EDA
B
4
O
and △ABF.
(6)
(5)
x
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TOPIC
2
9
Analytical geometry
Unit 1: Equations of a circle
You learnt these analytical formulae in Grade 11:
_________________
• The distance between points A(x1; y1) and B(x2; y2) = √(x2 − x1)2 + ( y2 − y1 )2
y −y
2
1
• The gradient between points A(x1; y1) and B(x2; y2) = ______
x2 − x1
1
2 ______
2
; 1
• The midpoint between points A(x1; y1) and B(x2; y2) = ( ______
2
2 )
• You can find the equation of a line using the formula:
* y = mx + c when you know the c value (y-intercept)
* y − y1 = m( x − x1 )for any line passing through point ( x1;y1 )
• You find the inclination of a line (angle θ) using the formula: tan θ = m
x +x
y +y
Now consider the general formula for a circle.
Circles that are centred at the origin
Let the origin be the centre of the circle, and point (x; y) be any point
on the circumference of the circle with radius r.
According
to the distance formula:
______
_______________
r = √( x − 0 )2 + ( y − 0 )2 = √ x2 + y2
y
______
∴√
∴ x2 + y2 = r2
x2 + y2 = r
Therefore, the equation of a circle with centre (0; 0)
and radius r is:
x2 + y2 = r2
( x;y)
r
x
( 0; 0)
Circles that are centred off the origin
Let the centre of the circle be at point (a; b), and point
(x;y) be any point on the circumference of the circle
with radius r.
According to the distance formula:
y
_______________
√( x − a )2 + ( y − b )2 = r
∴ ( x − a )2 + ( y − b )2 = r2
Therefore, the equation of the circle with centre (a; b)
and radius r is: (x − a)2 + (y − b)2 = r2
To find the equation of a circle, we must first know
the centre and the radius of the circle.
188
(x;y)
r
x
(a;b)
Topic 9 Analytical geometry
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WORKED EXAMPLES
Determine the equation of the circle(s) with:
1
the centre as the origin and passing through point (−5;−7)
2
centre (−3;5) and radius 7
3
diameter AB where A = (2;−4) and B = (−2;10)
__
4
an x-intercept at (2;0), a y-intercept at (0; 6) and radius = 2√5
SOLUTIONS
1
2
3
Substitute the point (−5;−7) into x2 + y2 = r2
∴ ( −5 )2 + ( −7 )2 = r2
∴ 25 + 49 = r2
∴ 74 = r2
∴ x2 + y2 = 74
Substitute centre (a;b) = (−3;5) and r = 7 into (x − a)2 + (y − b)2 = r2
∴ ( x − ( −3 ) )2 + ( y − 5 )2 = ( 7 )2
∴ ( x + 3 )2 + ( y − 5 )2 = 49
Note: Be careful
The centre of the circle is at the midpoint
to substitute the
1 diameter.
of the diameter and the radius = __
coordinates of the centre
( 2
)
2
__________________
2
−4 + 10
2 − 2 ; ________
= ( 0; 3 ) and
So centre = _____
1 √ ( −2 − 2 )2 + ( 10 + 4 )2
r = __
2
________
Any point that lies on a graph
can be substituted into the
equation of that graph, in
place of x and y.
REMEMBER
___
1 √ 16 + 196 = √ 53
= __
2
4
in place of a and b, and
not x and y.
REMEMBER
So the equation :
___
( x − 0 )2 + ( y − 3 )2 = (√ 53 )2
∴ x2 + ( y − 3 )2 = 53
Substitute the points (2;0) and (0;6) into ( x − a )2 + ( y − b )2 = r2
__
__
∴ ( 2 − a )2 + ( 0 − b )2 = ( 2√ 5 )2
and ( 0 − a )2 + ( 6 − b )2 = ( 2√ 5 )2
∴ 4 − 4a + a2 + b2 = 20
and a2 + 36 − 12b + b2 = 20
2
2
∴ a − 4a + b = 16 ➀
and a2 − 12b + b2 = −16 ➁
∴ ➀ − ➁: − 4a + 12b = 32
∴ −a + 3b = 8
∴ a = 3b − 8 ➂
Substitute ➂ into ➁: ( 3b − 8 )2 − 12b + b2 = −16
∴ 9b2 − 48b + 64 − 12b + b2 = −16
∴ 10b2 − 60b + 80 = 0
∴ b2 − 6b + 8 = 0
∴ ( b − 4 )( b − 2 ) = 0
∴ b = 4 or b = 2
Substitute into ➂ ∴ a = 4 or −2
∴ circle centre = ( 4; 4 ) or ( −2; 2 )
radius
diameter
So there are two possible circles, with equations:
( x − 4 )2 + ( y − 4 )2 = 20 and ( x + 2 )2 + ( y − 2 )2 = 20
Unit 1 Equations of a circle
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Although the standard form equation of a circle centred at (a; b) is ( x − a )2 + ( y − b )2
= r2, you can sometimes multiply out these equations.
So for example, the equation ( x − 1 )2 + ( y + 5 )2 = 9 could be given as
x2 − 2x + 1 + y2 + 10y + 25 = 9
∴ x2 − 2x + y2 + 10y = −17
If the equation is given in this multiplied out form, first write the equation in the
form ( x − a )2 + ( y − b )2 = r2 to determine the centre and the radius of the circle. You
do this using the process of completing the square.
So, taking the example given above and working in reverse order:
x2 − 2x + y2 + 10y = −17
x2 − 2x + 1 + y2 + 10y + 25 = −17 + 1 + 25
REMEMBER
The constant term of a perfect
square trinomial, where the
coefficient of the first term is 1,
1 the
is always the square of __
2
coefficient of the middle term.
( __12 coefficient of x )
1 × −2 = 1
= ( __
)
2
2
2
( __12 coefficient of y )
1 × 10 = 25
= ( __
)
2
2
Add 1 and 25 to the
right-hand side to
balance the fact that
those values were
added to the left-hand
side.
2
Now form both perfect squares, so
x2 − 2x + 1 + y2 + 10y + 25 = −17 + 1 + 25
same
__
same
√1
sign
___
√ 25
sign
∴ ( x − 1 )2 + ( y + 5 )2 = 9
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WORKED EXAMPLES
Determine the radius and centre of the following circles:
1
x2 + y2 = 12
( x − 6 )2 + y2 = 100
2
3
x2 + 8x + y2 − 3y = −6
4
2x2 + 2y2 + 6x − 5 = 0
SOLUTIONS
1
2
3
This is in the form x2 + y2 = r2, so the centre is the origin and
___
__
radius = √ 12 = 2√3
( x − 6 )2 + (y − 0)2 = (10)2, so the centre is (6; 0) and radius = 10
x2 + 8x + y2 − 3y = −6
−3 2
−3 2
∴ x2 + 8x + 16 + y2 − 3y + ___
= −6 + 16 + ___
(2)
9 = −6 + 16 + __
9
∴ x2 + 8x + 16 + y2 − 3y + __
4
4
3 2 = ___
49
∴ ( x + 4 )2 + y − __
4
2
3 and radius = __
7
So centre is −4; __
2
2
2x2 + 2y2 + 6x − 5 = 0
5=0
∴ x2 + y2 + 3x − __
| Coefficient of x2 and y2 must be 1 so divide
2
(
4
(2)
(
)
)
by 2.
(
) (
)
9 + y2 = __
5 + __
9
9
1 coefficient of x 2 = __
1 × 3 2 = __
∴ x2 + 3x + __
| __
4
4
4
2
2
2
3 2 + y2 = ___
19
∴ x + __
4
2
___
√ 19
3
____
;
0
and
the
radius
is
So the centre is at −__
2
2
(
)
(
)
You need to recognise that an equation of the form cx2 + dx + cy2 + ey = f will
represent a circle.
Notice that:
• the coefficient of x2 and y2 must be the same.
• if d = e = 0 then the centre of the circle will be the origin.
f
d 2 + __
e 2 must be positive, as it represents r2.
the value of _ + __
•
c
( 2c ) ( 2c )
Unit 1 Equations of a circle
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EXERCISE 1
192
1
State whether the following equations represent a circle or not.
1.1
5x2 + 5y2 = 10
1.2
3x2 + 2x + 3y2 = 7
1.3
x2 − 5y + y2 + 7x − 8 = 0
1.4
2x2 − 2y2 = 50
1.5
6y2 + 2x2 − 8 = 0
1.6
x2 − 2x + y2 + 4y + 16 = 0
2
Determine the equation of the circle:
2.1
with centre the origin and radius 7
2.2
with centre the origin and passing through point (8;−3)
__
2.3
with centre (3;−5) and radius √ 3
2.4
with diameter CD where C = (0;−3) and D = (2;9)
2.5
with x-intercepts (2;0) and (−4;0) and radius 5
__
2.6
passing through the points (2;3) and (2;−1), with radius √ 5
2.7
passing through points (2;−2); (0;1) and (1; 0)
2.8
passing through point (−6;1), with circumference 12π and with the centre
on the y-axis.
3
Determine the radius and centre of the following circles:
3.1
x2 + y2 = 121
3.2
3x2 + 3y2 − 9 = 0
( x − 1 )2 + ( y − 5 )2 = 16
3.3
3.4
x2 + ( y + 15 )2 = 17
3.5
x2 + 4x + y2 − 8y = 5
3.6
3x2 + 6x + 3y2 − 18y + 12 = 0
3.7
x2 − 4 + y2 = 3x − y
3.8
8y2 − 4y + 8x2 = 16x
3.9
x2 + 5x + y2 = 0
3.10 2x2 − y + 2y2 − 10 = 0
3.11 x2 − 3x = y − y2
3.12 5x2 = 10y − 5y2 + 45
Topic 9 Analytical geometry
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Unit 2: Equation of a tangent to a circle
You should now have enough knowledge to determine the equation of a tangent to
any circle:
• A tangent is a straight line.
• A tangent is perpendicular to a radius of a circle at the point of contact.
• You can determine the equation of the tangent passing through a given point (x1; y1):
− using the formula y − y1 = m( x − x1 ) where ( x1; y1 ) represents the point of
contact of the tangent and the circle (or any other point that is given on the
tangent)
− substituting the given point of contact into the formula y = mx + c to solve for c,
then writing the equation in the form y = mx + c
− remembering to look out for horizontal or vertical tangents, where the tangent is
parallel to the x or y axis. The equation of the tangent will then be y = y1 or x = x1.
REMEMBER
The theorem:
WORKED EXAMPLE 1
y
Determine the equation(s) of the tangent(s)
to the circle where:
1
x2 + y2 = 20 at the point (2;− 4)
2
x2 − 5x + ( y + 3 )2 = 1 at the
y-intercept(s) of the given circle.
Tangent perpendicular to
radius
SOLUTIONS
1
4
= −__ = −2 ∴ m
x
O
1
= __
mradius
tangent
2
2
y − y1 = m( x − x1 ) using point ( 2;−4 )
1 (x − 2)
∴ y − ( −4 ) = __
2
1
__
y= x−5
(2;– 4)
2
2
First find the centre of the circle:
x2 − 5x + ( y + 3 )2 = 1
5 2 + y + 3 2 = 1 + __
5 2
∴ x2 − 5x + __
(
)
2
2
5 2 + y + 3 2 = ___
29
∴ x − __
(
)
tangent B
4
2
5 ;−3
∴ centre = __
(
)
( )
( )
y
(2 )
tangent A
x
Next find the co-ordinates of the
y-intercepts:
y-intercept (let x = 0): ( y + 3 )2 = 1
∴ y + 3 = ± 1 ∴ y = −2 or − 4
−2
We can now find the gradient of the radii
−4
from the centre to each y-intercept.
This will then enable us to know the
gradient of each tangent touching
the circle at each y-intercept, and thus the equation of the tangent at each
y-intercept.
2
−3 + 2
−3 + 4
2
∴ mradius = _______
= −__ or mradius = _______
= __
5
5
5
5
__
__
2
∴ mtangent A
2
5 and m
= __
2
tangent B
5
= −__
2
5x − 2
∴ y = __
| Tangent A
5
or y = −__
x−4
2
| Tangent B
2
Unit 2 Equation of a tangent to a circle
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REMEMBER
A tangent touches the circle
in one point only, thus there
will be only one point of
intersection between the two
graphs if it is a tangent.
KEY WORDS
secant − a line which passes
through a circle, intersecting
the circle at two points.
WORKED EXAMPLE 2
Determine whether the line y = x − 5 is a tangent to the circle
x2 − 4x + y2 = 5.
SOLUTION
We need to find the intersection of the two graphs:
x2 − 4x + ( x − 5 )2 = 5 ⇒ x2 − 4x + x2 − 10x + 25 = 5
∴ 2x2 − 14x + 20 = 0 ⇒ x2 − 7x + 10 = 0
∴ ( x − 5 )( x − 2 ) = 0 ⇒ x = 5 or x = 2
∴ the graphs intersect at ( 5; 0 ) and ( 2;−3 )
Since there are two points of intersection, the line is a secant to the circle and not
a tangent.
WORKED EXAMPLE 3
Show that the circle ( x + 3 )2 + ( y − 4 )2 = 4 touches the circle x2 + y2 = 9 .
y
SOLUTION
Circle ( x + 3 )2 + ( y − 4 )2 = 4 has centre (−3;4) and
circle x2 + y2 = 9 has centre (0;0).
(– 3; 4)
2
The __________
distance from (−3;4) to (0; 0)
___
= √ (−3)2 + ( 4 )2 = √ 25 = 5
3
x
(0;0)
Radius of circle ( x + 3 )2 + ( y − 4 )2 = 4 is 2
and radius of circle x2 + y2 = 9 is 3
The distance between the centres of the circles is
2+3=5
Note: Circles will touch when the
distance between their centres is
equal to the sum of their radii.
So the circles will touch.
WORKED EXAMPLE 4
y
Q
Given circle x2 + y2 − 4x + 10y = 7 with
tangent PQ touching the circle at point P.
P
x
If Q = (8; 3), determine the length of PQ.
6
SOLUTION
10
M (2;– 5)
x2 + y2 − 4x + 10y = 7
10 2 = 7 + 4 + 25
4 2 + y2 + 10y + ___
x2 − 4x + __
2
2
( x − 2 )2 + ( y + 5 )2 = 36
∴ centre = M = ( 2;−5 ) and radius = 6
( )
(8;3)
( )
_______________
_______
____
Distance MQ = √ ( 8 − 2 )2 + ( 3 + 5 )2 = √36 + 64 = √ 100 = 10
∴ PQ = 8 units | Pythagoras’ Theorem
194
Topic 9 Analytical geometry
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y
WORKED EXAMPLE 5
Determine the equation of the tangent(s) to the circle x2 + y2 = 10
from the point T (5;−5) outside the circle.
SOLUTION
N
There are two possible tangents from point T, as shown in the diagram.
x
O
Let N = (p, q).
As N is a point on the circle, substitute (p;q) into x2 + y2 = 10
2 = 10
∴ p2 + q______
∴ q = √ 10 − p2 (q is positive at N)
2
2
Let M = (a;b). Substitute (a;b) into x + y = 10
∴ a2 + b2 = 10
_______
M
But mTM
T(5;– 5)
_______
−5 + √ 10 − a2
= ______________
5−a
_______
√
2
∴ b = −√ 10 − a2 (b is negative at M)
_______
−5 + 10 − a
a
_______
∴ _________
= ______________
5−a
√ 10 − a2
Similarly to the working below, this simplifies to:
Notice that the calculations for a and p gave the same
answers, so we need only have done one calculation for
both results.
1
−5 + 3
−5 − 1
∴ mTM = _______
= −__ and mTN = _______
= −3
3
√
2
− 10 − a
a
_______
∴ mTM = _________
∴ mOM = __________
a
√ 10 − a2
∴ a2 − 2a − 3 = 0
∴ ( a − 3 )( a + 1 ) = 0
∴ a = 3 or a = −1
After checking a = −1 only, so M = (−1;−3)
5+1
5−3
Thus, the equations of the tangents at M and N respectively, are:
1
y + 1 = −__
(x + 3) and y − 1 = −3(x − 3)
3
1
y = −__
x − 2 (tangent MT) and y = −3x + 10 (tangent NT)
3
REMEMBER
a = __
c then ad = bc
If __
b
d
_______
Note: Since O = (0; 0) and
_______
N = (p; q) = (p;√ 10 − p2 ),
then
_______
_______
√10 − p2 − 0 _________
√10 − p2
____________
=
mON =
p
p−0
and perpendicular gradients are negative reciprocals,
thus
−p
_______
mTN = _________
√10 − p2
y2 − y1
− 5 − √10 − p2
_______________
But using _______
x2 − x1 , mTN =
5−p
_______
2
−
5
−
10
−
p
√
−
p
_______ = _______________
∴ _________
5−p
√10 − p2
_______
∴ − 5p + p2 = −5√10 − p2 − ( 10 − p2 )
_______
⇒ 5√10 − p2 = 5p − 10
∴ 25( 10 − p2 ) = 25p2 − 100 p + 100 ∴ 50p2 − 100p − 150 = 0
∴ p2 − 2p − 3 = 0 ⇒ ( p − 3 )( p + 1 ) = 0
∴ p = 3 or p = −1
After check p = 3 only.
So N = (3; 1)
Unit 2 Equation of a tangent to a circle
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A normal to a curve at a given point is the line
through that point perpendicular to the tangent at
that point. Thus mtangent × mnormal = −1
centre
tangent
Note: As the tangent to a circle is perpendicular to the
radius at the point of contact, the normal to a circle
will pass through the centre of the circle.
normal
WORKED EXAMPLE
Determine the equation of the normal to the circle 2x2 − 2x + 2y2 + 6y = 8
at the point (1;1).
y
SOLUTION
normal
2x2 − 2x + 2y2 + 6y = 8
⇒ x2 − x + y2 + 3y = 4
9
1 + y2 + 3y + __
⇒ x2 − x + __
4
tangent
(1;1)
4
9
1 + __
= 4 + __
x
4
4
3 2 = ___
26
1 2 + y + __
∴ x − __
4
2
2
3
1
__
__
∴ centre = ;−
2 2
3
−5
___
−__ − 1
2
2
___
=
=5
mnormal = mradius = _______
1−1
1
__
−__
) ( )
( )
(
2
2
∴ y − y1 = m( x − x1 ) ⇒ y − 1 = 5 (x − 1)
∴ y = 5x − 4
EXERCISE 2
1
Find the equation of the tangent and the normal to each circle at the given point
on the circle.
1.1
x2 + y2 = 13 at ( −3; 2 )
1.2
3y2 = 15 − 3x2 at ( 2; 1 )
1.3
( x − 2 )2 + ( y + 3 )2 = 4 at ( 2;−1 )
1.4
1.5
2
196
13 at ( 1; 4 )
( x − __12 ) + ( y − __32 ) = ___
2
2
2
x2 + 4x + y2 − 6y = 91 at ( 0;−7 )
1.6
x2 + 3x + y2 − 4y = −5 at ( − 1; 3 )
Find the equation(s) of the tangent(s) to the circle:
2.1
with the centre at the origin and passing through the point (5;−3), a point
on the circle
2.2
with centre (−4;2) and passing through the point (1;0), on the circle
( x + 2 )2 + y2 = 4 at the x-intercepts of the circle
2.3
( x − 1 )2 + ( y + 4 )2 = 17 at the origin
2.4
2.5
x2 + y2 = 7 which is parallel to the x-axis
( x + 3 )2 + ( y − 2 )2 = 25 which is parallel to the y-axis
2.6
2.7
x2 + y2 = 8 which is parallel to the line y = −x + 3
Topic 9 Analytical geometry
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3
4
5
6
Determine whether the given line is a tangent, secant or neither to the given
circle in each case:
3.1
x2 + y2 = 9 and 2y − 4x = 6
3.2
x2 + 6x + y2 − 2y = 22 and y + 4 = x
( x − 2 )2 + ( y − 1 )2 = 4 and y + 2x = −2
3.3
Determine whether the following circles touch each other, intersect, or
never cross.
4.1
x2 + y2 = 81 and x 2 + ( y + 10 )2 = 1
( x − 3 )2 + ( y − 1 )2 = 12 and ( x + 4 )2 + ( y + 5 )2 = 4
4.2
4.3
x2 + 4x + y2 + 2y = 4 and x2 − 4x + y2 = 0
Given circle ( x + 3 )2 + ( y + 2 )2 = 20 with tangent KL touching the circle
at K. Determine:
5.1
the length of KL, if point L = (7;−2)
5.2
the coordinates of point K, the point of contact of the tangent to the circle
5.3
the equation(s) of the tangent(s) KL.
The diagram below represents the
y
circle ( x − 2 )2 + ( y + 2 )2 = 20
with centre M.
AC and BC are tangents to the circle at A and B
respectively. Determine:
A
x
6.1
the coordinates of A, an x-intercept of
circle M
M
6.2
the coordinates of B, a y-intercept of circle M
C
6.3
the equations of the tangents AC and BC
B
6.4
the coordinates of C, the point of
intersection of AC and BC
6.5
the type of quadrilateral ACBM is, giving
reasons.
Unit 2 Equation of a tangent to a circle
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Revision Test Topic 9
Total marks: 237
__
1
Determine the equation of the circle centre O and passing through (−3; √ 7 ). (3)
2
Determine the equation of the circle with centre (−3;5) and passing through
3
4
5
the point (2;−9).
(4)
Given the circle x2 + y2 − 4x + y = 7.
3.1
Determine the equations of the tangents to the circle at the points
where x = −1.
3.2
Determine the point of intersection of these two tangents.
(10)
(4)
The line y + 3x = 2 intersects the circle x2 + y2 = 2 in two points A and B.
4.1
Determine the coordinates of A and B.
4.2
Then determine the length of AB.
4.3
Give the coordinates of M, the midpoint of AB.
4.4
Show that OM is perpendicular to AB.
(6)
(3)
(2)
(3)
Determine the equation of the tangent at A(3;−2) to the circle centred on
the origin which passes through A.
6
7
y
A circle centre O cuts the x-axis at A and
__
passes through C(√3 ;1). The tangent to
the circle at C meets the y-axis at B and
B
the x-axis at D.
C ( 3;1)
6.1
Determine the equation of the
circle.
(2)
0
6.2
Write down the coordinates of A.
O
A
D
(2)
6.3
Calculate the gradient of OC
(1)
6.4
Determine the equation of the
tangent BCD.
(4)
6.5
Determine the size of angle θ. (2)
6.6
Determine the coordinates of
B and D.
6.7
Calculate the area of △BOD. (Leave your answer in surd form.)
6.8
Calculate the length of BC. (Leave answer in surd form.)
6.9
If E(3;m) is a point so that BE = 5, determine m.
Given circle x2 − 2x + y2 − 16y + 39 = 0 with centre A and y-intercepts
B(0,p) and C(0;q) where p < q.
7.1
Determine the values of p and q.
7.2
Show that point D(2;13) lies on circumference of the circle.
7.3
Prove that points B, A and D are collinear.
(5)
x
(4)
(3)
(3)
(5)
(4)
(2)
(3)
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8
The diagram shows a circle with centre at the origin O,
y
passing though A(−10;0), B, D and E.
C(−8;4) is the midpoint of AB. BOD is
B
a diameter.
8.1
Determine the coordinates of B. (4)
C(– 8; 4)
8.2
Determine the equation of AB.
(3)
E
x
8.3
Write down the coordinates of D. (2) A(– 10;0)
O
8.4
Determine the equation of AD.
(3)
8.5
Prove analytically that OC ⊥ AB. (3)
8.6
Determine the area of △ AOB.
(6)
D
^
8.7
Find the size of BOE.
(4)
8.8
Determine the equation of the
tangent to the circle at point E.
(2)
8.9
Determine the coordinates of the point where the tangent found
in question 8.8 intersects with line AD.
(4)
8.10 Determine the equation of the tangent to the circle at point D.
(5)
9
The points A(−5;3), B(−3;−3) and C(5;3)
are the vertices of a triangle.
y
C(5;3)
A(– 5; 3)
Determine:
9.1
the length of BC (Leave your
answer in surd form.)
(3)
x
9.2
the midpoint of BC
(2)
9.3
the equation of the median of
triangle ABC drawn from A
(4)
B(–3 ;–3 )
9.4
the equation of the circle with
diameter BC
(3)
9.5
whether point A lies inside, outside or on the circumference of the circle
found in question 9.4
(3)
9.6
the coordinates of point D, so that ABCD is a parallelogram
(2)
^
9.7
the magnitude of BAC.
(5)
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REVISION TEST TOPIC 9 CONTINUED
10
A circle with equation x2 + y2 − 4x + 2y + k = 0 passes through the point (1;3).
Determine:
10.1 the centre of the circle
(4)
10.2 the value of k
(2)
10.3 the radius of the circle (in surd form)
(2)
10.4 the equation of the normal to the circle at (1;3).
(4)
11
Prove that the circles x2 + y2 = 16 and ( x − 4 )2 + ( y + 3 )2 = 1 touch
each other.
12
13
14
In the diagram alongside, the origin O is
the centre of the circle. A(x;y) and B(−3;4)
are points on the circle, C is an x-intercept and AOB is a diameter of
the circle. D is the point (p;−1) and BD is a tangent to the circle at B.
Determine:
12.1 the equation of the circle
12.2 the coordinates of A
12.3 the coordinates of C
12.4 the equation of AB
12.5 the equation of the tangent BD
12.6 the value of p
12.7 the length of BC (in surd form).
The equation of a straight line l is 2y − 5x = 6.
13.1 If P(4;a) is a point on l, prove that a = 13.
13.2 Q has coordinates (b;9) and PQ is perpendicular to l.
Calculate the value of b.
13.3 Determine the equation of a circle centred at P and passing
through Q.
P(−32) and Q(1;1) are two points on a circle with centre A.
The equation of the tangent to the circle at P is 2y + x − 7 = 0.
Determine:
14.1 the equation of PA
14.2 the equation of the perpendicular bisector of PQ
14.3 the coordinates of A
14.4 the length of AQ
14.5 the equation of the circle.
(5)
y
B(– 3; 4)
(3)
(2)
C
(2)
D(p; – 1)
(2)
(3)
(2)
(3)
x
O
A
(2)
(4)
y
(5)
P(– 3; 5)
Q(1;1)
(4)
(5)
(4)
(3)
(3)
x
A
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REVISION TEST TOPIC 9 CONTINUED
15
A, B(−10;2) and C(12;−5) are the vertices of △ABC. M is the midpoint
1 ) is the midpoint of AC.
of BC and N(7;2__
2
15.1 Write down the coordinates of M.
(2)
15.2 Determine the coordinates of A.
(4)
15.3 Determine the equation of the median of △ABC from A.
(4) B(– 10;2)
15.4 Show that AB ∥ MN.
(3)
15.5 Determine the length of MN in simplified surd form.
(3)
15.6 Without further calculations write down the length of AB.
(1)
^.
15.7 Determine the magnitude of B
(5)
15.8 Show that AB ⊥ AC.
(3)
15.9 Determine the area of △ ABC.
(4)
15.10 Determine the equation of the circumcircle of △ABC.
(5)
16
A(0;2), B(−3;−5), C(9;1) and D(x; y) are the vertices of parallelogram ABCD with
D in the first quadrant.
16.1 Determine the coordinates of D.
(4)
16.2 Show that the diagonals of ABCD bisect each other.
(3)
16.3 Determine the equation of the perpendicular bisector of AB.
(5)
16.4 Determine the equation of the altitude of △ABC from point A.
(4)
16.5 Determine the equation of line CA.
(4)
16.6 Determine the angle of inclination of AC.
(3)
16.7 If B lies on the circumference of a circle with centre at A, find:
16.7.1 the equation of the circle
(4)
16.7.2 the equation of the tangent to this circle at B
(4)
16.7.3 the point of intersection of this tangent and CA produced.
(5)
17
Determine whether the straight line x + y = 5 cuts the circle with centre (2;5)
and radius 1 unit.
(6)
18
A circle touches the x-axis and has its centre on the
line y = 2x. If it passes through the point (−1;2),
determine:
18.1 the centre of the circle
(6)
18.2 the radius of the circle.
(2)
y
A
1
N(7;2 2)
x
M
C(12;– 5)
KEY WORDS
circumcircle of a triangle
− the circle that passes
through all of the vertices of
that triangle
y
y = 2x
(– 1; 2)
x
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Mid-year Exam practice: Paper 1
Time: 3 hours
Total marks: 150
Question 1
1.1 Solve for x, correct to two decimal places where necessary:
1.1.1 ( 2x − 3 )( x − 1 ) = 0
10
1.1.2 3x + 1 = ___
x
1.1.3 ( x − 3 )( 3x + 5 ) > x − 3
1.2 Show that x2 − 6x + 10 = 0 has no real roots.
1.3
A large photograph is placed onto a rectangular
backing sheet. The dimensions of the backing sheet are
100 cm by 50 cm. The border is the same width
all the way around. Determine the width of the border
if the area of the border is 1 400 cm2.
(2)
(5)
(5)
(3)
(6)
[21]
Question 2
2.1 Consider the quadratic pattern 0; − 6; −14; −24; −36 and determine the nth term
in the sequence if the pattern continues in the same way.
(5)
2.2 The second term of an arithmetic sequence is 13 and the sum of the first and
fifth terms is 16. Determine the first term and the common difference of the
sequence.
(5)
2.3 Consider the geometric sequence given by 5x + 1; 4x − 4; 3x − 5.
2.3.1 Solve for x.
(5)
2.3.2 If x = 7, determine the sum to infinity.
(3)
n
2.4 Determine the value of n if
∑( 2k − 5 ) = 525.
k=1
202
(5)
[23]
Mid-year Exam practice: Paper 1
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Questions 3
Taku plans to buy a house for R2 000 000. He can afford monthly payments of
R16 000 and has enough money saved to pay a deposit of 15%. The interest rate
offered to him by Investor’s Bank is 9,5% per annum, compounded monthly.
3.1 If he pays R16 000 per month, how long will it take him to pay off the loan?
3.2 Determine the amount of his final payment.
3.3 If he missed the 200th, 201st, 202nd and 203rd payments and wants to pay
off the loan in 220 months, determine the monthly payment required from
the 204th payment on.
Question 4
3 − 1, g( x ) = − __
1 ( x − 3 )2 + 2 and
Consider the functions f( x ) = _____
x−2
2
x
−
3
h( x ) = 4 × 2
− 2.
4.1 For which values of x and y is f( x ) = g( x )?
4.2 Solve for x if h( x ) = 2.
4.3 Draw f, g and h on the same set of axes, clearly indicating the x- and
y-intercepts, the turning point(s), the asymptotes and the symmetry lines.
Question 5
Consider the functions f( x ) = 2x2 and g( x ) = 2 x .
5.1 How must the domain of f be restricted so that f − 1( x ) is a function.
5.2 State f − 1( x ) for each option stated in 5.2.
5.3 State g− 1( x ) in the form y = ...
5.4 h is the reflection of f in the x-axis. State the equation of h.
5.5 Sketch f, g, g − 1 and h on the same system of axes. Show the intercepts with
the axes and the coordinates of any turning points. Show at least 3 points
on each graph.
Question 6
6.1 Determine f ’( x ) from first principles if f( x ) = −x2 + 3.
__
dy
6.2 Determine ___ if y = 3x3 − 4x + √ x + 5.
dx
x2 − 3x − 4 .
6.3 Determine Dx __________
x+1
[
(5)
(5)
(8)
[18]
(6)
(3)
(9)
[18]
(2)
(4)
(2)
(1)
(8)
[17]
(5)
(4)
]
(3)
6.4 The function f( x ) = ax3 + bx2 + cx + d is sketched below. The stationary points
of f( x ) are P( −2;32 ) and Q( 2;0 ) and the y-intercept is R(0;16).
y f(x) = ax³ + bx² + cx + d
P(– 2;32)
R(0;16)
Q(2;0)
x
6.4.1 Determine the values of a, b, c and d.
6.4.2 For which value(s) of k will f( x ) = k have only one solution?
6.4.3 For which values of x is f( x ) a decreasing function?
6.4.4 State the equation of the tangent to f at x = −2.
6.4.5 Determine the average gradient between x = −2 and x = 2.
(8)
(2)
(3)
(1)
(2)
[28]
Mid-year Exam practice: Paper 1
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Question 7
Right-angled triangle ABC is rotated around its vertical side AB in such
__
a way that a cone is formed. The hypotenuse of the triangle is 4√ 3 units.
REMEMBER
The volume of a cone is given
1 π​r2h
by V = __
A
3
h
4 3
r
B
C
7.1 Express the volume in terms of h, the vertical height.
7.2 Determine the radius and height of the cone with maximum volume.
7.3 What is the maximum volume?
Question 8
Eighty Grade 12s were asked to fill in a survey form about their plans for the
following year.
20 planned to travel (T)
30 planned to work (W)
40 planned to study (S)
6 planned to work and travel
9 planned to study and travel
7 planned to study and work
8 had no plans to study, work or travel
Let the number of Grade 12s who planned to work, study and travel be x.
8.1 Draw a Venn diagram to represent the information about the learners.
8.2 Solve for x.
8.3 Determine the probability that one of these learners, selected at random:
8.3.1 will work and study, but not travel
8.3.2 will travel, but neither work nor study
8.3.3 will neither work, study nor travel.
204
(3)
(5)
(2)
[10]
(6)
(3)
(2)
(2)
(2)
[15]
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Mid-year Exam practice: Paper 2
Time: 3 hours
Total marks: 150
Question 1
The individual masses of 30 Grade 12 boys are recorded below:
59
63
66
66
67
67
68
69
72
72
74
74
75
75
78
79
80
84
84
85
85
86
89
92
92
98
99
101
104
108
1.1
1.2
1.3
Calculate the mean mass of the boys.
Determine the number of boys who lie within one standard deviation
of the mean.
Complete the table below:
Mass in kg
Frequency
(3)
(5)
Cumulative frequency
50 ≤ x < 60
60 ≤ x < 70
70 ≤ x < 80
80 ≤ x < 90
90 ≤ x < 100
100 ≤ x < 110
(4)
1.4
1.5
1.6
Draw an ogive (cumulative frequency curve) of the information in
the table.
Use your ogive to determine an estimate for the median mass and
indicate on the ogive where the reading was taken.
Use your ogive to determine the mass which should be used to
determine the heaviest 20% of the boys. Indicate on the ogive where
you took your reading.
(4)
(2)
(2)
[20]
Mid-year Exam practice: Paper 2
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Question 2
y
B
M
O
A
D
C
E
x
M is the centre of the circle which passes through A(−6;−2), B(0;6) and D.
BC is a tangent to the circle at B and AO passes through the origin.
2.1 State the coordinates of M.
2.2 Determine the equation of the circle.
2.3 Determine the length of BD.
2.4 Determine the equation of BC.
2.5 Determine the coordinates of C.
2.6 Show that AB = BC.
2.7 Determine the area of △ ABC
2.8 Determine the equation of AC.
2.9 Determine the coordinates of E.
^ O, correct to one decimal place.
2.10 Determine the size of BA
2.11 Determine the equation of the circle passing through A, B and C.
(2)
(4)
(3)
(3)
(2)
(4)
(4)
(3)
(7)
(5)
(4)
[41]
Question 3
3.1
A
C
D
B
A, B and C lie on the circumference of the circle. CD ⊥ AB and AD = BD.
Prove the theorem which states that DC passes through the centre of the circle.
(6)
206
Mid-year Exam practice: Paper 2
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3.2
N
V
G
H
K
Z
T
J
L
M
P
W
H, J, K, L, M, N, V and W lie on the circumference of the circle.
HK and LP intersect at G.
^ = LP
^ N = 90ο, LP = PM and VZ = ZW.
JK crosses NP at T and VW at Z. H
LM = 14 units, JK = 50 units and ZK = 10 units.
3.2.1 Prove that T is the centre of the circle.
3.2.2 Determine, with reasons, the length of TP.
3.2.3 Determine, with reasons, the length of VW.
Question 4
4.1
(6)
(3)
(3)
[18]
D
O
F
E
D, E and F are points on the circumference of the circle with centre O.
^ F = 2ED
^ F.
Prove the theorem which states that EO
4.2
C
B
2
1
3
2
1
O
1
(5)
E
2
1
D
x
A
Two circles intersect at B and D. A and E lie on the larger circle with centre O.
^ = x.
C and O lie on the smaller circle. A
4.2.1 Prove that BC = CE.
(7)
4.2.2 Is ABCD a cyclic quadrilateral? Justify your answer.
(3)
[15]
Mid-year Exam practice: Paper 2
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Question 5
A
12
E
B
C
1 2
3
20°
G
34
2
40° 1 D
F
A, B, C, D and E are points on the circumference of the circle. FDG is a tangent
^ = 20ο. Determine, with reasons,
^ = 40ο and C
to the circle at D and FDG || CE. D
1
2
the values of:
^
5.1 D
4
^
5.2 A
1
^
5.3 B
(4)
(3)
(3)
[10]
Question 6
cos ( 360° + x ) − tan ( 180° − x ) sin ( 360° − 2x ) cos ( −x )
6.1 Consider ______________________________________________
= cos 2x
(
)
sin 90° + x
cos 360° + x − tan 180° − x sin 360° − 2x cos −x
= cos 2x
6.1.1 Show that ____________________________________________
(
)
(
)
(
)
(
)
(
)
sin 90° + x
(7)
6.1.2 Then, without the use of a calculator, calculate the value of
cos 390° − tan 150° sin 300° cos( −30° ]
_________________________________
sin 120°
6.2 If sin 24∘ = p, determine the following in terms of p:
6.2.1 cos 24°
6.2.2 tan 66°
6.2.3 cos 33°
(2)
(3)
(2)
(3)
[17]
Questions 7
sin 3x − ______
cos 3x
7.1 Simplify the expression fully: ______
cos x
(4)
7.2 Without using a calculator, determine the value of:
sin 80° sin 110° − cos 100° sin ( −20° )
________________________________
(7)
sin x
cos 40° sin 70° + sin 220° sin 20°
7.3 f( x ) = cos x + 1 and g( x ) = sin ( 2x − 90∘ )
7.3.1 Solve for x if f( x ) = g( x ) and x ∈ [ −90°; 270° ]
7.3.2 Sketch f and g on the same system of axes for x ∈ [ −90°; 270° ].
7.3.3 For which values of x will f( x ) ≥ g( x )?
208
(8)
(6)
(4)
[29]
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Term 2 summary
Topic 6
Trigonometry: Problem solving in
two and three dimensions
The basic definitions: Use when you have a 90°
triangle
opposite
adjacent
opposite
sinA = __________ cos A = __________ tan A = ________
hypotenuse
hypotenuse
When the triangle is not
right-angled, choose
between:
The sine rule: Use this rule
when you have SSA
or ASA.
B
a = _____
c
b = _____
_____
adjacent
A
b
c
a
sin A
sin B
sin C
sin
A
sin
B
sin C
____
_____
or a =
= _____
c
b
C
The cosine rule: Use this rule when you have SAS
or SSS.
a2 = b2 + c2 − 2bc cos A or b2 = a2 + c2 − 2ac cos B
or c2 = a2 + b2 − 2ab cos C
b2 + c2 − a2 or cos B = __________
a2 + c2 − b2 or
cos A = __________
2ac
2bc
a2 + b2 − c2
cos C = __________
2ab
The area rule: Use this rule when you have SAS.
1 ab sin C = __
1 ac sin B = __
1 bc sin A
Area △ABC = __
2
2
angle of elevation
horizontal
2
Remember:
angle of depression
Topic 7
Functions: Polynomials
• Polynomials:
For a linear or 1st degree polynomial the standard
form is ax + b.
For a quadratic or 2nd degree polynomial the
standard form is ax2 + bx + c.
For a cubic or 3rd degree polynomial the standard
form
is ax3 + bx2 + cx + d.
• Polynomial functions:
y = ax + b is a linear function or straight line
graph.
y = ax2 + bx + c is a quadratic function or
parabola.
y = ax3 + bx2 + cx + d is a cubic function.
• Factorising cubic polynomials − this can be done
using:
• common factors, grouping or sum/difference of
cubes
• the remainder or factor theorem.
• The Remainder Theorem: If a polynomial f (x) is
divided by ax + b , then the remainder will be
b
f (− __
a ).
• Or more simply, when you divide a polynomial
f (x) by (x − a) the remainder is f(a).
• The Factor Theorem: If a polynomial is divided by
ax + b,and the remainder = 0, then ax + b is a
factor of the polynomial.
b
Or if f (− __
a ) = 0, then ax + b is a factor of f (x)
or if f (a) = 0 then x − a is a factor of f (x).
You have seen that a cubic polynomial has three
factors. A cubic equation will therefore have three
solutions and its graph will have three roots.
• You may need to use some identities that you studied
in Grade 11 as well as those from Topic 6 to simplify
the expression obtained in algebraic problems.
• Three-dimensional problems: You work in more
than one plane − the vertical plane, the horizontal
plane and sometimes the oblique plane. Remember:
• Right angles do not always look like right angles.
• Do not add or subtract angles that seem
adjacent, but are not in the same plane.
Term 2 summary
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Term 2 summary continued
Topic 8
Differential calculus
• When a function is undefined for a particular
value x it is important to know how the function
behaves near to the value of x for which it is
undefined.
A function has a limit if it approaches the same value
on either side of the undefined value.
A function does not have a limit if it approaches
different values on either side of the undefined value.
• The average gradient between points A and B on
y2 − y1
any curved graph = mAB = ______
x −x
2
1
• You can find the gradient of a curve from first
f( x + h ) − f( x )
principles using: f ’( x ) = lim ____________
h
h→0
• The short rule for the gradient, or derivative of
the function f( x ) = axn is given by f ’( x ) = anxn−1,
for n ∈ ℝ.
It is important that you simplify all functions before
applying this rule.
Note:
d k f( x ) = k___
d f( x ) ;___
d f( x ) ± g( x ) = ___
d f( x ) ± ___
d
___
[
]
[
] dx
[
] dx
[
] dx
dx
dx
[ g( x ) ]
d f( x ) × g( x ) ≠ ___
d g( x )
BUT ___
[
] d [ f( x ) ] × ___
[
]
dx
dx
dx
• Finding the equations of tangents to functions:
Find the gradient using calculus, and then
substitute the given point to find the equation of
the tangent (which is a straight line).
• Sketching cubic graphs: Determine the stationary
point (solve f ’( x ) = 0), the roots (f(x) = 0),
the y-intercept (x = 0)
and the point of inflection (f ”( x ) = 0).
• Optimisation: Maximum or minimum values are
optimal.
Create an equation of whatever needs to be
optimised.
Solve for the derivative of this equation = 0.
Ensure that the answer obtained is maximum or
minimum using the second derivative test, or
considering the shape of the function.
210
Term 2 summary continued
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Term 2 summary continued
Topic 9
Analytical geometry
• Formulae from Grade 11:
_________________
The distance formula: √ (x2 − x1)2 + ( y2 − y1 )2
( x +2 x y +2 y )
1
2 ______
2
Midpoint formula: ______
; 1
y2 − y1
Gradient = ______
x −x
2
1
Inclination: tan θ = m
Equation of line: y = mx + c or
y − y1 = m( x − x1 ) for any line passing through
point ( x1;y1 )
• Circle centred at the origin: x2 + y2 = r
• Circle centred at (a; b): (x − a)2 + (y − b)2 = r2
Use completing the square to get the equation into
this standard form.
• Equation of a tangent to a circle:
Remember that a radius is always perpendicular to
the tangent at the point of contact.
First find the gradient of the radius (from centre to
the point of contact); the perpendicular gradient
will be the gradient of the tangent. (Perpendicular
gradients have product = −1.)
Then substitute the coordinates of the point of
contact, and the gradient into y − y1 = m( x − x1 )
• A normal is the line perpendicular to the tangent at
the point of contact.
Term 2 summary continued
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Term
3
212
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Topic 10
Euclidian geometry
Unit 1
Unit 2
Unit 3
Unit 4
Unit 5
Revision: Grade 11 Geometry
214
Similar polygons
218
Parallel lines and proportionality theorem 221
Equiangular triangles and similarity
227
Triangles with proportional sides
and similarity
231
Unit 6
Pythagoras’ Theorem and similarity
235
Revision Test
237
Topic 11
Statistics
Unit 1
Unit 2
Revision of skewed and symmetric data 240
Bivariate data: scatter plots, regression
lines and correlation
243
Revision Test
253
Topic 12
Counting principles and probability
Unit 1
Revision of rules for independent,
mutually exclusive and
complementary events
Use Venn diagrams, tree diagrams and
contingency tables to solve problems
The Fundamental Counting Principle
Applications of the counting principle
to solve probability problems
Unit 2
Unit 3
Unit 4
Revision Test
Preliminary exam practice: Paper 1
Preliminary exam practice: Paper 2
Term summary
256
261
267
275
280
283
286
290
213
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TOPIC
2
10
Euclidean geometry
Unit 1: Grade 11 geometry revision
O
A
O
B
K
A
AK = KB | OK ⊥ AB
OR ⊥ AB
| line from centre to midpoint chord
E
F
A
B
0
C
^=F
^ | Equal chords AB & CD
E
D
D
B
Q
0
C
^ C = 2A
^
| ∠ at centre
BO
^ =D
^
| ∠s on chord BC
A
^
^
Reflex O = 2R | ∠ at centre
D
G
O
R
P
D
A
E
B
R
E
^ = 90° | ∠ on diameter
D
G
F
^ = 180°
^+G
E
| opposite ∠s cyclic quadrilateral
D
G
E
F
O
H
^H = D
^ | exterior ∠ cyclic quadrilateral
GF
A
^ C = 90°
OB
214
B
C
| Radius ⊥ tangent
Topic 10 Euclidean geometry
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D
F
G
E
1
A
E
D
2
DEFG is a cyclic quadrilateral
| DE subtends equal ∠s at F & G
3
B
C
^ =E
^ | tan AB, chord BD
B
1
^
^ | tan BC, chord BE
B3 = D
T
M
A
x
P
180°– x
N
B
V
PA = PB | tangents from common point P
MNVT is a cylic quadrilateral
| opposite ∠s supplementary
A
C
A
D
B
D
B
^ B = 90°
AB is the diameter | AC
C
E
ABCD is a cyclic quadrilateral
^ E = interior opposite A
^
| exterior DC
EXERCISE 1
O is the centre of the circle in the first 12 questions. Determine, with reasons, w, x, y
and z.
You must determine the angles alphabetical order. Determine only w in 4, 5, 6, 8 and 10.
In question 10, x, y and z indicate the lengths of the sides of ∆OPR.
1
2
D
A
3
E
126°
N
x
C
x
z
y
O
Oz
O
110°
x
F
v
G
H
J
M
y
40°
z
K
L
Unit 1 Grade 11 geometry revision
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5
4
A
6
D
w
70°
Q
F
y
P
w
P
w
S
z O
x
O
G
64°
O
y
x
y
R
Q
H
C
B
T
x
57°
z
78°
R
PQ and PR are tangents to the circle
at F and H.
7
8
9
A
51°
x
C
w
D
C
E
O
O
D
DE is a tangent to the circle at B.
DE is a tangent to the circle at A.
10
11
F
G
H
12
E
B
J
A
z
23°
w
y
P
y
F
G
x
O
z
25
R
S
y
z
O
H
62°
E
x
110°
B
O
67°
x
B
x
z
227°
z
67°
y
y
56°
z y
D
E
A
35
x
34°
E
M
Q
C
D
OB and OD are tangents to the circle
M at B and D.
216
Topic 10 Euclidean geometry
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13
O is the centre of the circle which passes through
A, C, D and E. AE and CD produced meet at B.
AD, BG and CE intersect at F.
^ = x.
H is the midpoint of EC and B
1
13.1 Prove that EBDF is a cyclic quadrilateral.
13.2 State, with reasons, 2 other angles equal to x.
13.3 Prove that OH || AE.
^ C in terms of x.
13.4 Determine the size of BA
13.5 Prove that OGFH is a cyclic quadrilateral.
A
G
2
O
C
2
1
1
2 1
5 6
4
3
E
H
1 F
3 2
2
1
2
1
D
3
x
1 2
B
14
RT and NW are tangents to the smaller and larger
^ = x. The circles intersect at P and
circles at V, with V
1
V, so PV is a common chord. QPN is a straight line, Q
and R lie on the larger circle and N lies on the smaller
circle.
14.1 Determine, with reasons, 4 other angles equal
to x.
14.2 Prove that QR || NQ.
Q
P
3
R
N
1
2
2
1
3 2
x
1
4
5
V
T
W
15
NT is a tangent to the circle at M, ON || KM and J, K, M and P
are points on the circle. MJ and ON intersect at R and O is the
centre of the circle.
15.1 Prove that R is the midpoint of JM.
^ =M
^ .
15.2 Prove that M
1
2
15.3 Prove that JOMN is a cyclic quadrilateral.
N
M
1
1
P
2
3
T
4
K
2
3
2
1
3
4
R
2
1
O
1 2
J
Unit 1 Grade 11 geometry revision
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Unit 2: Similar polygons
Similar polygons have their corresponding angles equal and corresponding sides in
proportion.
J
114°
E
F 124°
114°
A 124°
82° D
B
130°
C
82° I
130°
G
H
BC = ___
CD = ___
AB = ____
DE = ___
AE .
^ =F
^, C
^ =H
^, B
^=G
^, D
^ = ^I , E
^ = ^J and ___
ABCDE ||| FGHIJ ⇒ A
FG
GH
HI
IJ
FJ
Two polygons are similar if and only if their corresponding sides are in proportion
and their corresponding angles are equal.
WORKED EXAMPLE
Consider each sketch below and state whether the polygons are similar or not.
Justify your answers with suitable reasons.
1
4
E
H
3
A
D
2
B
3
3
2
C F
3
G
4
2
E
y
A
4
y
4
3
218
H
10
y
C
x
F
y
G
8
P
KEY WORDS
similar polygons – polygons
which have the same shape
as each other, but may be
different in size
corresponding sides – the
sides which join equal angles
corresponding angles – the
angles which join sides which
are in proportion
x
5
x
B
D 10
x
5
8
2
2
Q
T
S
3
5
5
3
U
W
4
R
4
V
Topic 10 Euclidean geometry
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4
A
68°
D
68°
51°
51°
C E
B
F
SOLUTIONS
1
2
3.
4.
3 so the sides are not in
AB = __
2 and = ___
AD = __
ABCD and EFGH are not similar. ___
EF
EH
4
3
proportion.
BC = ____
CD = ___
AB = ___
AD = __
1 and A
^ =E
^ =G
^
^, B
^=F
^, C
ABCD ||| EFGH because ___
EF
EH
FG GH
2
^
^
and D = H, so the corresponding sides are in proportion and the
corresponding angles are equal.
PQ 2
QR 5
PQRS and TUVW are not similar. ___ = __
and ___ = __
, so the sides are not in
TU 3
UV 4
proportion.
^ = 44ο, B
^ = C = 68ο and D
^ = 78ο, E
^=F
^ = 51ο,
△ABC is not similar to △DEF. A
so the corresponding angles are not equal.
EXERCISE 2
1
Consider each figure below and state whether the polygons are similar or not.
Justify your answers with suitable reasons.
57
1.1
E
H
A 38 D
51
B
x
34
y
x
80
120
F
1.2
20
A
D
B
C
30
C
y
G
E
H
60°
3
1.3
F
Q
23°
Two polygons are similar if:
• the corresponding angles
are equal
• the corresponding sides are
in proportion.
A parallelogram is a
quadrilateral with both pairs
of opposite sides parallel.
A rhombus is a parallelogram
with equal sides.
A rectangle is a parallelogram
with a right angle.
A square is a rectangle with
equal sides.
G
4
S
P
T
REMEMBER
23°
157°
23°
157°
157°
R
V
157°
U
Unit 2 Similar polygons
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1.4
E
D
H
G
A
F
C
B
1.5
2x
E
x
I
117°
2k
154°
B
2
220
H
50°
3y
2y
F 117°
86°
50°
154°
3k
A
D
133°
133°
G
C
State whether the statements below are true or false. Justify your answers.
2.1
If two polygons are equiangular, they will be similar.
2.2
Two rectangles are always similar to each similar.
2.3
Two squares are always similar to each other.
2.4
Two parallelograms are always similar to each other.
2.5
Two rhombuses are always similar to each other.
2.6
If the sides of two quadrilaterals are proportional, the quadrilaterals
will be equiangular.
Topic 10 Euclidean geometry
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Unit 3: Proportionality theorem:
A line drawn parallel to one side of a triangle divides the other two
sides proportionally.
Two parallel lines are the same distance apart.
h1
h2
h3
1 base × perpendicular height.
The area of a triangle = __
2
Triangles which have the same base and same height are equal in area.
P
R
Q
h1
h3
h2
V
S
A
T
U W
B
In the sketch above, AB is the base of three different triangles, that is,
△PAB, △QAB and △RAB.
The triangles have equal heights since P, Q and R all lie on the line PR, which is
parallel to base AB.
1 AB × h , Area△QAB = __
1 AB × h and Area △RAB = __
1 AB × h .
Area △PAB = __
2
1
2
2
3
2
Area △PAB = Area △QAB = Area △RAB | common base AB, same height, AB || PQR
Triangles with different bases, but equal heights are not equal in area. The ratio of
their areas is equal to ratio of the lengths of their bases.
△PVA, △PAB and △RBW have bases VA, AB and BW on the line VW, which is parallel
to PR. The third vertex of each of these triangles lies on PR, so they have equal
heights.
1
1 AB × h :__
1 BW × h = VA: AB: BW
VA × h : __
Area △PVA : Area △PAB : Area △RBW=__
2
In the sketch alongside, A
is the common vertex of
△ABC and △ADC.
1
2
2
2
3
A
If two triangles have a
common vertex and their
bases lie on the same
C
B
straight line, then the ratio
of their areas is equal to the ratio of their bases.
KEY WORDS
E
D
In the sketch alongside, A is a common vertex and BC and CD lie on the same straight line.
1 BC × AE
__
Area △ABC = _________
BC
2
___________
= ___
Area △ACD
1 CD × AE
__
2
CD
| The height AE is common
vertex – the point at which
two straight lines meet to
form an angle
common vertex – the point
at which three or more lines
meet to form two or more
angles
Unit 3 Proportionality theorem:
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Area △ABC : Area △ACD = BC : CD
Theorem: A line drawn parallel to one side of a triangle divides the other two sides proportionally.
A
Given: △ABC with DE || BC, such
that D lies on AB and E lies on AC.
Proof:
1 AD × h
__
2
Area △ADE = __________
2
AD
__________
= ___
Area △DBE
1 DB × h
__
2
DB
2
h1
| Common vertex E, same height h 2
D
1 AE × h
__
1
Area △AED = __________
2
AE
__________
= ___
Area △ECD
1 EC × h
__
2
1
h2
E
EC
| Common vertex D, same height h 1
C
B
Area △DBE = Area △ECD
| Common base DE, same height, DE || BC
Area △ADE = __________
Area △AED
__________
| Area △ADE is common and Area △DBE = Area △ECD
Area △DBE
AD = ___
AE
⇒ ___
DB
EC
Area △ECD
AD = ___
AE
Using the same sketch, we can also prove that____
AB
Proof:
1 AD × h
__
2
Area △ADE = _________
2
AD
__________
= ___
Area △ABE
AB
1 AB × h
__
2
2
1 AE × h
__
1
Area △AED = _________
2
AE
___________
= ___
1 AC × h
AC
__
Area △ACD
1
2
AC
| Common vertex E, same height h 2
| Common vertex D, same height h 1
Area △DBE = Area △ECD
| Common base DE, same height, DE || BC
Area △ADE = ___________
Area △AED
__________
| Area △ADE is common and
Area △ABE = Area △ACD
Area△ABE = Area △ACD
Area △ABE
| Area △ADE is common and
Area △DBE = Area △ECD
Area △ACD
AD = ___
AE
⇒ ___
AB
AC
AC
AB = ___
Finally, we can also prove that ___
REMEMBER
When we divide two numbers
the result is a ratio, so
4 = __
1 = 1:3.
4 ÷ 12 = ___
12 3
When two ratios are equal,
they are in proportion,
8 is an example of a
4 = ___
so __
5 10
simple proportion.
When two ratios are unequal,
they are not in proportion.
DB
1 AB × h
__
2
Area △ABE = _________
2
AB
__________
= ___
Area △DBE
1 DB × h
__
2
DB
2
1 AC × h
__
1
Area △ACD = _________
AC
2
___________
= ___
1 EC × h
EC
__
Area △ECD
1
2
EC
| Common vertex E, same height h 2
| Common vertex D , same height h 1
Area △DBE = Area △ECD
| Common base DE, same height, DE || BC
Area △ABE = ___________
Area △ACD
__________
| Area △ADE is common and
Area △ABE = Area △ACD
Area△ABE = Area △ACD
Area △DBE
Area △ECD
| Area △ADE is common and
Area △DBE = Area △ECD
AC
AB = ___
⇒ ___
DB
EC
222
Topic 10 Euclidean geometry
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A line drawn through the midpoint of one side of a triangle, parallel to
the second side, passes through the midpoint of the third side.
P
Consider △PQR with ST || QR and PS = SQ.
PS = ___
PT = 1
___
| Proportionality theorem, ST || QR and PS = SQ
SQ
S
T
TR
⇒ T is the midpoint of PR | PT = TR
Now construct a line through T, parallel to PQ. TU || PQ with U on QR.
QU
PT = ____
___
=1
| Proportionality theorem, UT || PQ and PT = TR
TR
R
Q
UR
P
⇒ U is the midpoint of QR | QU = UR
SQUT is a parallelogram
| Both pairs of opposite sides parallel, SQ || TU and
ST || QU.
ST = QU
1
QR
ST =__
| Opposite sides of parallelogram SQUT
TU = SQ
1
PQ
TU =__
| Opposite sides of parallelogram SQUT
S
T
| U is the midpoint of QR
2
R
U
Q
| T and U are midpoints of PR and QR respectively.
2
A line drawn from the midpoint of one side of a triangle, parallel to a second side of the triangle,
will pass through the midpoint of the third side of the triangle and its length will be half the
length of the side of the triangle to which it is parallel.
Three or more parallel lines divide the sides proportionally.
Consider the sketch below, with BG || CH || DJ.
A
F
G
B
C
F
A
G
B
H
C
H
J D
D
J
K E
E
Figure 1
K
Figure 2
Figures 1 and 2 are identical, but different triangles are shaded.
Area △BCH = Area △GHC
Area △ CDH= Area △ HJC
| Same base CH, same height, BG || CH
| Same base CH, same height, DJ || CH
Figure 1:
Area △BCH = ___
BC | Same height, common vertex H
___________
Figure 2:
Area △GHC = ____
GH | Same height, common vertex C
___________
Area △CDH
Area △HJC
Area △GHC = ___________
Area △BCH
___________
Area △HJC
Area △CDH
CD
HJ
| Proved above
GH = ___
BC
____
HJ
CD
Unit 3 Proportionality theorem:
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WORKED EXAMPLE 1
A
DE || BC, with D on AB and E on AC.
Determine, with reasons, the length of:
1.1 AD if AE = 15 units,
EC = 6 units and DB = 4 units.
1.2 AB if AD = 9 units,
AE = 12 units and EC = 4 units.
1.3 EC if AE = 16 units,
DB = 3 units and AB = 11 units.
D
E
C
B
SOLUTIONS
AD = ___
AE
1.1 ___
| Proportionality theorem, DE || BC
DB
EC
15 ⇒ AD = ___
60 = 10 units
AD = ___
___
4
6
6
AC
AB
___
___
=
| Proportionality theorem, DE || BC
1.2
AD
AE
16
AB
___ = ___ ⇒ 12AB = 144 and AB = 12 units
9
12
AD = ___
AE
| Proportionality theorem, DE || BC
1.3 ___
DB
EC
AD = AB – DB = 11 units – 3 units = 8 units
8 = ___
16 ⇒ 8EC = 48 and EC = 6 units
__
3
EC
WORKED EXAMPLE 2
P
Z
PQRS is a parallelogram.
Diagonals PR and QS intersect at T.
WR = 6 units. V is a point on QR
T
U
PZ = 25 units, RS = 40 units and
V
Q
S
W
R
such that QV : VR = 2 : 1 .
Determine, with reasons, the length of:
2.1 TZ
2.3 PR
2.2 QR
PU
2.4 ___
UV
SOLUTIONS
2.1 T is the midpoint of PR
Z is the midpoint of PS
1 RS
TZ = __
2
= 20 units
2.2 QR = PS
= 2PZ
= 50 units
QV
TW = ___
2.3 ____
WR
VR
TW = __
2 ⇒ TW = 12 units
____
6
1
| Diagonals of parallelogram PQRS bisect each other at T
| TP = TR and TZ || RS, midpoint theorem
| T and Z are midpoints of PR and PS respectively
| Opposite sides parallelogram PQRS
| Z is the midpoint of PS
| VW || QT, proportionality theorem
PR = 36 units
| T is the midpoint of PR
3
PU
18
PT
___
____
___
__
2.4
=
=
= units | UT || VW, proportionality theorem
UV
224
TW
12
2
Topic 10 Euclidean geometry
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WORKED EXAMPLE 3
A
D
F
E
G
B
C
E, D and G are points on AC and BC of △ABCsuch that GE || BD. AG intersects
BD at F.
3
BG = __
5 and ___
AD = __
___
GC
2
7
DC
AF
3.1 Determine ___
FG
Area △AFD
3.2 Determine __________
Area △AGE
SOLUTIONS
A
3y
3z 0
F
B
Use x, y and z to indicate the
ratio relationships.
D
5y
G
3
AD = __
3.1 ___
DC 7
BG
DE = ___
___
EC GC
5
= __
2
2x
3y
7y
3 = ___
BG = __
5 = ___
5x and ___
AD = __
Note: ___
GC
E 2y
5z
5x
7y
2
2x
DC
7
C
| Given
| GE || BD, proportionality theorem
| Given
AD : DE : EC = 3 : 5 : 2
AF = ___
AD
___
| FD || GE, proportionality theorem
FG
3
=__
5
DE
^D = θ
3.2 Let FA
REMEMBER
1 AF×AD × sinθ
__
Area △AFD = ______________
2
__________
Area △AGE
1 AG×AE × sinθ
__
2
AF × ___
AD
= ___
AG
AE
3 × __
3 = ___
9
= __
8 8 64
Make sure that you use common
^D
angle FA
You can determine the area of
a triangle with the area rule:
1 ab sin C
Area △ABC = __
2
1
= __
bc sin A
2
1
= __
ac sin B
2
= 0,36
Unit 3 Proportionality theorem:
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EXERCISE 3
1.
D
6
F
4
A
8
E
C
B
FE || DC and DE || BC. AF = 4 cm, DF = 6 cm and AE = 8 cm.
1.1
Determine, with reasons:
1.1.1
the length of EC
1.1.2
the length of AB
1.2
Determine Area △AFE: Area △ABC.
2
P
T
A
W
S
V
Q
R
PQ 7
W is the midpoint of PR. ST || QW, WA || RS and ___ = __
.QW and SR intersect
SQ
2
at V.
2.1
Determine, with reasons:
PT
____
2.1.1
TW
2.2
2.3
226
2.1.2
PT
___
2.1.3
PA
___
2.1.4
QV
____
TR
SQ
VW
If SR = 36 cm, determine:
2.2.1
the length of AW
2.2.2
the length of VR
Area △PSR
Determine: ___________
Area △WVR
Topic 10 Euclidean geometry
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Unit 4: Equiangular triangles and similarity
Triangles are similar if their:
• corresponding angles are equal
• corresponding sides are in proportion.
If you know that the triangles are equiangular, then show that the sides are in
proportion.
Theorem: If two triangles are equiangular, their corresponding sides are in proportion.
G
D
x
x
y
w
M
K
H
N
y
w
E
F
^, E
^ =G
^=H
^ and F
^=K
^
Given: △DEF and △GHK with D
DE = ___
DF = ___
EF
To prove: ____
GH
GK
HK
Construction: On DE and DF mark points M and N so that DM = GH and DN = GK.
Join MN.
Proof:
In △DMN and △GHK
1)
DM = GH
| Construction
2)
DN = GK
^
^ =G
D
| Construction
3)
| Given
∴ △DMN ≡ △GHK
| SAS
^ N = GH
^K
∴ DM
^
=E
| Congruency
| Given
^ N = corresponding ^E )
| ( DM
MN || EF
DE
DF
____
∴
= ____
∴
DM
| ( Proportional intercepts, MN || EF )
DN
| Construction
But DM = GH and DN = GK
DE = ___
DF
∴ ____
GH
GK
G
D
x
x
You do not
have to show
these sketches.
They are here
to help you
understand the
final step of
the proof.
A
y
w
K
H
y
E
w
B
F
Similarly, by marking off points A and B on ED and EF respectively so that EA = HG
DF = ___
EF
and EB = HK, we can prove that ____
GH HK
DE = ___
DF = ___
EF
∴ ____
GH GK HK
⇒ △DEF ||| △GHK
| Corresponding angles equal and corresponding sides in proportion
Unit 4 Equiangular triangles and similarity
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WORKED EXAMPLE
B
A
1
1 2
2 x
4
E
2
4
1
3
2
8
1
C
2
x
1
2
12
D
^ =C
^ = x, AE = 2 cm, BE = 4 cm, EC = 8 cm
AD and BC intersect at E. A
2
2
and CD = 12 cm.
1
Prove that △AEB ||| △CED.
2
Determine, with reasons, the lengths of AB and DE.
3
Prove that ABDC is a cyclic quadrilateral.
AC
4
Determine ___
BD
SOLUTIONS
1
In △AEB and △CED:
^ =C
^ =x
1)
A
2
2
^
^
2)
E1 = E3
^ =D
^
3)
B
1
2
3
4
2
△AEB ||| △CED
AE = ___
AB = ___
EB
___
| AAA
| △AEB ||| △CED
CE CD ED
2 = ___
AB = ___
4
__
ED
8
12
AB = 3 cm and ED = 16 cm
^ D = BC
^D = x
BA
| Given
ABDC is a cyclic quadrilateral | BD subtends equal angles at A and C
In △AEC and △BED:
^ =B
^
1)
A
| ∠ s on chord CD
1
2
^ =E
^
2)
E
| Vertically opposite ∠ s
4
2
^
^
3)
C1 = D 1
| ∠ sum △
△AEC ||| △BED
| AAA
AC = ___
EC = ___
8 = __
1
___
| △AEC ||| △BED
BD
228
| Given
| Vertically opposite ∠ s
| ∠ sum △
ED
16
2
Topic 10 Euclidean geometry
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EXERCISE 4
1
MNPQ is a parallelogram.
NPR and MTR are straight lines.
MN = 24 units and TP = 9 units.
M
Q
T
N
2
1.1
Prove, with reasons, that △RPT ||| △RNM.
1.2
RT
Determine ____
1.3
Area △RPT
Determine ___________
1.4
Prove, with reasons, that △RTP ||| △MTQ.
1.5
RT
Determine ____
1.6
NP
Determine ___
1.7
Area △RTP
Determine ___________
RM
Area △RNM
TM
PR
Area △MTQ
QR is the diameter of the circle. P and T
are points on the circle and UV ⊥ QR.
2.1
Prove that △RTU ||| △QPU.
2.2
Prove that:
2.2.1
△RVU ||| △RPQ
2.3
3
R
P
2.2.2
VU
RV = ___
___
2.2.3
RP2 × VU2
QR = RP + _________
RP
P
T
1
2
2
Q
RV2
3
2
PQ
2
U
1
4
1
O
V
2
R
Prove that PQVU is a cyclic
quadrilateral.
O is the centre of the circle, OT || QR and OT = 8 units.
R
T
P
O
Q
Determine, with reasons:
PR
___
3.1
PT
3.2
the length of RQ
3.3
the radius if PR = 30 units
3.4
Area △PTO : Area △PRQ.
Unit 4 Equiangular triangles and similarity
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4
A, B and C lie on the circle. DBE is a tangent to the circle at B.
E
C
1
2
A
2
1
3
B
D
4.1
4.2
4.3
Prove that △EBA ||| △ECB.
Prove that BE2 = AE.CE.
Determine CE if BE = 12 units and AE = 18 units.
^ = 90ο. GH ⊥ DF, DE = 27 cm, HF = 80 cm
△DEF is right-angled, with E
and GH = 18 cm.
5
D
H
27
80
18
G
E
230
F
5.1
5.2
5.3
Prove that △FHG ||| FED.
Determine, with reasons, the length of EG.
Area FHG
Determine _________
5.4
5.5
Determine the area of DEGH.
What type of quadrilateral is DEGH? Justify your answer.
Area FED
Topic 10 Euclidean geometry
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Unit 5: Triangles with proportional sides and
similarity
Triangles are similar if their:
• corresponding angles are equal
• corresponding sides are in proportion.
If you know that the sides of the triangle are in proportion, show that the triangles
are equiangular.
U
P
Q
x
x
y
y
T
x
y
W
V
R
Theorem: If two triangles have their corresponding sides in proportion, then they are equiangular.
PQ
UV
QT
VW
PT = ____
Given: △PQT and △UVW with ___ = ____
^,
^=U
To prove: P
^ =V
^ and
Q
UW
^=W
^
T
^R = V
^, QT
^R = W
^ and R and P are on opposite
Construction: Draw △ QRT so that TQ
sides of QT.
Proof:
In △ RQT and △ UVW:
^T = V
^ =x
1)
RQ
| Constructi on
2)
| Constructi on
3)
^ =y
^Q = W
RT
^
^ =U
R
△ RQT ||| △ UVW
RQ ____
QT
___
∴
= RT = ____
∴
UV UW VW
PQ
QT
PT = ____
But ___ = ____
UV UW VW
∴ RQ = PQ and RT = PT
| ∠ sum △
| AAA
| △ RQT ||| △ UVW
QT
VW
RQ
UV
PQ
UV
RT = ____
PT
| ___ is common, so ___ = ___ and ____
UW
UW
| Given
In △ RQT and △ PQT:
1)
RQ = PQ
| Proved
2)
RT = PT
| Proved
3)
QT is common
∴ △ RQT ≡ △ PQT
∴
^ T = PQ
^T
RQ
^
=V
^Q
^ Q = PT
RT
^
=W
^
∴ ^P = U
| SSS
| △ RQT ≡ △ PQT
^T = V
^
| Construction, RQ
| △ RQT ≡ △ PQT
^
^Q = W
| Construction, RT
| ∠ sum △
Unit 5 Triangles with proportional sides and similarity
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WORKED EXAMPLE 1
K
VP intersects MG at S and KG at T.
GM = 60 units, KM = 30 units,
GK = 80 units, GS = 40 units,
TS = 15 units and GT = 30 units.
1
Prove that △GST ||| △GKM.
2
Prove that KMST is a cyclic quadrilateral.
^ T.
3
Determine the size of SG
4
Determine the area of KMST.
M
P
T
S
V
G
SOLUTIONS
1.1 In △GST and △GKM:
GS = ___
40 = __
1
___
1)
GK
80
| Given
2
2)
30 = __
GT = ___
1
____
| Given
3)
ST = ___
15 = __
1
____
| Given
GM
KM
60
30
2
2
GS = ____
GT = ___
ST = __
1
△GST ||| △GKM
| Sides in proportion, ___
GK GM KM 2
^ G = KM
^G
1.2 ST
| △GST ||| △GKM
KMST is a cyclic quadrilateral | Exterior angle = interior opposite angle
302 + 402 − 152 = ___
91 ⇒ SG
^ T = _____________
^ T = 18,57°
1.3 cos SG
)
)(
(
96
2 30 40
1.4 Area KMST = Area △GKM – Area △GST
1 GM × GK sin MG
1 GT × GS sin MG
^ K − __
^K
= __
2
2
1 ( 60 )( 80 ) sin 18,57° − __
1 ( 30 )( 40 ) sin 18,57°
= __
2
2
^ K gives 573,33 units squared)
= 573,23 units squared (unrounded value of MG
D
WORKED EXAMPLE 2
KEY WORD
cyclic quadrilateral – a
quadrilateral with all four
vertices on a circle
REMEMBER
There are three ways to prove
a cyclic quadrilateral:
• opposite angles are
supplementary
• an exterior angle is equal to
an interior opposite angle
• a line subtends equal angles
on the same side.
232
A
AC and ED intersect perpendicularly at B. BE = 6 cm,
AB = 8 cm, BD = 12 cm and CD = 15 cm.
2.1 Prove that △ABE ||| △DBC.
2.2 Prove that AECD is a cyclic quadrilateral.
15
B
6
SOLUTIONS
________
___
2.1 BC = √______
152 − 122 = √ 81 = 9
____
AE = √82 + 62 = √100 = 10
In △ABE and △DBC
10 = __
AE = ___
2
___
1)
2)
3)
12
8
DC 15 3
8 = __
AB = ___
2
___
DB 12 3
6 = __
BE = __
2
___
BC 9 3
AE = ___
AB = ___
BE = __
2
___
DC DB BC 3
| Pythagoras’ Theorem
| Pythagoras’ Theorem
E
| Pythagoras’ Theorem and given
| Given
C
Use the area
rule with
^ T, the
SG
common
angle.
| Given and Pythagoras’ Theorem
Use the cosine rule because
you have all 3 sides.
△ABE ||| △DBC
| Sides in proportion
^
^
| △ABE ||| △DBC
2.2 A = D
AECD is a cyclic quadrilateral | EC subtends equal angles at A and D
Topic 10 Euclidean geometry
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EXERCISE 5
1.
AB = 24 cm, BC = 21 cm, AC = 27 cm,
DF = 9 cm, DE = 8 cm and EF = 7 cm.
1.1
Prove that △ABC ||| △DEF.
Area △DEF
1.2
Determine __________
1.3
2
4
Area △ABC
^ , correct to
Determine the size of E
one decimal place.
D
24
QT and PR intersect at K,
with PK : KR = 7: 2.
PQ = 24 units, TR = 16 units, TK = 14 units
and PR = 27 units.
2.1
Determine the length of KR.
2.2
Prove that △PQK ||| △TRK.
Area △TRK
2.3
Determine ___________
2.4
2.5
3
A
9
8
E
27
T
F
7
B
C
21
14
P
K
24
Area △ PQK
9
Prove that PQRT is a cyclic quadrilateral.
^ T, correct to one decimal place.
Determine the size of PK
A
E
Area △ABD
Area ABCE
Determine __________
3.6
Without substituting the lengths of the sides:
3.6.1
Prove that ED.DA = DE2 + DE.EA
3.6.2
Then prove that 2BC2 = DE2 + DE.EA
Area △ABD
△ABC and △DEF are right-angled triangles.
AB = 7 cm, AC = 25 cm, DF = 21 cm and
DE = 72 cm.
4.1
Prove that △ABC ||| △FDE.
4.2
Prove △FGD ||| △FDE
4.3
Determine the length of FG.
4.4
Prove that △ABC ||| △FGD.
Area △FGD
4.5
Determine __________
R
Q
AB = 24 units, BC = 20 units and AE = 7 units.
BC = CD, EC : BD = 3 : 8 and ED : BC = 5 : 4.
3.1
Prove that △ABD ||| △CED.
3.2
Prove that ABCE is a cyclic quadrilateral.
3.3
Prove that △ABD is a right-angled triangle.
Area △CED
3.4
Determine __________
3.5
16
B
D
D
C
F
21
A
G
72
25
7
C
B
Area △ABC
5
A
E
AE = 4 cm, AB = 5 cm, BE = 6 cm,
4
5
CE = 16 cm, CD = 20 cm and ED = 24 cm.
E
5.1
Prove that △ABE ||| △CDE.
6
B
5.2
Prove that AB || DC.
5.3
Determine the length of:
5.3.1
BC
5.3.2
AD (Leave your answer in simplified surd form.)
5.4
Is △AED ||| △CEB? Justify your answer by making use your answers in 5.3.
D
24
20
16
C
Unit 5 Triangles with proportional sides and similarity
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6
7
AC = 41 units, BC = 40 units, EF = 27 units
and ED = 120 units.
6.1
Prove △ABC ||| △FED.
Area △ABC
6.2
Determine __________
Area △FED
6.3
Determine the length of:
6.3.1
GH
6.3.2
JK
6.3.3
KH
6.3.4
JG.
6.4
Determine the area of JGFE.
PT = 9 units, QT = 12 units,
PQ = 15 units, TR = 16 units and
QR = 20 units.
F
A
7.4
7.5
8
9
120
B
C
40
H
K
D
P
9
T
16
12
Prove that △PQR ||| △PTQ ||| QTR.
Prove that △PTQ is a right-angled triangle.
20
Name two other right angles and justify your
Q
answer.
Determine Area △PQR:Area△PTQ:AreaQTR.
If a circle is drawn through points P, Q and R
which side of△PQR will be a diameter of the circle. Justify your answer.
BC = 12 units, AB = 18 units, AC = 22 units,
A
AD = 27 units and BD = 33 units.
8.1
Which triangle is similar to △ABC? Fully
18
22
justify your answer.
8.2
Is △ABC a right-angled triangle. Fully
B
E
justify your answer.
Area
△ABC
12
__________
8.3
Determine
Area △ABD
C
8.4
Determine, correct to one decimal place,
^.
the magnitude (size) of D
8.5
Is ABCD a cyclic quadrilateral? Fully justify your answer.
8.6
Is AD || BC? Fully justify your answer.
DH = 2 units, GH = 3 units, DG = 4 units, FH = 8 units,
EH = 12 units and EF = 16 units.
9.1
Prove that DEFG is a trapezium.
9.2
Is DEFG a cyclic quadrilateral?
Justify your answer.
R
27
D
33
D
G
4
3
2
H
8
12
E
234
J
G
41
15
7.1
7.2
7.3
E
27
16
F
Topic 10 Euclidean geometry
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Unit 6: Pythagoras’ Theorem and similarity
Theorem: The perpendicular line drawn from the vertex of the right angle of a right-angled
triangle to the hypotenuse divides the triangle into two triangles that are similar to each other and
similar to the original triangle.
^ R = 90° and PT ⊥ QR
Given: △ PQR with QP
x
90
°–
x
P
x
90° – x
T
Q
R
To prove: △ PQR ||| △T PR ||| △ TQP
^ =x
Proof: Let Q
^ P = 90° − x
∴ QR
| ∠ sum △ PQR
^ T = 90° − x
∴ QP
^T = x
∴
RP
| ∠ sum △ PQT
| Adjacent complementary
In △PQR and △TP R and △TQP
^T = x
^R =
^T =
RP
PQ
PQ
^P =
QR
^R =
QP
^P =
TR
^R =
PT
^ T = 90° − x
QP
^ P = 90°
QT
△PQR ||| △TPR ||| △TQP
| Proved
| Proved
| Given
| AAA
Pythagoras’s theorem states that the square on the hypotenuse is equal to the sum of the squares
on the other two sides.
We can prove this theorem by using △ PQR ||| △TPR ||| △ TQP as follows:
PQ
QR
PR = ____
___
= ___
PR
| △PQR ||| △TPR
PQ
QR
PR = ____
___
= ___
| △PQR ||| △TQP
TP
TR
⇒ PR2 = TR × QR
TQ
TP
QP
➀
⇒ PQ2 = TQ × QR
➁
PR2 + PQ2 = TR × QR + TQ × QR
| from ➀ and ➁
= QR( TR + TQ )
= QR2
In any right-angled triangle, the line drawn through the
right angle, perpendicular to the hypotenuse, subdivides
the hypotenuse in such a way that the perpendicular
distance squared is equal to the product of the two lengths
of the subdivided hypotenuse. In this sketch, PT2 = QT.TR. Q
TP = ___
TR = ___
PR ⇒ PT2 = QT.TR
___
TQ
TP
QP
P
T
R
| △T PR ||| △ TQP
Unit 6 Pythagoras’ Theorem and similarity
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WORKED EXAMPLE
AC is a diameter and BCD is a tangent. AED is a straight line with E on the circle.
1
Prove that CE2 = AE.ED
2
Prove that CD2 = AD.ED
A
O
1
E
2
1
2 3
C
B
D
SOLUTION
1
2
^ D = 90°
AC
^ = 90°
E
| Radius OC ⊥ tangent CD
|
∠ on diameter AC
1
2
CE = AE.ED
| Right-angled △ACD with CE ⊥ AD
△ACD ||| △CED | Right-angled △ACD with CE ⊥ AD
CD
AD = ___
___
| △ACD ||| △CED
CD
This is also known as the
tangent/secant theorem
ED
CD2 = AD.ED
EXERCISE 6
1
2
^ C = 90ο, BD ⊥ AC and AF || DE. CE = 64 units,
AB
EF = 36 units and DA = 90 units.
Determine, with reasons, the length of:
1.1
CD
1.2
BD
1.3
FB
1.4
AF
1.5
ED
1.6
DG
1.7
GB
1.8
FG.
C
PR is a chord of the circle with centre O.
Diameter TS is perpendicular to PR at V.
TV = 4VS and TS = 10 units.
2.1
Determine, with reasons,
the length of PR:
2.1.1
using similarity
2.1.2
using Pythagoras’ Theorem.
2.2
What type of quadrilateral is OPSR?
Give a reason for your answer.
2.3
Is OPSR a cyclic quadrilateral? Justify
your answer.
A
90
D
G
64
E 36 F
B
T
O
P
R
V
S
236
Topic 10 Euclidean geometry
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Revision Test Topic 10
Total marks: 160
Clearly state your reasons in all questions.
1
P
T
W
V
Z
R
Q
PW = __
1 and ___
PV = __
2
TV || WR , ____
WQ
1.1
1.2
1.2.1
1.2.2
1.2.3
2
2
VR
3
If PT = 2x, determine TW in terms of x.
Then, or otherwise, calculate the numerical value of:
(3)
QZ
___
(3)
ZV
Area
△PQV
__________
Area △PQR
Area △TQV
__________
Area △PQR
(3)
(5)
BEH is a common tangent to both circles. BAG is a tangent to the smaller circle.
^ =x
BCD, FEC and AED are straight lines and D
D
H
x
G
F
E
3 4
2
5
1 6
1 2
A 3
1
2 C
1 2
2.1
2.2
2.3
2.4
2.5
Name three other angles equal to x.
Prove that △FEA ||| △CED.
Prove that FE.ED = CE.EA.
Prove that BAEC is a cyclic quadrilateral.
^ = AB
^ C.
Prove that A
1
B
(5)
(4)
(2)
(5)
(2)
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REVISION TEST TOPIC 10 CONTINUED
3
A, B, C and D lie on the circle with centre O.
FDG is a tangent to the circle and FDG || AEC.
^ = x.
AC and BD intersect at E, GC ⊥ DC and B
3.1
Determine FOUR other angles equal to x.
3.2
Prove that BD bisects AC.
CD = ____
DG .
3.3
Prove that ___
3.4
3.5
4.
5
EA
1
1
C
O
(3)
x
(6)
B
The circles intersect at A and E. B and F lie on the circumference of the smaller
circle and C and D lie on the circumference of the larger circle. BFD and CFE are
straight lines.
^ B = x and ED
^ B = y.
CD
4.1
Determine, in terms of x and y:
^
4.1.1
A
(2)
^
4.1.2
F4
(2)
4.1.3
F2.
(2)
4.2
Prove that CD is a tangent to circle EFD.
(4)
4.3
Prove that △CDF ||| △CED.
(4)
2
4.4
Prove that CD = CE.CF.
(3)
PMRN is a cyclic quadrilateral. MN is the diameter of the circle and
PT ⊥ MN. MN and PQ intersect at T. O is the centre of the circle, MTN,
MSR and PTSQ are straight lines.
5.1
Prove that TSRN is a cyclic quadrilateral.
(5)
P
^ .
^ =Q
5.2
Prove that P
(4)
2
1
2
1
5.3
Prove that △PSM ||| △RSQ.
(4)
5.4
Prove that PS.SQ = RS.SM.
(2)
5.5
Prove that MP2 = MT.MN.
(5)
5.6
Prove that MP is a tangent to the circle through P, T and N.
(3)
5.7
Prove that △MTS ||| △MRN.
(5) M 1 2
3
2
5.8
Prove that MP = MR.MS.
(3)
MT.TN .
5.9
Prove that PT = ______
(4)
TQ
5.12
2
1
2 3
1 4E
(5)
AD
Area △MRN
Area △MSQ
Determine ___________
Area △MST
4
2 3
2
A
(8)
(4)
Prove that CD2 = EC.DG.
If AE = 12 units and AE = 2DE,
determine the length of BE.
5.10 If MN 75 units, TN = 48 units and NR = 21:
5.10.1 determine the length of PT
5.10.2 determine the length of MR
5.10.3 determine TS : SQ
5.10.4 determine MS : SR
5.10.5 is △PMN ||| △RNM? Justify your answer.
Area △MTS
5.11 Determine ___________
G
D
F
A
E
2
B
1
2
F
1
4
3
y
2
D
x
1 2
C
N
1
2
O
T
2 3
1 4
1
4 3
1 2
2
S
2
(3)
1
R
1
Q
(2)
(5)
(5)
(3)
(4)
(4)
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REVISION TEST TOPIC 10 CONTINUED
6
^ D. BED and AEC are straight lines.
BA || EF || CD and EF bisects AE
A
1 2
1
B
F
2
E1 2 3
2
1
1
2
5 4
2
D
1
C
6.1
6.2
^ = x, find five other angles equal to x.
If A
1
Prove that:
AF = ___
BE
___
6.2.1
6.2.2
6.2.3
7
FD
EC
EF
FD
(10)
(4)
ABCD is a cyclic quadrilateral
BC .
AB = ___
___
(2)
(7)
O is the centre of the circle. COD, BED and AEO are straight lines. B, C and D lie
on the circle and AO ⊥ CD.
7.1
Prove that:
7.1.1
ABOD is a cyclic quadrilateral.
(5)
7.1.2
△AOC ||| △DBC
(4)
(2)
7.1.3
DC.OC = BC.AC
2
7.1.4
2OC = BC.AC
(2)
B 2
7.2
Draw AD and BO.
^ D.
1
7.2.1
Prove that AO bisects CA
(4)
7.2.2
State the relationship between AB
and AC if BO || AD.
A
3
4
(2)
E
2
1
1 2
C
D
O
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2
11
Statistics
Unit 1: Revision of symmetric and skewed
data
Skewed to the right
-
6-
4-
2-
-
-
-
-
0
2
4
6
8
mode
median
mean
Consider the histograms below:
20 -
30
10
x
-
27,5 25 15 -
10 -
0
100
200
300
400
Ratio
Median
12,5 10 7,5 5
Mean
Skewed to the right
500
600
-
-
2,5 0
45 46 47 48 49 50 51 52 53 54 55
–2,5 -
-
-
-
17,5 15 -
-
5-
22,5 20 -
-
Range = highest value –
lowest value
Interquartile range = upper
quartile – lower quartile
-
8-
-
REMEMBER
10 -
-
Skewedtotothe
theleft
lef t(or
Skewed
negatively skewed)
y
-
Skewedtotothe
theright
right(or
Skewed
positively skewed)
Skewed to the right
Although a box-and-whisker plot is useful
to show the spread of data, we cannot always
be sure whether data is symmetrical or skewed.
We must determine the mean of the data to
make sure. What is important is how the mean
Symmetrical
Skewed
to the right
compares
with the median. When
we have
perfect symmetry, the median (middle value),
mode (value with highest frequency) and
mean will all be the same, as shown in the
frequency polygon alongside.
Frequency
rical
Symmetrical
-
Skewed to the lef t
-
Symmetrical
Skewed data
values are more
spread out on one side than
on the other.
It is possible for three sets of data values to have the same range and the same
interquartile range, but for one to be skewed to the left, one to be symmetrical and
one to be skewed to the right. See the box-and-whisker plots below.
-
The median is the middle
value.
In Grade 11 you discovered that when you analyse statistical data, you get an idea of
how data values are distributed throughout the range. Symmetrical data values are
balanced on either side of the median, with the data being evenly spread on either
side as shown in the box-and-whisker diagram in the Remember box.
-
REMEMBER
Frequency
he lef t
TOPIC
Age
Mean
Median
Skewed to the left
We can conclude:
• If the mean = the median, then the data is symmetrical.
• If the mean > the median, then the data is skewed to the right.
• If the mean < the median, then the data is skewed to the left.
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Topic 11 Statistics
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WORKED EXAMPLE
The frequency table shows the results
achieved by Shivaan's class, in a
mathematics test that was out of 45.
Determine any relevant information
to discuss the distribution and
symmetry of the given data.
Percentage
obtained
Number of
learners
20 ≤ x < 30
12
30 ≤ x < 40
20
40 ≤ x < 50
14
50 ≤ x < 60
7
60 ≤ x < 70
10
70 ≤ x < 80
16
80 ≤ x < 90
20
90 ≤ x < 100
12
REMEMBER
SOLUTION
Determine the mean:
×
+
×
+
×
+
×
+
×
+
×
+
×
+
×
25 12 35 20 45 14 55 7 65 10 75 16 85 20 95 12
Mean = __________________________________________________________________
111
= 60,41
Note: This is continuous
Q3 + 1,5 IQR = 149,5
∴ no outliers
60 -
40 -
20 10
-
20°
Box-and-whisker plot:
-
∴ Q1 – 1,5 IQR = –30,5
-
-
∴ IQR = 45
80 -
-
Q3 = 82
-
-
Q2(median) = 61
-
-
Q1= 37
100
-
From this ogive curve we
can estimate:
Cumulative frequency
If we estimate the median score, we would know that
data, so the mid-class
value of the first interval
the middle value is the 56th value. The 56th value falls
1 (20 + 30).
is __
in the interval 60 ≤ x <70 because the cumulative
2
frequency to the end of the previous interval is 53.
If we want to be more accurate, we could draw an ogive curve:
To find the mean, multiply
each frequency by the
mid-class value and divide
this sum by the total
frequency. Or use the STATS
mode on the calculator,
but remember to have the
frequency option on (find this
via the set up key).
40°
60°
80°
100°
Q1
Q2
Q3
Percentage
The range of the data is
100 – 20 = 80. The results are widely spread and there are no outliers.
Using the calculator, we can determine that the standard deviation is 23,55, which
is fairly high. This means that it is not only a few values that are causing the range
to be large. Many data values deviate largely from the mean.
As the mean is almost equal to the median, we can say that the data is almost
symmetrical. We can also see this in the box-and-whisker plot. There are two modal
groups (30 ≤ x < 40 and 80 ≤ x < 90), which are fairly evenly distributed on either
side of the median. Thirty-two learners scored lower than 40 and 32 learners scored
higher that 80, creating symmetry on the highest and lowest parts of the range as
well. It is significant that, while the data is symmetrical, more than half the test
results (64 out of 111) were distributed far from the mean and median.
REMEMBER
Modal group is the group
with the highest frequency
Unit 1 Revision of symmetric and skewed data
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EXERCISE 1
1
y
Use the information given on the
histogram to discuss the distribution
and symmetry of the data given.
Determine any relevant information
that will assist in your discussion.
20
20 15 10 5-
10
6
5
4
3
2
1
-
-
9
0
2
5 10 15 20 25 30 35 40 45 50 55 60
x
Discuss and compare the distributions.
Data A
-
Data B
0
3
5
10
15
20
25
30
This data shows the weights of two groups of rugby players.
3.1
The weights in kilograms of the 1992 Springbok rugby players were:
78
3.2
78
85
88
89
90
92
93
101 105 106 106 108 112 112
3.1.1
Determine the mean weight of the players.
3.1.2
Determine the standard deviation of the weights of the players.
3.1.3
What percentage of players’ lies within 1 standard deviation of the
mean?
3.1.4
Draw a box-and-whisker diagram for the data.
A box-and-whisker diagram is given for the weights of the 2010 SA Schools
Rugby players.
–
–
–
–
–
–
–
–
–
–
The mean weight for these players is 97,66 kg and the standard deviation
is 11, 76.
70
75
80
85
90
95
100
105
110
115
120
Use the information above and your answers to 3.1 to discuss and compare
the weights of both teams with particular reference to their distribution
and symmetry.
242
Topic 11 Statistics
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Unit 2: Bivariate data: scatter plots,
correlation and regression lines
When we conduct a study on the relationship between two types of data we are
working with bivariate data (that is, ‘two variables’). For example, ice cream sales
compared with the temperature of the day.
It is very useful to display the bivariate data in graphical form. This is called a
scatter plot. The values of one variable are on the x-axis and the values of the other
variable are on the y-axis. If one variable is influencing or causing the other, we use
that variable on the x-axis (this would be the independent variable), with the other
variable (the dependent variable) on the y-axis. If there is a relationship between the
variables without a ‘cause and effect’, either variable can be on either axis.
The shape of the graph tells us a lot about the relationship between the two variables.
Ask these questions when you are analysing a scatter plot:
• Is there a positive or negative association?
positive association: y
as the one variable
positive association:
increases, so the
as the one variable
other variable also
increases, so the other
increases
y
KEY WORDS
bivariate data – two sets of
data values that both vary
scatter plot – the graphical
representation of bivariate
data in the form of points
plotted on a Cartesian plane
linear trend – to form a
pattern that closely follows a
straight line
negative association:
as the one variable
negative association:
increases, so the
as the one variable
other variable
increases, so the other
decreases.
variable also increases
variable decreases
x
x
• Is there a strong relationship or a weak relationship between the variables?
The above examples are both strong relationships, whereas the following examples
are weaker relationships, because the points are more scattered and do not follow as
clear a pattern:
y
y
x
x
• Does the relationship follow a linear trend or a non-linear trend?
The examples above all follow a linear trend, whereas the examples below do not.
y
exponential
y trend
y
x
x
parabolic trend
y
x
x
• Are there any outliers?
In this scatter plot most of the points show a strong positive
relationship between the two variables, except for the second
last point, which we can classify as an outlier.
y
x
Unit 2 Bivariate data: scatter plots, correlation and regression lines
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• Are there any groupings?
In this scatter plot the data values are grouped in
two groupings, with a gap between the groups.
y
x
WORKED EXAMPLE
Mr Naicker owns an ice-cream shop in Durban. He analysed his ice cream
sales over a randomly selected 14 days. He then compared his sales with the
temperature on the day of each sale. The results of his survey are shown in the
table below:
Temperature
in °C
Ice creams sold
per day
1
2
20
13
18
23
19
28
21
38
26
22
17
14
15
16
116
85
107
139
123
172
128
127
148
124
112
89
101
96
Draw a scatter plot to represent the relationship between the temperature
and the ice cream sales on each day for the two weeks given.
Discuss the relationship between these two sets of data. Say whether the
relationship is strong or weak, positive or negative, and whether there is a
linear trend or otherwise. Also say whether there is a possible outlier.
SOLUTIONS
1
Number of ice creams sold
KEY WORDS
correlation coefficient –
the measure of association
between two variables
negative correlation – the
measure that indicates that
the two variables move in
opposite directions, so as
one increases so the other
decreases
positive correlation – the
measure that indicates that
the two variables increase or
decrease together
244
y
200
150
100
50
0
10
20
30
40
x
Temperature in degrees Celsius
2
Besides the outlier at (38;127), the rest of the data points show a linear trend
with a strong positive relationship between the two variables.
We can further determine the strength of the relationship between two variables
by calculating the correlation coefficient. This value can vary between –1 (which
indicates a very strong negative correlation) and 1 (which indicates a very strong
positive correlation). If the correlation coefficient is zero, there is no relationship
between the two variables.
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We generally describe:
• a value of r = 1 as perfect positive correlation
• any value of 0,8 ≤ r < 1 as having a strong positive correlation
• any value of 0,5 ≤ r < 0,8 as having a moderately strong positive correlation
• any value of 0 < r < 0,5 as having a weak positive correlation
• a value of r = 0 as having no correlation
• any value of −0,5 < r < 0 as having a weak negative correlation
• any value of −0,8 < r ≤ −0,5 as having a moderately strong negative correlation
• any value of −1 < r ≤ −0,8 as having a strong negative correlation
• a value of r = −1 as perfect negative correlation.
Examples:
y
y
y
x
x
r = 1 (perfect line)
r =very
1 (perfect
(a
strongline)
positive
(a verycorrelation)
strong
r = –0,99
r = –0,99
(a strong
negative
correlation)
(a
strong negative
positive correlation)
correlation)
y
x
r = –0,09
r = 0,09
(as this
is close to zero,
almost
no
(as there
this is is
close
to zero,
correlation)
there
is almost no
x
r = –0,68
r = –0,68 strong
(a moderately
negative
correlation)
(a moderately
strong
negative correlation)
correlation)
y
y
x
It is meaningless to discuss a correlation
in each of these cases because we do not
have bivariate data. Bivariate data requires
two sets of data values to vary, and in
these cases one set of values remains
x
constant while the other varies.
When we have drawn a scatter plot, we can draw a line that best fits the data. This
line is called a regression line. The line is chosen so that it comes as close as possible
to the data points and passes through the mean values for each set of data. The
standard form equation of the line of regression is given as:
y = A + B × x where the regression parameters A and B are described below:
The regression coefficient, which is the
gradient of the regression line
A constant value, which is the y-intercept
of the regression line.
( standard deviation of x )
standard deviation of y
B = r × ____________________
The equation of the regression line is a useful way of predicting the value of one of
the variables based on given values of the other variable. If we substitute a value of
one variable into the equation of the regression line, we will determine the predicted
^ or ^y.
value of the other variable. We call this value x
The difference between this predicted value and the observed (actual) value is called
the residual. The regression line aims to minimise the sum of the squares between the
observed and the predicted values of y. This is why we also refer to the regression line
as the ‘least squares regression line’.
REMEMBER
_
_
x refers to the mean of the set
of x values.
KEY WORDS
regression line – the ‘line of
best fit’ for a set of plotted
data points; also called the
line of least squares
residuals – the deviations
from the line of best fit, that
is: y − ^y
interpolation – using a
regression equation to predict
values within the data range
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KEY WORDS
extrapolation – using a
regression equation to predict
values outside of the data
range
When we substitute values of the variable (x) that are within the range of observed
values, into the regression line to predict values for the other variable (y), we call this
interpolation. This should provide us with a reasonably accurate prediction. When
we substitute values of the variable x, which are outside of the range of observed
values, into the regression line to predict values for the other variable (y), we call this
extrapolation; the result is often inappropriate or not creditable.
Calculate the values of A and B using the STAT option on a calculator as follows:
• Go to MODE: choose the STAT option
• Choose the option A + Bx: You will now have a table on your screen.
• Type in each of the data values, pressing = after each entry.
• Press AC
• Go to SHIFT 1 (that is, the other STAT option)
• Choose the Reg option
• Choose the A option, followed by =
• Press AC
• Go to SHIFT 1 (that is, the other STAT option)
• Choose the Reg option
• Choose the B option, followed by =
WORKED EXAMPLE
1
Use the values given in the previous worked example (the analysis of Mr
Naicker's ice cream sales compared with temperature) to draw a regression
line to best fit the data. Determine the equation of this line by
1.1
including the outlier
1.2
removing the outlier.
Using the regression line which was calculated by removing the outlier,
determine and comment on the number of ice creams that the shop owner
predicts he will sell on a day when the temperature is
2.1
24 °C
2.2
5 °C
2
SOLUTIONS
Number of ice creams sold
1.1
The outlier causes the
gradient of the best-fit line
to be less steep than it would
otherwise have been, and
thus it does not best fit the
trend of the rest of the data
points.
y = 2,5598x + 66,047
r = 0,7135
y
200
150
100
50
10
20
30
40
Temperature in degrees Celsius
Note: When a point lies far from the others in
a horizontal direction, it is called an ‘influential
observation’.
246
Using a calculator, we find
x that the parameter values for
this regression line are:
A = 66,047 and B = 2,5598
∴ the equation of the
regression line is:
y = 2,5598x + 66,047
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1.2
Number of ice creams
sold
y
With the outlier removed, the
regression line fits the trend
of the rest of the data more
accurately.
y = 5,7625x + 6,2742
r = 0,9756
Using a calculator, we find
that the parameter values for
this regression line are:
A = 16,222 and B = 5,2742
Temperature in degrees Celsius
x
∴ the equation of the
regression line is:
y = 5,2742x + 16,222
2.1 Substitute x = 24 into y = 5,2742x + 16,222
y = 5,2742( 24 ) + 16,222 = 142,8
∴ it is predicted that he will sell 143 ice creams on the day when the
temperature is 24 °C.
All other variables would have to remain constant (such as the fact that ice
cream sales are often seasonal and are dependent on holidays and not only
weather. So we would have to be comparing days that are all either in holiday
season or all out of season). If we have taken this into account, then we could
take this prediction as reasonably accurate because:
• we can see that the data points closely fit the regression line
• the correlation coefficient calculated previously for this set of data (0,9756)
shows a strong positive correlation
• the value substituted into the regression line was within the known data
range (interpolation).
However, because there are only 13 data values taken into consideration in
this situation, it would not be accurate to make too many assumptions with
regard to the trend. See the note on the next page.
2.2 Substitute x = 5 into y = 5,2742x + 16,222
y = 5,2742( 5 ) + 16,222 = 42,593
∴ it is predicted that he will sell 43 ice creams on the day when the
temperature is 5 °C.
Do not take this as an accurate or even creditable prediction, because the
value substituted into the regression line was outside the known data range
(extrapolation). There could be other factors that make the regression line no
longer valid for data values outside of the known range. For example, it could
be that when the temperature drops below a certain temperature, there is a far
smaller probability of people choosing to buy an ice cream.
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Note:
• To make truly valid comments about the correlation between bivariate data values,
or to make accurate predictions based on a regression line, we require a large set
of data values. However, for learning and practising the techniques, we restrict the
number of data values in Exercise 2.
• Any predictions for the future based on a set of bivariate data where the
independent variable involves time, will involve extrapolation and will therefore
be questionable. Many things may change over time to alter the trend that was
present before.
We now consider the aspect of residuals in more detail to help us determine whether
the trend of a set of bivariate data is linear or not.
In this diagram the plotted points show
no pattern of being above or below the
regression line. The residual values will be
randomly positive and negative in no clear
order. This will always be the case when the
scatter plot suggests a linear trend.
y
x
Note: When drawing a regression line onto
a scatter plot it is not easy to be perfectly
accurate, nor is it essential. If you draw a close
approximation of the line of best fit, it should give
an indication on whether plotted points lie above
or below the line.
Now consider the scatter plots in the next two figures. The plotted points show a
non-linear trend as shown with the green curve. The purple line shows the linear
regression line in each case.
In each diagram the vertical distance between the plotted point and the regression
line represents the residual (that is, y − ^y), where y is from the plotted point and ^y is
from the point on the regression line. So, if the plotted point is above the regression
line then y − ^y will be positive, and if the plotted point is below the regression line
then y − ^y will be negative.
y
x
248
In this diagram the plotted points are first
below, then above, and then below the
regression line. This means that the residual
values will be negative, then positive and
then negative again. This suggests that the
scatter plot will show a parabolic trend (as
shown by the green curve). The opposite
would have been true if the plotted points
had suggested a negative parabola. The
residuals would have been negative for the
central values and positive at the high and
low ends.
Topic 11 Statistics
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In this diagram the plotted points are first
above, then below, and then above the
regression line. This means that the residual
values will be positive, then negative and
then positive again. This suggests that the
scatter plot will show an exponential trend
(as shown by the green curve). The opposite
would have been true if the plotted points
had suggested exponential decay.
y
x
In general, if there is a random nature to the residual values (positive and negative)
with no apparent pattern, this indicates that a linear trend is appropriate for the
bivariate data. If there is a pattern to the residual values, then it indicates that a
linear model would not be appropriate. The trend that exists between the variables
is possibly exponential or parabolic (or another trend). We would have to look more
carefully at a larger sample of data points to decide on the suggested trend of the
points if a linear trend does not seem appropriate.
EXERCISE 2
1
For the following sets of bivariate data, use your calculator to determine:
1.1
the value of r, the correlation coefficient, and what this indicates about
the relationship between the variables
1.2
the regression parameters and the equation of the regression line
1.3
the predicted value required, and whether it is interpolation or
extrapolation.
A
x
y
10
B
x
y
180
11
20
160
30
C
x
y
35
8
19
22
71
16
68
140
29
82
25
28
40
120
38
148
37
52
50
100
53
149
55
56
60
80
59
188
59
97
70
60
75
259
81
24
80
40
82
286
94
78
Predict y when x = 150
Predict y when x = 67
Predict y when x = 6
Unit 2 Bivariate data: scatter plots, correlation and regression lines
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D
Predict y when x = 10.
60 50 40 30 20 10 -
E
-
-
-
-
-
-
-
-
-
0
10
20
30
40
50
60
70
Predict y when x = 100.
90 80 70 60 50 40 30 20 10 -
F
2
-
-
-
-
-
-
-
0
20
40
60
80
100
x
20,2
35,5
41,7
52,9
63,4
75,1
89,3
98,0
28
y
32
200
10
135
89
100
225
150
predict
Candice's Grade 11 Mathematics class wrote two tests. The results achieved by
10 learners are recorded in a table as percentages.
Learner 1
2
3
4
5
6
7
8
9
10
Test 1
92
68
58
47
84
65
59
53
67
75
Test 2
89
70
55
50
80
63
62
57
61
78
2.1
2.2
2.3
2.4
2.5
2.6
250
Draw a scatter plot to represent the relationship between these two sets
of data.
Determine the correlation coefficient for this set of bivariate data.
Discuss the relationship between the results of these two tests.
Show the regression line on the diagram of the scatter plot. Discuss
whether a linear model is appropriate for this set of bivariate data.
Determine the equation of the regression line.
A Grade 11 learner, Candice, wrote the first test and her result was
76%. She missed the second test. Predict the result that she would have
achieved, and discuss whether this is an accurate prediction or not.
Topic 11 Statistics
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3
Nomakhushe found a chart showing her height in cm from 1 year to 3 years’ old,
recorded every 3 months, as shown below:
Age in months
12
15
18
21
24
27
30
33
36
Height/Length in cm
76
79
82
85
87
89
92
94
96
3.1
3.2
3.3
3.4
3.5
3.6
Draw a scatter plot to show her growth over time, using her age as the
independent variable. Start the x-axis at age 0 (that is, at birth), although the
first point to plot is (12;76). Continue your scale to the age of 48 months.
Use your calculator to determine the values of:
3.2.1
the correlation coefficient
3.2.2
the regression parameters.
Write the equation of the regression line, and use it to predict Rabia’s
height when she was 26 months old.
Nomakhushe was interested to know her length at birth. What would her
predicted birth length be if she used the equation of the regression line?
Why would this not be an accurate prediction of her birth length?
Nomakhushe then found a more detailed chart, which included with her
length in the first and 4th years. The additional information is shown
below:
Age in months
birth
3
6
9
39
42
45
48
Height/Length in cm
50
61
67
72
98
100
101
103
4
Add these points to the scatter plot drawn in 3.1.
3.7
What trend seems evident in the relationship between age and height
when considering the time period from birth to 48 months? Show on the
scatter plot how the residual values could assist in making this decision.
Scatter plots are drawn for each of the two rugby teams referred to in Exercise 1,
Question 3, page 242.
1992 Springboks: Height versus weight
120 –-110 –-100 –-90 –-80 –-70 –-60 –
–
–
–
–
–
–
–
Weight (kg)
y
150
160
170
180
190
200
x
210
Height (cm)
Unit 2 Bivariate data: scatter plots, correlation and regression lines
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2010 SA Schools Players: Height versus weight
120 –-110 –-100 –-90 –-–
80 -–
70 -60 –
–
–
–
–
–
–
–
Weight (kg)
y
150
160
170
180
190
200
x
210
Height (cm)
4.1
5
Which team appears to have a better correlation between height and
weight? Give a reason for your answer.
4.2
Draw the line of best fit for each team.
4.3
If reserve players are brought into the teams, use the lines of best fit to
estimate:
4.3.1
the weight of a 1992 Springbok reserve player with a height of
195 cm.
4.3.2
the height of a 2010 SAS reserve player with a weight of 100 kg.
Mr Kerbelker was not happy with some of the test results of his class. He decided
to do a survey to find out how much time they had spent watching TV each day.
The results of his survey are shown below, together with the test result of each
student.
Time spent watching
TV (in hours)
5.1
5.2
5.3
5.4
252
Test results (%)
3
49
1,5
78
3
50
1
72
2,5
63
3,5
47
0,5
83
4
48
2
75
5
36
Draw a scatter plot to show the relationship between these two variables.
Determine the equation of the line of best fit.
Determine the correlation coefficient for this set of data values (to 2
decimal places).
Describe the correlation between the time spent watching TV and the
results obtained.
Topic 11 Statistics
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Revision Test Topic 11
Total marks: 100
1
The results, out of 50, for two tests written by a Grade 12 class are given below:
Test A
39
33
35
44
37
40
24
31
30
5
42
12
46
16
35
Test B
41
45
48
40
47
42
37
44
43
24
39
45
49
43
48
1.1
Use a calculator to determine for each test, the value of:
1.1.1
the mean
1.1.2
the standard deviation.
Determine the five-number summary for each set of values. Then
draw a box-and-whisker plot for the results of each test.
Discuss the distribution of each set of test results.
Determine the correlation coefficient; and hence describe the
relationship between the two test results.
Draw a scatter plot and then discuss the relationship between the
results of the two tests.
Determine the equation of the regression line for this set of
bivariate data.
1.2
1.3
1.4
1.5
1.6
2
(4)
(4)
(12)
(3)
(2)
(4)
(2)
Four histograms are given below.
2.1 y
2.2
y
x
2.3
x
2.4
y
y
x
x
Match the correct box-and-whisker plot to each of the histograms above. State
whether the data is symmetrical, skewed to the right, or skewed to the left.
A
B
C
D
(2 × 4)
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Revision Test Topic 11
3
The heights of 90 randomly selected learners at a school were recorded.
3.1
3.2
3.3
120 < x ≤ 135
1
135 < x ≤ 150
15
150 < x ≤ 165
45
165 < x ≤ 180
28
180 < x ≤ 195
1
(3)
(4)
(5)
(3)
A company analyses the relationship between expenditure on advertising their
product and their sales over the previous 6 years.
4.1
Year
Advertising expenditure
(in Rand)
Annual sales figures
(in Rand)
1
21 300
221 300
2
23 800
248 200
3
19 800
204 150
4
29 540
242 530
5
40 600
327 500
6
24 100
235 180
Draw a scatter plot to represent the relationship between these two sets
of data.
(4)
Determine the correlation coefficient, and discuss what this tells us about
the correlation between the advertising expenditure and sales figures of
the company.
(2)
Discuss the trends that the company should notice with regard to the
relationship between money spent on advertising and sales figures.
(3)
Discuss whether a linear model is appropriate for this set of bivariate data. (2)
Determine the equation of the regression line.
(2)
The company would like to predict their sales figures for the next year if
they plan to spend R35 900 on advertising. Determine the predicted sales
figures based on the line of regression.
(2)
Would the result in 4.6 have made use of extrapolation or interpolation?
Then discuss whether the result should be considered accurate or not. (2)
4.2
4.3
4.4
4.5
4.6
4.7
5
Number of learners
Determine the mean height of this group of learners.
Draw an ogive curve to represent this data.
Use your ogive curve to estimate the median height and the
interquartile range.
Discuss the distribution and symmetry of the data.
3.4
4
Height in cm
A group of environmentalists monitored the levels of oil in a river after a spillage
as they cleaned up. The results shown in the table represent the percentage of oil
in the water as each day after the spillage passed:
Days
0
1
% oil
0,05
0,043 0,032 0,029 0,023 0,019 0,016 0,014 0,012 0,01
5.1
2
3
4
5
6
Draw a scatter plot to represent this data.
7
8
9
10
0,007
(4)
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5.2
5.3
5.4
Number of days since drug administered
5
10
15
20
25
30
Number of mice still affected by the virus
219
95
28
2
24
102
100 90 80 70 60 50 40 30 20 10 -
-
-
-
-
Determine:
7.1
an outlier
7.2
the equation of the regression line:
7.2.1
including the outlier
7.2.2
excluding the outlier
7.3
the correlation coefficient:
7.3.1
including the outlier
7.3.2
excluding the outlier.
7.4
Discuss the relationship between this set of
bivariate data.
-
-
Refer to the scatter plot and answer the questions
that follow.
-
6.4
Draw a scatter plot to represent this data.
(4)
Does this seem to follow a linear, exponential or parabolic trend? Suggest a
reason for why this might be the case.
(1)
Determine the equation of the regression line, and use this to confirm
your answer to Question 6.2.
(2)
Determine the correlation coefficient, and describe what this tells us
about the relationship between the drug being administered and
the virus surviving.
-
6.3
8
(4)
A group of scientist are studying the effects of a certain drug that is being tested
on mice who have a particular virus. Their results are recorded below:
6.1
6.2
7
(1)
-
6
Does this data seem to follow a linear, exponential or parabolic trend?
Determine the equation of the regression line, and use this to confirm
your answer to Question 5.2.
Determine the correlation coefficient for this set of data, and describe
the correlation between the number of days that have passed after the
spillage and the percentage of oil in the river.
0
10
20
30
40
50
60
70
(1)
(2)
(2)
(2)
(2)
(4)
A group of learners were asked to complete a form showing their height and
their most recent mathematics test result. The information gathered is shown in
the table below:
Height (cm)
159
145
152
163
149
156
168
155
150
Mathematics result (%)
75
95
52
48
82
84
98
74
59
Determine the correlation coefficient for this set of data values, and then discuss
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TOPIC
2
12
Counting and probability
Unit 1: Revision of rules for independent,
mutually exclusive and complementary
events
Note: P(A′) means the
probability of A not
happening, so
P(A′ ) = 1 – P(A) implies
P(A′) + P(A) = 1
which means that the
probability of an event not
happening, together with
the probability of the event
happening, is a total of 1.
(which is 100%).
Note: The symbol ⋂
represents ‘intersection’, and
implies the word ‘and’.
Note: The symbol ⋃
represents ‘union’, and
implies the word ‘or’.
Remember these important facts and notations that you learnt in Grade 11:
the number of ways an event can occur
• The probability of an event = ____________________________________________
the total number of possible outcomes for the event
• Probability based on experimental or observed data is called empirical probability.
• Probabilities range in value from 0 (impossible) to 1 (certain to occur).
∴ P(A′ ) = 1 – P(A)
• The sample space is all of the possible outcomes for an event.
• Notations to remember are:
– P(A) means__ ‘the probability of event A happening’.
– P(A′) or P(A) means ‘the probability of event A not happening’.
– P(A ⋂ B) means ‘the probability of event A and event B happening’
– P(A ⋃ B) means ‘the probability of event A or event B happening’
– P(A | B) means ‘the probability of event A happening, assuming that event B
is taken as given.
– n(A) means the number of possibilities of A happening.
• Events A and B are complementary events if P(A) = 1 – P(B).
This implies that P(not A) = P(B)
• Events are mutually exclusive if there is no overlap or intersection between
the two events, that is, P(A and B) = 0.
• Events are exhaustive if together they use up the whole sample space, that is,
there are no options in the sample space that do not belong to either event A or
event B, or both.
• Two events are independent when one result does not affect the result of the other.
The product rule for independent events is P(A and B) = P(A) × P(B).
When events are not independent, then P(A and B) = P(A) × P(B | A)
• The identity P(A or B) = P(A) + P(B) – P(A and B) applies to all events.
When events A and B are mutually exclusive P(A and B) = 0, which leads to the
addition rule for mutually exclusive events, P(A or B) = P(A) + P(B).
• Complementary events will be both mutually exclusive and exhaustive.
Conditional probability is for
enrichment purposes only
and is not examinable.
If events A and B are complementary, then P(A) + P(B) = 1.
• Conditional probability occurs when a condition is given that restricts the sample
from which you are choosing. In such a case, when the full sample space is not
being used, the probability of the event
the number of ways an event can occur
= _____________________________________________________________
the total number of possible outcomes that satisfy the conditions given
P(A ∩ B)
so P (B | A) = ________
P(A)
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Topic 12 Counting and probability
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These techniques are used to assist with problem solving:
• Venn diagrams:
We use circles to represent each event in Venn diagrams. If it is possible for both
events to take place, then the circles will intersect, with the area enclosed in the
overlap representing when the two events both happened. The circles are placed
inside a rectangle, which represents the sample space. This technique is a useful
method of providing a clear visual picture of each event, both individually, and how
they relate to each other. It is not necessary to be restricted to two events only.
You can enter data quantities or probability values into the various parts of a Venn
diagram. (For probability values the total value for the sample space will be 1). When
you enter data into the Venn diagram, start with the intersection values (use x for this
if it is unknown at that stage). Use the intersection to calculate the values in the rest
of the areas of the Venn diagram. If necessary, solve for x, knowing that the values of
each section must total to the sample space.
WORKED EXAMPLE
1
Sellmove estate agency has 31 clients to find suitable properties for. It also
has 23 clients whose properties they must sell. The company has 42 clients
altogether who fall into one or other of these categories. Draw a Venn
diagram and then answer the questions:
1.1
How many clients is the company both trying to find a suitable
property for and trying to sell for?
1.2
What is the probability that one of the clients for whom the company
is trying to find a suitable property is also one for whom they are trying
to sell a property?
SOLUTION
We can draw a Venn diagram where one circle
represents the clients who have properties to sell (S),
and the other circle represents clients who are looking
to buy a property (B).
1.1 We can now solve for x:
31 − x + x + 23 − x = 42
∴ x = 12
There are 12 clients for whom
the company is both trying to
find a property to buy and to
sell a property.
42
S
B
31 – x
x
23 – x
Note: As there are 31 buyers, but x of those are also
sellers, this means that 31 – x are buyers only. Similarly,
there are 23 – x who are sellers only.
1.2 We need to find the probability of the client being a seller as well as a buyer,
so:
12
2
P(S ∩ B) = ___
= __
7
42
Unit 1 Revision of rules for independent, mutually exclusive and complementary events
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• Tree diagrams:
When more than one event takes place consecutively or simultaneously, it is sometimes
useful to represent them as a tree diagram. Each event is represented by a column of
branches. The number of branches is determined by the number of possible outcomes
for that event. When the probability of each branch is equally weighted, each outcome
number of successful paths
from each path will be equally weighted. Probabilities will = _______________________ .
total number of paths
When the probabilities of each branch are not equal, the probability of any outcome
is the sum of each successful path. The probability of each successful path is
determined by multiplying the probabilities of the branches along the path. Tree
diagrams are also useful when the events are not independent.
WORKED EXAMPLE 1
Judy and her friends decide to take a break from studying and go for lunch. Judy
goes to a restaurant and has a choice of:
Starter: Soup or salad
Main course: Beef, pork or chicken
Dessert: Ice cream or trifle
If she is equally likely to choose any option, draw a tree diagram and determine
the probability that she will have ice cream and not beef.
SOLUTION
Beef
Soup
Chicken
Pork
Beef
Salad
Chicken
Pork
Ice cream
Trifle
Ice cream
Trifle
Ice cream
Trifle
Ice cream
Trifle
Ice cream
Trifle
Ice cream
Trifle
SoBIc
SoBT
SoCIc
SoCT
SoPIc
SoPT
SaBIc
SaBT
SaCIc
SaCT
SaPIc
SaPT
Key:
where
So = soup
Sa = salad
B = beef
C = chicken
P = pork
Ic = ice cream
T = trifle
The tree diagram shows all the options. The blue pathways (labelled SoCIc; SoPIc;
SaCIc and SaPIc) show the options that satisfy the criteria of the question.
REMEMBER
This means that the outcome
of the second draw will
depend on the first, so these
events are not independent.
Notice that the probabilities
of each colour being drawn in
the second draw is different
to the probabilities in the
first draw.
258
Having ice cream and not beef is 4 out of the possible 12 options, so the
4
1
= __
. (In this example each branch on each level is equally
probability is ___
12
3
weighted.)
WORKED EXAMPLE 2
A marble is drawn from a bag containing 5 blue, 3 red and 8 green marbles. A
second marble is then drawn, without replacing the first. Draw a tree diagram
and determine the probability that:
2.1 the marbles drawn will be the same colour
2.2 the marbles drawn will not both be green.
Topic 12 Counting and probability
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SOLUTION
4
15
Blue
3
15
8
15
5
16
3
16
5
15
Red
8
16
5
15
Green
2
15
8
15
3
15
7
15
Blue
Red
Green
Blue
BB 2.1 P(BB or RR or GG)
3
8
5
4
2
7
× ___ + ___
× ___ + ___
× ___
= ___
BR
16 15
16 15
16 15
41
BG
= ____
) (
(
) (
)
120
RB
Green
Blue
2.2 P(not GG) = 1 – P(GG)
RR
23
8
7
= 1 – ___
× ___
= ___
16
15
30
RG
GB Note: As there would be too many possible paths
Red
GR
Green
GG
Red
(
)
that satisfy this criteria, it is quicker to use the rule
of complementary events: P(not A) = 1 – P(A)
• Contingency tables:
These are statistical tables that represent the relationships between two or more
variables. The frequencies of each variable are shown in rows and columns.
WORKED EXAMPLE
Northcross High School has a total of 654 learners. Three hundred and thirty one
learners are boys. Nazamo's Grade 12 class organised a Charity Fun Run where
all proceeds would be donated to the Red Cross Children's Hospital. In total 602
learners attended the Fun Run, but 25 of the girls did not attend. All 654 learners
at the school were given a raffle ticket (no other raffle tickets were given out),
which was drawn on the day of the Fun Run. Draw a contingency table and
determine the probability that the winner of the raffle was:
1
at the Fun Run
2
a boy, given that the winner was not at the Fun Run.
SOLUTION
First enter the given data into
the table (those typed in red),
after which we can calculate the
missing data, knowing that the
totals of each column and row
must equal the values at the end
of each column and row.
Did attend Did not attend
Total
Boys
304
27
331
Girls
298
25
323
Total
602
52
654
1
There were a total of 602 learners at the Fun Run out of a possible
654 learners. So the probability of the raffle winner being at the Fun Run
602
is ____
≈ 92,05%.
654
2
There were 27 boys not at the Fun Run out of a total of 52 who did not
attend. So the probability that the winner was a boy, given that the winner
was not at the Fun Run
the number of ways an event can occur
= _____________________________________________________________
the total number of possible outcomes that satisfy the conditions given
27
= ___
52
Unit 1 Revision of rules for independent, mutually exclusive and complementary events
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EXERCISE 1
1
A spinner has four options: 1, 3, 5 or 9. After a fair spin,
determine the probability that the spinner will land on:
1.1
a prime number
1.2
an even number
1.3
a multiple of 5
1.4
a factor of 30.
5
9
1
2
A card is randomly drawn from a normal pack of 52 cards.
What is the probability that the card will be:
2.1
a queen
2.2
a red card
2.3
a picture card
2.4
a black king
2.5
a red card or a seven
2.6
not a jack.
3
A random sample of school children from Sunshine College was surveyed. The
results of the survey showed that the probability that a learner from the chosen
3
group came from a family where their parents were still married was __
(call this
4
event A). The probability that a learner from the group surveyed had at least one
sibling was 0,8 (call this event B), and the probability that a learner from the
group surveyed had lost a parent due to HIV Aids was 0,25 (call this event C). It
was further found that events A and C were mutually exclusive.
3.1
Are A and C are exhaustive?
3.2
Are B and C are complementary?
3.3
If A and B are independent, determine P(A ∩ B).
4
One hundred and eighty people attended a charity function in aid of the Nelson
Mandela Children’s Fund. Fifty percent of the people were women, and 10%
of the women were unmarried. There were 15 unmarried men. On arrival,
each person was given a ticket for a lucky draw. Draw a contingency table, and
determine the probability that the lucky winner will be a married man.
5
Pradesh has a choice of these items of clothing to wear to a certain event:
2 pairs of trousers (black or blue)
3 shirts (white, green or blue)
3 ties (striped, black or blue)
Draw a tree diagram. If he is equally likely to wear any of the items, determine
the probability that he will choose to wear:
5.1
the black trousers, white shirt and striped tie
60
5.2
a striped tie
C
5.3
nothing that is blue
13
17
5.4
trousers and tie of the same colour.
A
Given the Venn diagram alongside, determine:
16
B
5
6.1
P(A ∩ B)
6.2
P(B ∪ C)
6
6.3
6.5
6.7
7
260
3
P(C)′
P(B | A)
n(C′ ∪ B′)
6.4
6.6
6.8
n(A ∪ B ∪ C)′
P(C | A′)
n(A′| B)
9
In a school with 650 learners, the number of learners who play only soccer is
10 more than the number of learners who play both soccer and rugby. The number
of learners who play only rugby is twice the number of learners who play only
soccer. The number of learners who play neither rugby nor soccer is six times the
number of learners who play both soccer and rugby. Draw a Venn diagram, and
determine the number of learners who play neither rugby nor soccer.
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Unit 2: Use Venn diagrams, tree diagrams and
contingency tables to solve problems
When you have complex, wordy problems involving probabilities, it helps to
visualise the data as diagrams. Diagrams make the information easier to understand.
Using worked examples, we will discuss how to decide which type of diagram (Venn
diagram, tree diagram or contingency table) would be most useful in each situation.
WORKED EXAMPLE 1
Nadia did a survey where she asked a group of 85 learners whether they drank
fizzy cooldrink or fruit juice. Forty two said that they drank both, 21 said that they
drink fruit juice only, and
12 learners said that they drink neither.
1.1 What is the probability that a learner chosen at random drinks only fizzy
cooldrink?
1.2 Are the events ‘drinking fizzy cooldrink’ and ‘drinking fruit juice’
independent events?
SOLUTION
This example suits a Venn diagram well as there are two aspects to compare:
fizzy cooldrink (C) and fruit juice (J). We can represent each as a circle.
85
C
J
10
42
21
Fill in the intersection of 42 first, then
the 21 and 12 in the relevant areas.
We know that the total is 85, so we
subtract the total of 12, 42 and 21
from 85. This tells us that there were
10 learners who said they drink fizzy
cooldrink only.
12
10
2
1.1 ∴ P(C only) = ___
= ___
85
17
63
52
42 ___
1.2 P(C ∩ J) = ___
≠ 85 × ___
so not independent.
85
85
We could also have answered this question using a contingency table.
Fill in the red values first, then calculate the others by totalling each column and
row.
Note: These are
Fizzy
cooldrink
Not fizzy
cooldrink
Juice
42
21
63
Not juice
10
12
22
52
33
85
complementary
events, which is why a
contingency table or a
Venn diagram works well
in this example.
10
2
∴ P(C only) = ___ = ___
85
17
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WORKED EXAMPLE 2
Mrs Sachs asked her class of 30 learners how they researched information for their
history projects. Ten said they read books from the library, 25 said they had used
the internet, and 8 said they had interviewed people. Two said they had used
none of these three options. Three said they had interviewed people and used the
internet but not the library, and 7 said they had used the library and the internet.
One said he had used the library only. Determine the probability that a learner
chosen at random from the class had used all three options.
SOLUTION
30
A Venn diagram is the best option in
L
this example as there are three different
I
1
options. There is also much information
7–x
15
given that relates to intersections between
x
2
the three options. We will draw one circle
3
for library users (L), one for internet
3–x
users (I) and one for interviewing people
2
(P). There are intersections between all
P
three options, so we will draw the circles
intersecting. We start by filling in the
intersection of all three. As we do not know that value, we will call it x. Next fill
in the rest of the values shown in bold. Finally determine the rest of the values by
calculation, knowing the totals for each circle.
We can now create an equation because we know the total of the sample space to
solve for x:
1 + 7 − x + 15 + x + 2 + 3 + 3 − x + 2 = 30
∴x=3
3
1
= ___
So the probability of all three options = P(L ∩ I ∩ P) = ___
30
10
WORKED EXAMPLE 3
Three friends, Kate, Sechaba and Gatiep,
play two rounds of a computer game. From
past experience, the friends know that Kate
has a 30% chance of winning the game,
Sechaba has a 25% chance of winning
and Gatiep has a 45% chance of winning.
There can be only one winner each time.
Determine the probability that:
3.1 Sechaba wins both games
3.2 Kate wins neither of the games
3.3 Gatiep wins at least one game, and Sechaba did not win a game
3.4 Kate wins at least one game
3.5 Gatiep does not win both games.
262
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SOLUTION
0,3
As this example involves two events happening
consecutively with each event having more than
two options, a tree diagram would be the easiest way
of visualising the options. Each branch of the tree has
different weightings.
1
3.1 P(SS) = 0,25 × 0,25 =___
16
K
0,3
0,25 S
0,3
0,45
0,3
G
3.2 P(Kate neither) = P(SS or SG or GS or GG)
49
= (0,25 × 0,25) + (0,25 × 0,45) + (0,45 × 0,25) + (0,45 × 0,45) = ____
100
K
0,25 S
0,45
G
K
0,25 S
0,45
G
K
0,25 S
0,45
G
3.3 P(G and not S) = (0,3 × 0,45) + (0,45 × 0,3) + (0,45 × 0,45)
189
= ____
400
3.4 P(KK or KS or KG or SK or GK)
= ( 0,3 )2 + 2( 0,3 )( 0,25 ) + 2( 0,3 )( 0,45 )
51
= ____
100
3.5 P(not GG) = 1 – P(GG) = 1 – ( 0,45 )2
319
= ____
400
WORKED EXAMPLE 4
Researchers at Groote Schuur Hospital found that many people are carriers of a
certain disease. They tested 200 people and found that 5 of the 90 females tested
were carriers, and that 7 of the males tested were carriers. Determine:
4.1 the probability that a male from those tested is a carrier
4.2 the number of males in South Africa who are likely to be a carrier (assuming
that the population of South Africa is 50 million)
4.3 the probability that the person tested is a carrier, given that the person is
a female
4.4 the probability that the person tested is male, given that he is a carrier.
SOLUTION
You could use a contingency table for this example (because both the two factors
being considered, male/female and carrier/non-carrier are complementary events).
You could also use a tree diagram (because there are two events that are both
exhaustive, and happen simultaneously):
Fill in the red numbers first
Carrier
Not a carrier
Male
7
103
110
Female
5
85
90
12
188
200
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7
4.1 P(M ∩ C) = ____
200
7
4.2 ____
× 50 000 000 = 1 750 000
200
5
1
4.3 P(C | F) = ___
= ___
90
18
P(M ∩ C) ___
7
4.4 P(M | C) = ________
= 12
P(C)
OR:
Carrier
7
110
Male
110
200
90
200
103
110
Not
carrier
Carrier
5
90
Female
85
90
Not
carrier
Note: In a tree diagram when the first option
is given, ignore that probability. Look only at
the probability of the next branch.
110 ____
7
7
4.1 P(M ∩ C) = ____
×
= ____
200 110
200
7
4.2 ____
× 50 000 000 = 1 750 000
200
4.3 P(C | F)
5
1
= ___ = ___
90
18
110 ____
7
____
×
P(M and C) __________________
200 110
7
4.4 P(M | C) = __________
= 110
= ___
90
5
7
12
____ ____
____
___
P(C)
200
Note: P(C) has two
possibilities, each of
which must be reached
by following a path.
So each path must be
multiplied, and the
two options added.
× 110 + 200 × 90
The contingency table method is often easier when the
events are complementary as in this case: male/female
and carrier/not carrier. Tree diagrams are best used only
when there are more than two options.
WORKED EXAMPLE 5
Umfolozi School has 752 learners in Grades 8 to 12 and 387 are boys. There are
160 Grade 8s, 157 Grade 9s, 151 Grade 10s and 143 Grade 11s. Ninety Grade 8s and
73 Grade 11s are boys, while 82 Grade 9s and 78 Grade 10s are girls.
5.1 If a learner is selected randomly from the school, determine P(boy in Grade 10).
5.2 If a learner is selected randomly from Grade 8, determine P(boy)′.
5.3 If a learner is selected randomly from the school, determine P(Grade 9 | girl)
5.4 If a learner is selected randomly from the school, determine P(Grade 12)′.
SOLUTION
A contingency table will be most useful in this example. A Venn diagram is not
possible. A tree diagram is possible, but would require too many branches.
Grade 8
Grade 9
Male
90
75
73
73
76
387
Female
70
82
78
70
65
365
160
157
151
143
141
752
73
5.1 P(boy in Grade 10) = ____
752
82
5.3 P(Grade 9 | girl) = ____
365
264
Grade 10 Grade 11 Grade 12
70
7
5.2 P(M′| 8) = ____
= ___
160
16
13
5.4 P(Grade 12)′ = 1 – P(Grade 12) = ___
16
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EXERCISE 2
1
Ngwenya School’s soccer coach analyses the performance of his team after half
the season is over. He finds that if it rains, the probability that the team will win
is 0,4. If it does not rain, the probability that the team will win is 0,7. During the
month ahead, the probability that it will rain is 0,2. The coach considers various
outcomes of his team’s next match, which is in two weeks’ time. Determine the
probability that:
1.1
it will rain and his team will win
1.2
it will not rain
1.3
his team will win
1.4
his team will lose, given that it does not rain.
2
A cutlery drawer contains 12 knives, 10 forks and 8 spoons. If only the handles
are visible and Xhanti randomly selects three items without replacing them,
what is the probability that he:
2.1
selects a knife, fork and spoon
2.2
selects three of the same
2.3
selects three knives
2.4
selects a fork then two spoons
2.5
first selects two knives, and then does not select a fork.
3
A group of 100 people were asked which exercises they do to keep fit: running,
cycling or going to the gym. Fifteen said that they only run, 30 said that they go
to the gym only and 35 said that they do none of these options. The cyclists said
that they do not go to the gym or run, and none of those who run or go to the
gym said they go cycling. The same number of people who said they run and go
to the gym said that they go cycling.
3.1
Determine the probability that a person chosen randomly from the group
of 100:
3.1.1
runs and goes to the gym
3.1.2
does not cycle
3.1.3
goes to the gym, given that they run.
3.2
Prove that running and going to the gym are independent events.
Unit 2 Use Venn diagrams, tree diagrams and contingency tables to solve problems
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4
REMEMBER
__
A = A′
5
6
266
A survey was conducted on 58 women and 64 men. They were asked whether
they enjoyed fishing. Fifty people answered yes, and 15 men answered no.
4.1
Determine the probability that a person chosen randomly from the group
of 122 surveyed:
4.1.1
was a woman
4.1.2
had answered no
4.1.3
was a man who had answered yes
4.1.4
enjoyed fishing, given that she was a woman.
4.2
Are the events ‘being a man’ and ‘enjoying fishing’ independent events?
4.3
Are the events ‘being a woman’ and ‘enjoying fishing’ complementary
events?
Glen is a contestant in a lucky draw competition. Without seeing the numbers,
he randomly picks out a number from a box which contains 3 ones, 4 twos
and 2 threes. If he picks out a 1, then he is out of the lucky draw. If he picks
out anything other than a 1, he is allowed to pick another number. The process
continues three times without replacing the numbers drawn. If he is able to get
to the end of three rounds without picking out a 1, then he is a winner.
What is the probability that Glen will:
5.1
be out in the first round
5.2
be a winner
5.3
be a winner, given that he has made it past the first round
5.4
get to the third round?
Given two events A and B, such
that:
__
P(A only) = 2P(A ∩ B); P(B |A) = P(A ∩ B) + 0,3;
n(A ∪ B)′ = 60; P(A ∪ B) = 0,7
Determine:
6.1
n(A ∩ B)
6.2
P(A)
6.3
P(B | A)
__
6.4
P(A ∪ B)
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Unit 3: The fundamental counting principle
Consider the word HELP. If we want to know how many
possible ‘words’ we could create from the letters H E L P, we
could list all possibilities:
HELP; HEPL; HLPE; HLEP; HPEL; HPLE; EHLP; EHPL; ELHP;
ELPH; EPLH; EPHL; LHEP; LHPE; LEHP; LEPH; LPHE; LPEH;
PHEL; PHLE; PELH; PEHL; PLHE; PLEH
Note: For this example
we will assume that the
arrangements do not
need to create a real
word belonging to any
language.
This would be a tedious method if the original word had
more letters. Another approach would be to recognise that when choosing a letter
to use in each position, the number of options to choose from diminishes after each
previous choice has been made. So:
The first letter is a
choice out of 4 letters
The second letter is
a choice out of the 3
letters left
The third letter is a
choice out of the 2
letters left
The fourth letter has
only 1 option
4
3
2
1
So in total there are 4 × 3 × 2 × 1 = 24 possible ‘words’ that you can form.
Note: In the same way
that we multiply when
following pathways
in a tree diagram,
multiplication is used
in the fundamental
counting principle.
This is called the fundamental counting principle:
If there are a ways that one event can be performed, b ways that a second event can
be performed, c ways that a third event can be performed, and so on, then there are
a × b × c × … ways in total that the events can be performed successively.
In the example above, we assume that we can use each letter once only. What would
happen if we had been allowed to repeat any of the letters? Our choices for the
second letter will be any one of the 4 letters again, and so on, so the total number of
possibilities will be:
4 × 4 × 4 × 4 = 44 = 256
4
4
4
4
What would happen if we were told that the first letter had to be H, with repetitions
not allowed?
In this situation where conditions are given, always consider the condition/restriction
before continuing with options for the rest of the positions.
So, there is only one option for position 1 which leaves us with 3, then 2 then 1
choice thereafter:
1×3×2×1=6
1
3
2
1
If we were told that the last letter must not be E, we would start by knowing that
there are 3 possible choices for the last letter, then 3, 2, 1 for the other positions:
3 × 2 × 1 × 3 = 18
3
2
1
3
If we need to know how many five-digit numbers we can form using 1; 2; 3; 4 and 5,
we could apply the same principle: the first digit is chosen from 5 possibilities, the
second from the remaining 4 numbers , the third from 3 numbers, and so on, thus
the total number of possibilities is:
Unit 3 The fundamental counting principle
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5 × 4 × 3 × 2 × 1 = 120 possible numbers can be formed.
KEY WORDS
factorial (symbol !) –
multiplying a series of
descending natural numbers
Note: 0! = 1
In general, the total number of possible arrangements of n items, where no
repetitions are allowed, will be:
n × (n – 1) × (n – 2) × … × 1
This is called n factorial, and is written n!
So, 5! = 5 × 4 × 3 × 2 × 1 and
10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
What would happen if we had more values to choose from than positions to be filled?
If repetitions are allowed, then the total number of possible arrangements of
n items where only r positions need to be filled will be:
n × n × n × ……… × n (r times) = nr
Now consider the situation where repetitions are allowed: To determine how many
three-digit numbers we can create from 1; 2; 3; 4 and 5 with no repetitions, the first
digit would be chosen from 5 possibilities, the second from 4 and the third from 3.
Thus there would be:
5 × 4 × 3 = 60 possible numbers in total. This is the same
5×4×3×2×1
5!
as _______ = ________________
(5 − 3)!
2×1
120
= ____
=
60
2
In general, the total number of possible arrangements where repetitions are not
n!
_______
allowed, will be:
(n − r)!
n = the number of items that are available to choose from, and
r = the number of items chosen.
WORKED EXAMPLES
1
2
3
268
How many different outfits can Claire wear if she has 3 skirts, 4 blouses,
and 2 pairs of shoes from which to choose?
A four-character code is formed either by choosing from the numbers 1, 2, 3,
4, 5 or by choosing from the letters A, B, C, D, E. How many different codes
can be formed if no letter or number can be repeated?
How many five-digit codes can be formed from the numbers 1, 2, 3, 4 and 5, if:
3.1
no digit may be used more than once
3.2
no digit may be used more than once and the code must start with 5
3.3
there is no restriction on how often a digit is used, but the code must
end with 5
3.4
there is no restriction on how often a digit is used, but the code must
start with 5 and not end with 1
3.5
there is no restriction on how often a digit is used, but the code must
start with 5 or 1
3.6
there is no restriction on how often a digit is used, but the code must
start with 5 or end with 1?
Topic 12 Counting and probability
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SOLUTIONS
1
2
3.1
3.2
3.3
3.4
3.5
3.6
Claire will first have a choice of 3 skirts, then a choice of 4 blouses, followed
by a choice of 2 pairs of shoes, so in total she will have 3 × 4 × 2 = 24
possible outfits.
There are 5! possible arrangements of the numbers,
Note: 1,2,3,4,5 and
or 5! arrangements of the letters. Thus, using the
A,B,C,D,E are mutually
addition rule for mutually exclusive events: there
exclusive.
are 5! + 5! = 240 possible codes.
There will be 5! = 120 possible codes.
There will be 1 × 4 × 3 × 2 × 1 = 24 possible codes.
There will be 5 × 5 × 5 × 5 × 1 = 625 possible codes.
There will be 1 × 5 × 5 × 5 × 4 = 500 possible codes
We could answer this by considering that the first digit is a choice of 2,
followed by a choice of 5, then 5, then 5, then 5. So the total number of
codes will be 2 × 5 × 5 × 5 × 5 = 1 250. Alternatively, we could consider that
there will be 1 × 5 × 5 × 5 × 5 = 625 codes that start with 5, and 625 that
start with 1. Since ‘starting with 5’ and ‘starting with 1’ are mutually exclusive
events, we can apply the addition rule, so there will be
625 + 625 = 1 250 possible codes.
Since ‘starting with 5’ or ‘ending with 1’ are not mutually exclusive events
(some codes can start with 5 AND end with 1), we must subtract the number
of codes that can have both, so there will be
625 + 625 – (1 × 5 × 5 × 5 × 1) = 1 125 possible codes.
Arranging items where the order of the items matters, is called a permutation.
• When repetitions are allowed in a permutation, the number of ways of arranging r
items from a choice of n items will be nr.
n!
When repetitions are not allowed, instead of using the formula _______ given above,
•
(n − r)!
we could use the key for permutations on the calculator: nPr
REMEMBER
P( A or B ) = P( A ) + P( B )
− P( A ∩ B )
KEY WORDS
permutation – an
arrangement of items where
the position or order of the
items is important
Other notations for permutations are: nPr or P(n,r).
WORKED EXAMPLES
1
2
How many four-digit numbers can be formed from the digits 7, 6, 5, 4, 3, 2, 1,
with no repetitions?
How many of these numbers would:
2.1
be even
2.2
have a value greater than 6 000
2.3
be even and have a value greater than 6 000?
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SOLUTIONS
7!
= 840. Or we could key in: 7; nPr; 4 = and get the same
We could use _______
(7 − 4)!
answer.
Note: once the last digit has been chosen
2.1 For the number to be even, it
to ensure that it is even, there will then be
would have to end with a 2, 4,
6 numbers left from which to choose the
or 6. So, the last digit will be a
first digit.
choice of 3, with the other
digits then being a choice of 6, 5, 4.
Thus the total number of possibilities would be:
6 × 5 × 4 × 3 = 360.
2.2 To have a value greater than 6 000, the first digit would have to be either 7 or
6. Thus the total possibilities would be:
2 × 6 × 5 × 4 = 240.
2.3 The first digit must be either 7 or 6, and the last digit must be even. Because
6 is an even number we consider the two options for the first digit separately,
as this choice will make a difference to how many even numbers we could
choose for the last digit.
If the first digit is a 7, the possibilities will be: 1 × 5 × 4 × 3 = 60
If the first digit is a 6, the possibilities will be: 1 × 5 × 4 × 2 = 40
So the total number of possibilities will be 60 + 40 = 100
1
Now consider arranging items in such a way that some of them have to be next to
each other.
WORKED EXAMPLES
How many ways can Greta, Mavis, Ava, Gertrude, Desmond, Yaseen and Basil
sit in a row, so that:
1
There are no restrictions on who must sit where.
2
Greta and Mavis have to sit together.
3
The girls must sit together and the boys must sit together.
4
Only the girls have to sit together.
SOLUTIONS
1
2
270
This means that we can arrange 7 people in any order, so there will be 7!
(that is, 5 040 possible arrangements).
We consider Greta and Mavis as one unit, and so look at the number of
arrangements we can have with 6 units (the pair, and the other 5 individuals).
In an example like this, repetitions are not valid as none of these people
would be able to sit in more than one place at the same time. So, that means
that there are 6 × 5 × 4 × 3 × 2 × 1 = 720 possible arrangements. However,
the pair of Greta and Mavis can be seated in two different ways and still be
together (Greta then Mavis, or Mavis then Greta). So in total there are (6 × 5
× 4 × 3 × 2 × 1) × 2 = 1 440 different arrangements.
Topic 12 Counting and probability
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3
4
Here we consider the two groups of boys, and girls, and realise that there are
2! (that is, 2) arrangements of the groups, but each group can be arranged
within the group, so in total there are:
4! × 3! × 2 = 288 arrangements.
Note: Always think of
the group that needs
Now we consider girls as one unit and the boys as
to be together as being
individuals. Thus there are 4! arrangements.
one unit, which must be
But the girls can be arranged in 4! ways amongst
arranged together with
themselves, which means there are 4! × 4! = 576
the rest of the data.
arrangements.
Now consider situations like those at the start of this unit: Determine the number of
‘words’ that we can form from certain given letters. The example that we considered
used the letters in the work HELP. What would happen if the word that we started
off with had repeated letters? For example, if we want to determine the number of
‘words’ that we can form from CONSOLIDATION? It is important to know whether
the repeated letters will be treated as different letters or not. If the repeated letters are
treated as different letters, then there will be 13! (6 227 020 800) arrangements, as
CONSOLIDATION has 13 letters. If the repeated letters are treated as the same, then
some of those 6 227 020 800 ‘words’ that are created will be the same, thus there
would be fewer different arrangements. As there are 3 Os, 2 Ns, and 2 Is, the number
13!
of different arrangements would be __________
= 259 459 200 .
3! × 2! × 2!
In general, the number of different ways that n letters can be arranged, treating all
repeated letters as the same, where a of the letters are identical, b are identical, c are
n!
identical, and so on, will be: ________
a! b! c! ...
WORKED EXAMPLES
How many ‘words’ can be formed, using all of the letters in the word
CHARACTERISTIC, if:
1
the repeated letters are treated as different letters
2
the repeated letters are treated as identical
3
the repeated letters are treated as different, and the word starts with an H
4
the repeated letters are treated as identical, and the word starts with an A
and ends with a C
5
the repeated letters are treated as different, and the word starts with an A or
ends with a C
6
the repeated letters are treated as identical, and the word starts and ends with
the same letter?
Unit 3 The fundamental counting principle
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SOLUTIONS
1
2
3
4
5
There are 14 letters in the word, so therefore there would be
14! = 87 178 291 200 words.
14!
There are 3 Cs, 2 As, 2 Rs , 2 Ts and 2 Is, so there would be __________________
3! × 2! × 2! × 2! × 2!
= 908 107 200 words.
If the word must start with an H, there are 13 remaining letters that can be in
any position, which would mean 13! possible arrangements of those letters.
There would therefore be 13! = 6 227 020 800 words possible if the repeated
letters are treated as being different.
If the first and last letters are set and cannot be changed, there are 12
remaining letters that can be arranged in 12! ways. However, after using an
A at the start and a C at the end, there remain 2 Cs, 2 Rs, 2 Ts and 2 Is. Thus
12!
there will be ______________
= 29 937 600 words.
2! × 2! × 2! × 2!
It is important to note that ‘starting with an A’ and ‘ending with a C’ are not
mutually exclusive, thus we use the identity
n(A or B) = n(A) + n(B) – n(A and B).
n(starting with an A) = 1 × 13!; n(ending with a C) = 13! × 1
n(starting with A and ending with C) = 1 × 12! × 1
Thus, the number of words = 13! + 13! – 12!.
Alternatively, instead of using the identity, we
could have considered a Venn diagram of the
situation:
6
starting
with A
ending
with C
13! – 12! 12! 13! – 12!
It is then clear to see that n(starting with A or
ending with C) = 13! – 12! + 12! + 13! – 12!
= 11 975 040 000.
If the word has to start and end with the same letter, the options would be:
C
C
T
T
A
A
R
R
I
I
In the first option, the remaining 12 letters would have 2 As, 2 Rs, 2 Ts and
12!
2 Is. Thus there would be ______________ = 29 937 600 words. In the second
2! × 2! × 2! × 2!
option, the remaining 12 letters would have 3 Cs, 2 As, 2 Rs and 2 Is. Thus
12!
there would be ______________ 9 979 200 words.
3! × 2! × 2! × 2!
In the third option the remaining 12 letters would have 3 Cs, 2 Ts, 2 Rs and
2 Is. This would also result in 9 979 200 possible words. In the last option the
remaining 12 letters would have 3 Cs, 2 Ts, 2 As and 2 Is, thus also resulting
in 9 979 200 possible words. Similarly there would be 9 979 200 words.
So, in total there will be 29 937 600 + 4(9 979 200) = 69 854 400 possible
words.
272
Topic 12 Counting and probability
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EXERCISE 3
1
How many five-digits codes can be formed using the digits 1, 2, 3, 4, 5, 6, 7, 8,
and 9, if:
1.1
digits can be repeated
1.2
digits cannot be repeated?
2
Given three towns: A, B and
A
B
C
C. There are three roads to get
from A to B, and two roads
from B to C.
Assuming that you have to drive through B to get to C, determine:
2.1
the number of different ways that a person could drive from A to C via B
2.2
the number of different ways that a person could drive from A to C and
back to A, via B each way, assuming that he does not want to use the same
road more than once.
3
A cricket team has 11 players. How many
possible variations can there be for the choice
of captain and vice-captain from the 11 players?
4
How many two-digit or three-digit numbers can
we form using the digits 7, 8 and 9, if:
4.1
digits can be repeated
4.2
digits cannot be repeated?
5
How many different six-letter ‘words’ can we
form using the letters of the word TRAVEL if:
5.1
no repetitions are allowed
5.2
repetitions are allowed?
6
How many different three-letter ‘words’ can we
form using the letters of the word TRAVEL if:
6.1
no repetitions are allowed
6.2
repetitions are allowed?
7
How many different 13-letter ‘words’ can we form using the letters of the word
MATHEMATICIAN if no repetitions are allowed?
8
How many different 6-letter ‘words’ can we form using the letters of the word
MATHEMATICIAN if no repetitions are allowed?
9
Tony owns a shops that makes sandwiches.
His customers can choose between 3 different types of
bread, 4 different types of cold meat, 2 types of cheese,
and whether to have lettuce or no lettuce. How many
different types of sandwiches does he offer?
10
Emma has 8 different cards and places them in a row.
After taking note of the order of her cards, she shuffles them and then gives the
cards to her friend, Dirk, to guess the order that she had placed her cards. How
many possible arrangements would Dirk have to choose from?
Unit 3 The fundamental counting principle
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274
11
From Monday to Friday each week, Magda chooses either to bring her own
sandwiches from home for lunch or she buys something from the school tuckshop.
For example, in one week her lunch choices were: Monday: home sandwich;
Tuesday: home sandwich; Wednesday: tuckshop; Thursday: tuckshop; Friday:
tuckshop. How many possible ways can she arrange her lunch choices in any week?
12
How many different five-digit codes can we form using the digits 0, 1, 2, 3, 4, 5,
6, 7, 8 and 9, if:
12.1 the digits can be used in any order and can be repeated
12.2 there must be no repeated digits
12.3 the code must not start with 0, but can have repeated digits
12.4 the code must not start with 0, and must not have any repeated digits
12.5 the code must start with 1, and can have repeated digits
12.6 the code must start with 9 and end with 0, and must not have any
repeated digits
12.7 the code must start with two 5s, and although repeated digits are allowed,
there must not be another 5
12.8 the code must start with 9, then 8, and there must not be any repeated
digits in the code
12.9 the code must end with the same number that it started with, and there
must be no other repeated digits
12.10 the code must end with the same number that it started with, and digits
can be repeated.
13
A test consists of 8 multiple choice questions, each having 5 possible answers.
In how many possible ways can the multiple choice questions be answered,
assuming that all questions are answered?
14
Determine how many different codes can be formed from the digits 1, 2, 3, 4 and 5
if the code can be a three-digit or a four-digit number, and no digit may be repeated.
15
A four-character code can consist either of four digits chosen from the numbers
0, 1, 2, 3, 4 and 5, or it can consist of four letters chosen from V, W, X, Y and Z.
How many possible codes will there be if:
15.1 any digit or letter can be repeated
15.2 the digits can be repeated but the letters must not be repeated
15.3 the digits must not be repeated, but the letters can be repeated?
16
Bongani has 2 Mathematics books, 4 English books, 3 History books and
1 Geography book on his shelf. How many different ways can he arrange these
books on his shelf if:
16.1 the books can be arranged in any order
16.2 the books for each subject must be together
16.3 the English books must be together, but the rest of the books can be in
any order?
17
In how many different ways can the names Piet, Maria, Valmarie and Lara be
written, if:
17.1 the names can be in any order
17.2 Piet must be written first and the other names can be in any order
17.3 Maria must be written last and the other names can be in any order
17.4 Valmarie must be written first and Lara must be written second
17.5 Maria, Lara and Valmarie must be written together
17.6 Piet must not be written first.
Topic 12 Counting and probability
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Unit 4: Applications of the counting principle
to solve probability problems
Whenever we need to know the probability of a certain event taking place,
we can use any one of the techniques covered in this topic, including the
counting principle. As long as we remember that the probability of an event
the number of ways an event can occur
= ____________________________________________, we can apply whichever technique
the total number of possible outcomes for the event
best suits the situation in each case. We have discussed using Venn diagrams, tree
diagrams and contingency tables in depth, so we will now apply the counting principle.
WORKED EXAMPLE 1
Consider the word HEATHER. If the letters of this word are arranged in any order
in which the repeated letters are treated as identical, what is the probability that:
1.1 the ‘word’ formed will start with an H and end with an E
1.2 the ‘word’ formed will start and end with the same letter
1.3 the letters R and T are adjacent
1.4 of the arrangements where R and T are adjacent, the vowels are also adjacent?
SOLUTIONS
1.1 There are 7 letters in HEATHER with 2 Es and 2 Hs, so there are
7!
______
= 1 260 ways in total that the letters can be arranged.
2! × 2!
Now determine how many arrangements will start with an H and end with
an E. If the first and last letters are fixed as being H and E, there are
5 other letters that can be in any order (E, A, H, T and R). As none of these
are repeated letters, there will be 5! ways that these letters can be arranged.
Therefore, the probability of the arrangement chosen starting with an H and
5!
2
ending with an E is _____
= ___
.
1 260
21
1.2 If the arrangement must start and end with the same letter, then it must be
either
E
E
or
H
H
5!
In each case there are __
= 60 ways to arrange the remaining letters (5 letters,
2!
with 1 pair of repeated letters). So the total number of possible arrangements
that start and end in the same letter is 60 + 60 = 120. Therefore, the
probability that the arrangement chosen will start and end in the same letter
120
2
is _____
= ___
.
1 260
21
1.3 We consider R and T as one unit, and therefore we are looking at the number
of arrangements we can have with 6 letters (the pair, and the other 5 letters,
which contain 2 pairs of repeated letters). This would be
6!
______
= 180 arrangements. However, there are two ways in which
2! × 2!
the R and T can be together (either RT or TR), so there would be a total of
2(180) = 360 possible arrangements. Therefore the probability of the R and T
360
2
being adjacent = _____
= __
.
7
1 260
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1.4 Consider the R and T as one unit and the E, A and
4!
E as one unit. This would be __
= 12 arrangements
2!
(the remaining letters are repeated Hs, therefore the
division by 2!). However, there are two ways that
3!
the R and T can be together, and __
= 3 ways that
2!
Note: This means that
the arrangement where
R and T are adjacent
is the total number of
possible outcomes in this
situation.
the E, A and E can be together (3 letters, but 2 of
them are repeated). So, there will be 2(3)(12) = 72
possible arrangements that have the R and T together as well as the vowels
together. Therefore, the probability that this happens, given that the R and T
72
1
are together = ____
= __
.
5
360
WORKED EXAMPLE 2
A five-digit number is formed using the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9, with
no repetitions.
What is the probability that the number is:
2.1 an odd number
2.2 a number less than 50 000
2.3 a number less than 50 000 given that it is divisible by 5?
SOLUTIONS
2.1 The total number of ways of arranging the given numbers with no repetitions
9!
= _______ = 15 120 or 9P = 15 120
(9 − 5)!
5
To be an odd number, the final digit must be a 1, 3, 5, 7, or 9, which would
have 5P1 = 5 possibilities. The other 4 digits could be any of the digits except
for the final digit as there must be no repetitions. Thus there would be
8
P4 = 1 680 possibilities. In total, there would be 1 680 × 5 = 8 400 possible
ways to have an odd number.
Therefore, the probability that the number formed will be an odd number
8 400
5
= __
is: ______
15 120
9
2.2 For the number to be less than 50 000, the first digit must be a 1, 2, 3, or 4,
and the other 4 digits can be any of the 9 digits, except for the digit that was
first. Therefore, there will be 4 × 8P4 = 6 720 possible numbers that are less
6 720
4
than 50 000. The probability of getting such a number is _______
= __
15 120
9
2.3 To be less than 50 000 and divisible by 5, the first digit is a choice of
4 possibilities, the last digit must be a 5. The remaining 3 digits are chosen
from 7 digits. So there are 4 × 7P3 × 1 = 840 possible numbers, and the
probability of getting such a number, given that we are only choosing from
1
840
= __
numbers that are divisible by 5 = _______
8
P4 × 1
276
2
Topic 12 Counting and probability
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WORKED EXAMPLE 3
A code is chosen by using 3 letters of the alphabet followed by 5 digits. The letters
and digits can be repeated. Determine the probability that the code chosen:
3.1 has only one 7 in it
3.2 has at least one A in it
3.3 starts with an A and ends with a 9.
SOLUTIONS
3.1 The total number of ways of creating a code which has 3 letters and 5 digits
with repetitions will be 263 × 105. If we want a code with only one 7:
The 3 letters in the code can be any of 26 letters. These can be repeated, so
there will be 263 possibilities for the 3 letters. The 5 digits must have one 7, so
the remaining 4 digits can be arranged in 94 ways. But the 7 can be in any of
the 5 positions.
Therefore, the probability that the code chosen has only one 7 in it is
263 × 94 × 5
___________
= 0,32805
Note: There are 10 possible
digits to choose from in total:
0, 1, 2, 3, 4, 5, 6, 7, 8 and 9,
but there must only be one
7. That leaves 9 to choose
from in each of the remaining
positions.
263 × 105
3.2 For the code to have at least one A (it could have one, two or three As), it is
easiest to remember that P(A) = 1 – P(A′). So, the probability that there is at
253 × 105
1 951
least one A = 1 – the probability that there is no A = 1 – _________ = ______
263 × 105
17 576
3.3 If the code must start and end with a specific letter or digit, there is no choice
for those positions.
The number of possibilities for the remaining positions is 262 × 104, so the
262 × 104
1
probability of this happening is _________ = ____
263 × 105
260
WORKED EXAMPLE 4
Repeat Question 3, but assume that there are no repetitions of letters or digits.
SOLUTIONS
4.1 The total number of ways of creating a code which has 3 letters and 5 digits
26!
10!
with no repetitions will be ________
× ________ = 471 744 000
(26 − 3)!
(10 − 5)!
(or 26P3 × 10P5).
If we want a code with only one 7:
The 3 letters in the code can be any of 26 letters. These must not be repeated,
so there will be 26P3 possibilities for the 3 letters. The 5 digits must have
one 7, which can be in any of the 5 positions, so the remaining 4 digits must
be chosen from 9 digits. Therefore, in total, there will be
26P × 5 × 9P = 235 872 000 possible codes with one 7, and the probability of
3
4
235 872 000
1
= __
this will be ___________
2.
471 744 000
25!
10!
________
× ________
(25 − 3)!
(10 − 5)!
3
4.2 P(A) = 1 – P(A′) = 1 – _________________
= ___
471 744 000
26
4.3
There is no choice for the start and end position.
The number of possibilities for the remaining positions is
1 814 400
1
25P × 9P = 1 814 400 so the probability = ___________
= ____
2
4
471 744 000
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WORKED EXAMPLE 5
Amina, her brothers, Aaqib and Asheeq, and their parents, are posing for a
family photograph. Assuming that the photographer wants them to stand in a
line, what is the probability that:
5.1 Aaqib and Asheeq will be next to each other
5.2 Amina will be in the middle, given that her brothers are next to each other?
SOLUTIONS
Note: If the brothers must
be next to each other, this
position will have to be the
second brother (if the first
position was one of the
brothers), or the second
parent (if the first position was
a parent). So, once the first
position is filled this position
no longer has a choice, and
will be only one possible
option
5.1 If we consider Aaqib and Asheeq as one unit, there will be 4 units to arrange,
which will be possible in 4! ways. As there are two ways that Aaqib and Asheeq
can be next to each other, there will be 2 × 4! = 48 possible arrangements with
Aaqib and Asheeq next to each other. In total there would be 5! possible ways to
arrange the whole family in a row, so the probability of Aaqib and Asheeq being
2
48 __
next to each other = ___
= 5.
5!
5.2 For Amina to be in the middle, the number of other possibilities will be:
4
1
Amina
2
1
That is, 8 possibilities.
Therefore the probability of Amina being in the middle, given that her
8
1
= __
brothers are next to each other is ___
48
6
WORKED EXAMPLE 6
Hong-lee considers the dinner menu at a restaurant. The menu consists of a choice
from two starters, three different main courses, and three desserts, one of which is
malva pudding. Assuming that he is equally likely to choose any option, what is
the probability that he will have malva pudding with his meal?
SOLUTION
In the past we might have answered this
question by using a tree diagram:
From this diagram we can see that
6 out of the 18 meal options include
malva pudding, so the probability of
6
Hong-lee having malva pudding is: ___
18
1
__
=
3
Note: a choice of
2 starters followed by a
choice of 3 main courses,
followed by the one
option of malva
pudding.
278
Using counting principles, we know that
the total number of meal options
= 2 × 3 × 3 = 18. The number
of these meals that will include
malva pudding will be 2 × 3 × 1 = 6,
so the probability of Hong-lee having
6
1
malva pudding is: ___
= __
.
18
3
Main 1
Starter 1
Main 2
Main 3
Main 1
Starter 2
Main 2
Main 3
Dessert 1
Dessert 2
Malva
Dessert 1
Dessert 2
Malva
Dessert 1
Dessert 2
Malva
Dessert 1
Dessert 2
Malva
Dessert 1
Dessert 2
Malva
Dessert 1
Dessert 2
Malva
Note: a choice of 2 starters followed
by a choice of 3 main courses,
followed by a choice of 3 desserts.
Topic 12 Counting and probability
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EXERCISE 4
1
2
3
4
5
6
7
Determine the probability that a four-digit code chosen from the letters of the
alphabet starts with an A if:
1.1
letters can be repeated
1.2
letters cannot be repeated.
The cards ace, two, three, four, five, and six of hearts are shuffled and placed face
down on a table, and then turned over. Determine the probability that the order
of the cards:
2.1
starts with a 6
2.2
ends in a three or five
2.3
has the ace and two next to each other.
Valda has decided on the make and model for her new car, but has a choice
between four different colours, two different engine sizes and whether to have
power steering or not.
3.1
How many different possible cars is she choosing between?
3.2
Determine the probability that she will not have power steering in her car.
A code is created by choosing three digits from 1, 2, 3, 4 or 5, or three digits from
6, 7, or 8 with no repetitions. Determine the probability that the code will:
4.1
end with a 5
4.2
have no 8.
Five letters from the word SPECIAL are arranged randomly with no repetitions.
Determine the probability that:
5.1
the word SPICE will be chosen
5.2
the word will contain an E
5.3
the letters of the word SPICE will be chosen in any order.
Consider the word ADMINISTRATION. The letters of this word are arranged in
any order, where the repeated letters are treated as being identical, what is the
probability that:
6.1
the ‘word’ formed will start with an S
6.2
the ‘word’ formed will start with an M and end with an S
6.3
the letters D and O are adjacent?
Repeat Question 6, assuming that the repeated letters are treated as being
different.
Unit 4 Applications of the counting principle to solve probability problems
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Revision Test Topic 12
Total marks: 158
1
One person is randomly selected from this list: Rashaad, Terry, Sarah, Thandi,
Chris and Fadiel. Determine the probability that the person selected is:
1.1
Terry
(2)
1.2
not Thandi
(2)
1.3
a person whose name starts with a T
(2)
1.4
either Sarah or Fadiel.
(2)
2
A teacher puts the names of the 25 learners in her class in a bag. She randomly
selects 5 learners to assist her with a project on each day of the week. The learner
with the first name she draws from the bag assists her on Monday, the second
name she draws assists her on Tuesday, and so on, until she has 5 names to assist
her from Monday to Friday. How many different possible lists of helpers could
she have if after drawing a name she:
2.1
returns the name into the bag
(2)
2.2
does not return the name to the bag?
(2)
3
In a horse race there are ten horses
running. Ted would like to place a bet
where he has to choose the first three
places. How many possible variations
are there for the bet that he would
place, if:
3.1
he does not need to state the
order that the first three horses
will be placed
(2)
3.2
he has to state which horse will
finish first, which horse will finish second, and which horse will finish
third?
(2)
4
A test contains ten questions in which the learners have to state whether the
statement is true or false. Determine how many different answers are possible,
assuming that all ten questions are answered.
(2)
5
A four-letter code is formed with no repetitions using letters from the words
TEACHER and CLAP. Treat repeated letters in the words as different letters.
5.1
How many codes are possible if:
5.1.1
any letter from either of the words can be used
(2)
5.1.2
the code must either use four letters from the word TEACHER
or four letters from the word CLAP
(4)
5.1.3
the code must use one letter from the word TEACHER and the
other three letters from the word CLAP
(4)
5.1.4
the code must use one letter from the word CLAP and the other
three from the word TEACHER?
(4)
280
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5.2
6
7
8
What is the probability that the code:
5.2.1
uses no letters from the word CLAP, given that any letter from
either of the words can be used
(4)
5.2.2
the code starts with a T, given that the code must either use
four letters from the word TEACHER or four letters from the
word CLAP
(4)
5.2.3
the code contains an R, given that the code must use one letter
from the word TEACHER and the other three letters from the
word CLAP
(4)
5.2.4
the code starts and ends with an A, given that the code must
use one letter from the word CLAP and the other three from
the word TEACHER?
(4)
The menu at a seaside restaurant is:
Starter: Soup or Prawn cocktail
Main course: Lamb, Chicken or Fish
Dessert: Fruit salad, Cheesecake, Chocolate mousse or Crème brulè
6.1
How many different meals are possible?
6.2
What is the probability that a meal will include chicken?
Consider the numbers 1, 2, 3, 4, 5, 6 and 7.
7.1
How many even three-digit numbers can be formed from these
numbers, assuming that digits are not repeated?
7.2
What is the probability that the even three-digit number formed
will be greater than 6 00?
Consider the numbers 5, 6, 7, 8, 9, and 0. Assuming that the digits can be
repeated:
8.1
how many even three-digit numbers can be formed from these
numbers
8.2
what is the probability that the even three-digit number formed
will end with a 0?
(2)
(4)
(2)
(4)
(2)
(4)
9
A security code is formed so that the code consists of 3 digits chosen from 1, 2, 3,
4 or 5 followed by one letter chosen from A, B, C, D or
E. No repetitions are allowed.
For example:
9.1
How many such codes are possible?
(5) 352D; or 142C
9.2
What is the probability that the code formed
starts with a 1 and contains a vowel?
(2)
10
The letters from the word AEROPLANE are arranged in a line, and the repeated
letters are treated as different. Determine:
10.1 the total number of possible arrangements
(2)
10.2 the probability that the letters R and N will be adjacent
(4)
10.3 the probability that the vowels will be adjacent, given that R and N
are adjacent.
(2)
281
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REVISION TEST TOPIC 12 CONTINUED
11
12
13
14
15
16
Tsepho either walks to school or catches a bus. If he catches the bus, the
probability that he will be on time for school is 0,95. The probability that
he is late for school, given that he walks to school is 0,2. The probability
that Tsepho is on time for school on any day is 0,9.
11.1 What is the probability that he walks to school?
11.2 What is the probability that he will be on time for school, given that
he takes the bus?
11.3 Given that he is late for school, what is the probability that he took
the bus?
11.4 Are the events walking to school, and being on time independent
events?
(6)
(2)
(4)
(5)
Given that P(A) = 0,6; P(B′ ) = 0,15; P(A ∪ B)′ = 0,1 and P(A and B) = x:
12.1 determine the value of x
12.2 determine P(A′ ∩ B)
12.3 determine P(A | B)
12.4 determine whether A and B are independent events.
(5)
(4)
(2)
(5)
Given that P(A) = 0,75 and P(B) = 0,15, determine:
13.1 P(A ∩ B), given that A and B are independent
13.2 P(A ∩ B), given that A and B are mutually exclusive
13.3 P(A | B), given that A and B are independent
13.4 whether A and B are complementary events.
(4)
(1)
(3)
(4)
Ten learners are seated in a row. There are 3 from school A, 5 from school B
and 2 from school C.
14.1 In how many ways can the learners be seated if they can arrange
themselves in any way?
14.2 In how many ways can the learners be seated if each school’s learners
must sit together?
14.3 What is the probability that the learners from each school will be
together, given that the seating arrangement was randomly selected?
14.4 What is the probability that Ntsika, a learner from school A, will be
seated at the end, given that the schools sit together?
Eighty five people attended a conference. Sixty two of the attendees can
speak English, 39 can speak Afrikaans and 45 can speak Zulu. Ten people
can speak all three languages, 32 can speak both English and Afrikaans,
5 can speak Afrikaans and Zulu but not English, and 11 can speak Zulu only.
Determine:
15.1 n(English or Afrikaans or Zulu)′
15.2 P(English and Zulu)
15.3 P(English | Zulu)
15.4 P(Zulu | Afrikaans)
(2)
(4)
(3)
(5)
(6)
(2)
(2)
(2)
Tom has eight bags of marbles. Five bags contain 3 blue and 5 green marbles
and the other three bags each contain 4 blue and 6 green marbles. If 1 marble
is drawn at random from any bag, what is the probability that the marble
will be blue?
(6)
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Exam practice: Preliminary Paper 1
Time: 3 hours
Total: 150 marks
Question 1
1.1 Solve for x:
1.1.1 (x − 2)2 = 8 + x
−6
1.1.2 x − 4 = _____
x+3
(4)
(4)
1.1.3 (2x − 3)2 < 9
1.2 Solve simultaneously for x and y in:
y + 7 = 2x
x2 − xy + 3y2 = 5
1.3 Given f(x) =
f(x) if x = 2.
(5)
(7)
_________
x
, determine, without a calculator, the value of
√___________
x
+x
2013
2012
2009
(4)
1.4 If 2x−1 + x−2 = 3 and x < 0, determine, without the use of a calculator,
the value of (x2 − 3x)−1.
(5)
[29]
Question 2
2.1 A class of 30 learners wrote a mathematics tests. The lowest test score
was 30%. No two learners scored the same mark and every succeeding mark
in the class was 2% higher than the previous one.
2.1.1 Use the above information to write down the first 3 test scores as
a sequence.
2.1.2 What was highest mark in the class?
2.1.3 What was the total of all 30 tests?
2.1.4 Find the class average for the 30 learners.
(1)
(2)
(2)
(1)
2
2
2.2 Given the series 18 + 6 + 2 + __
+ .......... + __7
3
3
2.2.1 Express the series in sigma ( Σ ) notation.
2.2.2 If the series is continued indefinitely, find the sum to infinity.
(5)
(2)
2.3 Consider the sequence 5 ; 8 ; 15 ; 26 ; 41 …
2.3.1 Find the next two terms of the sequence if the pattern continues
in same manner.
(2)
2.3.2 Calculate a formula for the nth term of the sequence.
(5)
2.3.3 Use your formula from 2.3.2 to determine which term of the sequence
is equal to 176.
(3)
[23]
Exam practice: Preliminary Paper 1
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Question 3
3.1 Simone’s grandmother plans to give her R50 000 on her 21st birthday. She
takes out an annuity which offers interest of 8,5% p.a. compounded annually.
If she starts paying into the annuity on Simone’s 10th birthday, and stops her
contributions one month before Simone turns 21, what must her monthly
contributions be so that Simone receives R50 000 on her 21st birthday?
(5)
3.2 Calculate, correct to two decimal places, the effective interest rate on loans
with interest compounding monthly at an annual rate of 12,8%
(4)
3.3 The population of a large insect community is decreasing by 0,6 % each year.
If the population continues to decrease at this rate, how many years will it
take for this population to halve in size?
(4)
[13]
y
Question 4
The graphs of f(x) and g(x)are drawn alongside
(not to scale). A(2;5) is the turning point of
C(–2;4)
f(x) and B(0;1) is the y-intercept of f(x) and g(x).
C(–2;4) is a point on g(x).
4.1 Consider the graph of f(x) and write
B(0;1)
down:
4.1.1 the equation of f(x)
(4)
4.1.2 the turning point of f −1(x)
(2)
4.1.3 the turning point of f(x − 1).
(2)
4.2 Consider the graph of g(x) and write down:
4.2.1 the equation of g(x) in the form
y=…
(2)
4.2.2 the equation of g−1(x) in the form y = …
4.2.3 the equation of g(x) − 1 in the form y = …
A(2;5)
f
g
x
(2)
(2)
[14]
Question 5
−3
5.1 Draw neat sketch graphs of the functions f(x) = _____
+ 3 and g(x) = −2x + 3.
x+2
Show all intercepts with the axes and
any asymptotes.
5.2 Write down the domain of f(x) and the domain of f −1(x).
5.3 About which line is f(x) symmetrical?
5.4 Write down the equation of g−1(x) in the form y = …
(6)
(2)
(2)
(2)
[12]
Question 6
6.1 Determine from first principles the derivative of f(x) = −4x2 + 5.
dy
6.2 Find ___ if:
(4)
dx
6.2.1 y = (3x + 2)2
8x3 −
__ 1
6.2.2 y = _______
√x
6.3 For what value of m will the curve of y
1
?
at x = __
2
(3)
(3)
= mx − 2x3 have a local minimum
(4)
[14]
284
Exam practice: Preliminary Paper 1
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Question 7
7.1 Draw a neat sketch graph of f(x) = −x3 + 3x2 + 9x − 27 and g(x) = −x2 + 9 on the
same set of axes as g(x). Show clearly all intercepts with the axes and coordinates
of any local maximum and minimum points.
(10)
7.2 Use your graphs to determine for which value(s) of x:
7.2.1 f(x) is increasing
(2)
7.2.2 f’(x) and g(x) are both negative.
(2)
7.3 Determine the coordinates of the point of inflection of f(x).
(3)
7.4 Determine the equation of the tangent to f(x) at x = 1.
(4)
[21]
Question 8
In the figure, △ABE has a base of length x metres.
The base and the perpendicular height of the triangle add
B
up to 12 m. The triangle is mounted on a rectangle BCDE
which has a perimeter of 24 m.
8.1 Show that the area of the figure ABCDE is equal
3x
to ___
(12 − x) m2.
(4) C
2
8.2 Determine the value of x for which ABCDE has
a maximum area.
A
h
x
E
D
(3)
[7]
Question 9
9.1 Alexander creates several different 7 character screen names. He uses
the arrangements of the first 3 letters of his name (ALE), followed by
arrangements of the 4 digits, 1987, his date of birth. How many different
screen names can he arrange in this way?
(3)
9.2 A vet surveys 30 of his clients. Sixteen clients have dogs, 12 have cats,
and 6 have fish. Five clients have dogs and cats, 4 have dogs and fish,
1 has a cat and fish and no one has all three kinds of pets.
9.2.1 Draw a Venn diagram to illustrate the above information.
(4)
9.2.2 How many of these clients own none of these pets?
(1)
9.2.3 The vet is offering free de-worming tablets for dogs to these clients.
What is the probability that the client he chooses at random is a
dog owner?
(2)
9.3 The probability that a person picked from the general public is blonde
is 25% and that they are left-handed is 10%. These two events are independent.
9.3.1 What is the probability that a person picked at random from the
general public is right-handed and is not blonde?
(4)
9.3.2 What is the probability that a person picked at random from the
general public is either left handed or blonde?
(3)
[17]
Exam practice: Preliminary Paper 1
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Exam practice: Preliminary Paper 2
Time: 3 hours
Total: 150 marks
Question 1
The heights of 60 U15 Rugby players are recorded in the table.
Height in cm
Number of boys
150 ≤ x < 155
7
155 ≤ x < 160
10
160 ≤ x < 165
15
165 ≤ x < 170
12
170 ≤ x < 175
9
175 ≤ x < 180
5
180 ≤ x < 185
2
Cumulative frequency (cf)
1.1 Copy and complete the table.
1.2 Draw an ogive (cumulative frequency curve) on graph paper.
1.3 Use the ogive to estimate:
1.3.1 the median value
1.3.2 the interquartile range
1.3.3 the percentage of players who are taller than 173 cm.
1.4.1 Use the information from 1.3 to draw a box-and-whisker diagram.
1.4.2 Comment on the distribution of the data in light of what the
box-and-whisker diagram shows.
(2)
(4)
(1)
(2)
(2)
(2)
(2)
[15]
Question 2
The data below shows the temperature of sea water at different depths
in an ocean.
Water depth (x) in metres
25
50
75
100
125
150
200
Water temperature (y) in °C
18
15
12
10
7
4
1
2.1 Determine the least squares regression line y = a + bx for the data.
Round the values of a and b to three decimal places.
2.2 Use the line to predict the temperature to the nearest integer at
a depth of 240 m.
(4)
(2)
[6]
286
Exam practice: Preliminary Paper 2
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Question 3
y
Consider the points A(–2;3), B(5;2), and
C(2;–3).
Determine:
A(–2;3)
3.1 the length of AC in simplified
surd form
(2)
3.2 the coordinates of D if ABCD is a
parallelogram
(2)
3.3 the equation of the perpendicular line
from B to AC, giving your answer in
(5)
the form ax + by + c = 0.
3.4 whether or not point E, the midpoint
of AC, lies on the line in 3.3
(4)
3.5 the value of k if A, C and F(5;k) are
collinear
(3)
3.6 the value(s) of t if BC = BT and T is the point (0;t)
^ B.
3.7 the magnitude of AC
B(5;2)
x
C(2;–3)
(4)
(5)
[25]
Question 4
y
In the diagram Q is the centre of the circle and lies on
the line y − 2x = 5. P lies on the circle and the x-axis.
T lies on the circumference of the circle. PT is a diameter.
Determine:
4.1 the equation of the circle
4.2 the equation of the tangent to the circle at T
4.3 the centre and radius of another circle with
equation x2 − 2x + y2 + 6y + 6 = 0.
T
Q
(5)
(4)
(4)
P
x
[13]
Question 5
5.1 Given sin 40° = a, determine, without a calculator and showing all steps,
each of the following in terms of a:
5.1.1 cos (−50°)
5.1.2 cos 80°
5.1.3 sin 230°.
5.2 Without using a calculator, fully simplify the expression below. Show all
your working.
sin ( 180ο − x ).cos( − x)
__________________________________
tan ( 360ο − x ).cos(180° + x).sin(90° + x)
Question 6
cos x + sin x
1 + sin 2x ____________
= cos x − sin x
6.1 Prove the identity _________
cos 2x
(2)
(2)
(2)
(7)
[13]
(5)
cos 2x
sin 2x
6.2.1 Show that the equation _______
+ _______
= 2 can be written as:
sin 60°
sin 30°
__
√3
sin(2x + 60°) = ___
2
6.2.2 Then find the general solution for x.
(3)
(5)
[13]
Exam practice: Preliminary Paper 2
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Question 7
Given f(x) = sin 3x and g(x) = cos (x − 30°)
7.1 Sketch the graphs of f and g on the same system of axes for x ∈ [−180°;180°].
Indicate the x and y intercepts and the coordinates of the turning points.
7.2 Use the graphs to find for which values of x ∈ [−180°;180°] g(x) ≥ f (x).
(7)
(2)
[9]
Question 8
A
In the diagram, which is not drawn to scale, B, C and D
^ C = 40°. AB is a vertical
lie in the same horizontal plane. DB
pole. BD = 96 m and BC = 106 m. The angle of elevation
from C to A is 50°.
8.1 Calculate AC.
(2)
8.2 Calculate CD if AC = AD.
(3)
^ C.
8.3 Determine the size of BD
(3) B
[8]
24,41°
96 m
D
40°
106 m
50°
C
Question 9
9.1 In the figure, X, Y and Z are three points on
the circle with centre O. PZQ is a tangent to the
^ Y = 140°, calculate,
circle at Z. XY || PQ. If ZO
^ X.
with reasons, the magnitude of OZ
9.2 In the figure, ABCD is a cyclic quadrilateral.
O is the centre of the circle. Redraw the diagram
^ +C
^ = 180ο
and prove the theorem that A
X
Y
O
(7)
P
140°
Z
Q
A
B
O
D
C
(6)
9.3 In the figure, ABC is a tangent to circle BDEF
at B.
FE || ABC and FB || EDC .
^ F = x.
FB = FD and AB
^ and EF
^B are
9.3.1 Give reasons why C
both equal to x.
(2)
9.3.2 Find, with reasons, three other
angles equal to x.
(6)
^ C and DB
^ C in terms
9.3.3 Express FE
of x.
(4) A
[25]
288
F
E
D
x
B
C
Exam practice: Preliminary Paper 2
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Question 10
10.1 Complete the statements of the following theorems by writing down
the missing words:
10.1.1 A line drawn parallel to one side of a triangle divides …
10.1.2 If all the corresponding sides of two triangles are proportional,
the triangles are …
10.2 In the diagram, AP = 36 mm,
A
PC = 72 mm, BM = 45 mm and
36 mm
AB || PM .
P
10.2.1 Calculate, giving reasons,
72 mm
the length of MC.
(3)
y
x
x
10.2.2 Calculate the value of __y .
(2)
[7] B
45 mm
(1)
(1)
C
M
Question 11
PR is the diameter of the circle in the figure. SB is
perpendicular to PR and cuts PR in T. A point V on
the circumference of the circle is joined to P and R.
Prove:
11.1 TAVR is a cyclic quadrilateral.
11.2 △PSR ||| △PTS
11.3 PS2 = PR. PT
11.4 △PTA ||| △PVR
11.5 PS2 = PA. PV
(4)
P
(3)
(2)
(3)
(4)
[16]
S
2
T
1
1
2 3
12
1
2
R
A
B
V
Exam practice: Preliminary Paper 2
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Term 3 summary
Topic 10
Exponents and surds
Topic 10
Topic 11
Euclidean geometry
Equations and inequalities
Two polygons are similar if and only if their corresponding sides are in proportion and their corresponding angles
are equal.
• Triangles which have the same base and same height are equal in area.
• Triangles with different bases, but equal heights: the ratio of their areas is equal to ratio of the lengths of
their bases.
A
In the figure, A is the common vertex of △ABC and △ADC
Area △ ABC
BC
= ___
AE is the common height, so __________
CD
Area △ ADC
• A line drawn parallel to one side of a triangle divides
C
B
E
D
the other two sides proportionally
In the sketch, DE || BC
AD AE
∴ ___ = ___
DB
A
EC
When using this fact, state the proportionality theorem,
Topic
as well as the parallel
lines12
involved.
Number
patterns
A
line
drawn
through
the
midpoint of one side of a triangle,
•
parallel to the second side, will pass through the midpoint of
the third side.
CD
HJ
290
DF
R
G
B
H
T
Q
F
A
G
B
C
C
H
D
J D
J
E
K E
K
Figure 2
G
D
x
their corresponding sides are in proportion.
EF
C
B
F
A
Figure 1
DE
E
S
• If two triangles are equiangular,
^ =D
^ = ^F
^; H
^ = ^E and K
In the figure, G
GH HK GK
∴ ___ = ___ = ___ (△GHK ||| △DEF)
h2
P
In the figure, PS = SQ and ST || QR
∴ ST = TR
This is called the midpoint theorem.
It also follows that QR = 2 ST.
• Three or more parallel lines divide
the sides proportionally.
In the figure, BG || CH || DJ
BC DH
∴ ___ = ___
h1
D
x
A
y
H
w
K
y
E
w
B
F
Term 3 summary
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Term 3 summary continued
Topic 10
Euclidean geometry (continued)
When using this fact always mention the similar triangles involved in the order of the corresponding angles. Then
the proportionate sides are easy to find:
the first two letters of each triangle; the last two letters of each triangle, outer letters of each triangle.
• If two triangles have their corresponding sides in proportion, then they are equiangular.
This is the converse of the previous theorem.
• The perpendicular line drawn from the vertex of the right angle of a right-angled triangle to the hypotenuse
divides the triangle into two triangles, which are similar to each other and similar to the original triangle.
In the figure, △ PQT ||| △ RPT ||| △ RQP.
x
90
°–
x
P
x
90° – x
T
Q
R
Grade 12 Geomerty
Proportionality theorem
A line drawn parallel to one side of a triangle divides the other two sides
proportionally.
AD ___
___
= AE
| Proportional intercepts, DE || BC
DB EC
AD ___
___
= AE
AB AC
| Proportional intercepts, DE || BC
EC
DB ___
___
=
| Proportional intercepts, DE || BC
AB
AC
A
D
E
C
B
Converse proportionality theorem
QV ____
QW __
VW || PR
| Converse proportionality theorem, ___
=
=k
VP WR h
Q
kx
VW || PR
PQ RQ h + k
| Converse proportionality theorem, ___ = ____ = ____
VW || PR
PQ ___
RQ ____
h+k
| Converse proportionality theorem, ___
=
= h
PV RW
VQ
WQ
ky
V
k
W
R
P
Midpoint theorem
JK || GH
| Midpoint theorem, J and K midpoints of FG and FH
respectively
1
__
JK = 2 GH
| Midpoint theorem, J and K midpoints of FG and FH
respectively
hy
F
J
K
G
H
Term 3 summary
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Term 3 summary
Converse midpoint theorem
M
R
Q
N
P
MR = RP
| Converse midpoint theorem, Q midpoint MN and QR || NP
If two triangles are equiangular, their corresponding sides are in proportion
D
z
A
z
x
x
y
C
B
y
E
F
△ABC ||| △DEF
| AAA
AC ___
BC
AB ___
___
=
=
| △ABC ||| △DEF
DE
DF
EF
If two triangles have their corresponding sides in proportion, they are equiangular
M
G
kz
az
H
ky
ay
ax
J
△GHJ ||| △MNP
^ =M
^, H
^ =N
^ and ^J = P
^
G
N
kx
P
GJ ___
HJ __
a
GH ___
| Corresponing sides in proportion, ____
=
=
=
MN MP NP k
| △GHJ ||| △MNP
The perpendicular line drawn form the vertex of the right angle of a right-angled triangle to the
hypotenuse divides the triangle into two triangles that are similar to each other and similar to the
original triangle.
B
△ABC ||| △ADB ||| △BDC
BD2 = AD.DC
^ C = 90° and BD⊥AC
| AB
| △ADB ||| △BDC
A
292
D
C
Term 3 summary
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Term 3 summary continued
Topic 11
Statistics
• If the mean = the median, then the data is
symmetrical.
• If the mean > the median, then the data is skewed
to the right.
• If the mean < the median, then the data is skewed
to the left.
• Association between bivariate data is represented
as a scatter plot, and can be linear (positive or
negative), exponential, parabolic, or other.
• Correlation coefficient, r = the measure of
association between two variables. –1 ≤ r ≤ 1.
r = 0,5 → moderately strong positive
r = –0,5 → moderately strong negative correlation
0 < r < 0,5 → weak positive correlation
0 > r > –0,5 → weak negative correlation
r > 0,8 → strong positive correlation
r < –0,8 → strong negative correlation
• Line of best fit from scatter plot is called a
regression line: y = A + Bx
Find A and B using the STAT mode on the calculator.
• You can apply a regression line to predict values
using interpolation or extrapolation.
• The predicted value can differ from the observed
value, and this difference is the residual.
• Sometimes you can omit outliers to have a more
creditable regression line.
Topic 12
Counting and probability
A summary of all notations and definitions that were
studied in Grade 11 appear on the first page of this
topic.
• You can use Venn diagrams, tree diagrams and
contingency tables studied in Grade 11 to assist
with certain probability questions.
• The fundamental counting principle:
– If there are a ways that one event can be
performed, b ways that a second event can be
performed, c ways that a third event can be
performed, and so on, then there are
a × b × c × … ways in total that the events can
be performed successively.
Topic 12 continued
Counting and probability
– In general, the total number of possible
arrangements of n items, where no repetitions
are allowed, will be
n × (n – 1) × (n – 2) × … 3 … × 1
This is called n factorial, and is written n!
– If repetitions are allowed, then the total number
of possible arrangements of n items where only r
positions must be filled will be:
n × n × n × … × n (r times) = nr
• When there are more items to choose from than
positions to be filled:
– In general, the total number of possible
arrangements (permutations) where repetitions
are not allowed will be
n!
_______
where n = the number of items that are
(n − r)!
available to choose from and r = the number of
items chosen OR nPr on the calculator.
– When repetitions are allowed in a permutation,
then the number of ways of arranging r items
from a choice of n items will be nr.
• When arranging letters from a word that has
repeated letters:
– The number of different ways that n letters can
be arranged, treating all repeated letters as
the same where a of the letters are identical,
b are identical, c are identical, and so on, will
n!
be ________
a! b! c! ...
• When applying any of these techniques to solve
probability problems, remember:
probability of an event
the number of ways an event can occur
= _________________________________________
the total number of possible outcomes for the event
Term 3 summary
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Term
4
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Unit 1
Unit 2
Structure of final examination
Exam practice A
Paper 1
Paper 2
Unit 3 Exam practice B
Paper 1
Paper 2
Term summary
Examination tips
296
299
299
303
307
307
311
316
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Unit 1
Structure of final examination
The final Grade 12 Mathematics examination consists of TWO PAPERS.
Each paper is THREE HOURS long and totals 150 MARKS.
This examination is externally set, marked and moderated.
The final examination makes up 75% of your total mark for Mathematics.
Although Grade 10 lays a critical foundation for Grades 11 and 12, work covered in
Grade 10 is not directly examinable in the final examination.
All work covered in Grade 11 is examinable, although you will not cover it again in
Grade 12.
In the mid-year examination, trial examination and final examination, you will be
examined on the entire Grade 11 curriculum.
It is essential that you spend time revising Grade 11 work as it will form the greater
part of the mid-year examination.
The table below contains the mark distribution for the Mathematics NCS End-of-year
Grade 12 papers.
PAPER 1
Maximum 6 marks for bookwork
Description
Weighting of marks
Algebra, equations and inequalities
25 ± 3
Patterns and sequences
25 ± 3
Finance, growth and decay
15 ± 3
Functions and graphs
35 ± 3
Differential calculus
35 ± 3
Probability
15 ± 3
TOTAL
150
PAPER 2
Theorems and/or trigonometric proofs: maximum 12 marks
Description
Weighting of marks
Statistics
20 ± 3
Analytical geometry
40 ± 3
Trigonometry
40 ± 3
Euclidean geometry and Measurement
50 ± 3
TOTAL
150
Questions will not necessarily be compartmentalised in sections, as this table
indicates.
Various topics can be integrated in the same question.
Modelling as a process should be included in both papers, so expect contextual
questions in any topic.
A formula sheet is provided in the final examination.
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Cognitive levels
The final examination will test the four cognitive levels listed in the table below.
The weighting of these levels as well as the skills which will be tested, are indicated
for each category.
It is important that you familiarise yourself with these skills. They will form the
basis of all your assessments for the year and not only the final examination.
Cognitive levels
Weighting
Skills to be tested
Examples
Knowledge
20%
• Recall
• Identification of correct formulae
on the formula sheet, without
changing the subject of the
formula
• Use of mathematical facts
• Appropriate use of mathematical
vocabulary
1
State the domain and range of the
3 + 2.
function f( x ) = _____
2
Write down the first four terms of the
sequence with general term Tk = 3 × 2k−1.
Complete the statement: The exterior
angle of a cyclic quadrilateral …
State three ways to prove that a
quadrilateral is a cyclic quadrilateral.
• Estimations and appropriate
rounding of numbers
• Proofs of prescribed theorems and
derivation of formulae
• Identification and direct use of
formulae on the formula sheet,
with or without changing the
subject of the formula
• Perform well-known procedures
• Simple applications and
calculations which might involve
a few steps
• Derivation from given
information may be involved
• Procedures similar to those
encountered in class
1
30 marks
Routine
procedures
35%
52 marks
x−4
3
4
Solve for x if x2 − x = 12.
5
2
Calculate
∑( 2k − 1 )
k=1
3
4
5
Determine f ’( x ) if f( x ) = ( x − 3 )2
Determine the general solution of the
equation 2cos ( x + 30° ) + 1 = 0.
^ B, subtended by arc AB at
Prove that AO
the centre of the circle, is twice the size of
^ B, subtended by the same arc on the
AC
circumference of the circle.
C
O
A
B
Complex
procedures
30%
45 marks
• Solve problems involving
complex calculations and/or
higher order reasoning
• Solve problems not having an
obvious route to the solution
• Solve problems not based on a
real world context
• Make significant connections
between different representations
• Conceptual understanding
1
2
3
(
)3
x −__1
Determine f ’( x ) if f( x ) = _______
√
x
If cos ( α − β )
= cos α cos β + sin α sin β, derive a formula
for:
2.1
cos ( α+ β )
2.2
sin ( α+ β )
Determine the equation of the tangent
which touches the circle
x2 + 2x + y2 − 4y = 5 at the point (−4;1).
Unit 1 Structure of final examination
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Problem solving
15%
23 marks
• Non-routine problems, which are
not necessarily difficult
• Higher order reasoning and
processes
• Might require the ability to
break down a problem into its
constituent parts
1
2
Show that 0,˙
9=1
In △PQR, PT ⊥ QR.
cos Q r − q cos P
Prove that _____ = _________
P
q − r cos P
q
r
Q
cos R
R
T
p
This question is quite challenging and it is
important to consider options before rushing in
to solve it.
^ are used, whereas on the
^ and Q
On the LHS P
^
RHS only P is used.
^ +R
^ +R
^+Q
^ = 180∘ ⇒ P = 180∘ − ( Q
^)
P
| ∠ sum △PQR
r − q cos( 180∘ − ( Q + R ) )
q − r cos( 180 − ( Q + R ) )
RHS = ______________________
∘
r + q cos( Q + R )
q + r cos Q + R
= ______________
)
(
r + q[ cos Q cos R − sin Q sin R ]
q + r cos Q cos R − sin Q sin R
= __________________________
[
]
r + q cos Q cos R − ( q sin R ) sin Q
q + r cos Q cos R − r sin Q sin R
= ____________________________
)
(
r + q cos R cos Q − ( r sin Q ) sin Q
)
(
= ____________________________
q + r cos Q cos R − q sin R sin R
sin Q
sin R = _____ ⇒ r sin Q = q sin R
| _____
q
r
r + q cos R cos Q − r sin2 Q
q + r cos Q cos R − q sin R
= _______________________
2
r + q cos R cos Q − r ( 1 − cos2 Q )
= ____________________________
2 )
(
q + r cos Q cos R − q 1 − cos R
r + q cos Rcos Q − r + r cos2 Q
q + r cos R cos Q − q + q cos R
= __________________________
2
cos Q( q cos R + r cos Q )
cos R( r cos Q + q cos R )
= ____________________
cos Q
= RHS
= _____
cos R
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Exam practice A: Paper 1
Time: 3 hours
Total: 150 marks
Question 1
1.1 Solve for x, correct to two decimal places where necessary.
1.1.1 x2 = 5x
1.1.2 ( 2x − 3 )( x + 2 ) ≥ 22
1.1.3 43x = 8 x + 2
______
1.1.4 √ 2x + 3 = x
(2)
(5)
(3)
(5)
1.2 Show that x2 + 2x + 3 = 0 has no real roots.
(3)
__
__
_______
___
1.3 Determine a possible value of p2 + q2 if √p + √ q = √ 9 + √56 .
Question 2
A quadratic pattern has a third term equal to 2, a fourth term equal to –2
and a sixth term equal to −16. Calculate the second difference of this
quadratic pattern.
(7)
[25]
[5]
Question 3
3.1 The third term of an arithmetic sequence 11 and the sum of the second
and fifth terms is 26. Determine the constant difference and the value of the
first term.
(5)
4
3.2 Evaluate
∑ 12 × 2 1 n
−
(4)
n = −1
3.3 Consider the series: 125( x − 3 ) + 25( x − 3 )2 + 5( x − 3 )3 + ...
3.3.1 For which value(s) of x will the sequence be convergent?
3.3.2 Determine the sum to infinity of the series if x = 7.
3.4 1 + 2 + 3 + 4 + 6 + 7 + … + 79 + 81 is a series of natural numbers
from which all the multiples of 5 have been removed. Determine the
sum of this series.
(4)
(3)
(6)
[22]
Exam practice A: Paper 1
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Question 4
4.1 Siseko buys a house for R600 000. He pays a deposit of R42 000 and
his monthly payments are R7 500.
The interest rate is 11,25% per annum, compounded monthly. He pays
the deposit immediately and his first monthly repayment one month
after securing the loan.
4.1.1 Determine the deposit as a percentage of the house price?
4.1.2 How long will it take him to pay off the loan?
4.1.3 What is the amount of his final payment?
(1)
(4)
(4)
4.2 Mohammed borrows R300 000 to fund a new bu siness venture. He
agrees to pay back R50 000 after the first year, R100 000 after the second and
third years and a final payment at the end of the fourth year. Interest
is calculated at 13% per annum, compounded monthly for the first two years
and 12,5% per annum, compounded quarterly for the final two years.
What is the final payment?
(6)
[15]
Question 5
y
g
h
C(3;9)
f
2
A(1;1)
–2 B
5
y=– 3
x
k
x
The graphs of f( x ) = ax2 + bx + c, g( x ) = _____
x − p + q and h(x) = vw + z are sketched
above.
A(1;1), B(−1;−1) and C(3;9), the turning point of f, are given.
5.1 Determine the values of a, b, c, k, p, q, v, w and z. Show all working.
(12)
5.2 For which values of x is:
5.2.1 h( x ) < g( x )
5.2.2 f( x ) ≥ h( x )?
(3)
(2)
5.3 State one value of x for which f( x ) = g( x ) = h( x ).
(1)
3
5.4 Calculate
6
∑f x − ∑f x . Show all working.
(
x=0
300
(
)
x=4
)
(4)
[22]
Exam practice A: Paper 1
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Question 6
f( x ) = log__1 x
3
6.1 State f −1( x ).
(2)
6.2 If g is the reflection of f in:
6.2.1 the x-axis, determine the equation of g
6.2.2 the y-axis, determine the equation of g and state the domain of g.
(1)
(2)
6.3 Determine the equation of h if h(x) is obtained by shifting f 2 units
to the left.
(2)
6.4 Determine the average gradient of f between x = 3 and x = 9.
(4)
[11]
Question 7
2 , determine f ’( x ) by using the definition.
7.1 If f( x ) = − __
x
(5)
7.2 Determine:
d [ ( 2x − 3 )( 5x + 1 ) ]
7.2.1 ___
(3)
dx
7.2.2
( x + 3 )3
__
g ’( x ) if g( x ) = _______
7.3
y
E
(4)
√ x3
f(x) = x³ – 4x ² – 11x + 30
A
C
B
x
D
F
3
2
f( x ) = x − 4x − 11x + 30 has been sketched above.
7.3.1 Show that (x − 2) is a factor of f(x).
7.3.2 Determine the coordinates of A, B, C and D.
7.3.3 Determine the coordinates of the stationary points E and F.
7.3.4 Determine the equation of the tangent to f at x = 1.
(2)
(5)
(5)
(3)
[27]
Question 8
There are 1 000 Grade 12 learners at Fiseka High School. All of them take either
Mathematics or Mathematical Literacy as a subject. The table below is incomplete,
but shows that 300 girls take Mathematics, 120 boys take Mathematical Literacy
and there are 640 boys in total.
Contingency Table
Boys
Mathematics
Girls Total
300
Mathematical
Literacy
120
Total
640
1 000
8.1 Copy the table and fill in the missing values.
8.2 What is the probability that a randomly selected Grade 12 learner is a boy
who takes Mathematics?
(3)
(2)
[5]
Exam practice A: Paper 1
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Question 9
A
C
D
G
E
B
F
The cone has a height of 10 cm and a radius of 5 cm. The volume of a cylinder is
1 πr2h. The radius of the
given by V = πr 2h and the volume of cone is given by V = __
3
cylinder is x cm.
9.1 Determine the height of the cylinder in terms of x.
(2)
9.2 Show that the volume of the cylinder is given by
V = 10πx2 − 2πx3.
(3)
9.3 Determine the maximum volume of the cylinder which can be inscribed in this
cone.
(4)
[9]
Question 10
A bar code is designed with a mixture of letters and numbers. The first three values
must be letters from the alphabet and the last four values may be any numbers
from zero to nine.
10.1 If there are no restrictions, how many bar codes are possible?
(3)
10.2 If no letter or numbers may be repeated, how many bar codes are possible?
(3)
10.3 What is the probability that a randomly selected code has no vowels,
contains only prime numbers and has no repeated letters or numbers?
302
(3)
[9]
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Exam practice A: Paper 2
Time: 3 hours
Total: 150 marks
Question 1
The following 19 consecutive batting scores were recorded for the top run scorer at
Thembisa High School during the cricket season:
67
11
28
42
49
61
52
40
42
60
53
31
76
47
71
79
41
1.1 Give the five-number summary of the data.
(4)
1.2 Represent his scores using a box-and-whisker diagram.
(3)
1.3 Briefly discuss the distribution of the data.
(1)
54
59
[8]
Question 2
The response time to a stimulus was measured for 13 people of varying ages.
The results are summarised below in the table.
Age (years) x
17
20
22
23
25
31
33
38
40
45
53
59
72
Time (seconds) y
0,8
0,9
1,0
0,8
1,3
1,1
1,2
1,5
1,4
1,6
1,7
1,9
2,3
Give your answers correct to four decimal places for this question.
2.1 Draw a scatter plot for the data.
(3)
2.2 Calculate the equation of the least squares line for this data.
(4)
2.3 Calculate the correlation coefficient.
(1)
2.4 Comment on the correlation of the data.
(2)
2.5 If Vinnie is 65 years old, estimate his response time to the stimulus.
(2)
[12]
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Question 3
y
B
A
M
C
O
0
x
D
A is the centre of the circle which touches the x-axis at D(4;0) and crosses
the y-axis at C(0;2) and B(0;8). AM ⊥ BC.
3.1 Determine the equations of these straight lines:
3.1.1 AD
3.1.2 AM
3.1.3 BD
3.1.4 CD.
(1)
(2)
(3)
(3)
3.2 Determine, without reasons, the lengths of:
3.2.1 AM
3.2.2 MC
3.2.3 AD.
(1)
(2)
(2)
3.3 Determine the equation of the circle.
(3)
3.4 Determine θ, correct to one decimal place.
(5)
3.5 Determine the equation of the tangent to the circle at C.
(3)
3.6 Determine the equations of the two tangents to the circle which are
perpendicular to the tangent at C.
(8)
3.7 Determine the area of:
3.7.1 quadrilateral AMCD
3.7.2 △BCD
(4)
(3)
[40]
Question 4
Do not use a calculator in any part of this question.
___
4.1 Given 3 tan β + 5 = 0, β ∈ [ 180°;360° ], evaluate √ 34 ( cos β − cos ( 90° + β ) ).
(4)
4.2 If sin 17° = p, express the following in terms of p:
4.2.1 cos 343°
4.2.2 sin 62°.
(2)
(4)
4.3 Calculate the value of ( sin 105° + cos 105° )2.
(3)
(
)
cos 2θ − sin 90° + θ
cos θ − 1
4.4 Prove the identity __________________
= ________
)
(
sin 2θ + cos 90° − θ
304
sin θ
(5)
[18]
Exam practice A: Paper 2
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Question 5
5.1 5.1.1 Determine the general solution of cos 2x = sin ( x + 60° ).
5.1.2 Hence, solve for x if cos 2x = sin ( x + 60° ) and x ∈ [ − 90°;180° ].
(6)
(2)
5.2 Draw, on the same set of axes f( x ) = cos 2x and g( x ) = sin ( x + 60° ) for
x ∈ [ − 90°;180° ]. Show clearly all the intercepts on the axes and the
coordinates of the turning points.
(6)
5.3 Write down the period of g.
(1)
5.4 For which values of x is:
5.4.1 f( x ) ≥ g( x )
5.4.2 f( x ).g( x ) < 0?
(2)
(2)
[19]
Question 6
K
h
D
x
G
x
y
E
F
The figure shows the boundaries of a sports field DEFG. DG || EF and DE ⊥ EF.
KG is a vertical pylon for a floodlight. The angle of elevation of K from F is x.
^ F = x, DF
^G = y and KG = h m.
ED
^ F in terms of x and y.
6.1 Express DG
(3)
h cos ( y − x )
6.2 Prove that DF = ___________
sin x
(6)
[9]
Question 7
7.1
P
T
R
^
△PQR is a right-angled triangle with QRP = 90° and RT ⊥ PQ.
Prove Pythagoras’ Theorem by similar triangles.
Q
(8)
Exam practice A: Paper 2
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7.2
A
C
B
1
2
1 3 4
2
F
G2
12
34
1 O
1
D
23
4
E
AC and AE are tangents to the circle at B and D respectively. F is a point
on the circle with centre O.
7.2.1 Prove that BG = GD.
7.2.2 Prove that ABOD is a cyclic quadrilateral.
^ =F
^
7.2.3 Prove that O
1
^
7.2.4 If OD bisects BDF, prove that AE || BF.
7.2.5 Prove that BG2 = AG.GO
7.2.6 If GO = 9 units and AO = 25 units, determine the length of OD.
(4)
(4)
(3)
(6)
(3)
(4)
[32]
Question 8
A
E
F
D
C
B
D is a point on AB, E and F are points on AC. DE || BF, DF || BC and
AD : DB = 2 : 1
Determine, with reasons:
8.1 AE : EF : FC
(4)
8.2 DE : BF
(4)
Area △ADE
8.3 __________
(4)
Area △ABC
306
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Exam practice B: Paper 1
Time: 3 hours
Total: 150 marks
Question 1
1.1 Solve for x: Leave your answers in simplified surd form where necessary.
1.1.1 x2 – 2x – 4 = 0
_____
1.1.2 √ x + 7 + 1 = 2x
1.1.3 (2x − 3)(x + 1) ≥ 33
2
__
1.1.4 4x 3 = 9
1 = ___
1
1.1.5 8x.____
x−1
4
1.2
32
(3)
(4)
(5)
(2)
(3)
If 2x2 − 5xy − 12y2 = 0 and xy > 0:
x
1.2.1 find the value of __
y
1.2.2 if x + y = 4 , use this and your answer from 1.2.1 to solve for x and y .
(3)
(3)
[24]
Question 2
2.1 Pattern 1 = number of squares
Stage 1
Stage 2
Stage 3
Stage 4
Pattern 2 = number of squares
Stage 1
Stage 2
Stage 3
Stage 4
2.1.1 Write down the first four terms of pattern 1 and pattern 2.
2.1.2 Which pattern is linear and which is quadratic?
Give a reason for your answer.
2.1.3 Write down a formula for Tn, the nth term of the pattern 1.
2.1.4 Write down a formula for Tn, the nth term of the pattern 2.
(2)
(2)
(5)
2.2 In a geometric progression the third term is 24 and the sixth term is 3.
Find the sum to 10 terms.
(5)
24 + 12 + 6x + 3x2 + ... is an infinite geometric series.
2.3 ___
x
2.3.1 Find the values of x for which the series converges.
2.3.2 Find the sum to infinity in terms of x.
(4)
(3)
n
2.4 Determine the value of n for which
∑( 2r − 3 ) = 120.
r=1
(2)
(5)
[27]
Exam practice B: Paper 1
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Question 3
Jano invests R5 000 at a rate of 7,2% per annum compounded monthly.
3.1 Write down an expression for the value of the investment after n full years.
(2)
3.2 What will be the value of the investment at the end of 8 years?
(2)
3.3 If the value of the investment exceeds R10 000 after n full years, calculate
the minimum value of n.
(2)
3.4 Jano takes out an annuity of R55 per month, instead of investing the
lump sum of R5 000. If his payments start in one month’s time, what
would his investment be worth after 8 full years if he receives the
same interest rate of 7,2% p.a. compounded monthly?
(3)
3.5 How long will it take for the investment in 3.4 to exceed R10 000?
(3)
[12]
Question 4
4.1 Sketch the graph of the function h(x) = 2−x.
(2)
4.2 Determine the equation of q(x), the graph obtained by reflecting h(x)
in the y-axis.
(1)
4.3 Determine the equation of h−1(x), the graph obtained by reflecting h(x),
in the line y = x.
Write your answer in the form y = …
(2)
4.4 Write down the range of h(x).
(1)
4.5 Sketch the graph of h–1(x) on the same set of axes as h(x) and use it
to determine the values of x for which is h–1(x) ≥ −3.
(4)
[10]
Question 5
4 + 1 and g(x) = 2x + 3.
The figure shows the graphs of y = f(x) = _____
x+2
A is the y-intercept of both graphs, C is the point of intersection of the
asymptotes of f(x) and D and E are the x-intercepts of the two graphs.
B is a point of intersection of f(x) and g(x).
5.1 Write down the coordinates of C.
(2)
5.2 Write down the coordinates of A and B.
(5)
5.3 For which values of x is f(x)g(x) ≤ 0?
(3)
5.4 Write down the domain of f –1(x).
(2)
[12]
308
y
g
A
f
C
D
E
x
B
Exam practice B: Paper 1
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Question 6
6.1 If f(x) = −x2 + 3x, determine the derivative, f ’(x), from first principles.
(4)
dy
6.2 Determine ___ for these functions:
dx
__
1
3
x − ___
6.2.1 y = √
3x
5x3 − 5
6.2.2 y = _______
x−1
(3)
(3)
6.3 The figure shows the curve of f(x) = ax3 + 1.
Point P(2;5) lies on the curve of f.
1.
6.3.1 Show that a = __
2
6.3.2 Determine the average gradient of the curve between x = 2
and x = 4.
6.3.3 Determine the equation of the tangent to f(x) at point P.
6.3.4 The graphs below represent f(−x) or f −1(x) or −f(x).
Match the graph with its function and write down the letter
corresponding to each function on your answer sheet.
y
y
P(5;2)
y
(2)
(3)
(3)
P(2;5)
x
(3)
y
P(–2 ;5)
x
x
x
P(2;– 5)
A
B
C
[21]
Question 7
The graphs of the functions f(x) = −x3 + 6x2 − 9x + k and
g(x) = ax2 + bx + c are drawn below.
y
7.1 Show that k = 2 if (x − 2) is a factor of f(x).
(2)
7.2 Write down the coordinates of A.
(2)
(
)
1
7.3 If the coordinates of B and C are −__
;0 and (2;0) respectively,
2
determine the values of a, b and c.
(4)
A
B
7.4 Determine the coordinates of D and E, the x-intercepts of f(x),
correct to two decimal places. State the nature of these roots. (6)
7.5 Determine the coordinates of T and P, the turning points of
f(x).
7.6 For which values of x is f(x) increasing?
P
C
D
T
E
x
f
g
(5)
(2)
[21]
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Question 8
A window in the shape of a rectangle with a semicircle on top is designed
to let in the maximum amount of light by maximising the area of the window.
The rectangle measures 4x m by h m.
h metres
8.1 If the perimeter of the whole window is 16 m, express h in terms of x.
(2)
8.2 Hence show that the area of light L, let in by the window is given
by A = 32x − 8x2 − 2πx2.
(3)
8.3 Find the value of x, correct to two decimal places, that should be used
to allow this design to let in the maximum amount of light.
4x metres
(4)
[8]
Question 9
9.1 If you are given 2 yellow cards, 1 red card, l green card and 1 blue card,
determine the number of different ways in which you can arrange these
cards in a single row.
(2)
9.2 There are 75 boys in Grade 12 in a school. Forty eight play rugby,
8 play rugby and hockey, 12 play squash and 2 play hockey
and squash. If 1 boy plays all three sports and 4 boys do not play any sport,
what is the probability that a Grade 12 boy chosen at random plays only
hockey and no other sport.
(Hint: Draw a Venn diagram.)
(6)
9.3 A team of 3 learners is chosen at random to take part in a debate. The team
is chosen from a group of 7 girls and 3 boys. Find the probability that:
9.3.1 only girls are chosen
9.3.2 two girls and one boy is chosen.
310
(4)
(3)
[15]
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Exam practice B: Paper 2
Time: 3 hours
Total: 150 marks
Question 1
1.1 If the following set of numbers (5; 7; 8; x; 15) has a mean of 9, calculate
the standard deviation.
(4)
1.2 A scientist has 100 female rats and 100 male rats. She measured their lengths
(excluding the tail) to the nearest cm and represented her results in the
box-and-whisker diagrams below.
-
-
-
-
-
-
Female rats
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
-
-
-
-
-
-
Male rats
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Use the diagrams to determine:
1.2.1 the interquartile range for the female rats
1.2.2 the range of lengths for all 200 rats.
(2)
(2)
[8]
Question 2
A number of bacterial cultures were grown in a laboratory for a Life Sciences
experiment. The results showing the number of bacteria in millions and their
ages in days are recorded in the table.
Age (x) in days
Number of bacteria (y) in millions
1
36
2
108
3
137
4
183
5
195
6
233
7
8
270
300
2.1 Draw a scatter plot with a scale up to 15 days on the x-axis and a scale up
to 415 million on the y-axis.
(3)
2.2 Determine the equation of the least squares regression line and the
correlation coefficient, r. Use r to comment on the results.
(4)
2.3 Readings were taken a few days later and recorded in the table below.
Age (x) in days
13
14
15
Number of bacteria (y) in millions
402
404
408
2.3.1 Plot these on your graph and describe what they show.
2.3.2 Would you use the least square regression line to predict the number
of bacteria for 16 days? Explain whether you would you be using
interpolation or extrapolation.
(2)
(2)
[11]
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Question 3
3.1 A(3;2), B(1;a) and C(4;5) are points in the Cartesian plane. Determine
the value(s) of a if :
3.1.1 A, B and C are collinear
___
3.1.2 the length of AB = √20 units.
(4)
(5)
3.2 Consider the points A(−2;8), B(2;4), C and D that are the vertices of
a parallelogram. The equation of line BD is 3y − x = 10.
A(–2;8 )
P
D
y
B(2;4)
x
C
Determine:
3.2.1 the coordinates of D and C
3.2.2 the equation of AC
3.2.3 the coordinates of P, the point of intersection of the two diagonals
3.2.4 by calculation that ABCD is a rectangle
^ B.
3.2.5 the magnitude of AD
(4)
(4)
(2)
(4)
(5)
[28]
Question 4
x2 + y2 + 4x + 6y − 12 = 0 defines the circle with centre M.
y
B
A
M
O
x
4.1 Determine the coordinates of M and the radius of the circle.
(5)
4.2 Find the coordinates of A and B, the points of intersection of the circle
and the x-axis.
(4)
4.3 Determine the equation of the tangent to the circle at the point A.
312
(4)
[13]
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Question 5
__
√5
2 , A < 90° and cos B = ___
5.1 Given sin A = __
, B > 90° determine, without
3
6
solving for A or B, the value of sin (B − A).
(6)
5.2 Without using a calculator, fully simplify the expression below.
Show all your working.
2 sin 165°.cos 345°
____________________________
(4)
[10]
cos 45° cos 15° + sin 45° sin 15°
Question 6
cos 2x
cos x − sin x
Given _____________
= ___________
3
1 + sin 2x
(cos x + sin x)
6.1 Prove the identity.
(5)
6.2 For which values of x in the interval [0°;360°] is the identity undefined?
(5)
[10]
Question 7
Given: f(x) = cos 2x and g(x) = sin(x − 45°)
7.1 Find the general solution to the equation cos 2x = sin(x − 45°).
(5)
7.2 Sketch the graphs of f and g on the same system of axes for x ∈ [−180°;180°].
Indicate the x- and y-intercepts and the coordinates of the turning points.
(6)
7.3 Use the graphs to find values of x ∈ [−90°;90°] for which f(x) > g(x)?
(2)
[13]
Question 8
In the figure below, AB and CD are two vertical towers of equal height 50 m
standing on a horizontal plane BDE. From E the angle of elevation of A is 42°
and the angle of elevation of C from E is 23°.
8.1 Find the lengths of EB and ED.
(4)
A
50 m
C
B
42°
E
50 m
23°
D
^ D = 55°, calculate the area and perimeter of the horizontal plane BED.
8.2 If BE
(4)
[8]
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Question 9
B
E
9.1 In the figure alongside, ABCD is a cyclic quadrilateral and TC is a tangent
to the circle centre O.
^ O= 30° is the angle between the chord AC and the radius OC.
AC
^
TCD = 70°.
A
Determine the following angles, giving reasons:
^C
9.1.1 AO
(2)
^
9.1.2 ADC
(2)
^
9.1.3 ABC
(2)
^D
9.1.4 CA
(2)
^C
9.1.5 AE
(2)
C
30°
70°
O
T
D
9.2 In the diagram, O is the centre of circle ABC.
C
O
A
B
Redraw the diagram to prove the theorem which states:
^ B = 2AC
^ B.
If O is the centre of the circle, then AO
(6)
9.3 In the diagram below O is the centre of the circle BCD. CD is produced
^ = 35°.
to A so that AO ⊥ BC. AO and BD intersect at E. B
1
B
2 1
35°
A
E
1
2
2
1
O
3
1 2
3
D
C
9.3.1 Calculate, with reasons, the size of:
^
(a)
O
2
^
(b)
D
1
9.3.2 Prove that COED is a cyclic quadrilateral.
9.3.3 Prove that ADOB is a cyclic quadrilateral.
314
(4)
(2)
(3)
(3)
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Question 10
10.1
A
D
E
B
G
F
C
In the diagram D is a point on AB and E is a point on AC such that DE ∥ BC.
G is the midpoint of BC. F is another point on BC such that AF ∥ EG
AD = __
1.
and ___
DB
5
Calculate, with reasons, the numerical value of the following:
AE
10.1.1 ___
(2)
10.1.2
(4)
AC
FG
___
FB
10.2 Complete the statement of the theorem that states:
If △ABC and △PQR are equiangular, then …
A
C
B
(2)
P
R
Q
10.3 In the diagram below MN and MQ are tangents to the circle PQN and
NM || PQ.
N
1 2
P
1
2
Q
10.3.1
10.3.2
10.3.3
10.3.4
M
^ = x, name, with reasons, three more angles equal to x.
If N
2
Prove that: △QNP ∣∣∣ △QMN.
Prove that: QN2 = QP.QM
If PQ = 35 and QN = 26, calculate the numerical value of QM.
(6)
(3)
(2)
(2)
[21]
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Term 4 summary
Exam Tips
How to prepare for final examinations
Here are suggestions to help you prepare for the final matriculation Mathematics papers:
• Do not cram! Start your preparations early. Remember that you have been preparing for these examinations since the start
of Grade 11. The curriculum for the Grade 12 examination papers covers everything that was taught in Grades 11 and 12.
You may have forgotten some of the details about the work done in Grade 11, so it will be necessary to revise it.
• Plan a carefully structured revision schedule that is realistic, and then keep to it.
• First ensure that you are fully aware of all of the definitions, laws, formulae and rules for each topic. Learn the proofs
that are required for examination purposes. Bookwork will make up part of the marks in the papers.
• Go through each topic and make a summary of the essential rules and methods, as well as reminders of when to use
each method. Write these out on clearly presented pages, using bold headings and colour. Put this somewhere where
you will often see it. Read it over many times.
• Use the internet if there is a topic you are feeling unsure of, or if there is a question that you are battling to understand.
There are videos that teach certain topics and websites that take you through step-by-step explanations of how to
tackle certain topics. The internet may also provide you with answers to questions that trouble you.
• The best way to revise mathematics is to do mathematics. Mathematics is a subject that requires lots of practice; it is
best that this practice is consistent and regular. You will improve your mathematics only when you start working through
problems. The more you practise, the more you will start getting the answers correct. The more answers you get correct,
the more your confidence will grow. Having confidence in your mathematics is the key to doing well in this subject,
especially when you are in examination situations. Start by practising the work topic by topic. First ensure that you have
remembered all the concepts and methods covered in each topic, and then move on to essential practice doing mixed
exercises or examination papers. This will help you recognise when to use each method.
• Make sure that you practise calculator work. The finance and statistics sections, in particular, require a thorough
knowledge of your calculator. If you do not have enough practice at this skill, you may find that you are either
unfamiliar with the process, or very slow at doing calculator work in the examinations.
• As you work through revision exercises, make sure you learn from your mistakes. Look back at tests and examinations
that you have written during Grades 11 and 12. These will remind you of the mistakes that you made in the past.
As you work through past tests and papers as well as revision exercises, make a list of your common errors as well
as concepts or methods that you had forgotten. Keep this list on hand and keep adding to it. Read through this list
regularly to remind you not to make the same errors in the next revision exercise or examination that you practise.
• Do not give up and be too quick to look at a memorandum when you are working through a revision exercise. Spend
time working at a problem before looking at the memorandum. If you keep at it and finally master the problem
yourself, you will have made far more progress in your learning that if you give up and look at the memorandum. If
you do need to look at the memorandum, analyse what it was that you had forgotten; add this to your list of points to
remember.
• It is sometimes useful to work together with a friend or group of people. Discussing a mathematics problem can be
beneficial in terms of hearing other people’s perspectives and thinking patterns. Telling others about your thoughts also
helps consolidate your own understanding. Explaining how to do a mathematics problem to someone else is one of the
best ways of ensuring that you have fully understood it.
• As the final examination date gets closer, practise revision papers under the same conditions that you will have in
the final examination. Do the paper in one sitting and complete it in the correct time allocation, without looking
at memoranda or notes. This will enable you to get used to concentrating for three hours without a break, as well
as ensure that you can complete the paper under the pressure of time constraints. This process of providing an
examination simulation assists in building your confidence for the final examination. It is very important that you
do not enter an examination feeling worried and stressed. If you have examination anxiety, start this examination
simulation process earlier to put yourself in an examination situation often enough for you to feel more confident in the
final examinations.
• Finish your examination preparations by looking at your summary sheets, the list of your common errors and the
bookwork required for the examination.
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Term 4 summary
Exam Tips
How to write examinations
Here are tips to help you cope with the final matriculation mathematics papers:
• Make sure that you have enough sleep the night before an examination! You need a clear, alert and calm mind, which
is only possible when you are fully rested.
• Have all necessary stationery (including a ruler, pencil, eraser and spare pen), and remember to check that your
calculator is set on deg and not rad or grad.
• Avoid last minute cramming and talking through the work with friends immediately prior to an examination. This often
causes you to panic.
• Read the instructions carefully (for example, take note of how many decimal places answers require). Make sure you
answer all parts of a question, and do not forget to look at the back page of the question papers.
• Use the reading time before the examination starts to plan your approach. Take note of the questions that you can see
will be difficult and leave them to the end. Look for the easiest questions and make sure that you do them early in the
examination time. You can answer your examination in any order, as long as you label your questions correctly.
• Do not panic! If you have prepared yourself well by revising and practising, you do not need to stress. If you come
across a question that you cannot do, leave it out and try not to think about it. There will always be questions in an
examination that are meant to extend you. Firstly complete all the questions that you know how to do when you are in
a calm state of mind. Then go back and tackle the more challenging questions.
• As soon as you are allowed to start writing, jot down the rules and identities that you think you may forget so you can
refer to them when you need them. However, do not spend too long writing out all the rules; this might cause you
time pressures towards the end of the examination.
• Do not spend too long on any one question. As you have 180 minutes in which to complete 150 marks, spend
1,2 minutes on each mark. If you have been battling with a question for too long, leave it out and move on. Once you
have finished the examination and if you have time, return to the questions that you did not complete.
• Show all your working. The mark allocation will be an indication of how much working is required.
• If a question requires you to use an answer from a previous question that you were unable to answer, estimate an
answer for the previous question so you can show that you know the method required for the new question. Tell the
examiner that your first answer is an estimate, and you will score part marks for using the correct method.
• Do not forget to write the units with answers that require units and to write degree symbols for angles.
• In questions where you are asked to prove, or show, a statement to be true, do not use what you have been asked to
prove. Remember to set out your argument clearly and to include all steps of working, giving reasons.
Term 4 summary
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Term 4 summary
Exam Tips
What bookwork do you need to know?
Here is a list of the derivations and proofs that you must learn for the final matriculation mathematics papers.
PAPER 1
Sequences
• Derivations of the formulae for the sum of arithmetic and geometric series, including S∞ .
PAPER 2
Trigonometry
sin θ ; sin2 θ + cos2 θ = 1
• Derivations of basic identities: tan θ = _____
cos θ
• Derivations of all compound and double angle formulae, accepting that cos ( α − β ) = cos α cos β + sin α sin β
• Derivations of the sine, co sine and area rules
Euclidean Geometry
Proofs of circle theorems:
• The line drawn from the centre of a circle perpendicular to a chord bisects the chord.
• The angle subtended by an arc at the centre of a circle is double the size of the angle subtended by the same arc
at the circle (on the same side of the chord as the centre).
• Angles subtended by a chord of the circle on the same side of the chord are equal.
• The opposite angles of a cyclic quadrilateral are supplementary.
• Two tangents drawn to a circle from the same point outside the circle are equal in length.
• The angle between the tangent to a circle and the chord drawn from the point of contact is equal to the angle in
the alternate segment.
Proofs of triangle theorems:
• A line drawn parallel to one side of a triangle divides the other two sides proportionally.
• Equiangular triangles are similar, and triangles with sides in proportion are similar.
• Pythagoras' Theorem using similarity
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Answers
TOPIC 1: EXERCISE 1
1.3
T40 = 268
1.1
d = −6
2.1
7; 12; 17
1.2
−13;−19
2.2
Tn = 5n + 2
1.3
Tn = 17 − 6n
2.3
100 is not a term in the sequence
1.4
T20 = −103
3.1
11; 8; 5
1.5
43 is not a term in the sequence because n ∉ ℕ
3.2
Tn = 14 − 3n
2.1
d=7
3.3
T17 = −37
2.2
30; 37
3.4
−72 is not a term in the sequence.
2.3
Tn = 7n – 5
4
T31 = 8
2.4
T12 = 79
5.1
Tn = 15 − 2n
2.5
n = 20
5.2
2.6
4
n = 28__
; No, n ∉ ℕ
T15 − n = 15 − 2 ( 15 − n ) = −15 + 2n
Tn + T15−n = ( 15 − 2n ) + ( ( −15 + 2n ) ) = 0
3
Sequence A:
6.1
3
7
__
Tn = __
2n + 2
3.1
x = −1
6.2
93 is not a term in the sequence.
3.2
Tn = 5 − 3n
7.1
Row 11 will have 43 Ks
3.3
T19 = −52
7.2
The letter T appears 79 times in row 20.
7
Sequence B:
3.1
x=4
3.2
Tn = 7n − 11
3.3
T19 = 122
Sequence C:
3.1
x = 1,5
3.2
Tn = −7,5 + 2,5n
3.3
T19 = 40
Sequence D:
TOPIC 1: EXERCISE 3
Sequence A:
1.1
1 875
1.2
Tn = 3 × 5n−1
1.3
Tn = 46 875
1.4
n=8
Sequence B:
1.1
2
1.2
a=2
3.1
x = − 0,5
3.2
Tn = − 6,5 + 1,5n
1.3
T7 = 8
3.3
T19 = 22
1.4
n – 1 = 7 and n = 8
Tn = 2n − 4
Sequence C:
TOPIC 1: EXERCISE 2
1.1
162
1.1
1.2
a = 2 and r = −3
−5; 2; 9
Tn = 2(– 3)n –1
Tn = 7n − 12
1.2
T10 = 58
1.3
T7 = 1 458
1.4
n=9
Answers
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Sequence D:
3.2
1.1
15
___
1.2
1
a = 60 and r = __
2
4
If x = 12 and y = 18, AP 6; 12; 18, GP 12; 18; 27
3
1
__
If x = __
4 and y = − 4 2 ,
3
3
1
1
__
__
__
AP 6; __
4 ;− 4 2 , GP is 4 ;− 4 2 ; 27
Tn = 15 × 23−n
15
T7 = ___
16
4.1
1.3
If x = 9, y = 12
If x = 1, y − 4
1.4
7n = 10
4.2
If x = 9 and y = 12, AP 6; 9; 12, GP 9; 12; 16
If x = 1 and y = − 4, AP 6; 1;− 4, GP is 1;− 4; 16
9
11
_____
; _____
Sequence A:
2.1
x = 10 or x = −2
5.1
2.2
If x = −2, Tn = −10
1 n−1
If x = 10, Tn = 98 × __
7
5.2
2.3
If x = −2, T10 = −10
1 9 _______
2
If x = 10, T10 = 98 × __
7 = 823 543
( )
TOPIC 1: EXERCISE 5
( )
Sequence B:
2.1
5
x = 3 or x = __
7
2.2
5
1(
__
)n−1
If x = __
7 , Tn = − 7 −5
(3)
1 n−1
If x = 3, Tn = 9 __
= 33−n
2.3
1 953 125
1(
__
_________
)9
10 = − 7 −5 =
7
5
If x = __
,T
7
1
( ) = _____
2 187
1
If x = 3, T10 = 9 __
3
9
1.1
Sn = − 4 650
1.2
Sn = 1 653
1.3
Sn = 1 090
2.1
40 [ ( )
(
)( ) ]
S40 = ___
2 2 2 + 40 − 1 8 = 6 320
2.2
Sn = – 4 040
2.3
Sn = 1 210
3.1
T1 = 37
T2 = 25
T3 = 13
Tn = 49 − 12n
3.2
T1 = 14; T2 = 11; T3 = 8
Sequence C:
2.1
1
x = − __
3 or x = 7
2.2
1
4(
__
)n−1
If x = − __
3 , Tn = − 3 −2
( )
1 n−1
If x = 7, Tn = 50 × __
5
2.3
Tn = 2n – 1
2 048
1
_____
If If x = − __
3 , T10 = 3
( )
1 9 ______
2
If x = 7, T10 = 50 × __
5 = 78 125
3.3
Tn = 17 – 3n
4
T1 = 2; T2 = 8; T3 = 14;
Tn = 2 + ( n − 1 ) ( 6 ) = 6n − 4
5
S55 = 5 445
6
First three terms: 2; 5; 8
Sequence D:
2.1
x = 49 or x = 4
2.2
If x = 4, Tn = 3 × ( −2 )n−1
7 n−1
If x = 49, Tn = 48 × __
4
2.3
If x = 4, T10 = −1 536
( )
Tn = 3n – 1
( )
121 060 821
7 9 ___________
If x = 49, T10 = 3 × __
4 =
16 384
TOPIC 1: EXERCISE 4
( )
1
1 n−1
T1 = 96; T2 = 24; T3 = 6; Tn = 96 × __
4
2
If r = −5, T1 = 3; T2 = −15; T3 = 75 and
Tn = 3 × ( −5 )n − 1
If r = 4, T1 = 3; T2 = 12; T3 = 48 and
Tn = 3 × 4n−1
3.1
320
1 024 4 096
2n − 1 ______
2n − 1
Tn = ______
4n = 22n
TOPIC 1: EXERCISE 6
1
S 12 = 107,17
2
S 10 = 557,89
3
6 560
S8 = _____
81
4
S10 = 3 906,2496 ≈ 3 906,25
5
T5 = 1
If x = 12, y = 18
3
1
__
If x = __
4, y = − 42
Answers
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TOPIC 1: EXERCISE 7
3.2
R255
1.1
S∞ = 54
3.3
17 weeks
1.2
80
S∞ = − ___
4
a
17
________
____
___
S∞ = _____
1 − r = 1 − 0,01 = 0,99 = 99
5
15,5 m
1.3
80
S∞ = ___
6.1
16 cm × 32 cm
1.4
81
S∞ = − ___
5
6.2
Unshaded area = 243,92 cm2
2.1
−2 < x < 6
7
280 m
2.2
4
S∞ = − __
5
3.1
1
1
It converges r = __
and −1 < __
<1
3.2
S∞ = 67,5
3
3
0,17
0,17
TOPIC 1: EXERCISE 10
3
3
TOPIC 1: EXERCISE 8
1.1
Tn = 7n − 5
1.2
∑( 7n − 5 )
1.1
–57 and –81
1.2
Tn = −2n2 + 2n + 3
1.3
T17 = −541
2
If x = −20 and y = 20
GP: 20; – 20; 20
AP: – 20; 20; 60
If x = 30 and y = 45
GP: 20; 30; 45
AP: 30; 45; 60
3
1
p = __
2 and q = 5
48
n=1
1.3
Sn = 7 992
2.1
Tn = 2 × 9n−1
2.2
n=8
p = 8 and q = 20
8
∑2 × 9n 1
4.1
4 + 8 + 12 + … + 732
4.2
3 + 9 + 27 + … 729
4.3
67 344 + 1 092 = 68 436
∑ ( 3)
n 1
5.1
5 + 10 + 15 + 20 + … + 100
5.2
3 + 6 + 12 + … + 96
3.3
Sn = 21,6
5.3
1 050 + 189 = 1 239
4
n=8
6.1
67 and 92
6.2
Tn = 2n2 − n + 1
7.1
1 ( )( )
Area of Triangle 1 = __
2 2 1 =1
2.3
−
n=1
2.4
Sn = 10 761 680
3.1
2
Tn = 36 − __
3
∞
3.2
( )
n−1
2 n−1
36 − __
=
5
5
∑r 3 = 224
r=2
1 ( )( )
Area of Triangle 2 = __
2 4 2 =4
TOPIC 1: EXERCISE 9
1
S∞ = 4 915,2 cm2
2.1
7th week
2.2
138 km
2.3
S10 = 840 km
3.1
He will not be able to afford the 9th payment.
The final payment he can afford is R128.
1 ( )( )
Area of Triangle 3 = __
2 6 3 =9
1 ( )( )
Area of Triangle 3 = __
2 8 4 = 16
1 ( )( )
2
Area of Triangle n = __
2 2n n = n
7.2
This is a quadratic sequence or pattern.
Answers
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TOPIC 1: REVISION TEST
5.1
The second difference is 6.
Sequence A:
5.2
The first term is – 4
1.1
Arithmetic sequence
5.3
Tn = 3n2 − 2n − 5
1.2
Tn = 5n − 4
6.1
−7 < x < 7
1.3
n = 20
6.2
196
S∞ = − ____
11
1.4
T308 = 1 536
7.1
n = 19
1.5
T20 = 96
7.2
S10 = 150
50
∑( 5n − 4 )
8.1
x = −11
1.6.1
n=1
8.2
Tn = −2n2 + n + 4
50 [ ( )
( )]
⇒ S50 = ___
2 2 1 + 49 5 = 6 175
9.1
She cycled 90 km on the 36 day
Sequence B:
9.2
777 km
1.1
geometric sequence
9.3
1.2
Tn = 3 × 2n−1
On the 90th day she will cycle 282 km.
This is an excessive distance, it is unlikely that
she will be able to maintain this programme!
1.3
n=6
10.1
–3
1.4
T10 = 1 536
10.2
1.5
T20 = 1 572 864
Tn = n2 − 14n + 45
1.6.2
11.1
50
1.6.1
Sn = a + ar + ar 2 + … + ar n−2 + ar n−1
∑3 × 2
n−1
➀
n=1
1.6.2
r Sn =
S50 = 3 ( 250 − 1 )
Sequence C:
1.1
quadratic sequence
1.2
Tn = n2 + 2n − 3
1.3
T9 = 96
1.4
1 536 is not a term in the sequence as n must be
a natural number.
2
ar + ar + … + ar
+ ar
n−1
0
−ar n
+ ar
n
➁
➀ − ➁: Sn − r Sn = a 0
0
0
a ( 1 − rn )
a ( rn − 1 )
________
Sn ( 1 − r ) = a ( 1 − rn ) ⇒ Sn = ________
(1 − r) = r − 1 , r
≠1
3
11.2.1 r = __
5
( __35 ) → 0 as n → ∞
n
∞
11.2.2
∑15 × ( __35 )n 1
−
1.5
T20 = 437
1.6
Not applicable as this is a quadratic sequence.
11.2.3 S∞ = 37,5
2.1
x = 25
11.3.1 S12 = 1 062 880
2.2
x = ± 24
11.3.2 T12 = 708 588
3
A = 2 441 406,25
4.1
45 and 80
11.3.3 T1 = 4
T2 = 12
4.2
5 242 880 − 205 = 5 242 675
4.3
3 125 + 167 772 155 = 167 775 280
4.4
Tn = 5( 2n − 1 ) and Tn = 5 × 2n−1
Both formulae are divisible by 5, so all the terms
in the series will be divisible by 5.
322
n−2
n=1
r = 3, Tn = ar n−1 = 4 × 3n−1
12.1
9
11
___
and ___
12.2
2n − 1
Tn = ______
n
13.1
p = 10 and q = 30
32
64
2
Answers
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13.2
Tn = n2 + 4n − 2
9.2.3
Tn = −2n − 3 + 3n+1
14.1
1,88 m
10.1
T11 = S11 − S10 = 29
14.2
Maximum height is 1,89 m
10.2
Tn = 3n − 4
15.1
d=2
10.3
n = 28
15.2
1; 3; 9
11.1
1
1
__
r = − __
2 or r = 2
16
14; 28; 56; 112
11.2
1
If r = − __
2 , 30; –15; 7,5
17
13; 4;−5
1.1
x=4
1.2
Tn = 22 − 5n
1.3
T29 = −123
1.4
S20 = − 610
1.5
S20 − S10 = −555
2.1
x = 25
2.2
1
1
__
Converging series: r = __
3 , −1 < 3 < 1
2.3
S∞ = 243
3.1
Tn = 3n + 2
3.2
n = 100
1
If r = __
2 , 10; 5; 25.
12
x=7
7; 21 and 56
13.1
76
13.2
−34 450
13.3
10
14
225 m
15.1
15.2
1
1
− __ < x < __
2
2
16
___
x = − 33
16.1
T6 + T7 = − 4 + 97 = 93
16.2
n = 11
100
∑( 3n + 2 )
16.3
79
n=1
17
n = 6 or n = 15
3.4
S100 = 15 350
18.1
33
27
___
and ___
3.5
S100 − S50 = 11 425
4
3
765 __
96 − ____
8 =8
18.2
6n − 3
Tn = ______
2n
5
n = 32
19
531 441π
S∞ = ________
= 98 210,07 units
17
6
Refer to page 10 in your book for the proof.
20.1
8x; 4x; 2x; ...
7
32 + 34 + 36 + … is a geometric series with
a = 9, r = 9 en n = 20
S20 = 9 + 92 + 93 + ... + 920
➀
9S20 =
92 + 93 + ... + 920 + 921
➁
20.2
4√ 3 x; 2√3 x; √ 3 x
20.4
16√ 3 x2; 4√ 3 x2; √ 3 x2
20.5
S∞ =
20.6
x = 30
3.3
S20 − 9S20 = 9 0
0
0
−921 | ➀ − ➁
−8S20 = 9( 1 − 920 )
− 9 ( 920 − 1 )
64
__
__
__
9 ( 920 − 1 )
S20 = _________
= ___________ = _________
−8
−8
8
__
__
__
64√3 x2
_______
3
1.1
One-to-one
1.2
Many-to-one
8
1
r = __
3 , a = 162
1.3
One-to-many
9.1
The third term = 72
1.4
One-to-one
9.2.1
Tn = −2n − 3
1.5
Many-to-one
9.2.2
= 3n+1
2.1
Yes
Tn
__
TOPIC 2: EXERCISE 1
Take out a common factor of 9
9 ( 1 − 920 )
32
Answers
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TOPIC 2: EXERCISE 2
2.2
No
2.3
Yes
2.4
No
3.1
y ≥ 2 or y ≤ 2
3.2
y ≥ − 1 or y ≤ − 1
3. 3
y ≥ 2 or y ≤ 2
4.1
f(–1) = 9, f (0)= 6
f (2) = 0
4.2
Domain: {−1; 0; 2} Range: {9; 6; 0 }
1.1
4.3
a) and b)
y
f ˉ¹
(0;1)
(0;–3)
f
y
(2;0)
c)
x+3
y = _____
3
d)
Domain f: x ∈ ℝ
Range f: y ∈ ℝ
Domain f −1: x ∈ ℝ
Range f −1: y ∈ ℝ
x
1.2
a) and b)
y = –2x – 6
4.5
f(x) is a one-to-one function
5.1
g(−1) = 0
g(0) = 3
g(1) = 0
1.3
c)
−x + 6
x
y = _______
= −__ + 3
2
2
d)
Domain f: x ∈ ℝ; Range f: y ∈ ℝ
Domain f −1: x ∈ ℝ Range f −1: y ∈ ℝ
a) and b)
(–3;1)
x
(–3;–1)
(–1;–3)
(1;0)
f
c)
5.4
g(x) is a function because a vertical line cuts it
once. This means for every x there is one
y value.
5.5
g(x) is a many-to-one function
4
9
16
d)
1.4
0
1
25
4
5 or 6 or 7 or 8 or 9 or
3
2
1
0
-1
It is not a function because for every x, except
x = 0, there are two y values, which shows it is a
one-to-many relation.
324
(1;–3)
x
g
y
y= x
y
f ¯¹
y
x
x -
f ¹
(3;0)
(0;3)
6
(6;0)
f
Domain: {−1; 0; 1}
Range: {0; 3; 0 }
(–1;0)
y= x
(0;3)
f(x) is a function because for every x there is one
corresponding y value.
5.3
y
(0;6)
f
5.2
x
(1;0)
(–3;0)
(0;6)
4.4
y = 3x – 3
y = –3x ²
___
−x
___
y=±√ 3 ,x≤0
Domain f: x ∈ ℝ
Range f: y ∈ ℝ, y ≤ 0
Domain f −1: x ∈ ℝ, x ≤ 0
Range f −1: y ∈ ℝ
a) and b)
f
(–2;5)
y y = x² + 1
y=x
(2;5)
(5;2)
(0;1)
(1;0)
f
–1
x
(5;–2)
Answers
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_____
c)
y = ± √x − 1
d)
Domain f: x ∈ ℝ
Range f: y ∈ ℝ, y ≥ 1
Domain f −1: x ∈ ℝ, x ≥ 1
Range f −1: y ∈ ℝ
5.4
(0;0)
2.2
3
1
y = − __
x − __
5.5
2.3
___
y = ± √ −x , x ≤ 0
6.1
2.4
f –1 is not a function. f is a many-to-one function
and the inverse is f –1. Therefore this is a one-tomany relation and not a function.
Restriction: x ≥ 0 or x ≤ 0
2
1
1
__
LHS = −2x − 3 + ____
2 + 3 = −2x − 2
−x
x
2
RHS = ____
− (−2(−x) − 3) −3
−x2
___
f(x) and g(x) are functions.
f and g are inverses.
6.3
f and g are inverses.
6.4
f and g are inverses.
TOPIC 2: EXERCISE 3
a)
y
x2
a) and b)
6
5
4
3
2
1
y
f –1
(3;–3)
____
: y = ± √ −3x x ≤ 0
b)
f
f
1
2
1
y = ______
−2
(x − 3)
a)
y
f
g −1: y = −3x
f
8
f
6
y=x
3.4
See graph in question 3.1
1
4
2
−1
f (x) is not a function because for for every x
there is more than one y value
3.6
If the domain of f(x) is restricted to x ≥ 0 or
x ≤ 0 then f −1(x) will be a function.
4.1
f(x) is a function because for every x, there is
one corresponding y value
4.2
x ≥ 0 or x ≤ 0.
4.3
x−4
y = _____
2
4.4
y ∈ ℝ, y ≥ 0
5.1
5
a = __
4
5.3
5 6
−1
3.3
5.2
x
1 2 3 4
–3
–4
–5
–6
x
g
(–3;–3)
3.5
f
–6 –5 –4 –3 –2 –1
–2
(3;–1)
3.2
1
4
y = + √__
5x
∴ LHS = RHS
3.1
f
6.2
1
2
2
= ____ − 2x + 3 − 3 = −2x − __
−x2
x
(5;–2)
A(−1;−1) B(3;−9)
2.5
y
A(–2;5)
2.1
2
f
– 4 –2
–2
–4
b)
3
_____
y = 3 ± √x + 1
a)
y
f
6
5
4
3
2
1
–6 –5 –4 –3 –2 –1
–2
___
4
y = − √__
5x
Domain of f : x ∈ ℝ, x ≤ 0
Range of f −1: y ∈ ℝ, y ≤ 0
x
6 8
2 4
(3;–1)
1 2 3 4
5 6
–3
–4
–5
–6
b)
1
y = ______
+4
(x − 2)
Answers
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4
a)
2.3
y
a)
y
2
1
(– 4;1)
–7 –6 –5 – 4 –3 –2 –1
1; 9
4
1;
9
4
1 2
9
;1
4
(1;–4)
9 ;–1
4
–2
–3
–4
–5
–6
b)
c)
f –7
b)
_____
y = − 4 ± √1 − x
2.4
TOPIC 2: REVISION TEST
1.1
1.2
1.3
1.4
1.5
1.6.1
1.6.2
2.1
True. x and y have been swapped around.
False. It is a one-to-many relation.
False. It is a many-to-one function. The inverse
is therefore one-to-many.
False. They are reflections in the x-axis but not
inverses, which are reflections in the line y = x.
True. The vertical line test shows it cuts the
graph more than once.
False. The range must be restricted to y ≥ 1, for
it to be a function.
False. The maximum valve of the inverse graph
is 2.
a)
y
___
4
2√
__
y = ± √ __
9x = ± 3 x
Range of f : y ∈ ℝ; y ≥ 0
Domain of f −1: x ∈ ℝ; x ≥ 0
__
a)
y
(O;2)
(–8,0)
(2;0)
(–2;0)
(0;–2)
2.5
____
x+8
_____
√ 2
y=±
c)
Range of f : y ∈ ℝ; y ≥ − 8;
Domain of f −1: x ∈ ℝ; x ≥ − 8
a)
y
(4;5)
2.2
(4;0)
(0;–1)
1x + 2
y = −__
2
Range of f: y ∈ ℝ
Domain of f −1: x ∈ ℝ
a)
(–2;0)
(4;0)
(0;–2)
326
4
y = _____
x−4
c)
Range of f: y ∈ ℝ; y ≠ 4;
Domain of f − 1: x ∈ ℝ; x ≠ 4
A and C
B and D
3.2
A and D
E and C
4.1
q(x) = y = 2x − 6
4.2
p(x) = y = (x − 1)2
4.3
y = (x + 3)2 + 2
5.2
)
3
9 __
C(2;−1) and D ( __
2;2 )
5.3
y=x−3
5.1
y = 2x + 4
Range of f: y ∈ ℝ
Domain of f −1: x ∈ ℝ
b)
3.1
y
(0;4)
b)
c)
(5;4)
(–1;0)
(2;0)
b)
c)
(0;–8)
b)
(0;4)
(0;2)
(7)
5.4
(
3 __
9
A( −1;2 ) and B __
2; 2
__
x
y = ± √__
2
Answers
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5.5
y
TOPIC 3: EXERCISE 1
g
39
B 2; 2
1.1
x=4
1.2
3
x = __
5
1.3
x = −6
1.4
x=6
2.1
a)
f : y = 5x
1 x
g: y = 5 − x = __
5
b)
Domain f = Domain g: x ∈ ℝ
Range f = Range g: y ∈ ℝ, y > 0
c)
a=2
a)
f: y = 2x
g(x) = y = −2x
(
)
1
A(1;1) and D ( −5;−__
5)
b)
Domain f = Domain g: x ∈ ℝ
Range f: y ∈ ℝ, y > 0
Range g: y ∈ ℝ, y < 0
1
f −1: y = __
x
f and f −1 are the same graphs because
1
the graph of y = __
x is symmetrical in the line
y=x
c)
1
a = ___
16
D 9; 3
A(–1;2)
g
22
C(2;–1)
5.6.1
x>0
5.6.2
x < −3
5.6.3
x ≥ 0 or x ≤ 0
5.7.1
3
5.7.2
−2
6.1
6.2
6.3
6.4
2.2
1 ;−5
A(1;1) and B −__
5
2.3
y
g
1
A(1;1)
1
5
B – ; –5
6.6.1
0 < x ≤ 1 or x < −5
6.6.2
4
0 < x ≤ __
7.1
f(−x) = B, f −1(x) = C, −f(x) = A
7.2
C is not a function.
The range must be restricted to y ≥ 2 or y ≤ 2
8.1
4
y = _____
x−1
8.2
y = 1 ± √x − 1
8.3
3
y=√
x−1
5
_____
__
( )
x
b)
Domain f = Domain g: x ∈ ℝ
Range f : y ∈ ℝ, y > 0
Range g: y ∈ ℝ, y < 0
c)
9
a = − __
4
g
y=x
C –5 ; 5
( )
2 x
f: y = __
3
a)
2
g(x) = y = − __
3
x+4
y = _____
5
6.5
( )
TOPIC 3: EXERCISE 2
1.1
25 = 32
1.2
103 = 1 000
1.3
1
3−4 = ___
81
1.4
53 = 125
1.5
( __12 ) 6 = 64
2.1
log2 16 = 4
2.2
1
log3 ___
27 = −3
2.3
loga 5 = 2
2.4
log3 8 = x
2.5
log2 x = y
2.6
logc a = 3
3.1
x=5
3.2
1
x = ____
100
3.3
x = −3
3.4
x=2
−
Answers
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3.5
x=7
4.1
x = 2,32
4.2
x = 1,93
4.3
x = 8,83
4.4
x = 0,56
4.5
x = 3,95
1.3
a) to c) See graph
y y= 5
1
–2;25
4.7
x=4
4.8
x=4
4.9
x = 2 or x = −1
4.10
x = 3 or x = −1
y = log x
5
x
(1;0)
1
;–2
25
TOPIC 3: EXERCISE 3
f(x) = 4x
1.4
a) to c)
y y = 4x
d)
f − 1(x): y = log 5 x
e)
Range of f(x): y ∈ ℝ, y > 0
Domain of f −1(x): x ∈ ℝ, x > 0
f)
f(x) and f −1(x) are both increasing
functions.
a) to c) See graph
y=x
x
y= 1
2
(2;16)
(0;1)
–3; 1
64
(16;2)
y = log x
x 4
(1;0)
(–1;2)
(0;1)
e)
Range of f(x): y ∈ ℝ, y > 0
Domain of f −1(x): x ∈ ℝ, x > 0
f)
f(x) and f −1(x) are both increasing
functions
a) to c)
(1;0)
(2;–1)
1 ;–3
64
y = log4 x
4
y= x
y
(–3;8)
d)
x
y= 1
2.1
x
y = log 1 x
2
(8;–3)
d)
f −1(x): y = log__1x
e)
Range of f(x): y ∈ ℝ, y > 0
Domain of f −1(x): x ∈ ℝ, x > 0
f)
f(x) and f −1(x) are both decreasing
functions.
2
a) to c) See graph
x
y y = 2 +1y = x
(2;5)
y=x
y
(−2;16)
(–1;4)
(0;1)
(0;2)
(1;0)
(4;–1)
(5;2)
y = log (x –1)
2
(2;0) x
x
y = log 1 x
(16;–2)
d)
f −1(x): y = log __1 x
e)
Range of f(x): y ∈ ℝ, y > 0
4
d)
f −1(x): y = log2(x − 1)
e)
Range of f(x): y ∈ ℝ, y > 1
Domain of f −1(x): x ∈ ℝ, x > 1
f)
f(x) and f −1(x) are both increasing
functions.
4
Domain of f −1(x): x ∈ ℝ, x > 0
328
y=x
(25;2)
(0;1)
x=1
1.2
x
(2;25)
4.6
1.1
f(x) and f −1(x) are both decreasing
functions
f)
Answers
9780636143319_plt_mat_g12_lb_eng_zaf.indb 328
2013/05/31 11:19:01 AM
2.2
1.2
a) to c) See graph
x
x = –2 y y =3 – 2
2.3
y=x
(2;9)
y =x
(2;7)
(–1;0) (1;1)
(0;–1)
y y = 3x
(7;2)
y = log3( x + 2)
x
y = –2
(9;2)
(0;1)
1
–3; 27
d)
f −1(x): y = log3(x + 2)
e)
Range of f(x): y ∈ ℝ, y > −2
Domain of f −1(x): x ∈ ℝ, x > −2
f)
f(x) and f −1(x) are both increasing
functions.
(1;0)
1
27 ;–3
Symmetrical about or reflections in the line
y=x
1.3
y = 3x
y
a) to c) See graph
x
y
y = 13 + 3
(1;3)
y=x
_1
–3; 27
(0;1)
x
(–1;6)
(0;–1)
(0;4)
y= 3
(4;0)
(6;–1)
(1;–3)
x
x
y = –3
y = log 1 (x – 3)
Symmetrical about the x-axis or reflections in
the line y = 0
3
x =3
1.4
y
−1
d)
f
e)
Range of f(x): y ∈ ℝ, y > 3
Domain of f −1(x): x ∈ ℝ, x > 3
(x): y = log __1(x − 3)
3
−1
f)
f(x) and f (x) are both decreasing
functions.
TOPIC 3: EXERCISE 4
1.1
y = log 3x
x
1
4 ;2
(1;0)
1
4 ;–2
(8;3)
y = log 2 x
x
y = – log 2 x = log 1 x
(8;–3)
2
y
1
;3
64
(1;0)
1
;–3
64
(16;2)
y = log 4 x
Symmetrical about the x-axis or reflections in
the line y = 0
x
y = log x
1
4
(16;–2)
Symmetrical about the x-axis or reflections in
the line y = 0
Answers
9780636143319_plt_mat_g12_lb_eng_zaf.indb 329
329
2013/05/31 11:19:01 AM
1.5
y
y=5
1.2
x
a)
f −1: y = log6 x
b)
See graph
y =6 x
y
(1;5)
1 (0;1)
–3; 125
(1;0)
y=x
(2;36)
x
(36;2)
y = log6x
x
(0;1)
(5; –1)
y = log 1 x
1
(1;0)
–2; 36
1
;–2
36
5
No symmetry
1.6
y
y = –log1x
3
1
– –;
2
9
1
–;
2
(–1;0)
(9;–2) y = log x
9
(1;0)
x
1.3
1
3
c)
g: y = − 6x
d)
1 x
h: y = __
6
a)
f −1: y = 3x
( )
b)
Symmetrical about the y-axis or reflections in
the line x = 0
y y =3 x
(2;9)
TOPIC 3: EXERCISE 5
1.1
a)
b)
(9;2)
y = log3x
x
(0;1)
f −1: y = log__1 x
1
y
(1;0)
–2; 9
5
x
y= 1
5
y=x
1
;–2
9
y=x
(–2;25)
(0;1)
(1;0)
xx
y = log 1 x
5
(25;–2)
1.4
c)
g: f (x) = log __1 x
d)
h: y = log 3(−x), x < 0
a)
2 x
f −1: y = __
3
d)
h: y = 5x
( )
b)
y= 2
x
y
3
9
–2; 4
(5)
c)
1 x
g: y = − __
3
y=x
(0;1)
x
y = log 2 x
(1;0)
9
;–2
4
330
c)
g: y = −log __2 x = log __3 x
d)
h: y = log __2(−x), x < 0
3
3
2
3
Answers
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2013/05/31 11:19:01 AM
2.1
a)
b)
1
a = __
3
2.4
( )
1 x
f −1: y = __
3
c)
y= 1
1
a = __
8
a)
c)
x
y
3
y = 18
y=x
(–2;9)
x
y
y=x
(1;0)
x
y = log 1 x
(–2;64)
(0;1)
3
(9;–2)
d)
g(x) = log3 x
e)
x ∈ ℝ, x > 0
a)
a=2
b)
f −1: y = 2x
c)
(0;1)
x
y = log 1 x
(1;0)
2.2
( )
1 x
f −1: y = __
8
b)
(64;–2)
d)
e)
8
g(x) = log8 x
x ∈ ℝ, x > 0
TOPIC 3: EXERCISE 6
y y =2
1
x
y
y=x
(3;8)
(0;1)
–2;1
4
Q(2;9) y = 2
(8;3)
(1;0)
1 ;–2
4
y= log x
2
x
x
y = log x
3
2.3
d)
g(x) = log__1 x
e)
x ∈ ℝ, x > 0
a)
a=5
b)
−1
log3 x ≤ 2 when 0 < x ≤ 9
2
f
2
y
y = log 1 x
4
: y = 5x
c)
y
y = 5x
y=x
x
y = –1
Q(4;–1)
(0;1)
(1;5)
(5;1)
(1;0)
y= log5 x
x
log__1 x ≥ −1 when 0 < x ≤ 4
4
d)
g(x) = log__1 x
e)
x ∈ ℝ, x > 0
5
Answers
9780636143319_plt_mat_g12_lb_eng_zaf.indb 331
331
2013/05/31 11:19:02 AM
3
y = ± √4x
3.6
it is a one to many relation
3.7
x > 0 or x < 0
3.8
y = log__1 x
4.1
k=4
4.2
At Q: y = 0
x = 40 = 1
4.3
0 < x ≤ 16
4.4
y = 4x
5.1
y=1
5.2
x=1
x
5.3
y = log2(x − 1)
y = log 1x
6.1
2
b = __
3
6.2
y = log__2 x
6.3
y=x
y = log x
5
y=1
x
Q(5;1)
log5 x ≥ 1 when x ≥ 5
4
y
y=x –1
P(1;0)
2
log__1 x ≥ x − 1 when 0 < x ≤ 1
2
5
___
3.5
y
4
3
6.4
y
y
f
f
–3; 27
y = log x
8
4
9
–2; 4
x
P(1;0)
–2;– 9
(0;1)
(0;–1)
4
y = – 2x+2
TOPIC 3: REVISION TEST
3
(3)
6.5
2 x
y = − __
6.6
Graph of h
6.7
y ∈ ℝ, y < 0
6.8
h is an increasing function because as x
increases the corresponding values of y increase.
x = −3
1.2
x=7
1.3
x = 16
1.4
x = 3,11
7.1
y = −1
1.5
x = 3,8
7.2
x = −1
1.6
x = 1,66
7.3
2
31
x
y = ± √__
2
7.4
( )
3.1
1 2
__
a is g(x) = __
4 x , b is f(x) = 4
3.2
P = (0;1)
3.3
1
k = __
4
3.4
x<0
332
Answers
x
9 ;–2
4
27 ;–3
8
y = log 2 x
1.1
1 x
(1;0)
h
log4 x ≤ −2x + 2 when 0 < x ≤ 1
9780636143319_plt_mat_g12_lb_eng_zaf.indb 332
-1
__
x≥0
8.1
y = b0 = 1
P = (0;1)
8.2
b=4
a = −3
a = −3, b = 4, p = 1, q = 4
2013/05/31 11:19:02 AM
8.3
( )
x
1
f(−x) = 4−x = __
4
reflection in the y-axis
x
8.4
h(x) = logb x and f(x) = b
f(x) → h(x) is a reflection in the line y = x
8.5
Range of f: y ∈ ℝ, y > 0
Range of g: y ∈ ℝ, y ≤ 4
8.6
x<1
8.7
One-to-many
8.8
x ≥ 1 or x ≤ 1
9.1
a = −2, y = 5
9.2
y ∈ ℝ, y > 1
9.3
f −1(x) = log3 x
9.4.1
9.4.2
x > 2,5
0 < x ≤ 1 or x ≥ 2,5
10.1
10.2
5.2
Interest = R31 492,60
6
F = R30 163,88
7
x = R11 000
8
x = R18 252,73
TOPIC 4: EXERCISE 2
1
R130 672,27
2.1
R63 756,22
2.2
R18 756,22
3.1
R4 213,94
3.2
R1 988 654,40
4.1
R388 606,50
4.2
R1 897,08
a=k=2
4.3
R146 904,32
f −1(x) = log2 x
5
R71 691,21
6
x = R608,41 (using ANS key for more accurate
i(12))
10.3 and 10.4
y
f
y=x
f –1
TOPIC 4: EXERCISE 3
1
R11 796,86
2
R96 785,36
3.1
R2 045,05
3.2
R310 812
4
R16 222,04
5.1
R10 120,81
10.5.1 0 < x ≤ 1
10.5.2 0 < x ≤ 2
5.2
R55 266,20
6.1
i ( 12 ) ≈ 11,39%
TOPIC 4: EXERCISE 1
6.2
R3 124,24
7.1
9,8%
7.2
R43 418,94
7.3
R51 081,11
7.4
R65 262,17
7.5
R14 181,06
Q(1;2)
P
T(2;1)
g
x
1.1
n ≈ 8,3%
1.2
i = 14,3%
2
n ≈ 5,45 years
3.1
x = R334 283,73
3.2
ieff = 10,47%
4
i = 10,09%
He would need an interest rate of more than
10,09%p.a.
5.1
F = R3 117,85
TOPIC 4: EXERCISE 4
1.1
R91 714,08
1.2
R1 055 456,98
1.3
R963 742,90
Answers
9780636143319_plt_mat_g12_lb_eng_zaf.indb 333
333
2013/05/31 11:19:02 AM
1.4
R7 517,56
TOPIC 4: EXERCISE 9
2.1
12%
1.1
i ( 12 ) ≈ 10,11%
2.2
8%
1.2
F = R634 814,49
2.3
R846 190.81
1.3
Total value = R3 920 763,57
2.4
R4 318,06
1.4
Barry is better off by R155 249,12
2.1
Option 1: R11 092,36
Option 2: R14 619,15
2.2
Option 1: R1 682 166,40
Option 2: R1 231 447
Option 2
TOPIC 4: EXERCISE 5
1
R17 519,54
2.1
R246 036,80
2.2
R1 782 866,51
2.3
2.3
R1 536 829,71
TOPIC 4: REVISION TEST
2.4
R11 536,39
1.1
20 years
3
R2 610,03
1.2
11,53 years
4
R154 718,82
2.1
R80 926,01
TOPIC 4: EXERCISE 6
2.2
R38 432,49
1.1
R4 506,38
3
11,35%
1.2
R141 711,25
4
R27 792,35
2
R717 406,04
5.1
R1 005 421,52
3
R436 756,04
5.2
R191 360,30
TOPIC 4: EXERCISE 7
5.3
R814 061,22
1
R37 392,54
5.4
R7 261,06
2.1
8,8%
6.1
R1 200,38
2.2
R5 189,33
6.2
R36 931,99 or R36 932,04
2.3
R179 886,29
7.1
R56 225,03
2.4
R135 634,45 short
7.2
R65 945,66
3
R11 212,94
8.1
4
Susan has R605,07 more than Pieter.
R900 will not cover interest on loan
∴ bank will not grant the loan
5
R45 248,62
8.2
85 payments (84 of R2 000 and one smaller
payment)
TOPIC 4: EXERCISE 8
8.3
R1 306,11
1
8.4
R44 306,11
n = log1,1 1,61051 = 5 years
176 344,70
__________
≈ 8 years
2
n = log0,91 375 000
9.1
12,5% p.a.
3.1
R318 848,35
9.2
R2 525,36
3.2
91 payments of R5 000
10.1
10,11%
3.3
R1 718,41
10.2
R634 814,49
3.4
Impossible to pay back the loan with payments
of R1 000
10.3
R3 920 763,57
334
Answers
9780636143319_plt_mat_g12_lb_eng_zaf.indb 334
2013/05/31 11:19:02 AM
10.4
Barry is better off by R155 249,12
3
Identity proofs
11
x = R76 897,68
3.1
cos(90° + θ) = −sinθ
LHS = cos 90° cos θ − sin 90° sin θ
= 0.cos θ − 1 sin θ
= −sin θ
= RHS
3.2
sin( 360° − θ ) = −sin θ
LHS = sin 360° cos θ − cos 360° sin θ
= 0.cos θ − 1.sin θ
= −sinθ
= RHS
3.3
cos( A + B ) + cos(A − B) = 2cos A cos B
LHS = cos A cos B − sin A sin B + cos A cos B
+ sin A sin B
= 2 cos A cos B
= RHS
3.4
sin ( A+B ) + sin ( A − B ) = 2 sin A cos B
LHS = sin A cos B + cos A sin B + sin A cos B
− cos A sin B
= 2 sin A cos B
= RHS
3.5
cos ( A + B )cos(A − B) = cos2A – sin2B
LHS = (cos A cos B − sin A sin B)(cos A cos B
+ sin A sin B)
2
= cos A cos2 B − sin2 A sin2 B
= cos2 A(1 − sin2 B) − (1 − cos2 A)sin2 B
= cos2 A − cos2 A sin2 B − sin2 B + cos2 A sin2 B
= cos2 A – sin2 B
3.6
sin ( A + B )sin(A − B) = cos2 B – cos2 A
LHS = (sin A cos B + cos A sin B)(sin A cos B
− cos A sin B)
2
= sin A cos2 B − cos2 A sin2 B
= (1 − cos2 A)(cos2 B) − cos2 A(1 − cos2 B)
= cos2 B − cos2 A cos2 B − cos2 A + cos2 A cos2 B
= cos2 B – cos2 A
= RHS
3.7
cos(A − 30°) = sin(60° + A)
LHS = cos A cos 30° + sin A sin 30°
TOPIC 5: EXERCISE 1
1.1
___
√ 61
1.2.1
5___
____
1.2.2
37
2.1
–32
2.2
4
− __
7
3
1
√ 61
4.1–4.2
1.2.3
5
__
6
Identity proofs
5.1
cos x
1
______
= − _____
5.2
sin x
− _____
sin x = −1
6.1
−k
7.1
− √3
8.1
15,24° or 124,76°
8.2
x = – 30°; 150°; 120°; – 60°
8.3
θ = ± 104,48° + n.360° or θ = 0° + n.360°, n ∈ ℤ
8.4
θ = 12°; 192°; –168°; –348°
8.5
n = 27,5° + n.90° or θ = −55° + n.180°, n ∈ ℤ
− sin x
__
tan x
_____
6.2
√ 1 + k2
7.2
1
− __
2
6.3
_____
√ 1 + k2
TOPIC 5: EXERCISE 2
1.1
cos θ cos 2 β − sin θ sin 2β
1.2
cos 3x cos y + sin 3x sin y
1.3
sin 3θ cos 2β + cos 3θ sin 2β
1.4
sin x cos 4y − cos x sin 4y
2.1
2.2
2.3
2.4
1
__
√3
___
2
cos 80°
__
√3
___
2
2.5
cos x
2.6
cos 2x
2.7
1
__
2.8
2 __
√3
___
2
__
√3
1
= ___ cos A + __ sin A
2
2
RHS = sin 60° cos A + cos 60° sin A
__
√3
1
= ___ cos A + __ sin A
2
2
∴ LHS = RHS
Answers
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335
2013/05/31 11:19:02 AM
3.8
__
√3 cos x − sin x
cos(30° + x) = ______________
2
LHS__= cos 30° cos x − sin 30° sin x
√3
1
= ___ cos x − __ sin x
2
2
5.4.2
7
___
5.4.3
−24
____
=
TOPIC 5: EXERCISE 3
__
√3 cos x − sin x
______________
2
= RHS
3.9
sin(x + 30°) − sin(x − 30°) = cos x
LHS = sin x cos 30° + cos x sin 30°
− (sin x cos 30° − cos x sin 30°)
= sin x cos 30° + cos x sin 30°
− sin x cos 30° + cos x sin 30°
__
__
√3
√3
1
1
= ___ sin x + __ cos x − ___ sin x + __ cos x
2
= cos x
2
2
2
= RHS
3.10
2__
cos(A − 45°) − sin(A − 45°) = ___
cos A
√
2
LHS = cos A cos 45° + sin A sin 45°
− (sin A cos 45° − cos A sin 45°)
1__
1__
1__
1__
= ___
cos
A + ___
sin A − ___
sin A + ___
cos A
√
√
√
√
2
2__
___
=
cos A
√2
2
2
4.2
4.3
4.4
4.5
4.6
__
__
__
√ 2 (1 + √ 3 )
1+√
_______
__3 or ___________
2√2 __
__ 4
__
√ 2 (1 + √ 3 )
1+√
_______
__3 or ___________
2√2 __
__ 4 __
√2 − √6
1−√
_______
__3 or _________
2√2 __
__ 4
__
√2 − √6
1−√
3
_______
_________
__ or
2__√2
__ 4
__
√3 − 1
√6 − √2
_______
_________
__ or
4
2__√2
__
√3 − 1
_______
__
or 2 − √3
√3 + 1
5.1
−16
= ____
65
5.2.1
−96
____
5.2.2
−220
_____
5.2.3
−171
_____
5.3.1
−33
____
5.3.2
−56
____
5.3.3
56
___
5
221
221
65
65
33
5.4
y
5
3
x
–4
5.4.1
336
−24
____
25
x
7
1.1
2 sin 2A cos 2A
1.2
1 − 2 sin2 6A of 2cos2 6A − 1
1.3
2 cos2 3A − 1 of cos2 3A − sin2 3A
1.4
2 cos2 4A − 1
1.5
4 sin A√ 1 − sin2A (1 − 2 sin2 A)
1.6
1 − 2 sin2 5A
1.7
sin x
1.8
cos B
1.9
2 sin 6A
1.10
sin 2x
______
2.1
= RHS
4.1
2
25
________
2
__
√3
___
2
2.2
0
2.3
1
__
2.4
1__
___
2.5
−√3
_____
2.6
2.7
2
√ 2 __
2
__
√3
___
2__
√3
___
2
2.8
1
3.1
cos 2x
3.2
cos 2x
3.3
1 + sin 2 x
3.4
tan 2x
3.5
3 sin2 x
3.6
cos2 x
3.7
cos 2x
3.8
sin2 x
4.1
a
4.2
a
4.3
a
_______
_____
√1 − a2
4.4
4.5
_____
2a√1 − a2
2a
Answers
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2013/05/31 11:19:02 AM
4.6
4.7
4.8
a
__ _____
a + √3 √1 − a2
______________
2
__ _____
√3 √1 − a2 − a
______________
6
Identify proofs
6.1
sin 3x ______
cos 3x
______
+
=2
sin x
cos x
sin 3x cos x − cos 3x sin x
LHS = _______________________
2
5
Identify proofs
sin(3x − x)
= __________
5.1
sin 3x = 3 sin x – 4 sin3 x
sin(2x)
= __________
LHS = sin(2x + x)
= sin 2x cos x + cos 2x sin x
= 2 sin x cos x cos x + sin x(1 − 2 sin2 x)
= 2 sin x(1 − sin2 x) + sin x − 2 sin3 x
= 2 sin x − 2 sin3 x + sin x − 2 sin3 x
= 3 sin x − 4 sin3x
= RHS
5.2
5.3
cos 3x = cos x(1 − 4 sin2 x)
LHS = cos(2x + x)
= cos 2x cos x − sin 2x sin x
= cos x(1 − 2 sin2 x) − 2 sin x cos x sin x
= cos x(1 − 2 sin2 x) − 2 cos x sin2 x
= cos x(1 − 2 sin2 x − 2 sin2 x)
= cos x(1 − 4 sin2 x)
= RHS
cos 4x = 8 cos4 x − 8 cos2 x + 1
LHS = 2 cos2 2x − 1
= 2(2 cos2 x − 1)2 − 1
= 2(4 cos4 x − 4 cos2 x + 1) − 1
= 8 cos4 x − 8 cos2 x + 2 − 1
= 8 cos4 x − 8 cos2 x + 1
= RHS
5.4
1 – cos 4x = 2sin2 2x
LHS = 1 − (1 − 2 sin2 2x)
= 2 sin2 2x = RHS
5.5
sin 4x = cos x( 4 sin x − 8sin3 x )
LHS = 2 sin 2x cos 2x
= 2(2 sin x cos x)(1 − 2 sin2 x)
= 4 sin x cos x − 8 sin3 x
= cos x( 4 sin x − 8 sin3 x )
= RHS
5.6
sin 4x = 4 sin x cos3 x − 4 cos x sin3 x
LHS = 2 sin 2x cos 2x
= 2(2 sin x cos x)(cos2 x − sin2 x)
4 sin x cos3 x − 4 cos x sin3 x
sin x cos x
sin x cos x
sin x cos x
2 sin x cos x
= ___________
sin x cos x
= 2 = RHS
6.2
(cos x + sin x)(cos x − sin x)
cos 3x ______
sin 3x ________________________
______
−
=
cos x
sin x
sin x cos x
cos 3x cos x − sin 3x sin x
LHS = _______________________
cos(3x − x)
= __________
sin x cos x
sin x cos x
cos(2x)
= __________
sin x cos x
co
s2 x − sin2 x
____________
= sin x cos x
(cos x + sin x)(cos x − sin x)
= ________________________
sin x cos x
= RHS
TOPIC 5: EXERCISE 4 EXTENSION WORK
1
1 + tan β
tan(45° + β) = ________
1 − tan β
sin(45° + β)
LHS = __________
cos(45° + β)
sin 45° cos β + cos 45° sin β
______________________
= cos 45° cos β − sin 45° sin β
___
1__
1__
cos β + ___
sin β
√2
√2
________________
= ___
___
1__
1__
cos β − √ sin β
√2
2
___
1__
1
(cos β + sin β) _____
cos β
√2
______________
____
= ___
× 1
1__
(cos β − sin β) _____
cos β
√2
1 + tan β
_______
= 1 − tan β
= RHS
or using tan( A + B) formula
1 + tan β
tan(45° + β) = ________
1 − tan β
tan 45° + tan β
LHS = _______________
1 − tan 45° tan β
1 + tan β
= ________
1 − tan β
= RHS
Answers
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2
__
tan 75° = 2 + √3
4
sin(45° + 30°)
sin(45° + x)
sin 45° cos 30° + cos 45° sin 30°
= ____________________________
cos 45° cos 30° − sin 45° sin 30°
____________________
sin 45° cos x + cos 45° sin x ____________________
sin 45° cos x − cos 45° sin x
= cos
45° cos x − sin 45° sin x + cos 45° cos x + sin 45° sin x
__
√3
1__ ___
1__ __
___
__
.
+ ___
.1
√2 2
√2 2
2√__
2
_____________
____
__
=
×
√3
1__ ___
1__ __
1
2√ 2
___
.
+ ___
.
√2 2
√2 2
___
1__
1__
(cos x + sin x) ___
(cos x − sin x)
√2
√2
______________
______________
+
___
1__
1__
(cos x − sin x) ___
(cos x + sin x)
√2
√2
___
1__
__
Note: sin 45° = cos 45° = √
√3 + 1
√3 + 1
__
__
= _______
× _______
√3 − 1
√3 + 1
=
(cos x + sin x)
___________
__
3 + 2√ 3 + 1
___________
(cos x + sin x)2 + (cos x − sin x)2
__________________________
__
4 + 2√ 3
= ________
(cos x − sin x)(cos x + sin x)
2
2
2
2
_________________________________________
= cos x + 2 sin x cos x + sin 2x + cos2 x − 2 sin x cos x + sin x
__
= 2 + √3
cos x − sin x
______
2
= cos 2x ( sin2x + cos2x = 1 )
= RHS
= RHS
or using tan(A + B) formula
__
tan 75° = 2 + √3
Alternative solution:
LHS = tan(45° + 30°)
2
tan(45° + x) + tan (45° − x) = ______
cos 2x
LHS = tan(45° + x) + tan (45° − x)
tan 45° + tan 30°
= ________________
1 − tan 45° tan 30°
tan 45° + tan x
tan 45° − tan x
= _______________ + _______________
1 − tan 45° tan x
1 + tan 45° tan x
1__
__
1 + ___
√3
√3
___
__
= __________
×
1__
√3
1 − (1)(___
)
√3
1 + tan x ________
1 − tan x
________
+
1 − tan x
1 + tan x
(1
+
tan
x)(1
+ tan x) + (1 − tan x)(1 − tan x)
= _______________________________________
(1 − tan x)(1 + tan x)
__
__
√3 + 1
√3 + 1
__
__
× _______
= _______
√3 − 1
√3 + 1
__
3 + 2√3 + 1
___________
=
1 + 2 tan x + tan2x + 1 − 2 tan x + tan2x
= ___________________________________
1 − tan2x
2 x)
2(1
+
ta
n
= ___________
1 − tan2x
3−1
__
4 + 2√3
= ________
2
__
= 2 + √3
= RHS
3
p
1
_____ and cos x = _______
_____
tanx = p; sin x = _______
2
2
√1 + p
sin 2x
tan 2x = ______
cos 2x
p
1
_____ _______
_____
2_______
√1 + p2 √1 + p2
____________________
=
2
p
2
1
_______
_____
_____
− _______
√1 + p2
√1 + p2
2p
______
1 + p2
1 + p2
_____
______
=
2 ×
1
−
p
1 + p2
______
1 + p2
) (
√1 + p
sin2 x
2(1 + _____
)
cos2 x
cos2 x
___________
=
× _____
sin2 x
_____
cos2 x
1−
cos2 x
2(cos2 x + sin2 x)
= _______________
cos2 x − sin2 x
2
= ______
cos 2x
2 sin x. cos x
= ____________
cos2 x − sin2 x
(
2
(cos x − sin x)
___________
(cos x − sin x) + (cos x + sin x)
3−1
2
sin (45° − x)
__________
LHS = __________
cos(45° + x) + cos(45° − x)
LHS = ____________
cos(45° + 30°)
__
2
tan(45° + x) + tan (45° − x) = ______
cos 2x
= RHS
)
2p
= ______2
1−p
or using tan2x formula
tan x = p
2 tan x
tan 2x = _________
2
2p
= ______
1 − tan x
1 − p2
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Answers
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5
a
tan x = __
b
TOPIC 5: EXERCISE 5
sin 2x = 2 sin x cos x
a
b
________
______
______
= 2 ________
2
2
2
2
1.1
cos 2x = cos4 x − sin4 x
RHS = cos4 x − sin4 x
= (cos2 x − sin2 x)(cos2 x + sin2 x)
= cos 2x × (1)
= cos 2x = LHS
1.2
1 − sin 2x
___________
= sin x − cos x
( √a + b ) ( √a + b )
2ab
= ______
a2 + b2
sin 2x
tan 2x = ______
cos 2x
2 sin x cos x
= ____________
cos2 x − sin2 x
a
b
________
______
______
2 ________
√ a2 + b2 √ a2 + b2
_____________________
=
2
a
2
b
a2 + b2
sin x − cos x
n2 x + cos2 x − 2 sin x cos x
_________________________
LHS = si
sin x − cos x
si
n2 x − 2 sin x cos x + cos2 x
_________________________
=
sin x − cos x
(sin x − cos x)(sin x − cos x)
________________________
=
(sin x − cos x)
b −a
______
= sin x − cos x = RHS
(
)(
(
________
______
√ a2 + b2
) (
)
______
− ________
√ a2 + b2
)
2ab
______
= ______
2
2
a2 + b2
1.3
2ab
= ______
b2 − a2
Alternative solution using tan 2x formula
2 tan x
tan 2x = _________
2
a
2 ( __
b)
1 − tan x
b2
__
= _______
a 2 × b2
1 − ( __
)
b
1.4
2ab
= ______
b2 − a2
sin(A + B)
sin A cos B + cos A sin B
cos A cos B
_________
= ______________________ × __________
1
__________
cos A cos B − sin A sin B
cos(A + B)
cos A cos B
sin
A
cos
B
cos
A
sin
B
__________ + __________
cos A cos B
cos A cos B
= _____________________
C = (180° – (A + B))
cos A cos B __________
sin A sin B
__________
cos A cos B − cos A cos B
RTP: tan A.tan B.tan C
A
= tan A + tan B + tan C
tan C = tan(180° − (A + B))
tan C = − tan(A + B)
(
tan A + tan B
tan C = − _____________
1 − tan A tan B
)
tan C(1 − tan A tan B) = − tan A − tan B
tan C − tan A tan B tan C = − tan A − tan B
∴ tan A.tan B.tan C = tan A + tan B + tan C
(co
s2 x − sin2 x)2
______________
= cos 2x
cos4 x − sin4 x
(cos2 x − sin2 x)(cos2 x − sin2 x)
(cos x − sin x)(cos x + sin x)
cos2 x − sin2 x
= ____________
1
= cos 2x = RHS
1.5
tan A + tan B
= _____________
1 − tan A tan B
6.2
1 + cos x + cos 2x
sin x + 2 sin x cos x
LHS = ____________________
1 + cos x + 2 cos2 x − 1
sin x(1 + 2 cos x)
= _______________
cos x(1 + 2 cos x)
sin x
= _____
cos x
= tan x = RHS
LHS = __________________________
2
2
2
2
1
__________
6.1
sin x + sin 2x
________________
= tan x
B
1.6
1 − cos 2x
_________
= tan x
sin 2x
1 − (1 − 2 sin2 x)
LHS = _______________
2 sin x cos x
2 sin2 x
= ___________
2 sin x cos x
sin x
= _____
cos x
= tan x = RHS
sin 2x − cos x
cos x
_______________
= _____
1 − cos 2x − sin x
sin x
2 sin x cos x − cos x
LHS = _____________________
2
1 − (1 − 2 sin x) − sin x
cos x(2 sin x − 1)
= _______________
sin x(2 sin x − 1)
cos x
_____
= sin x = RHS
Answers
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1.7
1.8
1.9
1.10
1 − cos 2x + sin 2x
________________
= tan x
1 + cos 2x + sin 2x
1 − (1 − 2 sin2 x) + 2 sin x cos x
LHS = ___________________________
1 + (2 cos2 x − 1) + 2 sin x cos x
2 sin2 x + 2 sin x cos x
___________________
=
2 cos2 x + 2 sin x cos x
2 sin x(sin x + cos x)
= __________________
2 cos x(cos x + sin x)
sin x
= _____
cos x
= tan x = RHS
2.3
( )
sin x − cos 2x ________
sin x + 1
_____________
=
cos x
sin 2x − cos x
sin x − (1 − 2 sin2 x)
__________________
LHS = 2 sin x cos x − cos x
2 sin2 x + sin x − 1
= _________________
2 sin x cos x − cos x
(2 sin x − 1)(sin x + 1)
= ___________________
cos x(2 sin x − 1)
sin x + 1
= ________
cos x = RHS
(cos x + sin x)2 − 1
________________________
= tan 2x
(cos x + sin x)(cos x − sin x)
2
s x + 2 sin x cos x + sin2 x − 1
____________________________
LHS = co
(cos x + sin x)(cos x − sin x)
1 + 2 sin x cos x − 1
_________________
=
cos2 x − sin2 x
sin 2x
______
=
cos 2x
= tan 2x = RHS
2.4
+ sin 60° sin x)
( )
2.5
2 cos 6x cos 4x − cos 10x + 2 sin2 x = 1
LHS = 2 cos 6x cos 4x − cos(6x + 4x) + 2 sin2 x
= 2 cos 6x cos 4x − (cos 6x cos 4x − sin 6x sin 4x)
+ 2sin2x
= 2 cos 6x cos 4x − cos 6x cos 4x + sin 6x sin 4x
+ 2 sin2 x
= cos 6x cos 4x + sin 6x sin 4x + 2 sin2 x
= cos(6x − 4x) + 2 sin2 x
= cos 2x + 2 sin2 x
= cos2 x − sin2 x + 2 sin2 x
= sin2 x + cos2 x
= 1 = RHS
2.6
2 sin 5x.cos 4x − sin 9x = sin x
LHS = 2 sin 5x.cos 4x − sin(5x + 4x)
= 2 sin 5x.cos 4x − (sin 5x cos 4x + cos 5x sin 4x)
= sin 5x.cos 4x − cos 5x sin 4x
= sin(5x − 4x)
= sin x = RHS
2.7
tan x − tan y = __________
cos x cos y
2
sin x + 2 cos x
(1 − cos2x) + sin(2x)
LHS = __________________
sin x + 2 cos x
(sin2 x) + 2 sin x cos x
___________________
=
(sin x + 2 cos x)
sin x(sin x + 2 cos x)
__________________
=
(sin x + 2 cos x)
sin(x + y)
__________
= tan x + tan y
= RHS
2.2
cos 2x _____
sin 2x ______
1
______
−
=
sin x
cos x
cos x
sin 2x cos x − cos 2x sin x
_______________________
sin x cos x
sin(2x − x)
= __________
sin x cos x
sin x
__________
= sin x cos x
1
= _____
cos x = RHS
__
( )
√3 2
1 2
= __ cos2 x − ___ sin2 x
2
2
3
1
__
2
2
= 4(1 − sin x) − __
4 sin x
3
1 __
1
__
2
2
= __
4 − 4 sin x − 4 sin x
1
2
= __
4 − sin x = RHS
1 − cos x − sin(−2x)
__________________
= sin x
cos x cos y
sin x cos y + cos x sin y
LHS = ____________________
cos x cos y
sin x cos y __________
cos x sin y
__________
= cos x cos y + cos x cos y
= tan x + tan y
( )
1__ 2
1__ 2 2
= ___
cos2 x − ___
sin x
√2
√2
1
1
= __(1 − sin2 x) − __ sin2 x
2
2
1 __
1
1
__
__
2
= − sin x − sin2 x
2 2
2
1
__
2
= − sin x = RHS
2
1
2
cos(60° + x)cos(60° − x) = __
4 − sin x
LHS = (cos 60° cos x − sin 60° sin x)(cos 60° cos x
= (cos2 60° cos2 x − sin2 60° sin2 x)
= sin x = RHS
2.1
1
2
sin(45° + x)sin(45° − x) = __
2 − sin x
LHS = (sin 45° cos x + cos 45° sin x)(sin 45° cos x
− cos 45° sin x)
2
= (sin 45° cos2 x − cos2 45° sin2 x)
sin(x − y)
sin y
sin x _____
LHS = _____
cos x − cos y
sin x cos y − cos x sin y
= ____________________
cos x cos y
sin(x − y)
= __________
cos x cos y
= RHS
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Answers
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2.8
sin 3x − sin x
____________
= cos 2x
2 sin x
sin(2x + x) − sin x
LHS = ________________
2 sin x
sin 2x cos x + cos 2x sin x − sin x
_____________________________
=
2 sin x
2 sin x cos x cos x + cos 2x sin x − sin x
_________________________________
=
2 sin x
sin x(2 cos2 x + cos 2x − 1)
_______________________
=
2 sin x
sin x(cos 2x + cos 2x)
___________________
=
2 sin x
sin x(2 cos 2x)
_____________
=
2 sin x
= cos 2x = RHS
2.9
2.10
cos 2x + 2 sin 2x + 2 = (3 cos x + sin x)
(cos x + sin x)
LHS = cos2 x − sin2 x + 2(2 sin x cos x) + 2(sin2 x
+ cos2 x)
2
2
= cos x − sin x + 4 sin x cos x + 2 sin2 x + 2 cos2 x
= 3 cos2 x + 4 sin x cos x + sin2 x
= (3 cos x + sin x)(cos x + sin x) = RHS
2 sin(x − y)
cos x __________
sin x _____
_____
sin y − cos y =
sin 2y
sin x cos y − cos x sin y
LHS = ____________________
sin y cos y
sin(x − y)
2
__
= _________
sin y cos y × 2
TOPIC 5: EXERCISE 7
1
2
TOPIC 5: EXERCISE 6
x = 0°; ± 180°; ± 360°;−63,43°; 116,57°;
−243,43°; 296,57°
3
4
x = ± 70,53° + n.360° of x = ± 120° + n.360°, n ∈ ℤ
x = ± 60°; ± 300°; 56,31°; 236,31°;−123,69°;
−303,69°
5
x = 90° + n.180° of x = ± 120° + n.360°, n ∈ ℤ
6
x = 0° + n.180° of
x = 30° + n.360° of 150° + n.360°, n ∈ ℤ
7
x = ± 131.81° + n.360° of x = 0° + n.360°, n ∈ ℤ
8
x = 90° + n.360°, n ∈ ℤ
9
x = 90° + n.180° of x = 45° + n.180°, n ∈ ℤ
10
x = ± 60° + n.360°; x = ± 60°
11
x = −90° + n.360° of x = ± 60° + n.360°, n ∈ ℤ
12
x = 146,31°; 326,31°; 45°; 135°
13
x = 90° + n.180°
of x = 14,48° + n.360° of 165,52° + n.360°
of x = −30° + n.360° of 210° + n.360°
14
x = 0° + n.180° of x = 45° + n.180°, n ∈ ℤ
x = 0°;180°;−180°;45°;−135°
2 sin(x − y)
= __________
2 sin y cos y
2 sin(x − y)
= __________ = RHS
sin 2y
x = ± 60° + n.360° of x = ± 180° + n.360°, n ∈ ℤ
TOPIC 5: EXERCISE 8
1
x = 42,33° + n.360° of x = 177.67° + n.360°, n ∈ ℤ
1
x = 25°; 51,67°; 171,67°; − 68,33°
2
x = 24,46° + n.360° of x = −74,46° + n.360°, n ∈ ℤ
2
3
x = −18,70° + n.360° of x = 78,7° + n.360°, n ∈ ℤ
x = 40° + n.120°, n ∈ ℤ
of x = 60° + n.360°, n ∈ ℤ
4
x = 35° + n.180° of x = 5° + n.180°, n ∈ ℤ
3
x = 30°; 10°; 130°; 250°
5
x = −12,42° + n.180°, n ∈ ℤ
x = 102,42° + n.180°, n ∈ ℤ
4
x = 45° + x + n.360°, n ∈ ℤ
x = −15° + n.120°, n ∈ ℤ
6
x = 15° + n.180°, n ∈ ℤ
x = 75° + n.180°, n ∈ ℤ
5
x = 25° + n.180°, n ∈ ℤ
6
x = 24,35° + n.180°, n ∈ ℤ
x = −24,35° + n.180°, n ∈ ℤ
x = 39° + n.120°, n ∈ ℤ
x = 63° + n.360°, n ∈ ℤ
7
8
x = 69,39° + n.180°, n ∈ ℤ
x = − 69,39° + n.180°, n ∈ ℤ
x = −37,5° + n.180°, n ∈ ℤ
of x = − 82,5° + n.120°, n ∈ ℤ
8
9
x = 66,03° + n.180°, n ∈ ℤ
x = − 66,03° + n.180°, n ∈ ℤ
x = 13,33° + n.120°, n ∈ ℤ
of x = 140° + n.360°, n ∈ ℤ
10
x = 30° + n.120°, n ∈ ℤ
x = −30° + n.120°, n ∈ ℤ
7
Answers
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TOPIC 5: REVISION TEST
1.1
C
1.2
B
4.5
2
A
2.1
√1 − m
2.2
√1 − m2
2.3
−2√1 − m2 (m)
2.4
2m2 − 1
2.5
2m√1 − m2
3.1
−56
____
______
2
______
Identity proofs
4.1
1
__
[cos ( A + B )+ cos (A − B)] = cos A cos B
2
1
__
5.2
5.4
1
−__
2
__
√3
___
2__
√3
___
2
5.5
0
__
√3
6.1
tan 120°. sin 390° cos 156°.cos(−135°) ___
√3
________________________________
=
= cos A cos B = RHS
6.2
342
sin 105° cos 75°.tan 135°
1
_____________________
= −__
sin(−330°)
2
sin 75° cos 75°.(−tan 45°)
2
LHS = _____________________
× __
2
sin(30°)
__
2 sin 75° cos 75°.(−tan 45°)
= _______________________
2 sin(30°)
1 + √3
sin(30° + x) + cos(30° − x) = _______
2 (sin x + cos x)
LHS = sin 30° cos x + cos 30° sin x + cos 30° cos x
+ sin 30° sin x
sin 2(75°).(−tan 45°)
= __________________
2sin(30°)
__
sin ( A + B )sin(A − B) = sin2 A − sin2 B
LHS = (sin A cos B + cos A sin B)(sin A cos B
− cos A sin B)
= sin2 A cos2 B − cos2 A sin2 B
= sin2 A(1 − sin2 B) − (1 − sin2 A)sin2 B
= sin2 A − sin2 A sin2 B − sin2 B + sin2 A sin2 B
= sin2 A − sin2 B = RHS
2
2
√3
√3
1
1
= __ cos x + ___ sin x + ___ cos x + __ sin x
4.4
sin 315°.sin 66°
__
1
1__
−√3 .(__)(−cos 24°).(− ___
)
2
√2
______________________
=
1__ ).cos 24°
(−___
√2
__
√3
= ___ = RHS
2
2
2
2
__ 2
√3
1
= __(sin x + cos x) + ___(sin x + cos x)
2
2 __
1 √3
= (sin x + cos x)(__ + ___)
2
2
__
1 + √3
_______
=
2 (sin x + cos x) = RHS
__
−tan 60°.sin 30°(−cos 24°).(−cos 45°)
LHS = _______________________________
(−sin 45°).cos 24°
1
= __[2 cos A cos B]
__
2
5.6
1
LHS = __
2[ cos A cos B − sin A sin B + cos A cos B
+ sin A sin B]
cos (90° + A) = −sin A
LHS = cos 90° cos A − sin 90° sin A
= (0)cos A − (1)sin A
= −sin A = RHS
__
5.1
5.3
4
2
2 sin(30° − x) = cos x − √3 sin x
LHS = 2(sin 30°__cos x − cos 30° sin x)
√3
1
= 2(__ cos x − ___ sin x)
2
__ 2
= cos x − √3 sin x = RHS
______
3.3
4.3
2
4.6
______
65
−33
____
65
56
___
33
4.2
2
= 0 = RHS
1.3
3.2
cos(45° − x) − sin (45° + x) = 0
LHS = cos 45° cos x + sin 45° sin x − (sin 45° cos x
+ cos 45° sin x)
1__
1__
1__
1__
___
cos x + ___
sin x − ___
cos x − ___
sin x
√
√
√
√
sin 150°(−1)
= ___________
1
2(__
2)
1
= −__ = RHS
2
6.3
sin 80° − sin 40° = sin 20°
LHS = sin(60° + 20°) − sin(60° − 20°)
= sin 60° cos 20° + cos 60° sin 20° − (sin 60° cos 20°
− cos 60° sin 20°)
= 2(cos 60° sin 20°)
1
= 2(__) sin 20°
2
= sin 20° = RHS
Answers
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6.4
3
sin2 20° + sin2 40° + sin2 80 = __
2
2
2
2
8.2
2
LHS = sin 20° + sin 40° + sin 80 sin 20° +
[sin(60° − 20°)]2 + [sin(60° + 20°)]2
2
= sin 20° + (sin 60° cos 20° − cos 60° sin 20°)2
+ (sin 60° cos 20° + cos 60° sin 20°)2
= [sin2 20° + sin2 60° cos2 20°
− 2 sin 60° cos 20° cos 60° sin 20°
+ cos2 60° sin2 20° + sin2 60° cos2 20°
+ 2 sin 60° cos 20° cos 60° sin 20°
+ cos2 60° sin2 20°]
= sin2 20° + 2(sin2 60° cos2 20°)
+ 2(cos2 60° sin2 20°)
__
7.1
√3
1
= sin2 20° + 2(___)2 cos2 20°) + 2(__)2 sin2 20°)
2
2
3
1
__
__
2
2
2
= sin 20° + cos 20° + sin 20°
2
2
3
__
2
2
= (sin 20° + cos 20°)
2
3
__
= = RHS
2
1
sin(45° + x).sin(45° − x) = __
2 cos 2x
8.3
7.2
( )( )
1__
= ___
= cos 45° = RHS
√2
8
Identity proofs
8.1
1 − sin 2x = (sin x − cos x)2
LHS = 1 − sin 2x
= sin2 x + cos2 x − 2 sin x cos x
= (sin x − cos x)(sin x − cos x)
= (sin x − cos x)2
1 + sin x − cos 2x 1 + 2 sin x
sin x + 2 sin x cos x
LHS = _____________________
1 + sin x − (1 − 2 sin2 x)
sin x(1 + 2 cos x)
= _______________
2 sin2 x + sin x
sin x(1 + 2 cos x)
= _______________
sin x(2 sin x + 1)
8.4
cos2 3x − cos 6x = sin2 3x
LHS = cos2 3x − cos 2(3x)
= cos2 3x − (cos2 3x − sin2 3x)
= sin2 3x = RHS
8.5
sin 2x tan x
1
1
________
+ ________ = ___________
8.6
cos 2x − cos x _____
1
1
_____________
=
− _____
1 + cos x 1 − cos x
sin4 x
1
−
cos
x
+
1
+
cos
x
LHS = __________________
(1 + cos x)(1 − cos x)
2
= _________
1 − cos2 x
2
_____
=
sin2 x
sin x
2 sin x cos x( _____
cos x )
RHS = ________________
sin4x
2 sin2 x
_______
=
sin4 x
2
= _____
sin2 x
∴ LHS = RHS
sin 105° + cos 105° = cos 45°
LHS = sin(45° + 60°) + cos(45° + 60°)
= sin 45° cos 60° + cos 45° sin 60° + cos 45° cos 60°
− sin 45° sin 60°
1__
1__
1__
= ___
cos 60° + ___
sin 60° + ___
cos 60°
√2
√2
√2
1__
− ___
sin 60°
√2
1__
= 2___
cos 60°
√2
1__ __
1
= 2 ___
√2 2
1 + 2 cos x
sin x + sin 2x
_______________
= __________
1 + 2 cos x
= __________ = RHS
1 + 2 sin x
1
1
= __ cos2 x − __ sin2 x
2
2
1
__
2
= (cos x − sin2 x)
2
1
__
= cos 2x = RHS
2
1
= __
4
cos2 x − sin2 x
sin 2x
LHS = ______
cos 2x
= tan 2x = RHS
LHS = (sin 45° cos x + cos 45° sin x).(sin 45° cos x
− cos 45° sin x)
= (sin2 45° cos2 x − cos2 45° sin2 x)
∴ sin 75°.sin 15°
2 sin x cos x
____________
= tan 2x
tan x
sin x
sin 2x + sin x
2cos2x − 1 − cos x
_________________
LHS = 2 sin x cos x + sin x
(2 cos x + 1)(cos x − 1)
= ____________________
sin x(2 cos x + 1)
(cos x − 1)
_________
=
sin x
cos x _____
1
RHS = _____
sin x − sin x
(cos x − 1)
= _________
sin x
∴ LHS = RHS
8.7
co
s4 x + sin2 x cos2 x
_________________
= 1 + sin x
1 − sin x
cos2 x(cos2 x + sin2 x)
LHS = __________________
1 − sin x
(1 − sin2 x)(1)
= ____________
1 − sin x
(1 − sin x)(1 + sin x)
= __________________
(1 − sin x)
= 1 + sin x = RHS
Answers
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8.8
(
)
1
(1 − cos 2x) 1 + ______
=2
2
tan x
(
11
2
cos x
LHS = (1 − (1 − 2 sin2 x)) 1 + _____
2
(
2
2
sin x + cos x
= 2 sin2 x ____________
sin2 x
2 sin2 x
_______
=
sin2 x
)
sin x
)
1 − sin x − (1 − 2 sin2 x)
2 sin2 x − sin x
= _________________
2 sin x cos x − cos x
sin x(2 sin x − 1)
= _______________
cos x(2 sin x − 1)
sin x
= _____
cos x
= RHS
8.10
cos 2x + 2 sin 2x + 2 = (3 cos x + sin x)(cos x + sin x)
LHS = cos2 x − sin2 x + 2(2 sin x cos x) + 2(sin2 x
+ cos2 x)
= 3 cos2 x + 4 sin x cos x + sin2x
= (3 cos x + sin x)(cos x + sin x)
= RHS
The identity in 8.2 is undefined if:
x = 45°; 135°; 225°; 315°
The identity in 8.3 is undefined if:
x = 0°; 180°; 360°; 210°; 330°
The identity in 8.6 is undefined if:
x = 0°; 180°; 360°; 120°; 240°
9.1
x = 9,74°; 80,27°; −170,26°; −99,74°
9.2
x = 70,53°; 289,47°; −70,53°; −289,47°
9.3
x = 90°; 270°
9.4
x = 0°; ± 180°
9.5
x = ± 60°; ± 300°; 33,69°; 213,69°;−326,31°;
−146,31°
10.1
x = 43° + n.180°, n ∈ ℤ
of x = −16,5° + n.90°, n ∈ ℤ
10.2
x = −19,07° + n.360°
of x = 39,07° + n.360°, n ∈ ℤ
sin 2x − cos x
LHS = _____________________
2 sin x cos x − cos x
=2
8.9
1 − sin x − cos 2x
_______________
= tan x
= tan x = RHS
1 − sin x − cos 2x
Hence solve for _______________
sin 2x − cos x = −1
tan x = −1
∴ x = − 45° + n.180°, n ∈ ℤ
12
x = 0° + n.180°, n ∈ ℤ
or x = 30° + n.180°, n ∈ ℤ
TOPIC 6: EXERCISE 1
1
ED = AE sin 21° = 15,85 m
2.1
BC2 = x2 + x2 − 2x.x cos 120°
= 3x2
__
BC = √ 3 x
2.2
300
__
x = ____
√
2.3
A = 48,59° or 131,41°
2.4
Area ABDC = 42 307,48 m2
3–6
Proofs. Refer to your teacher for details
3
TOPIC 6: EXERCISE 2
2
1
c
a2 = __________
2(1 − cos C)
2
a = 2c cos C
3–10
Proofs. Refer to your teacher for details
10.3
x = 22,5° + n.90°
10.4
x = 71,57° + n.180° or x = −45° + n.180°, n ∈ ℤ
TOPIC 6: EXERCISE 3
10.5
x = 30° + n.360°, n ∈ ℤ
of x = 50° + n.120°, n ∈ ℤ
1
^ C = 8,45°
FA
2.1
46,98 cm2
2.2
11,47 cm
2.3
9,62 cm
3.1
XA = 16,55 m
3.2
XC = 14,62 m
3.3
AC = 12,31 m
344
Answers
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4
CD = 169,52 m
8.2
^ C = 67,34°
AD
5.1
AB = 100,35 m
9.1
DC = 3,36 km
5.2
Area △BCD = 22 629,73 m2
9.2
Kite D is higher by 0,67 m.
6.1
△ABD ≡ △ABC
10
Refer to your teacher for details
6.2
AC = 102,16 m
6.3
CD = 81,47 m
6.4
AB = 34,94 m
7.1
Area △CBD = 582,23 m2
7.2
AB = 28,75 m
| SAS
TOPIC 7: EXERCISE 1
TOPIC 6: EXERCISE 4
1–3
Refer to your teacher for details
4.1
4p sin y
__
DC = _______
√3
4.2–4.3 Refer to your teacher for details
1
−13
2
17
3
6a − 6
4
2a2 + a + 2
5
0
6
16p3 + 24p2 + 4p
7
8xh + 4h2 − 5h
8
3a2p + 3ap2 + p3 − 8p
9
−2
10
2−a
_____
5.1
^ C = 180° − (x + y)
BD
5.2
Refer to your teacher for details
5.3
EF = 25,31 m
6
Refer to your teacher for details
7.1
^ V = 82,63°
TQ
1
(x + 1)(x + 2)(x − 2)
7.2
Area △TQV = 96,695 cm2
2
x(x + 4)(x − 2)
3
8(x − 2)(x2 + 2x + 4)
TOPIC 6: REVISION TEST
4
(x − 5)(x2 + 10x + 25)
1
^ M = 60,5°
VA
5
9x(x + 2)(x2 − 2x + 4)
2
^ B = 22,2°
ED
6
(x − 3)(x − 2)(x + 2)
3
Refer to your teacher for details
7
(x − 2)(x − 7)(x + 7)
4.1
BC = 26,97 m
8
− 4x(x − 4)(x − 5)
4.2
AB = 10,90 m
9
(x + 1)(x + 4)(x − 4)
4.3
AD = 37,76 m
10
−(x − 2)(x − 1)(x + 1)
4.4
Area △BDC = 450,14 m2
11
(3x − 1)(x2 + 1)
5–6
Refer to your teacher for details
12
5(x + 2)(x − 2)(x + 1)
7.1
△ABD ≡△ ABS
7.2
x
AC = _____
cos α
7.3
Refer to your teacher for details
7.4
AB = 58,92 m
8.1
^ C = 53,83°
AD
| SAS
a
TOPIC 7: EXERCISE 2
TOPIC 7: EXERCISE 3
1.1
f(− 4) = 0
1.2
f(2) = 54
2.1
k=5
2.2
k=1
Answers
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3.1
15
1
___
f(__
2) = 8
3.2
1
f(__
2) = 0
9
p = __
8
4.1
f(3) = 0
This means that (x − 3) is a factor of f(x)
4.2
f(3) = 4
This means that when f(x) is divided by (x + 1)
the remainder is 4
2
x = 0 or x = 0 or x = 4
3 real rational roots,
2 of which are equal.
3
x = 0 or x = 1 or x = 1
3 real rational roots,
2 of which are equal.
4
3
x = 2 or x = __
2 or x = −3
3 real rational roots
5
1
x = 1 or x = __
2 or x = − 6
3 real, rational roots
5
k=4
6
1
k = − __
2
6
7
f(−1) = 0 and f(2) = 0
b = 6 and a = 2 − 6 = − 4
x = 2 or x = 2 + √ 7 or 2 − √ 7
3 real roots: 1 rational and 2 irrational
7
x = 1 or x = −2 or x = −3
3 real, rational roots
8
x = 1 or x = − 4,30 or x = − 0,70
3 real roots: 1 rational, 2 irrational
9
x = 2 or x = −3 or x = 3
3 real, rational roots
10
x = 1 or x = 3 or x = −2
3 real, rational roots
11
x = 2 or x = 0,27 or x = 3,73
3 real roots: 1 rational and 2 irrational
12
x = −1 or x = 3 ± √ −3
1 real root and 2 non-real or imaginary roots
TOPIC 7: EXERCISE 4
__
__
1.1
(x + 1)(x − 4)(x + 2)
1.2
(x + 2)(x2 + x − 5)
1.3
– (x + 2)(x + 4)(x − 3)
1.4
(x − 3)(2x − 3)(x + 1)
1.5
(x − 1)(x + 4)(x − 3)
1.6
– (x + 3)(x − 1)(x + 1)
1.7
2(x + 3)(x − 3)(x + 1)
1.8
(2x − 3)(2x + 1)(x + 1)
2.1
(x − 2)(2x − 1)(x + 3)
TOPIC 7: REVISION TEST
2.2
(x − 1)(x − 1)(x − 1)
1.1
3
2.3
2(x − 1)(x + 1)(x + 3)
1.2
−9
2.4
(x − 1)(2x2 + 5x + 1)
1.3
(3a − 2)
_______
2.5
(x − 1)(4x + 5)(x − 2)
1.4
2.6
1
− __(x + 2)(2x2 − 7x + 2)
5t 2 − 11t + 6
1.5
0
2.7
(x − 3)(x − 6)(x + 2)
1.6
−2a3 − 5a2 + a + 5
2.8
– (x − 1)(3x − 1)(x + 1)
1.7
0
2.9
(x − 1)(4x + 1)(x − 5)
1.8
0
2.10
– (x + 1)(2x + 1)(3x − 1)
1.9
−6x − 3h + 1
1.10
6a2 + 6a + 2
1.11
−a
_______
1.12
4+b
_____
2.1
(x + 1)(2x − 3)(2x + 3)
2
TOPIC 7: EXERCISE 5
1
3
x = −1 or x = 2 or x = __
3 real rational roots
346
2
___
a
h(a + h)
−b
Answers
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__
2.2
2x(x + 7)(x − 2)
5.2.1
x = 2 or x = 1 ± √ 6
2.3
3(x − 3)(x2 + 3x + 9)
5.2.2
2.4
(x + 6)(x2 − 6x + 36)
x=1
__
or x = −2 ± √ 6
2.5
5x2(x + 2)(x2 − 2x + 4)
5.2.3
x = −2
2.6
(x − 2)(x + 2)(x + 3)
___
1 ± √ 33
or x = ________
4
___
1 ± √ 29
5.2.4
1
________
x = __
2 or x =
2
5.3.1
x = 3 or x = −2,30
or x = 1,30
5.3.2
x=1
2.7
(x − 3)(x − 2)(x + 2)
2.8
−5x2(x − 4)(x − 5)
2.9
(x + 1)(x + 4)(x − 4)
2.10
(x − 2)(−x2 + 3)
2.11
(5x − 1)(x − 2)(x + 2)
2.12
−3(x − 1)(x − 3)(x2 + 3x + 9)
3.1.1
1
f(__
2 ) = 0 ∴ (2x − 1) is a factor
TOPIC 8: EXERCISE 1
3.1.2
18
3.2.1
k = −6
It may be necessary to show the learners how to set up
these tables.
3.2.2
k = −13
3.3.1
R = 20
3.3.2
k = 10
3.4.1
a = −8
3.4.2
a=5
3.4.3
a=3
3.4.4
a=6
Function D:
˙
2
2,3
3.5.1
a = −5
b=2
Function E:
1
−1,9999
3.5.2
a = −5
b=5
Function F:
2
−3
4.1
(x − 1)(x − 2)(x − 4)
4.2
(x − 1)(x + 2)(x − 3)
4.3
(x − 3)(x + 2)(x + 4)
4.4
(x − 2)(2x + 1)(3x − 1)
4.5
(x + 1)(x − 3)(2x − 3)
4.6
(x + 1)(2x − 1)(x − 2)
5.1.1
x = −1 or x = −2 or x = 4
5.1.2
1
x = −2 or x = __
2 or x = 1
5.1.3
1
−3
___
x = __
2 or x = 2 or x = 1
5.1.4
3
x = −3 or x = __
2 or x = 4
____
3 ± √ −23
or x = __________
4
5.3.3
____
− 3 ± √ −27
x = 3 or x = ____________
2
Function A:
2
−2
Function B:
2
0,2
Function C:
2
6
Function G:
4
__
2
3
Function H:
2
12,5
Function I:
2
− 0,125
Function J:
2
2
Answers
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TOPIC 8: EXERCISE 2
1.5.10 g ’ ( 2,5 ) = 0
1
mAB = h − 5
1.5.11 gradient of the tangent at the turning point
mCD = − 4 − h
1.5.12 y = −12,25
2
h = −6
2.1
3
mCD = −2
4
1
h = __
5
mAB = −5
y f(x) = x² – 4x – 5
A(–2;7)
2
–1
5
–5
TOPIC 8: EXERCISE 3
y
1.1
g(x) = x² – 5x – 6 y = 2x – 12
4
P
–2
(2;–9)
2
–1 –2
–4
1
2
3
4
5
6
7
8
2.2
x
x
P(3;–8)
A ( −2;7 )
P ( 3;− 8 )
mAB = −3
2.3
y = −3x + 1
2.4.1
mAB = h − 8
2.4.2
mAB = − 4
2.5.1
mDE = 2x + h − 4
2.5.2
m=1
P ( −1;0 )
Q ( 2;−12 )
2.5.3
y = x − 11
2.5.4
f ’( 2 ) = 0
1.3
mPQ = − 4
2.5.5
1.4.1
mPB = h − 7
gradient of the tangent to f at the turning
point, ( 2;−9 )
1.4.2
mPB = − 6
2.5.6
y = −9
1.4.3
B ( 0;− 6 ); see graph for line through P and B.
3.1
m=3
1.4.4
y = − 6x − 6
3.2
y = 3x + 2
1.5.1
mDE = 2x + h − 5
3.3.1
mAB = 3 − 3h + h2
1.5.2
mDE = 2
3.3.2
mAB = 7
1.5.3
mDE = 2
3.3.3
f ’ ( −1 ) = 3
1.5.4
The gradients are identical.
If x = 1 and h = 5, D and E have the same
coordinates in 1.5.2 and 1.5.3.
3.3.4
y = 3x + 2
3.4.1
m = 3x2 + 3xh + h2
3.4.2
m==7
3.4.3
f ’ ( −2 ) = 12
3.4.4
the gradient of the tangent to f at x = −2 is 12.
3.4.5
y = 12x + 32
4.1
x = −1 and x = − 4
4.2
f( 0 ) = 4
–6
–8
–10
–12
–14
–16
–18
B
D
Q (2,5;–12,25)
y = – 6x – 6
1.2
1.5.5
See graph
1.5.6
y = 2x − 12
1.5.7
g ’ ( 1 ) = −3
1.5.8
g ’ ( 1 ) represents the gradient of the tangent to g
at D ( 1;−10 ).
1.5.9
y = −3x − 7
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Answers
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4.3.1
f ( −3 ) = 0
4.3.2
f ’ ( −1 ) = 0
4.4
y=4
4.5
A ( 2;54 ) and B ( −1;0 )
4.6
54 − 0
54
mAB = ________
= ___ = 18
(
)
3
4.7
C(2 + h; h3 + 12h2 + 45h + 54)
4.8
mAC = h2 + 12h + 45
4.9
f ’ ( 2 ) = 45
4.10
the gradient of the tangent to f at x = 2.
4.11
y = 45x − 36
2 − −1
f ’ ( −1 ) = −15
4
g ’( x ) = 2
g ’( 4 ) = 2
5
f ’( x ) = 0
f ’ ( −5 ) = 0
6
− 6x
7
f ’ ( x ) = −12x2
8
g ’ ( x ) = −18x
9
f ’( x ) = 0
10
g ’ ( x ) = −12x
11
9
f ’ ( x ) = − __
2
TOPIC 8: EXERCISE 4
12.
g ’ ( x ) = − 6x2
1
f ’ ( x ) = 2x − 1
f ’( 2 ) = 3
13
f ’ ( x ) = −5 − 6x
14
g ’ ( x ) = − 4x + 1
2
1
f ’ ( x ) = − __
3
4
x2
1
(
)
f ’ −2 = − __
4
g ’ ( x ) = 15x2
g ’ ( 4 ) = 240
3
f ’ ( x ) = − __
2
x
3
(
)
f ’ −5 = − ___
25
5
g ’ ( x ) = −10x + 2
g ’ ( −1 ) = 12
6
g ’ ( x ) = −2x
TOPIC 8: EXERCISE 5
1.1
f ( −3 + h ) − f ( −3 )
________________
= h − 11
h
1.2
lim ( h − 11 ) = −11
x→0
1.3
f ’ ( x ) = 2x − 5
f ’ ( −3 ) = −11
1.5
y = −11x − 9
2.1
g ’( x ) =
2.2
2.3
7
__
x2
1
g ’ ( −7 ) = __
7
g ’ ( −7 ) gives us the gradient of the tangent to g
at x = −7
2.4
1
y = __
7+2
3
f ’ ( x ) = −15x2
x
TOPIC 8: EXERCISE 6
1
dy
2
___
= 6x + __
2
5
2x − __
x2
3
1
4
dy
8
___
= − ___
5
50x − 30x2 − 4
6
6t − 4
7
3
__
8
dy
___
= 50x − 20
9
24x2 − 72x + 54
10
2__ _____
1__
_____
+
3
3√x2
2√x3
11
4x − 1
12
10x − 15
13
2
3
−___2 − __x
3
2x
14
2
1
___
− ___
15
dx
dx
x2
3x3
1
__
x2
dx
1
__
3x3
2
___
1
__
3x3
1
__
−
2x2
3
___
2
__
5x5
3
2 __ _____
= ____
− 5 __2
3
3√
x
5√ x
16
2
__
8 __53
18 __
9
__
t − 15t 3 + ___
+
3
1
__
t3
4
__
t3
Answers
9780636143319_plt_mat_g12_lb_eng_zaf.indb 349
349
2013/05/31 11:19:05 AM
__
17
√3
2
−___
− 8x − __3
x
x2
18
3
2
__
1 − ___
1 − 2
__
19
2−
20
f ’(x) = 10x4 − 12x3 + x2
21
g ’( x ) = 3
22
24x2
23
1
3t 2 − 6t + __
2
2.3.1
d ( )
d ( )
d ( )
___
[ f x + g ( x ) ] = ___
[ f x ] + ___
[g x ]
x
2x2
3
___
dx
1
__
2x2
t
__
25
5 __ ____
1__
3√x − ____
2√ x − √ x3
__
3__
3√x − 8 + ___
26
3
__
dy
___
= 36x3 − 15x2 + 1
27
3x2
28
1 __
1 + ____
2√ x
24
LHS = 2x − 2
RHS = 2x − 2
√x
dx
dx
dx
2.3.2
LHS = 2x − 6
RHS = 2x − 6
d ( )
d ( )
d ( )
___
f x − g ( x ) ] = ___
f x ] − ___
g x ]
dx [
dx [
dx [
2.3.3
LHS = 20x − 40
RHS = 20x − 40
d
d ( )
___
kf ( x ) ] = k___
f x ]
dx [
dx [
2.3.4
LHS = 6x2 − 32x + 32
RHS = 4x − 8
d ( )
d ( )
d ( )
___
f x × g ( x ) ] ≠ ___
f x ] × ___
g x ]
dx [
dx [
dx [
2.3.5
1
LHS = __
2
RHS = 2x − 2
d ( )
___
f x ]
f( x )
d ____
dx [
___
_______
=
d ( )
dx g ( x )
___
g x ]
dx [
[ ]
TOPIC 8: EXERCISE 7
1.1
f ( x )g ( x ) = 6x5 + 6x4 − 6x3 − 6x2
1.2
LHS = 36x2 + 24x2
RHS = 30x4 + 24x3 − 18x2 − 12x
1.3
f( x )
3
3
____
= ___ − ___
1.4
1
LHS = __
g( x )
2x
2
2x
x
3
3
__
RHS = − ___
2 + 3
2x
x
3.1
2x − 4
3.2
p ’( x ) = 2 ⇒ p ’( 3 ) = 2
3.3
2x5 +4x4 + 8x3 − 16x2 − 32x − 64
3.4
18
3.5
10x4 + 16x3 + 24x2 − 32x − 30
4.1
x+2
_____
4.2
2
h ’ ( x ) = − __
2
x
x
LHS = 10x + 6x2
TOPIC 8: EXERCISE 8
RHS = 10x + 6x2
1.1
y = 24x − 30
2.1.1
f ( x ) + g ( x ) = x2 − 2x − 8
1.2
Increasing, f ’ ( 2 ) > 0
2.1.2
f ( x ) − g ( x ) = x2 − 6x + 8
1.3
y=9
2.1.3
10f ( x ) = 10x2 − 40x
2.1.4
f ( x )g ( x ) = 2x3 − 16x2 + 32x
1.4
2.1
Stationary, f ’ ( −1 ) = 0.
4
f ’ ( x ) = − __
2
2.1.5
f( x )
x( x − 4 )
x
____
= _______ = __
2.2
m = f ’ ( 2 ) = −1
2.2.1
(
2.3
Decreasing, f ’ ( 2 ) < 0.
2.2.2
( f ( x ) − g( x ) )’ = 2x − 6
2.4
y = −x + 2
2.2.3
( 10f ( x ) )’ = 20x − 40
2.5
f ’ ( −1 ) = − 4
2.2.4
( f ( x )g( x ) )’ = 6x2 − 32x + 32
2.6
Decreasing, f ’ ( −1 ) < 0.
y = 4x − 10
2.2.5
( ) 2
2.7
2.8
4
No, f ’ ( x ) = − __
2 < 0 for all real values of x, x = 0.
3.1
2
f ’ ( x ) = 2x − 3; g ’ ( x ) = −2x − 1; k ’ ( x ) = − __
2
1.5
350
g( x )
2( x − 4 )
2
)
f ( x ) + g( x ) ’ = 2x − 2
’
f ( x)
1
____
__
g( x ) =
x
x
x
Answers
9780636143319_plt_mat_g12_lb_eng_zaf.indb 350
2013/05/31 11:19:05 AM
3.2.1
f ’ ( −2 ) = −7
f ( −2 ) = 6
y = −7x − 8
g ’ ( −2 ) = 3
g ( −2 ) = 4
y = 3x + 10
2.1
x = −1,1
2.2
17
f ’ ( −1,1 ) = − ___
5
f ( −1,1 ) = − 4,68
17
y = − ___
5 x − 8,42
17
g ’ ( −1,1 ) = − ___
5
1
k ’ ( −2 ) = − __
2
g ( −1,1 ) = − 0,63
k ( −2 ) = − 4
3.2.2
x + 2y + 10 = 0
2.3
f ’ ( 1 ) = = −1
f( 1 ) = − 6
y = −x − 5
g ’ ( 1 ) = −3, g ( 1 ) = 4
y = −3x + 7
g ’ ( 1 ) = −16
g ( 1 ) = −21
y = −16x − 5
2.4
f ’( x ) = 0
k ’ ( 1 ) = −2, k ( 1 ) = −1
y = −2x + 1
3.2.3
17
y = − ___
5 x − 4,37
f ’ ( 5 ) = 7, f ( 5 ) = 6
y = 7x − 29
g ’ ( 5 ) = −11, g ( 5 ) = −24
y = −11x + 31
13
2
___
( )
k ’ ( 5 ) = − ___
25 , k 5 = − 5
2x + 50y + 55 = 0
TOPIC 8: EXERCISE 9
( )
1
f − __
4 = − 6,125
1
y = − 6__
8
2.5
f ’ ( −3 ) = 4 ( −3 ) + 1 = −11 and f ( −3 ) = 2 ( −3 )2
+ ( −3 ) − 6 = 9 ⇒ ( −3;9 )
Substitute m = −11 and ( −3;0 ) into y − y1 = m
( x − x1 )
y − 9 = −11 ( x + 3 ) ⇒ y = −11x − 24
2.6
f ’ ( x ) = −3
f ( −1 ) = −5
y = −3x − 8
3
y=0
4.1
y = −x − 5
y = −x + 3
1.1
2
f ’ ( x ) = − __
2
1.2
1
x = ± __
2
1.3
y = −2x − 4
4.2
x + 4y = 4
1.4
176
4
14
____
___
If x = __
3 , y = − 27 = − 6 17
5.1
y = 12x − 21
If x = −2, y = 12
5.2
12x − 4y + 15 = 0
1.5
(
)
x
310
5
g − __
= ____
3
27
g( 1 ) = − 6
y = −3x − 3
TOPIC 8: EXERCISE 10
1.1
( −5;184 ) and ( 3;−72 )
1.2
f ” ( −5 ) < 0 concave down at ( −5;184 ) local
maximum
f ” ( 3 ) > 0 concave up at ( 3;−72 ) local minimum
( −1;56 )
1.6
g ’ ( 2 ) = 8, g ( 2 ) = − 4
y = 8x − 20
1.7
g ’ ( − 3 ) = 13
7
x = __
3
1.3
7
14
g __
= − ___
2.1
(3)
27
833
y = 13x − ____
27
1.8
m=1
1.9
m = −4
2.2
( − __32;− 44,75 ), (4;288)
3
3
__
g ” ( − __
2 ) > 0 concave up ( − 2 ;− 44,75 ) it a local
minimum
g ” ( 4 ) < 0 concave down (4;288) local maximum
2.3
(1,25;116,9375)
Answers
9780636143319_plt_mat_g12_lb_eng_zaf.indb 351
351
2013/05/31 11:19:05 AM
3.1
( −1;−2 ) and (1;2)
3.2
f ” ( −1 ) > 0 concave up at ( −1;−2 ) local
minimum
f ” ( 1 ) < 0 concave down local maximum
3.3
(0;0)
4.1
( 7;−179__23 ), ( − 5;108__13 )
4.2
6.10
(
)
2
g ” ( 7 ) > 0 concave up at 7;−179__
3 local
minimum.
(
7.1
14
( − __73;18___
27 ) and (1;0)
7.2
f ’ ( −3 ) > 0 and f ’ ( −2 ) < 0
14
( − __73;18___
27 ) is a local maximum, the first
)
1
g ” ( −5 ) < 0 concave down at −5;108__
local
maximum
4.3
(
2
1;−35__
3
3
derivative changes from positive to negative.
f ’ ( 0 ) < 0 and f ’ ( 2 ) > 0
(1;0) is a local minimum because the first
derivative changes from negative to positive.
)
5.1
17
( −2;−10__23 ) and ( 2,5;19___
27 )
5.2
f ” ( −2 ) > 0 concave up local minimum.
(
7.3
)
17
f ” ( 2,5 ) < 0 concave down at 2,5;19___
27 local
maximum
5.3
25
( 0,25;4___
48 )
6.1
(0;0) and ( −2;− 4 )
6.2
g ’ ( − 1 ) > 0, g ’ ( 1 ) < 0
(0;0) is a local maximum as the first derivative
changes sign form positive to negative at x = 0.
g ’ ( −3 ) < 0, g ’ ( −1 ) > 0 first derivative
changes sign from negative to positive at x =
−2, ( −2;− 4 ) is a local minimum
6.3
(
( )
( −1;−2 )
6.5
x = 0 or x = −3
f ” ( 1 ) > 0, concave up at (1;0) local minimum
7.4
7
( − __23;9___
27 )
7.5
2
f changes concavity at x = − __
3.
(
2
14
– –7 ;18 27
3
–3
7
– –2;9 –
–2
0
–1
x
(–1;–2)
(–2;–4)
–6
g'(x) = –3x 2 – 6x
6.8
g ’ ( x ) > 0 if − 2 < x < 0
g ’ ( x ) < 0 if x < − 2 or x > 0
6.9
The x-coordinates of the stationary points of g
are the same as the x-coordinates of the roots
of g ’.
27
3
–4
7.9
(–1;–3)
)
y
f '(x) = 3x + 4x – 7
2
y
g''(x) = –6x –6
)
7.6, 7.7 & 7.8
3
3
(
2
f is concave down from x ∈ − ∞ ;− __
3 and
2
__
concave up from − 3; ∞ .
2
f(x) = x + 2x – 7x + 4
6.6, 6.7 & 6.10 g(x) = –x – 3x
)
7
7
14
__
___
f ” − __
3 < 0, concave down − 3 ;18 27 local
maximum
g ” ( −2 ) > 0 so g is concave up at ( −2;− 4 ) local
minimum.
g ” ( 0 ) < 0 so g is concave down at (0;0) local
maximum.
6.4
352
When g ” ( x ) > 0 g is concave up, when
g ” ( x ) = 0, g has a point of inflection and when
g ” ( x ) < 0, g is concave down. When g ” changes
sign, g experiences a change in concavity and
the point of inflection indicates the point at
which the change in concavity takes place.
– 2–
3
4
– –7
3
1
x
–
– –2 ;–81
3
3
f ''(x) = 6x + 4
The x-coordinates of the roots of f ’ are the same
as the x-coordinates of the stationary points
of f.
The x-coordinate of the turning point of f ’ is
the same as the x-coordinate of the point of
inflection of f.
When f ’ ( x ) > 0, f is an increasing function and
when f ’ ( x ) < 0, f is a decreasing function.
When f ” ( x ) < 0, f is concave down, when
f ” ( x ) > 0, f is concave up and when f ” ( x ) = 0
there is a point of inflection.
Answers
9780636143319_plt_mat_g12_lb_eng_zaf.indb 352
2013/05/31 11:19:06 AM
TOPIC 8: EXERCISE 11
1
5
Roots: (2;0) and ( −3;0 )
1
3
–1 ;18,52
f(x) = – x3 – 2 x2 + 7x – 4 y
y
1
12
–4
1 9
;
3 26
(
4
14
___
Stationary points: (2;0) − __
3 ;18 27
(
1 ___
7
Point of inflection: __
3 ;9 27
)
(
)
2
7
__
___
Point of inflection: ( − 3 ;−927 )
1
Roots: ( − __
2 ;0 )
)
6
y
Roots: ( − 6;0 ), ( 1,5;0 ) and ( 6;0 )
1
11
;11 –
12
2
–6
1,5
6
1 3 1 2
f(x) = x – 2 x – 12x +18
3
x
−
(
7
Stationary point and point of inflection: __
2 ;512
(
)
7
)
2
20 –
3
2
–1 5
–
2 ;20 6
–2
4
–2
f(x) = 34
–x3 – 2 x 2 + x + 62
–
3
Roots: (1;0) or ( −2;0 )
(–1;2)
x
( __12;20__56 ) is a stationary point and a point of
x
1
)
Roots: ( −2;0 )
y
f(x) = – x – 3x – 4 y
inflection.
8
Roots: (0;0)
y
Stationary points: ( 0;4 ) and ( −2; 0 )
Point of inflection: ( −1;2 )
4
x
1
2
(
2
3
2
Stationary points: ( −3;40,5 ) and 4;−16__
3
1
11
___
Point of inflection: __
2 ;11 12
7
;512
2
169
4;–16
3
f(x) = (2x + 1) 4x2 − 44x + 169
y
(– 3;40,5)
3
x
1
14
___
Stationary point(s): −2__
3 ;−18 27 and ( 1;0 )
2
f(x) = (x – 2) ( x + 3)
–4
7
–
– 3–2 ;–923
–
–2 31
– ;–1814
27
x
2
–3
2
Roots: (1;0) and ( − 4;0 )
Roots: ( −2;0 ), ( −1,25;0 ) and ( 0;0 )
x
f(x) = – 4x3 – 13x – 10x y
1
– 2– ;2,25
2
(–1,08;0,66)
–2
–1,25
0
1
1;– –
3
f(x) = – 3–1 x3 + x2 – x
x
( 1;− __13 ) is a stationary point and a point of
(–1,67;0,93)
inflection.
(
2 ___
25
Stationary points: ( − 0,5;2,25 ) and −1__
3 ;− 27
1
___
Point of inflection: 112;0,66
(
)
)
Answers
9780636143319_plt_mat_g12_lb_eng_zaf.indb 353
353
2013/05/31 11:19:06 AM
9
( )
7
Roots: ( 1;0 ), __
4 ;0 and ( 3;0 )
15
Roots: ( −5,72;0 ), ( 1,97;0 ) and (3;0)
y
y f(x) = 1–(x – 3)(4x2 + 15x – 45)
6
(–3;54)
•
21
f(x) = – 4x3 + 23x2 – 40x + 21
– 1–;26,27 •
4
(2,5;2,25)
1
1,75
4 ;– 25
––
–
3 27
x
3
–5,72
(
(
)
10
Roots: (2;0)
f(x) = – x3 + 8 y
Stationary point and
point of inflection: (0;8)
11
(0,8)
Roots: (1;0)
y
3
16
f(x) = 4x 3 + 37x 2 + 110x + 104
–4
(
)
17
y
–4
f(x) = (x – 2)
–1
–1
(
1
Point of inflection: __
3 ;−2,81
Roots: ( −2;0 ), ( 0;0 ) and ( 2;0 )
– (x + 2) 4x2 + 13x – 50 y
f(x) = 1
6
(0,5;17,71)
–5,52
(–1,75;2,52)
–2
2,27
x
– 4;12 2
3
(– 1,15;3,08)
Stationary points: ( 0,5;17,71 )
( − 4;−12__23 )
x
(1,15; –3,08)
__
√
)
( −5,52;0 ), ( −2;0 ) and ( 2,27;0 )
f(x) = x3 – 4x
Stationary points:
x
Stationary points: ( − 0,64;0,86 ) and
( 1,31;− 6,49 )
18
2
2,22
1 ;–2,81
3
(1,31;–6,49)
x
Stationary point and point of inflection: (2;0)
__
√
–0,22
x
2
–8
)
(– 0,64;0,86)
(–3;–2)
(–2;–4)
3
12
)
y f(x) = 2x 3 + 2x 2 – 5x – 1
–1
0
(
Roots: ( −1;0 ), ( − 0,22;0 ) and ( 2,22;0 )
16
Roots: (2;0)
–2
x
–2
(–2,5;–2,25)
1
;− 0,66
Point of inflection: −3___
Roots: ( − 4;0 ) and ( −2;0 )
Stationary points: ( − 4;0 ) and ( −1;− 4 )
Point of inflection: (−3;−2)
y
(– 3,08;0,– 66)
37 ___
25
Stationary points: ( −2,5;−2,25 ) and − ___
12 ; 27
f(x) = x + 9x + 24x + 16
14
)
Roots: ( − 4;0 ), ( −3,25;0 ) and ( −2;0 )
(– 3,67;0,93)
2
(
104
x
x
1 __
7
Point of inflection: __
2 ;− 2
y
)
y
2
Stationary point(s): None
13
(
2
(
x
(2,5;–1,46)
1
11
___
Stationary points: 2__
2 ;−1 24 and ( −3;54 )
13
1
___
Point of inflection: − __
4 ;26 48
–1 ;– –7
2 2
–7
12
)
f(x) = 2x – 3x + 8x – 7
1
2,97 • • 3
•
•
25
4 ___
Stationary points: ( 2,5;2,25 ) and __
3 ;− 27
11
Point of inflection: 1___
12 ;0,66
• 22,5
Point of inflection: ( −1,75;2,52 )
__
√
__
√
2 3 _____
16 3
2 3 _____
16 3
____
( − ____
3 ; 9 ) and ( 3 ;− 9 )
Point of inflection: (0;0)
354
Answers
9780636143319_plt_mat_g12_lb_eng_zaf.indb 354
2013/05/31 11:19:07 AM
19
7
Roots: ( 3;0 ), ( 4,94;0 ) and ( −10,94;0 )
y
The x-coordinate of the turning point of f ’ is
the same as the x-coordinate of the point of
inflection of f.
(– 6;162)
(1;78,67) 53
3
4,94
2
–10,94
4;– 4 3
x
When f ’ ( x ) > 0, f is an increasing function and
when f ’ ( x ) < 0, f is a decreasing function.
1
– x 2 – 24x + 54
f(x) = 3
(
14
Stationary points: ( − 6;162 ) and 4;− ___
Point of inflection: ( −1;78,67 )
20
3
– 3;82
When f ” ( x ) < 0, f is concave down, when
f ” ( x ) > 0, f is concave up and when f ” ( x ) = 0
there is a point of inflection.
)
Roots: ( −3,82;0 ), ( 1,62;0 ) and ( −2;0 )
8
y f(x) = 5x3 + 21x2 – 9x – 62
(– 3;19)
–2
9
x
1,62
–62 (0,2;–62,92)
lim f ( x ) = −∞ and lim f ( x ) = ∞
x→ −∞
x→ ∞
1
f ’ ( x ) = 3x2 − 18x + 24 = 0 | Divide through by
3
x2 − 6x + 8 = 0
( x − 2 )( x − 4 ) = 0
x = 2 or x = 4
If x = 2, f ( 2 ) = 0 ⇒ ( 2;0 )
If x = 4, f ( 4 ) = − 4
Stationary points: (2;0) and (4;− 4)
f ” ( x ) = 6x − 18
f ” ( 2 ) = − 6 < 0 so f is concave up and ( −2;0 ) is
a local minimum.
f ” ( 4 ) = 6 > 0 so f is concave down and ( 4;− 4 ) is
a local maximum.
2
( 3;−2 )
3
f ” ( 3 ) = 0, f ” ( 2 ) = − 6 < 0 and f ” ( 4 ) = 6 > 0 so,
f changes from concave down to concave up
at ( 3;−2 ).
4
x = 2 or x = 5
Stationary Points: ( −3;235 ) and ( 0,2;− 62,92 )
( −1,4;−21,96 ) is a point of inflection
TOPIC 8: EXERCISE 12
Function A:
1
Stationary points: (1;0) and ( 3;− 4 )
f ” ( 1 ) = − 6 < 0, so f is concave down at (1;0)
and this point is a local maximum.
f ” ( 3 ) = 6 > 0, so f is concave up at ( 3;− 4 ) and
this point is a local minimum.
2
( 2;−2 )
3
f ” ( 1 ) < 0 and f ” ( 3 ) > 0
( 2;−2 ).
4
x = 1 or x = 4
y
f ( −5 ) = −324 and f ( 5 ) = 16
The minimum value of f( x ) for x ∈ [ −5;5 ] is −324.
The maximum value of f ( x ) for x ∈ [ −5;5 ] is 16.
Function B:
(–1,4;–21,96)
5
The x-coordinates of the roots of f ’ are the same
as the x-coordinates of the stationary points of f.
2
f '(x) = 3x – 12x + 9
5
2
f ' (x) = 3x – 18x + 24
(1;0)
2
3
(2;–2)
–4
4
y
x
(3;–2)
(2;–3)
(3;–4)
3
(2;0)
f ''(x) = 6x – 12
f(x) = x3 – 6x + 9x – 4
6
–18
See the sketch for f ’ and f ”.
f''(x) = 6x – 18
4 5
(3;–3) (4;–4)
x
–20
f(x) = x3 – 9x2 + 24x – 20
6
See the graph for f ’ and f ”.
Answers
9780636143319_plt_mat_g12_lb_eng_zaf.indb 355
355
2013/05/31 11:19:07 AM
7
The x-coordinates of the roots of f ’ are the same as
the x-coordinates of the stationary points of f.
The x-coordinate of the turning point of f ’ is
the same as the x-coordinate of the point of
inflection of f.
When f ’ ( x ) > 0, f is an increasing function and
when f ’ ( x ) < 0, f is a decreasing function.
When f ”( x ) < 0, fis concave down, when
f ”( x ) > 0, f is concave up and when f ”( x ) = 0
there is a point of inflection.
8
The minimum value of f( x )for x ∈ [ −5;5 ] is − 490.
The maximum value of f ( x )for x ∈ [ −5;5 ] is 0.
9
lim f ( x ) = − ∞ and lim f ( x ) = ∞
x→−∞
x→∞
Function C:
1
There is a stationary point at (3,5; 64)
f ” ( x ) = 6x − 21
f ” ( 3,5 ) = 6 ( 3,5 ) − 21 = 0, so the second
derivative test fails!
f ” ( 3 ) = −3 < 0 and f ” ( 4 ) = 3 > 0
There is a change in concavity at x = 3,5 and
so the stationary point ( 3,5;64 ) is a point of
inflection.
2
( 3,5;64 ) from 1
3
f ” ( 3,5 ) = 0, f ”( x ) < 0 when x < 3,5 and f ”( x ) > 0
when x > 3,5 so f changes from concave down
to concave up at ( 3,5;64 )
4
5
8
9
lim f ( x ) = −∞ and lim f ( x ) = ∞
x→−∞
x→∞
Function D:
1
(
)
2
f '(x) = 3x – 21x + 147
–
y
1
(2x + 1) (4x 2 + 44x + 169)
f (x) = –
8
( )
7
( −__14;− 4___
32 ) and this point is a local minimum.
2
f ” ( 3 ) = −13 < 0, so f is concave down at ( 3;18__
3)
and this point is a local maximum.
2
43
11
____
( ___
8 ;16 192 ) is a point of inflection.
3
11
11
___
( )
f ” ( x ) > 0 if x < ___
8 , f ” x < 0 if x > 8 and
11
f ” ___
8 = 0, so f changes from concave up to
11
concave down at x = ___
8.
4
x=1
x ≈ 4,48 or x ≈ 0,64
( )
5
3
4 x + 11
–
f (x) = – –
x2 + 3x – 23
–
6
2
3
y
f '(x) = – 8x + 11
1
––
4
–1
1
7
– – ;– 4–
4
32
– 21
See the sketch for f ’ and f ”.
7
f ’ ( x ) = f ” ( 3,5 ) = 0 so (3,5;64) is both a
stationary point and a point of inflection.
f ’ ( x ) ≥ 0 for all real values of x, so f is never a
decreasing function.
f ” ( x ) changes sign from negative to positive at
3,5, so there is a change of concavity in f from
concave down to concave up.
4,48
6
See graph
7
The x-coordinates of the stationary points of
f are same as the x-coordinates of the roots of
f ’ ( x ).
x
When f ’ ( x ) > 0 we can see that f is an increasing
function and when f ’ ( x ) < 0 we can see that f
is a
x
6
2
3;183
169
11
;
–
–
8 16
11
–
7,22
2
8;
0,64 11
3
–
8
– 23
–
6
2
f '(x) = – 4x + 11x + 3
f ''(x) = 6x – 21
(3,5;0)
)
1 = 13 > 0, so f is concave down at
f ” −__
4
(3,5;64)
36,75
21,125
(
1
7
2
___
__
Stationary points: − __
4 ;− 4 32 and 3;18 3
1
The only real solution is x = − __
2
4
356
The minimum value of f ( x ) for x ∈ [ −5;5 ] is
−550,125.
The maximum value of f ( x ) for x ∈ [ −5;5 ] is
67,375.
decreasing function. When f ” ( x ) < 0, fis
concave down, when f ” ( x ) > 0, f is concave
up and when f ” ( x ) = 0 there is a point of
inflection.
8
9
The minimum value is −18 and the maximum
value is 285,33.
lim f ( x ) = ∞ and lim f ( x ) = −∞
x→−∞
x→∞
Answers
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2013/05/31 11:19:08 AM
Function E:
1
3
4
5
f ” ( x ) < 0 when x < 1, f ” ( x ) > 0 when x > 1 and
f ” ( 1 ) = 0 so f changes concavity from concave
down to concave up at x = 1.
4
x = 1 or x = 3 or x = −1
5
y
3
2
f '(x) = 3x2 – 6x – 1 f ''(x) = 6x – 6 f(x) = x – 3x – x + 3
( − 4;40,5 )
( 3;16,67 )
f ” ( −4 ) = 7 > 0 so f is concave up and ( − 4;40,5 )
is a local minimum
f ” ( 3 ) = −7 < 0 so f is concave down and
(3;16,67) is a local maximum.
2
3
(– 0,15;3,08)
3
( −__12;−11,92 )
1
1 , f ” ( x ) < 0 when x > −__
f ” ( x ) > 0 when x < −__
2
2
1
and f ” − __
2 = 0, so f changes concavity from
1
concave up to concave down at x = − __
2.
–1 –0,15
( )
(1;– 4)
2
3;16 –
3
–
–1
2 ;12,25
–4
–7
1
–
–
–1
22
–1
1 ;11,92
––
2
5
–5 –
6
3
5
See graph
7
The x-coordinates of the roots of f ’ ( x )are the
same as the x-coordinates of the stationary
points of f. f ’ ( x ) > 0 when f is an increasing
function, f ’ ( x ) < 0 when f is a decreasing
function.
f ” ( x ) < 0 when x < 1, f ” ( x ) > 0 when x > 1 and
f ” ( 1 ) = 0, so f changes concavity from concave
down to concave up at x = 1.
8
f ( −5 ) = −192 and f ( 5 ) = 48
Minimum value is −192 and maximum value is
48
x
(–4;–40,5)
f '(x) = – x 2 – x + 12
See graph
7
The x-coordinates of the roots of f ’ ( x ) are the
same as the x-coordinates of the stationary
points of f. f ’ ( x ) > 0 when f is an increasing
function, f ’ ( x ) < 0 when f is a decreasing
function.f.
1 , f ” ( x ) < 0 when x > −__
1
f ” ( x ) > 0 when x < −__
2
2
1 = 0, so f changes concavity from
and f ” −__
2
1
concave up to concave down at x = − __
2.
( )
8
9
maximum.
f ” ( 2,15 ) = 6 ( 2,15 ) − 6 = 6,93 > 0, so f
is concave up and ( 2,15;−3,08 ) is a local
minimum.
2
1
x→∞
13
( − __23;9___
27 ); ( 2;0 )
2
f ” ( − __
3 ) = −8 < 0 so f is concave down at
13
( − __23;9___
27 )
and this is a local maximum. f ” ( 2 ) = 8 > 0
so f is concave up at (2;0) and this is a local
minimum
x→∞
( − 0,15;3,08 ); ( 2,15;−3,08 )
f ” ( − 0,15 ) = 6 ( − 0,15 ) − 6 = −6,9 < 0, so f
is concave down and ( − 0,15;3,08 ) is a local
lim f ( x ) = − ∞ and lim f ( x ) = ∞
x→−∞
Function G:
lim f ( x ) = ∞ and lim f ( x ) = −∞
Function F:
1
9
Minimum: −40,5
2
Maximum value is 16__
3
x→−∞
(2,15;–3,08)
6
f ''(x) = –2x –1
6
x
2,15 3
–6
1
x = −7 or x = __
2 or x = 5
1 (x – 5)(2x – 1)(x + 7)
f(x) = – –
6
y
–1 1
2
( __23;4,74 )
3
2
2
__
( )
f ” ( x ) < 0 when x < __
3 , f ” x > 0 when x > 3 and
2
f ” __
3 = 0 so f changes concavity from concave
2
down to concave up at x = __
3.
4
x = 2 or x = −2
( )
(1;0)
Answers
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357
2013/05/31 11:19:08 AM
y
5
f ’(x) = 3x² – 4x – 4
–2
–;913
–
3 27
–2
7
f ’ ( x ) has no roots and f ( x )has no stationary
points.
f ’ ( x ) < 0 for all real values of x and f is a
decreasing function for all real values of x.
The x-coordinate of the point of inflection is
the same as the x-coordinate of the turning
point of f ’ ( x ) and the root of f ” ( x ).
8
minimum −304 and maximum 36.
8
– –2
–2
3
2
x
3
2
–;–5–1
f(x) = x³ – 2x² – 4x + 8 3 3
f“(x) = 6x – 4
9
6
See graph
7
The x-coordinates of the roots of f ’ ( x ) are the
same as the x-coordinates of the stationary
points of f.
1
( )
( )
7
22
___
and ( __
3 ;−14 27 ) is a local minimum.
7
7
__
f ” __
3 = − 6 3 + 28 = 14 > 0 so f is concave up
( )
9
lim f ( x ) = − ∞ and lim f ( x ) = ∞
x→−∞
x→∞
f ” ( 7 ) = − 6 ( 7 ) + 28 = −14 < 0 so f is concave
down and ( 7;36 ) is a local maximum.
2
2
maximum value of f ’ ( x ) is − __
3
f is always a decreasing function and it has no
stationary points.
2
3
16
( 4__23;10___
27 )
3
2
2
__
( )
f ” ( x ) > 0 if x < 4__
3 , f ” x < 0 if x > 4 3 and
14
___
f ” 3 = 0, so f changes concavity from concave
2
up to concave down at x = 4__
3.
4
x = 1 or x = 4 or x = 9
5
y
Function H:
1
22
( __73;−14___
27 )
( 7;36 )
2
2
__
( )
f ” ( x ) < 0 when x < __
3 , f ” x > 0 when x > 3 and
2
f ” __
3 = 0, so f changes concavity from concave
2
down to concave up at x = __
3.
Minimum value is −147 and maximum value is 63
x→∞
Function I:
f ’ ( x ) > 0 when f is an increasing function, f ’ ( x )
< 0 when f is a decreasing function.f.
8
lim f ( x ) = ∞ and lim f ( x ) = −∞
x→−∞
( )
7
( −__53;−3___
27 )
2
3
f(x) = –x + 14x – 49x + 36
5
(
)
f ” ( −2 ) = 2 > 0 if x < − __
3 , f ” −1 = − 4 < 0
5
5
__
__
if x > − 3 and f ” − 3 = 0 so f changes from
5
concave up to concave down at x = − __
3.
( )
4
x = −3
5
f(x) = –x – 5x – 9x – 9 y
f ''(x) = – 6x + 28
28
36
1
2
–2
3
•
7
– ;–1422
–
– 49
2
– 5– ;– 2–
3
– 5–
3
3
–3
•
•• 4 2– ;1016
–
27
3
•
4 2•
•
3
3
(7;36)
1
–
4 –3 ;16 3
7
4–
3
9
27
2
f '(x) = –3x + 28x – 49
6
See graph
7
The x-coordinates of the roots of f ’ ( x )are the
same as the x-coordinates of the stationary
points of f. f ’ ( x ) > 0 when f is an increasing
function, f ’ ( x ) < 0 when f is a decreasing
function.f.
2
2
__
( )
f ” ( x ) > 0 when x < 4__
3 , f ” x < 0 when x > 4 3
14
___
and f ” 3 = 0, so f changes concavity from
x
7
–
– 5– ;–327
–9
3
–10
x
( )
f '(x) = –3x 2 – 10x – 10
f ''(x) = – 6x – 10
6
358
See graph
14
concave up to concave down at x = ___
3.
8
9
22
Minimum −14___
27 and maximum 756
lim f ( x ) = ∞ and lim f ( x ) = − ∞
x→−∞
x→∞
Answers
9780636143319_plt_mat_g12_lb_eng_zaf.indb 358
2013/05/31 11:19:08 AM
5
Function J:
1
( 3
1
1__
;14,81
)
y
f (x) = – x3 – 3x2 – 3x – 2
f''(x) = –6x – 6
( 6;−36 )
–1
( )
4
f ” __
3 = −14 < 0 so f is concave down at
–2
( 1__13;14,81 ) and this is a local maximum.
–3
f ” ( 6 ) = 14 > 0 so f is concave up at ( 6;−36 ) and
this is a local maximum.
2
11 ____
286
( ___
3 ;− 27 )
3
2
2
__
( )
f ” ( x ) < 0 if x < 3__
3 .f ” x > 0 if x > 3 3 and
11
f ___
3 = 0 so f changes concavity from
2
concave down to concave up at x = 4__
3.
4
x = 0 or x = 3 or x = 8
( )
5
y
f ''(x) = 3x – 22x + 24
–6
f'(x) = –3x 2 – 6x – 3
6
See graph
7
f ’ ( −1 ) = 0, but f ’ ( x ) < 0 for all other real values
of x, so f will be stationary at x = −1, but
decreasing on all other real values of x.
f ’ ( −1 ) = f ” ( −1 ) = 0 and ( −1;−1 ) is a stationary
point and a point of inflection.
8
minimum of −217, maximum of 63
2
24
f '(x) = –6x – 22
0
–22
9
4
– ;14,81
3
11
3
3
6
2
– ;10,59
33
2
1
1
3
3 –3 ;–16 –3
2
f(x) = –x – 11x + 24x
(6;–36)
6
See graph
7
The x-coordinates of the roots of f ’ ( x )are the
same as the x-coordinates of the stationary
points of f. f ’ ( x ) > 0 when f is an increasing
function, f ’ ( x ) < 0 when f is a decreasing
function.
11
11
___
( )
f ” ( x ) < 0 when x < ___
3 , f ” x > 0 when x > 3
11
and f ” ___
3 = 0, so f changes concavity from
11
concave down to concave up at x = ___
3.
9
( )
( −1;−1 )
3
f ” ( x ) > 0 if x < − 1, f ” ( x ) < 0 if x > − 1 and
f ” ( −1 ) = 0, so f changes concavity from
concave up to concave down at x = −1.
4
x = −2
2
3
2
( )
(3
)
1
down at __
;0,93 and this point is a local
maximum.
( ) ( )
3
at ( __
2 ;−2,25 ) and this point is a local minimum.
3
3
__
f ” __
2 = 24 2 − 22 = 14 > 0, so f is concave up
x→∞
2
3
1
1
__
f ” __
3 = 24 3 − 22 = −14 < 0, so f is concave
lim f ( x ) = −∞ and lim f ( x ) = ∞
There is only one stationary point at ( −1;−1 )
f ” ( −1 ) = 0, f ” ( −2 ) = 6 > 0 and f ” ( 0 ) = − 6 < 0
f changes concavity from concave up to concave
down at x = −1 so ( −1;−1 ) is a point of inflection.
| Divide through by
f ” ( x ) = 24x − 22
Function K:
1
f ’ ( x ) = 12x2 − 22x + 6 = 0
2.
6x2 − 11x + 3 = 0
( 2x − 3 ) ( 3x − 1 ) = 0
3
1
__
x = __
2 or x = 3
25
( ) ( ) − 11( __13 ) + 6( __13 ) = − ___
27 ≈ 0,93
1
⇒ ( __
3 ;0,93 )
3
3
3
3
__
__
__
f ( __
2 ) = 4 ( 2 ) − 11 ( 2 ) + 6 ( 2 ) = −2,25
3
⇒ ( __
2 ;−2,25 )
Minimum −520 and maximum 14,81
x→−∞
x→∞
1
1
__
f __
3 =4 3
( )
8
lim f ( x ) = ∞ and lim f ( x ) = −∞
x→−∞
Function L:
x
8
x
–2
(–1;–1)
2
11
f ” ( x ) = 24x − 22 = 0 ⇒ x = ___
12
( )
(
143
11
11
____
___
f ___
12 = − 216 ≈ − 0,66 ⇒ 12 ;− 0,66
( 12
)
)
11
___
;− 0,66 is a point of inflection.
3
11
11
___
( )
f ” ( x ) < 0 if x < ___
12 , f ” x > 0 if x > 12 and
11
f ” ___
12 = 0, so f changes concavity from concave
11
down to concave up at x = ___
12 .
( )
Answers
9780636143319_plt_mat_g12_lb_eng_zaf.indb 359
359
2013/05/31 11:19:09 AM
4
f ( x ) = 4x3 − 11x2 + 6x
4x3 − 11x2 + 6x = 0
x ( 4x2 − 11x + 6 ) = 0
x ( x − 2 ) ( 4x − 3 ) = 0
3
x = 0 or x = 2 or x = __
4
5
3
f ’ ( x ) < 0 for all real values of x so f is always
a decreasing function and has no stationary
points.
f ” ( x ) = 0 when f ’ ( x ) is has its maximum of − 6.
8
minimum value of −259 and a maximum value
of 81
2
f(x) = 4x –11x + 6x
y
2
f''(x) = 12x – 22x + 6
7
9
6
lim f ( x ) = ∞ and lim f ( x ) = −∞
x→−∞
x→∞
Function N:
1
– ;0,93
11
3
–
12
3 11
1
– ; – 143
– 1,5
–
–
3
4 12 216
0
2
x
(1,5;–2,25)
1
f has no stationary points.
2
( __13;−3,41 )
3
1
f ” ( 0 ) = −2 < 0 so f is concave down if x < __
3,
1
__
f ” ( 1 ) = 4 > 0 so f is concave up if x > 3 and
( )
11
–;– 4,08
12
1
1
__
f ” __
3 = 0 so f changes concavity at x = 3 from
concave down to concave up.
f''(x) = –24x – 22
4
6
See graph
7
The x-coordinates of the roots of f ’ ( x )are the same
as the x-coordinates of the stationary points of f.
f ’ ( x ) > 0 when f is an increasing function, f ’ ( x ) <
0 when f is a decreasing function.
11
11
___
( )
f ” ( x ) < 0 when x < ___
12 , f ” x > 0 when x > 12
11
and f ” ___
12 = 0, so f changes concavity from
11
concave down to concave up at x = ___
12 .
x=1
5
f '(x) = 3x 2 – 2x + 5
1 ;4–
2
–
3
5
–2
( )
8
9
lim f ( x ) = − ∞ and lim f ( x ) = ∞
x→∞
Function M:
6
See graph
7
f ’ ( x ) > 0 for all real values of x so f is always
an increasing function and has no stationary
points.
f ” ( x ) = 0 when f ’ ( x ) has its minimum value of
2
4__
3.
minimum value of −180 and a maximum value
of 120
2
( −1;−7 ).
3
f ” ( x ) > 0 if x < −1, f ” ( x ) < 0 if x > −1 and
f ” ( −1 ) = 0, so f changes concavity from
concave up to concave down at x = −1.
8
x = −2
9
f (x) = (x + 2)(–x 2 – x – 7) y
f '(x) = –3x 2 – 6x – 9
6
360
See graph
lim f ( x ) = − ∞ and lim f ( x ) = ∞
x→−∞
x→∞
TOPIC 8: EXERCISE 13
f ''(x) = –6x – 6
–2 –1
–6
(–1;–6)
(–1;–7) –9
–14
x
2
f has no stationary points
5
1
1
–
3
f(x) = (x – 1)(x + 5)
1
4
–5
3
f ''(x) = 6x – 2
Minimum value is −805 and maximum value is 255
x→−∞
y
x
1
a = −1, b = 6, c = 4 and d = −24
2
a = −1, b = 0, c = 21 and d = −20
3
a = 1, b = 3, c = 0 and d = −4
4
5
a = 1, b = −__
, c = 2 and d = −2
2
5.1
x = −1 and x = 3
Minimum at x = −1
Maximum at x = 3
Answers
9780636143319_plt_mat_g12_lb_eng_zaf.indb 360
2013/05/31 11:19:09 AM
5.2
5.3
x=1
At the point of inflection, f ’’ ( x ) = 0 and at the
turning point of f ’ ( x ), f ’’ ( x ) = 0.
7.2.4
y
f’
—
–;– 4 20
–2
27
3
f ’’ ( x ) > 0 if x < 1, f ’’ ( x ) = 0 if x = 1 and f ’’ ( x ) < 0
if x > 1, so f ( x ) changes concavity at x = 1 form
concave up to concave down.
5.4
–2
–
–2
y
f
3
–2
–1
1
x
3
8.1
y = −9x + 36
8.2
( −6; 90 )
9.1
(3; 4)
9.2
y = 9x − 23
6.2
y
2
–
3
(3;4)
0
–1
2
2,48
f
(0;0)
y = 9x – 23
10.1
(1;0)
10.2
y = 3x − 3
y
10.3
f
f
7.2.1
−592
y = 8x + _____
27
7.2.2
g
f”
(O;27)
1
a = __
2, b = −2, c = 4
(2;2)
f'
( 3 27 )
20
2 ; 4 ___
−__
is a local maximum
–0,73
0
(2; 0) is a local minimum.
7.2.3
x
2.55
(1,24;–3,92)
x
–3
7.1
x
(2;0)
g
f ’( 3 ) = 0, so f ( −3 ) = f ( 3 ) = 0 and there is a local
maximum at x = 3
a = b = 0 and c = −27
2
—
–; 2 10
3 27
y
9.3
6.1
4
f
2
–;–2–2
3 3
f'
–3
f"
1
2.73
x
(0,5;– 0,75)
10
( __23; 2 ___
27 ) is a point of inflection because g
–2
changes from concave down to concave up at
–3
2
x = __
3.
y = 3x – 3
Answers
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361
2013/05/31 11:19:09 AM
TOPIC 8: EXERCISE 14
1.1
8.3
r = 6,83 cm
h = 13,65 cm (13,66 using unrounded r)
1.2
878,76 cm2
2.1
429 − r2
h = _______
r
____
Area EBOD + Area HDE
1
1
= __ ( −x ) ( 8 + x3 + 8 ) + __ ( x + 2 ) ( x3 + 8 )
2
2
1 4 __
1 4
__
= − 8x − x + x + x3 + 4x + 8
2
2
= x3 − 4x + 8
|x < 0
__
2√ 3
8.4
2__
x = − ___
= − ____ | x < 0 from the sketch
3
√
3
r = √143 = 11,96 cm
____
h = 2√ 143 = 23,92 cm
8.5
Maximum area of HEBO = 11,08 units2
9.1
x, 2x and 4x
2.2
Maximum volume = 10 744,345 cm3
9.2
Dimensions: 21 × 14x
3.1
12500
A = 8x2 + ______
x
9.3
3.2
Dimensions: 9,21 × 36,84 × 14,74
Area rectangle – area sum of circles
= 294x − ( π​​( 4x )2 + 2 ( π​​( 2x )2 ) + 3π​x2 )
= 294x − 27π​x2
3.3
Minimum surface area = 2 035,81 cm2
9.4
49
Maximum area x ___
9π = 1,73 units
9.5
254,75 units2
10.1
s = 2r and h = √ 3 r
10.2
5 400π = 2π​r2 + 2πrH + πrs | Divide through by π
5 400 = 2r2 + 2rH + r ( 2r ) | s = 2r
2rH = 5 400 − 4r2
3
4.1
85x − 25x
V = __________
6
4.2
Dimensions: 1,06 × 5,32 × 1,77
4.3
Maximum volume = 1,06 × 5,32 × 1,77 = 9,98 m3
Maximum volume with unrounded dimensions
= 10,05 m3
5.1
h ( 2 ) = 42m
5.2
Average speed = 14,5 m/s
5.3
h ’ ( 2 ) = 9,5 m/s
5.4
Acceleration = h ” ( t ) = −10 m/s2
5.5
Maximum height when t = 2,95 seconds
Maximum height 46,51 m
5.6
The ball hit the ground after 6 seconds
5.7
At the moment of impact h ’ ( 6 ) = −30,5 m/s
6
a = −12 and b = 36
7.1
1
a = __
2 , b = 0, c = − 8 and k = 2
7.2
1 2
1 2
__
T ( p;2 ),Q ( − p;2 ),R − p;__
2 p − 8 and S p; 2 p − 8
7.3
QT and RS are horizontal lines, QR and TS are
vertical lines, angles in QRST are all right angles,
QRST is a rectangle.
7.4
Area QRTS = 20p − p3.
7.5
p=
7.6
Maximum area= 34,43 units2
2
(
5400 − 4r
2700
H = __________
= _____
r − 2r
2r
10.3
)
y = 2x − 10
8.1
B ( 0;− 8 ) and H ( −2;0 ),
8.2
DE = x3 + 8
2 3
1 2
__
2
V = __
3 π​r + π​r H + 3 π​r h
__
2700
2 3
1 2
__
2 ( _____
V = __
r − 2r ) + 3 π​r ( √ 3 r )
3 π​r + π​r
__
10.4
10.5
√ 3 π​r3
2
= __π​r3 + 2700πr − 2π​r3 + ______
3
3
__
π​r3 (
___
= 2 700πr −
4 − √3 )
______ 3
2700__
r = _______
≈ 34,5 cm
4 − √3
√
Maximum volume = 195113,6257 cm3
= 195,11l
TOPIC 8: REVISION TEST
(
)
1.1
x=4
1.2
f ( x ); 4,9; 4,99; 4,999; 4,9999; Undefined;
5,0001; 5,001; 5,01; 5,1
1.3
Yes, the limit is 5. Although f ( 4 ) is
undefined, lim f ( x ) = 5 and limf ( x ) = 5.
___
20
√___
3 ≈ 2,58 units, p < 0
7.7
362
__
x→4 −
x→4+
2.1
−6
2.2
6
__
3.1
1
Roots: − __
3 ;0 and ( 1;0 )
1
1
__
Turning point: __
3 ;−1 3
7
(
)
(
)
Answers
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2013/05/31 11:19:10 AM
3.2
y
f(x) = –3x 2 – 2x – 1
B(4;39)
P(–3;32)
1
––
3
(–1;4)
1 ;–1 1
3
3
y = 7x – 11
Q(1;0)
y = –8x + 8
x
10.2
dy
2
1__
___
= 4x + __ − ___
10.3
2
− __2
x
10.4
3 √__ ____
1
__
x − 1 __ + __
10.5
6
___
10.6
8x − 18x2 + 9
10.7
81x2 − 54x + 9
1
1
_____
+ ___
3.2.1
See graph
3.2.2
mPQ = −8
10.8
3.3.1
mAB = 3h − 8
10.9
3.3.2
m=7
3.3.3
See graph
3.3.4
mAB = 7
3.4.1
mDE = 6x + 3h − 2
10.10
dx
√x
x2
2
2√ x
x2
5x4
1
__
7
__
9
___
6x6
10x10
dy
2 − __35 __
1 − __2
2
1
___
__
= 5x
− x 3 = _____3 − _____2
3
dx
5x5
3x3
5
__
2
__
8t 3
8
8
___
__
___
3
3 + 10t + __13 − __43
3t
t
10.11 81x2
3 __12
1
___
10.12 __
2 x + 1 + __3
3.4.2
f ’ ( x )= 6x − 2
11.1
3.4.3
f ’ ( 2 ) = 10, gradient of the tangent to f at x = 2
is given, m = 10.
11.2
4.1
−h + 7
4.2
7
5
4
−__
x2
g ’ ( −2 ) = −1
6
(
2x2
(
)
325
179
−4; ____ and 8;− ____
3
3
)
g ’ ( x ) = 2x − 4
g ” ( 8 ) = 12 > 0 ⇒ g is concave up at
179
( 8;−____
3 )
g ” ( − 4 ) = −12 < 0 ⇒ g is concave down at
325
( − 4;____
3 )
11.3
1
(2;−35__
3)
f ’ ( x ) = 6x2
f ’ ( −3 ) = 54
12.1
400
( − __83;− ____
27 ) local minimum
7
g ’ ( x ) = −3
g ’ ( 4 ) = −3
12.2
16
( − __13;10___
27 )
12.3
x = −1 or x = 4 or x = − 4
8
f ’( x ) = 0
12.4
f(x) = –x3 – x 2 + 16x + 16 y
9.1
2
f ’ ( x ) = __
9.2
2
__
= 2 ⇒ x2 = 1 and x = ± 1
9.3
y = 8x − 8
9.4
833
23
___
y = − ____
27 = −30 27 and y = 45
9.5
12x − 27y + 1155 = 0
12x − y − 63 = 0
9.6
y = −13x + 12
9.7
175
y = −16x + ____
27 or 432x + 27y = 175
10.1
4
f ’ ( x ) = 7 − __
2
( 2;36 ) local maximum.
(2;36)
x2
– 1 ;10 16
27
3
x2
–4
–1
16
4
– 8 ;–14 22
3
27
x
Answers
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363
2013/05/31 11:19:10 AM
The maximum value is 336
18.1
After 2 seconds and again after 5,875 seconds.
lim f ( x ) = ∞
18.2
7,875 m
13
3
1
__
a = − __
2 , b = − 2 , c = 5, d = 12
18.3
After 3,875 seconds and again after 4 seconds.
14
a = 4, b = 1 and c = − 4
18.4
After 10 seconds.
15.1
y=0
18.5
63
h ’ ( t ) = −2t + ___
8 =0
15.2
y = −7x + 4
15.3
a = 1, b = 2, c = −7 and d = 4
16.1
Local maximum at x = −2
Local minimun at x = 4
16.2
f ” ( 1 ) = 0, so the x = 1 at the point of inflection.
16.3
f ” ( x ) < 0 when x < 1 and f ” ( x ) > 0 when x > 1
f is concave down when x < 1 and concave up
when x > 1.
12.5
12.6
x→−∞
16.4
63
63
−2t = − ___ ⇒ t = ___ = 3,9375 seconds
8
16
y
1
x
–2
18.6
Maximum height = 16,5038 m
18.7
2,875 m/s
18.8
It is an average speed.
18.9
15,625 m
18.10 Instantaneous speed.
18.11 1,875 m/s
f
4
Maximum height is reached after 3,9375
seconds.
18.12 h ” ( t ) = −2 m/s2
18.13 8 seconds
16.5
m = f ’( 0 ) = − 4
16.6
x=2
17.1
4 3
V = __
π​r + π​r2h and T.S.A. = 4π​r2 + 2πrh
17.2.1
4π​r2
18.14 h ’ ( 8 ) = − 8,125 m/s
3
+ 2πrh = 1,56π
0,78 − 2r2
h = _________
r
0,78
____
= r − 2r
0,78
4 3
2 ____
17.2.2 V = __
r − 2r
3 π​r + π​r
4
= __π​r3 + 0,78πr − 2π​r3
3
2
= 0,78πr − __π​r3
3
(
)
19.1
26,4 m
19.2
11,2 m/s
19.3
t = 2 or t = 2,8
19.4
2,4 seconds.
19.5
27,04 m
19.6
5 seconds
19.7
20,8 m/s
19.8
The ball is stationary at 2,4 seconds and the
velocity is zero.
20.1
A(4;7) and B ( 4;−18 )
17.2.3 V = 1,02 m3
9
4 3
__
2
17.3.1 ___
16 π = 3 π​r + π​r h
(
9
4r
h = ____
− __
3
16r2
( 16r
9
4r
__
17.3.2 A = 4π​r2 + 2πr ____
2 −
9π
4
= __π​r2 + ___
3
8r
3
__
17.3.3 r = 4 m
3
)
11
Minimum A = ___
π = 8,64 m2
364
(
In both 17.2 and 17.3 h = 0. A sphere is more
efficient than a capsule for minimising or
maximising the surface area.
)
20.2
x = 1,75
20.3
35,125 units
20.4
x = 1,75
20.5
24x + 16y − 289 = 0
24x + 16y + 273 = 0
20.6
k ∈ [ −20,25;16 ] or −20,25 ≤ k ≤ 16, k ∈ ℝ
4
17.4
)
1 ;13,75 and B −__
1 ;−11,25
A −__
2
2
27π = 64π​r3 + 48π​r2h
Answers
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2013/05/31 11:19:10 AM
20.7
x ∈ ( −3;1 ) ∪ ( 5; ∞ ) or
−3 < x < 1 or x > 5,x ∈ ℝ
y
15 • • (1;16)
+
+
1
20.8
21.1
5
x
x = 3,5
T ( 0 ) = 10
21.2
27 °C/s
21.3
45 °C/s
21.4
t = 8 seconds
21.5
266°C
22.
a = −15 and b = −33
23.1
A ( 2;6 )
23.2
12 units squared.
radius = √ 17 ; centre at ( 0;−15 )
3.5
radius = 5; centre = ( −2; 4 )
3.6
radius = √ 6 ; centre = ( −1; 3 )
3.7
radius =
3.8
1
__
radius = ____
4 ; centre = 1; 4
3.9
5
5
__
radius = __
2 ; centre = − 2 ; 0
3.10
9
1
radius = __
centre = 0; __
4 ;___
4
___
13
3 __
1
___
__
√___
2 ; centre = ( 2 ;− 2 )
√ 17
(
(
√ 10
____
(
(
)
)
)
3 __
1
__
2 ; centre = 2 ; 2
)
radius =
3.12
radius = √ 10 ; centre = ( 0; 1 )
___
TOPIC 9: EXERCISE 2
TOPIC 9: EXERCISE 1
1.1
Yes, with centre at the origin
1.2
Yes, with centre off the origin
1.3
Yes, with centre off the origin
1.4
No, (y2 term is negative)
1.5
No, (x2 and y2 coefficients different)
1.6
4 2
−2 2
No, −16 + ___
+ __ = −11 , which would
2
2
( )
__
3.11
1.1
(
___
3.4
( )
)
mean that r2 = −11 which is impossible.
Equation of tangent:
3(
)
y − 2 = __
2 x+3
Equation of normal:
2(
)
y − 2 = − __
3 x+3
3
13
___
∴ y = __
2x + 2
2
∴ y = − __
3x
1.2
Equation of tangent:
y − 1 = −2 ( x − 2 )
y = −2x + 5
Equation of normal:
1(
)
y − 1 = __
2 x−2
1
__
y = 2x
1.3
tangent is horizontal
∴ y = −1 (tangent) and x = 2 (normal)
1.4
Equation of tangent:
1(
)
y − 4 = − __
5 x−1
Equation of normal:
y − 4 = 5( x − 1 )
1
21
___
y = − __
5x + 5
y = 5x − 1
1.5
Equation of tangent:
1
y = − __
5x − 7
Equation of normal:
y = −5x − 7
1.6
Equation of tangent:
1(
)
y − 3 = − __
2 x+1
Equation of normal:
y − 3 = 2( x + 1 )
5
1
__
y = − __
2x + 2
y = 2x + 5
2.1
x2 + y2 = 49
2.2
x2 + y2 = 73
2.1
34
5
___
y = __
3x − 3
2.3
( x − 3 )2 + ( y + 5 )2 = 3
2.2
5
5
__
y = __
2x − 2
2.4
( x − 1 )2 + ( y − 3 )2 = 37
2.3
2.5
( x + 1 )2 + ( y − 4 )2 = 25 or ( x + 1 )2 + ( y + 4 )2 = 25
Tangent at ( 0;0 ): x = 0
Tangent at ( − 4;0 ): x = − 4
2.6
( x − 3 )2 + ( y − 1 )2 = 5 or ( x − 1 )2 + ( y − 1 )2 = 5
2.4
1
y = __
4x
2.7
65
7 2
7 2 ___
__
(x + __
2 ) + (y + 2 ) = 2
2.5
y = √ 7 or y = − √ 7
2.8
x2 + ( y − 1 )2 = 36
2.6
x = − 8 or x = 2
3.1
radius = 11 centre at the origin
2.7
3.2
radius = √3 ; centre at the origin
3.3
radius = 4; centre at ( 1; 5 )
Tangent at ( 2; 2 ):
y = −x + 4
Tangent at ( −2;−2 ):
y = −x − 4
__
__
__
Answers
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365
2013/05/31 11:19:10 AM
3.1
2 intersection points ∴ 2y − 4x = 6 is a secant
3.2
1 intersection point ∴ y + 4 = x is a tangent
3.3
Neither, the line does not intersect the circle.
4.1
sum of radii = 10
OA = 10 units (vertical distance)
∴ the circles touch
4.2
4.3
__
Sum of _______________
radii = 2√3 + 2 ≈ 5,46
___
AB = √ ( 3 + 4 )2 + ( 1 + 5 )2 = √85 > 5,46
∴ the circles do not intersect
Sum of __________________
radii = 5
___
(
√
AB =
−2 − 2 )2 + ( −1 − 0 )2 = √ 17 ≈ 4,12 < 5
∴ the circles intersect
__
4.4
3
__
−0
1
__
−1
5−0
5+1
5
5
1
= 3 and mAB = _____
= − __
mOM = _____
7
1
__
__
3
1 = −1 ∴ OM ⊥ AB
3 × −__
3
5
3
13
___
y = __
2x − 2
6.1
x2 + y2 = 4
6.2
A (–2; 0)
6.3
1__
___
6.4
y = − √3 x + 4
6.5
θ = 120°
6.6
4__
D = ___
;0
√
√3
__
( 3 )
B = ( 0;4 )
5.1
KL = 4√5
5.2
K = ( −1; 2 ) or ( −1;− 6 )
6.8
5.3
3
1
1
11
y = − __
x + __
; y = __
x − ___
__
BC = 2√ 3
6.9
m = 0 or m = 8
6.1
A = ( −2; 0 )
7.1
p = 3 and q = 13
6.2
B = ( 0;− 6 )
7.2
6.3
1
y = − __
x − 6; Equation of AC is y = 2x + 4;
Substitute x = 2 and y = 13:
LHS = 22 − 2 ( 2 ) + ( 13 )2 − 16 ( 13 ) + 39
= 0 = RHS
∴ ( 2;13 ) lies on the circle.
7.3
8−3
mBA = _____
1−0=5
2
| Pythagoras’ Theorem
2
2
2
6.7
2
1x − 6
equation of BC is y = −__
2
6.4
C = ( − 4;− 4 )
6.5
1
mAM = mCB = − __
2 ∴ AM ∥ CB
and MA = MB ( radii )
∴ ACBM is a square
TOPIC 9: REVISION TEST
2
2
1
x + y = 16
2
( x + 3 )2 + ( y − 5 )2 = 221
3
At ( −1;−2 ): y = −2x − 4
1
1
__
At ( −1;1 ): y = − __
2x+2
3.1
y = −2x + 4 or y = 2x + 3
3.2
( −3;2 )
4.1
1 __
7
A __
5 ; 5 and B ( 1;−1 )
4.2
AB =
4.3
3 __
1
M = __
5; 5
366
(
)
___
4√10
_____
(
5
)
3
13 − 8
mAD = ______
2−1 =5
mAC = mMB = 2 ∴ AC ∥ MB
1
2 × − __
2 = −1 ∴ AM ⊥ AC; AC ⊥ CB; CB ⊥ BM;
BM ⊥ AM
8__
Area △BOD = ___
units2
√
∴ B, A and D are colinear
8.1
B = ( − 6;8 )
8.2
y = 2x + 20
8.3
D = ( 6;− 8 )
8.4
1
y = − __
2x − 5
8.5
4
1
mOC = ___
= − __
−8
2
8
mAB = __
4=2
1
− __ × 2 = −1
2
∴ OC ⊥ AB
8.6
Area △ AOB = 40 units2
8.7
^ E = 126,87°
BO
8.8
Equation is x = 10
8.9
( 10;−10 )
8.10
3
25
___
y = __
4x − 2
9.1
BC = 10
Answers
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2013/05/31 11:19:10 AM
9.2
Midpoint BC = ( 1;0 )
9.3
1
1
__
y = − __
2x + 2
9.4
( x − 1 )2 + y2 = 25
9.5
(
15.4
∴ AB ∥ MN
−5;−1 )2 + ( 3 )2 = 45 > 25
∴ (−5;3) lies outside the circle
9.6
9.7
D = (3;9)
^ C = 71,57°
BA
10.1
Centre = ( 2;−1 )
10.2
k = −12
10.3
10.4
y = − 4x + 7
11
x2 + y2 = 16
Centre ( 0;0 ) and radius = 4
( x − 4 )2 + ( y + 3 )2 = 1
Centre ( 4;−3 ) and radius = 1
∴ sum of radii = 5
Distance between the centre s
_____
___
= √ 16 + 9 = √ 25 = 5
∴ the circles touch
| Equal gradients
15.5
___
MN = 2√ 13
15.6
AB = 4√ 13 | Midpoint theorem
15.7
^ C = 51,34°
AB
15.8
3
15
2
____
___
mAB = __
3 and mAC = −10 = −2
___
3
2 ___
__
×
= −1 ∴ AB ⊥ AC
3
15.9
___
r = √17
3
5 __
__
+
8
2
2
2
2
__
_____
__
mAB = ___
12 = 3 and mMN = 7 − 1 = 3
−2
Area △ABC = 130 units2
(
3
15.10 ( x − 1 )2 + y + __
2
533
) = ____
4
2
16.1
D = (12; 8)
16.2
3
3
9 __
9 __
__
Midpoint AC = __
2 ; 2 and midpoint BD = 2 ; 2
( )
∴ AC bisects BD
15
−3
___
y = ___
7 x− 7
16.4
y = −2x + 2
16.5
1
y = − __
9x + 2
Inclination = 173,66°
x2 + y2 = 25
16.6
12.2
A = (3;–4)
16.7.1 x2 + ( y − 2 )2 = 58
12.3
C = ( − 5;0 )
3
44
___
16.7.2 y = − __
7x − 7
12.4
4
y = − __
3x
16.7.3 ( − 26,1;4,9 )
12.5
3
25
___
y = __
4x + 4
12.6
29
p = − ___
3
12.7
13.1
__
BC = 2√5
13.2
b = 14
13.3
( x − 4 )2 + ( y − 13 )2 = 116
14.1
y = 2x + 11
14.2
y=x+4
14.3
A = ( −7;−3 )
14.4
AQ = √ ( 1 + 7 )2 + ( 1 + 3 )2 = 4√ 5
14.5
( x + 7 )2 + ( y + 3 )2 = 80
15.1
−3
M = 1;___
2
15.2
A = ( 2;10 )
15.3
23
y = ___
2 x − 13
(
)
17
No intersection
18.1
Centre = ( 1;2 ) or ( 5;10 )
18.2
Radius = 2a = 2 or 10
EXAM PRACTICE: PAPER 1
Substitute ( 4;a ) into 2y − 5x = 6
∴ 2( a ) − 5( 4 ) = 6
2a = 26
∴ a = 13
_______________
| Common midpoint
16.3
12.1
( )
1.1.1
x = 1,5 or x = 1
(2)
1.1.2
5
x = __
3 or x = −2
(5)
1.1.3
4 or x > 3
x < −__
3
(5)
1.2
b2 − 4ac = − 4
___
√ − 4 is non-real, roots are non-real
1.3
__
x ≠ 70 or x = 5
| x < 50
The border is 5 cm wide.
(3)
(6) [21]
2.1
Tn
= − n2 − 3n + 4
(5)
2.2
d = −5 and a = 18
(5)
2.3.1
x = 7 or x = 3
(5)
2.3.2
S∞ = 108
(3)
2.4
n = 25 only, n ≠ − 21 as n ∈ ℕ
(5) [23]
Answers
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2013/05/31 11:19:11 AM
3.1
n = 233,31
He will make 233 equal payment and a final
payment at the end of the 234 month. It will
take him 19 years and 6 months to pay off the
loan.
(5)
3.3
R30 231,23
4.1
( 0;−2,5 )
( 3;2 )
( 5;0 )
(6)
x=3
(3)
4.2
(8) [18]
3.2
Final payment = R4 981,67
4.3
3
f ( x ) = _____
x−2−1
Asymptotes: x = 2 and y = −1
Symmetry lines by substitution of ( 2;−1 ) into y = x + c and y = − x + c
Substitute into y = x + c ⇒ −1 = 2 + c, so c = −3 and y = x − 3
Substitute ( 2;−1 ) into y = − x + c ⇒ −1 = − ( 2 ) + c, so c = 1 and y = − x + 1
Symmetry lines by translation 2 units right and 1 unit down:
y = x becomes y = ( x − 2 ) − 1 ⇒ y = x − 3
y = − x becomes y = − ( x − 2 ) − 1 ⇒ y = − x + 1
x-intercept: (5;0), y-intercept ( 0;−2,5 )
1(
)2
g ( x ) = − __
2 x − 3 + 2: T.P. (3;2), y-intercept ( 0;−2,5 ) and x-intercepts: (1;0) or (5;0)
1(
)2
Calculations for x-intercepts: − __
| Multiply both sides by −2
2 x−3 +2=0
Difference of squares
Perfect square
Simplify and factorise
( x − 3 )2 − 4 = 0
( x − 3 )2 = 4
( x − 3 )2 − 4 = 0
[ ( x − 3 ) − 2 ][ ( x − 3 ) + 2 ] = 0 x − 3 = ± 2
x2 − 6x + 9 − 4 = 0
( x − 5 )( x − 1 ) = 0
x=3±2
x2 − 6x + 5 = 0
( x − 5 )( x − 1 ) = 0
x = 5 or x = 1
x = 5 or 1
x = 5 or x = 1
(5)
h( x ) = 4 × 2x − 3 − 2
Asymptote: y = −2
x-intercept: (2;0)
y-intercept: 4 × 2− 3 − 2 = −1,5 ⇒ ( 0;−1,5 )
3
f (x ) =
–1
y
x–2
h(x) = 4 × 2 x – 3 – 2
-
y= –x + 1
y= x – 3
( 3 ;2 )
-
y = –1
-
5
-
-
3
-
2
-
1
-
0
-
-
-
-
-
-
1-
x
–1
-
–1,5
y = –2
–2,5
-
–3 -
368
x=2
Answers
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2013/05/31 11:19:11 AM
Option 1: x ∈ ( − ∞;0 ) or x ≤ 0, x ∈ ℝ
Option 2: x ∈ [ 0;∞ ) or x ≥ 0, x ∈ ℝ
5.1
(2)
__
A
7.1
1
1
__
Option 1: x = 2y2 ⇒ y2 = __
2x and y = − √ 2x , x ≥ 0
5.2
___
1
1
__
Option 2: x = 2y2 ⇒ y2 = __
2 x and y = √ 2 x , x ≥ 0
(4)
5.3
y = log2 x
(3)
y f(x) = 2x ²
-
(1;2)
C
y=x
1
(–1;0,5)
-
-
( 2 ;1 )
(0;0)
-
1
2
__
| Pythagoras’ Theorem
1
= __π​​( 48 − h2 )h
3
1
= 16πh − __π​h3
3
(3)
7.2
h = 4, r = 4√ 2
(5)
7.3
Maximum volume
x
(0,5;–1)
B
r + h = ( 4√3 )2
r2 = 48 − h2
1 2
V = __
3 π​r h
2
g –1 (x) = log2 x
-
-
g(x) = 2
-
(–1;2)
r
x
-
5.4
h
4 3
__
128π
= _____
3
(–1;–2)
-
(1;–2)
≈ 134,04 units3
h(x)= –2x²
8.1
1
___
6.2
= 9x2 − 4 +
6.3
1
(3)
6.4.1
a = 1, b = 0, c = −12 and d = 16
(8)
6.4.2
Inequality notation: k < 0 or k > 32, k ∈ ℝ
Interval notation: k ∈ ( −∞; 0 ) ∪ k ∈ ( 32; ∞ )
(2)
Inequality notation: −2 < x < 2, x ∈ ℝ
Interval notation: x ∈ ( −2;2 )
(3)
6.4.3
(2) [10]
Diagram is solution for 8.1
(4)
1
__
2x2
6.4.4
y = 32
(1)
6.4.5
m = −8
(2) [28]
(6)
T20
W30
5+x
6–x
17 + x
x
9–x
7–x
24 + x
S40
8
8.2
8.3.1
8.3.2
8.3.3
x=4
(3)
3
___
= 0,0375
(2)
80
9
___
80 = 0,1125
8
___
80 = 0,1
(2)
(2)
Answers
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2013/05/31 11:19:11 AM
EXAM PRACTICE: PAPER 2
1.1
˙
2 411
x = _____
= 80,36 kg
30
1.2
19 boys
(5)
1.3
Mass in kg
Frequency
Cumulative
frequency
50 ≤ x < 60
1
1
60 ≤ x < 70
7
8
70 ≤ x < 80
8
16
80 ≤ x < 90
7
23
90 ≤ x < 100
4
27
100 ≤ x < 110
3
30
1.4
y
(4)
Masses of Grade 12 boys
30 28 26 24 -
Cumulative frequency
22 20 18 16 14 12 10 8642-
1.5
-
-
-
-
-
-
-
-
-
-
0
10
20
30
40
50
60
70
Mass in kilograms
80
90
100
110
120
The median mass is 79 kg (indicated on the
x-axis with an M)
(2)
1.6
The cut-off mass for the top 20% is 92,5kg
(indicated on the x-axis with a P)
(2)[20]
2.1
( −3;2 )
(2)
2.2
( x + 3 )2 + ( y − 2 )2 = 25
(4)
2.3
BD = 8 units
(3)
2.4
3
y = − __
4x + 6
(3)
2.5
C(8;0)
(2)
2.6
AB = BC = 10 units
(4)
370
P
-
-
M
-
-
x
2.7
Area △ABC = 50 units squared
(4)
2.8
8
1
__
y = __
7x − 7
(3)
2.9
E ( 1;−1 )
(7)
2.10
^ O = 34,7°
BA
(5)
2.11
( x − 1 )2 + ( y + 1 )2 = 50
(4)[41]
Answers
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2013/05/31 11:19:11 AM
3.1
3.2.1
Construction: Let E be a point on DC. Draw AE
and BE.
Proof: In △ADE and △BDE
1) AD = BD
| Given
2) DE is common
| Given
^ E = BD
^E
3) AD
| ED ⊥ AB
⇒ △ADE ≡ △BDE
| SAS
AE = BE
| △ADE ≡ △BDE
All points on DC are equidistant from A and B
and the centre of the circle is equidistant from
A and B, so DC passes through the centre of the
circle.
(6)
5.1
5.2
^ = 180° − x
^ +C
| From 4.2.1
A
ABCD is not a cyclic quadrilateral because the
opposite angles are not supplementary. (3)[15]
^ = 40°
E
^ = 40°
D
4
| tan FD, chord CD
| Alternate angles, CE ∥ FG
^ =C
^
A
1
3
= 40°
5.3
(6)
^ = 20°
^ =C
D
3
2
| Angles subtended by same
chord AE or angles in same
segment
3.2.2
TP = 24 units
(3)
3.2.3
VW = 40 units
(3)[18]
| ∠ on a straight line
^ = 100°
B
6.1.1
(4)
| equal chords BC and DE subtend
equal angles
| Alternate angles, CE ∥ FG
(3)
^ = 180°
D
2
JK passes through the centre of the circle
| JK is a diameter
PN passes through the centre of the circle
| NM ⊥ LM and LP = PM
T must be the centre of the circle
| NM and JK intersect at T
4.1
4.2.2
| opposite ∠ s cyclic
quadrilateral ABCD
(3)[10]
LHS
)
cos ( 360° + x ) − tan ( 180° − x ) sin ( 360° − 2x ) cos ( −x
______________________________________________
=
sin ( 90° + x )
(
(
)
cos
x
−
−
tan
x
−
sin 2x ) cos ( − x )
= _______________________________
cos x
sin x
( _____
cos x ) ( 2 sin xcos x ) ( cos x )
cos x ______________________
= _____
−
cos x
cos x
D
= 1 − 2sin2x
= cos 2x
1 2
6.1.2
= RHS
(7)
x = 30°
1
| by inspection ⇒cos 2x = cos 60° = __
2 (2)
O 4
3
2 1
60°
2
1
30°
E
K
Construction: Draw DO and extend the line
to K.
Proof: Let D1 = x and D2 = y
^=D
^ = x | Isosceles △DOE, OE = OD, radii
E
1
^ = 2x
O
| Exterior ∠ of △DOE
2
^=D
^ = y | Isosceles △DOF, OF = OD, radii
F
2
O1 = 2y
| Exterior ∠ of △DOF
^
EOF = 2x + 2y
= 2( x + y )
^F
= 2ED
(5)
4.2.1
3
F
^ =x
E
| Exterior angle cyclic quad ABED
2
^
O1 = 2x
| ∠ at centre = 2 ∠ at circumference
^ = 180° − 2x
C
| Opposite angles cyclic quad BODC
^ =x
B
| ∠sum △BCE
3
^ = ^E = x
BC = CE
|B
(7)
3
2
6.2.1
_____
cos 24° = √ 1 − p2
1
66°
p
24°
1 – p2
(3)
_____
6.2.2
tan 66° =
6.2.3
cos 33° =
7.1
2
√1 − p2
_______
(2)
p
____
1
+p
_____
√ 2
(3)[17]
(4)
7.2
__
√3
7.3.1
x = −90° or 90° or 270° or 120° or 240°
(7)
(8)
Answers
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2013/05/31 11:19:11 AM
g ( x ) = sin ( 2x − 90° ) = − cos 2x
7.3.2
(6)
y
–90
–45
(270;1)
(120;0,5)
(240;0,5)
45
–1
7.3.3
(90;1)
1-
-
-
(–90;1)
f(x) = cosx + 1
90
135
g(x) = –cos 2x
180
225
-
2
270
x
(180;–1)
x ∈ [ − 90°;90° ] ∪ [ 120°;240° ] or x = 270° (4)[29]
TOPIC 10: EXERCISE 1
13.5
^ O= 90˚
^ = 90˚, BG
H
2
GFHO is a cyclic quadrilateral.
14.1
14.2
^ = x, P
^ = x, P
^ = x, QR
^ V= x
V
4
2
1
^ V= V
^ =x
QR
| Proved in 14.1
1
1
x = 54˚, y = 108˚, z = 252˚
2
x = 55˚, y = 125, z = 250˚
3
x = 40˚, y = 80˚, z = 40˚
4
w = 78˚, x = 156˚, y = 102˚, z = 204˚
5
w = 90˚, x = 20˚, y = 33˚, z = 90˚
^ = 90˚
R
4
6
w = 64˚, x = 64˚, y = 128˚, z = 52˚
JR = RM
7
or
8
x = 67˚, y = 51˚, z = 62˚
^ C= 90˚, x = 34˚, y = 34˚, z = 112˚
w = 56˚, BA
9
x = 113,5˚, y = 66,5˚, z = 43˚
10
| Corresponding ∠s equal
QR || NW
15.1
^ = 90˚
M
3
| ∠ on diameter JK
| Corresponding ∠s, MK || RO
| OR ⊥ JM
O is midpoint JK
| O centre of circle, JK
diameter
w = 62˚, x = 62˚, OF = OG, y = 59˚, z = 82˚
RO || MK
| Given
11
x = 35 units, y = 37 units, z = 12 units
R is midpoint MJ
| Converse midpoint theorem
12
x = 90˚, y = 90˚, z = 34˚
13.1
^ = 90˚, BE
^ C= 90˚, D
^ = 90˚
E
3
1
15.2
1) PR is common
EBDF is a cyclic quadrilateral.
13.2
13.3
13.4
372
In △PJR and △PMR
^ =x
^ = x, C
D
2
2
2) JR = RM
| R midpoint JM
^ = 90˚
^ =R
3) R
2
| OR ⊥ JM
O is the midpoint of AC | AC diameter, O centre
of circle
H is the centre of EC | Given
OH || AE
| Midpoint theorem, O
and H midpoints of AC
and EC
or
H is the centre of EC | Given
^ = 90˚
H
| Line from centre circle to
1
midpoint chord
^ = 90˚
E
|
Proved in 13.1
3
OH || AE
| Corresponding ∠ equal
⇒ △PJR ≡ △PMR | SAS
^
^J = M
1
2
| △PJR ≡ △PMR
^
^J = M
1
1
^ =M
^
⇒M
1
15.3
2
^
^ =M
N
4
^
^J = M
2
4
^ = ^J
⇒N
| tan NM / chord MP
| Corresponding ∠s, NO || MK
| tan TM / chord MK
2
JOMN is a cyclic quadrilateral.
| MK subtends equal angles at N and J
^ C= 90˚
BA
Answers
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TOPIC 10: EXERCISE 2
1.1
1.2
1.3
2.1.1
P
Quadrilateral ABCD ||| Quadrilateral EFGH
| corresponding ∠s equal and corresponding
sides in proportion
5y
7x
ABCD is not similar to EFGH as the
corresponding angles are not equal.
W
BC
AB
___
> 1, but ___ = 1
FB
1.5
R
PS
5
PT
____
= ___ = __
TW
SQ
2
| ST || QW
2.1.2
PW = WR = 7y | W is the midpoint of PR
2.1.3
PA
___
=1
5
PT
___
= __
TR
BC
ABCDE and EBGHI are not similar because all
the corresponding sides are not in proportion:
3
AB
AE __
___
___
FB = 2, but FI = 2
V
Q
TV
ABCDE and FBCGH are not similar because all
the corresponding sides are not in proportion:
7y
2x
PS
PT
___
> 1 but ___ < 1
1.4
2y
S
PQRS and TQUV are not similar because all the
corresponding sides are not in proportion:
TQ
T
5x A
9
AS
| AW || SR
2,5
5
PA
___
= ___ = __
SQ
2
VW
4
= __
5
SA
QV
QS
____
= ___
4
2.1
False. The corresponding sides are not
necessarily in proportion.
2.1.4
2.2
False. The corresponding sides are not
necessarily in proportion.
2.2.1
AW = 18 cm
2.3
True. The corresponding sides are always in
proportion and all the angles are equal.
| Midpoint theorem, W midpoint
of PS, AW || SR
2.2.2
VR ____
WR
___
=
| VW || ST
2.4
2.5
2.6
| SA = AP, proved in 2.1.3
TR
VR = 28 cm
False. The angles may be different and the sides
are not necessarily in proportion.
2.3
False. The sides will be proportional, but they
may not be equiangular.
TOPIC 10: EXERCISE 4
False. A square and a rhombus always have
proportional sides, but unless the rhombus has
right angles, the square and rhombus will not
be equiangular.
1.1.1
EC = 12 cm
| FE || DC
1.1.2
AB = 25 cm
| DE || BC
20
AB ___
___
=
⇒ 25 cm
10
7
In △RPT and △RNM:
^ is common
1)
R
^ P = RM
^ N | Corresponding ∠s,
2)
RT
PT || NM
^
^
3)
TPR = N
| Corresponding ∠s,
PT || NM
⇒ △RPT |||△RNM
| AAA
1.2
3
RT
___
= __
1.3
8
8: 125
18
___
1.1
TOPIC 10: EXERCISE 3
1.2
SR
| SV || AW
1.4
RM
9
___
64
| △RPT |||△RNM
8
| From 1.2
△RTP ||| △MTQ
^ = QM
^R
1)
R
^
^Q
2)
RTP = MT
^
^R = Q
3)
QP
⇒ △RTP ||| △MTQ
1.5
3
RT
___
= __
1.6
5
NP __
___
= | PT || NM
MT
PR
| Alternate ∠s, MQ || PR
| Vertically opposite ∠s
| ∠sum of △
| AAA
5
3
Answers
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2013/05/31 11:19:12 AM
1.7
9
___
2.1
In △RTU and △QPU:
^
^ =Q
1)
R
| ∠s subtended by chord
2
1
PT
^=P
^ = 90°
2)
T
| ∠s subtended by
diameter QR
^ =U
^
3)
U
|
Vertically
opposite ∠s
4
1
⇒ △RTU ||| △QPU |AAA
2.2.1
25
In △RVU and △RPQ:
^ is common.
1)
R
1
^ = 90° | UV ⊥ QR and P
^ subtended
^U = P
2)
RV
by diameter QR
^R
^ = PQ
U
|
∠sum of△
3
⇒ △RVU ||| △RPQ
3)
5.1
In △FHG and △FED:
^ is common.
1)
F
^G = E
^
2)
FH
| ^E = 90 and GH ⊥ DE
^
^
3)
FGH = D
| ∠sum of △
⇒ △FHG ||| △FED
| AAA
5.2
FE = 120 cm
| △FHG ||| △FED
FG = 82 cm
| Pythagoras’ Theorem
EG = 120 cm – 82 cm = 38 cm
5.3
Area △FHG __
4
__________
=
5.4
Area DEGH = 900 units squared
5.5
Area △FED
9
^G = E
^ = 90°
FH
| Given
DEGH is a cyclic quadrilateral.
| Exterior ∠ = interior opposite ∠
2.2.2
VU
RV ___
___
=
2.2.3
RP2 × VU2
RHS = RP2 + _________
RV2
TOPIC 10: EXERCISE 5
RP2
= RP2 + ____2 × VU2
RV
1.1
RP
| △RVU ||| △RPQ
PQ
PQ2
= RP2 + ____2 × VU2
VU
2
| From 2.2.2
2)
2
= RP + PQ
3)
= QR2 | Pythagoras’ Theorem
2.3
3.1
3.2
3.3
^R = P
^ = 90o
UV
| Proved in 2.2.1
PQVU is a cyclic quadrilateral
| Exterior ∠ = interior opposite ∠
PQ
PR ___
___
=
| TO || RQ
=2
| PO =OQ, radii
PT
PO
| Midpoint theorem, PO = OQ and
OT || QR
= 16 units
RQ
DE
8
BC ___
21
___
EF = 7 = 3
1.2
1
__
1.3
2
__
4.1
In △EBA and △ECB:
^ is common.
1)
E
^ =B
^
2)
A
| tan EB, chord BC
3
^
^
3)
ABE = C
| ∠sum of △
2
⇒ △EBA |||△ECB
| AAA
EB
EA
___
= ___
EB
| △EBA |||△ECB
| Given
| Given
| Sides in proportion
9
7
^ = 73,4°
E
2.1
KR = 6 units
2.2
PK = PR – KR = 21
In △PQK and △TRK:
PQ ___
3
24 __
___
1)
| Given
TR = 16 = 2
2)
3)
3
PK
21
___
= ___ = __
TK
14
2
QK
3
9 __
____
__
RHS = 6 = 2
⇒ △PQK|||△TRK
1: 4
EC
AB
24
___
= ___ = 3
⇒ △ABC|||△DEF
^ = 90° | ∠ on diameter PQ
R
PQ = 34 units
| Pythagoras’ Theorem
Radius = 17 units
3.4
4.2
In △ABC and △DEF:
AC ___
27
___
1)
DF = 9 = 3 | Given
2.3
2.4
2.5
| Shown and given
| Given and from 2.1
| Sides in proportion
4
__
9
^=T
^ | △PQK|||△TRK
P
PQRT is a cyclic quadrilateral
| QR subtends equal angles at P and T
^ T = 81,8°
PK
BE2 = AE.CE
4.3
374
CE = 8 units
| From 4.2
Answers
9780636143319_plt_mat_g12_lb_eng_zaf.indb 374
2013/05/31 11:19:12 AM
3.1
A7
4.4
E
25
24
B
20
D
20
C
EC = 15 units
ED = 25 units
In △ABD and △CED:
40 __
8
BD ___
___
1)
| Shown above
ED = 25 = 5
2)
3)
32
8
AD
___
= ___ = __
5
CD
20
8
AB
24 __
___
___
CE = 15 = 5
⇒ △ABD ||| △CED
3.2
3.3
5.1
In△ABE and △CDE:
1)
2)
39
___
25
___
3.6.1
RHS = DE (DE + EA) = DE.DA = LHS
3.6.2
DA
BD
___
= ___
3.5
64
2)
3)
BC
24
1
___
= ___ = __
DE
72
AB
7
1
___
= ___ = __
FD
21
⇒ △ABC ||| △FDE
4.2
4.3
3
3
5.3.1
BC = 20
5.3.2
AD = 10√ 7
5.4
FG = 5,88 cm
9780636143319_plt_mat_g12_lb_eng_zaf.indb 375
__
__
__
longest side △AED _____
√7
10√ 7
________________
=
= ___
longest side △BEC
20
shortest side △BED
6
shortest side △AED __
4
2
_________________
= = __
2
3
The triangles are not similar because the sides
are not in proportion.
6.1
AB = 9 units
| Pythagoras’ in △ABC
FD = 123 units
| Pythagoras’ in △FED
In △ABC and △FED
AC ____
41
1
___
__
1)
FD = 123 = 3 | Given and Pythagoras’
Theorem
BC
40
1
___
____
__
2)
ED = 120 = 3 | Given
3)
9
AB ___
1
___
=
= __
FE
27
3
⇒ △ABC ||| △FED
1
__
6.3.1
GH = 41 units
6.3.2
JK = 40 units
6.3.3
HK = 9 units
6.3.4
JG = 18 units
| Given
6.4
900 units2
| Sides in proportion
7.1
In △PQR and PTQ:
In △FGD and △FDE
^ is common
1)
F
^ D = FD
^ E = 90° | Straight ∠ and given
2)
FG
^
^
3)
FDG = E
| ∠sum △
⇒ △FGD ||| △FDE
| AAA
| Given
| Sides in proportion
6.2
| Given and Pythagoras’
Theorem
| Given
^ =C
^ | △ABE ||| △CDE
A
AB || CD
| alternate angles equal
| △ABD ||| △CED
BC = 24
| Pythagoras’ Theorem in △ABC
EF = 75
| Pythagoras’ Theorem in △DEF
In △ABC and △FDE
AC ___
25 __
1
___
1)
FE = 75 = 3 | Given and Pythagoras’
Theorem
| Given
DE
4
24
5
AB
1
___
___
__
CD = 20 = 4
AE
4
1
___
___
__
CE = 16 = 4
5.2
ED.DA = DC.BD
= BC (2 BC)
| BC = DC
= 2BC2
ED.DA = DE2 + DE.DA
| From 3.6.1
2
2
⇒ 2BC = DE + DE.DA
4.1
6
BE
1
___
= ___ = __
⇒ △ABE ||| △CDE
| Sides in proportion
| △ABC ||| △FDE
| Given and straight ∠
| ∠sum of △
| AAA
625
| Shown above
3.4
ED
441
____
3)
BD2 = 402 = 1600
AB2 + AD2 = 242 + 322 = 1 600
BD2 = AB2 + AD2
^ = 90°
△ABD if a right angled triangle, with A
| Converse to Pythagoras’ Theorem
DC
4.5
| Shown above
^D = A
^
EC
| △ABD ||| △CED
ABCD is a cyclic quadrilateral
| Exterior ∠ = interior opposite ∠
64
In △ABC and △FDE
^ =F
^
1)
A
^ F = 90°
^
2)
B = DG
^
^
3)
C = FDG
⇒ △ABC ||| △FDE
| Pythagoras’ Theorem and
given
| Sides in proportion
9
1)
2)
3)
25
5
PR
___
= ___ = __
PQ
15
3
QR ___
20 __
5
___
TQ = 12 = 3
PQ ___
15 __
5
___
PT = 9 = 3
| Given
| Given
| Given
⇒ △PQR ||| △PTQ
| Sides in proportion
In △PQR ||| QTR
Answers
375
2013/05/31 11:19:12 AM
1)
2)
3)
7.2
25
5
PR
___
= ___ = __
4
QR
20
QR ___
20 __
5
___
=
=
TR
4
16
PQ
15 __
5
___
___
TQ = 12 = 4
| Given
| Given
| Given
TOPIC 10: EXERCISE 6
1.1
CD = 160 units
| Proportional intercepts, DE || FA
1.2
BD =120 units | In △ABC, B = 90° and BD ⊥ AC
PT2 + QT2 = 92 + 122 = 225
PQ2 = 152 = 225
PQ2 = PT2 + QT2
^ Q = 90°
PT
| Converse of Pythagoras’
Theorem
1.3
BC = 200 units | Pythagoras’ Theorem in △DBC
FB = 100 units
1.4
AB = 150 units | Pythagoras’ Theorem in △ABC
___
AF = 50√ 13 | Pythagoras’ Theorem in △ABF
1.5
In △CED and △CFA:
^ is common
1)
C
^ F = CF
^A
2)
CE
| Corresponding ∠s, DE || AF
^ F | ∠sum △
^
3)
CDE = CA
⇒ △CED ||| △CFA | AAA
CE
ED ___
___
| △CED ||| △CFA
FA = CF
^ R = QT
^ R = QT
^ R = 90°
PQ
7.4
25: 9: 16
7.5
PR – only a diameter can subtend a right angle
on the circumference.
8.1
In △ABC ||| △DAB:
1)
2)
3)
AC ___
22 __
2
___
=
=
BD
33
3
18 __
AB
2
___
___
DA = 27 = 3
BC ___
12 __
2
___
AB = 18 = 3
⇒ △ABC ||| △DAB
| △PQR ||| △PTQ ||| QTR
___
| Given
ED = 32√ 13
1.6
| Given
| Given
| Sides in proportion
8.2
AB2 + BC2 = 468
AC2 = 484
AC2 ≠ AB2 + BC2
△ABCis not a right-angled triangle.
8.3
4
__
8.4
D = 33°
8.5
No, it is not a cyclic quadrilateral.
^ ≠ AD
^ B |△ABC ||| △DAB
BCD
8.6
No, AB and CD are not parallel to each other.
It was proved in 8.2 that △ABC is not a rightangled triangle, and because △ABC ||| △DAB
^ B. The co-interior
^ C = DA
we know that AB
angles cannot be supplementary.
2)
3)
FE
4
16
3
HG ___
1
___
__
HE = 12 = 4
DH __
2 __
1
___
FH = 8 = 4
DB
| FG || ED
EB
1.7
540 _____
1500
GB = 120 − ____
17 = 17
1.8
In △BFG and △BED:
^ is common.
1)
B
^D
2)
B^FG = BE
| Corresponding ∠s, FG || ED
^
^ E | ∠sum of △
3)
BGF = BD
⇒ △BFG ||| △BED | AAA
FG
BF
___
___
| △BFG ||| △BED
ED = BE
___
400√13
FG = ________
17
In △DHG and △FHE:
DG ___
4
1
___
=
= __
DG ___
EF
___
=
540
DG = ____
17
9
1)
2.1
PV = VR
VS = 2 units
TV = 8 units
2.1.1
PV2 = TV.VS = 16
PV = 4 units
PR = 8 units
| OV ⊥ PR
^S = 90° and PV
| In △TPS, TP
⊥ TS
| PV = VR
2.1.2
TO = OP = OS = 5 units | Radii
OV = 3 units
PV = 4 units
| Pythagoras’ Theorem in
△OPV
PR = 8 units
| PV = VR
2.2
OPSR is a kite
| OS is the perpendicular bisector of PR
| Given
| Given
| Given
⇒ △DHG and △FHE | Sides in proportion
^ E = HE
^F
DG
| △DHG and △FHE
DG || EF
| Alternate angles equal
EFGH is a trapezium | DG || EF
376
^ E ≠ HF
^E
DG
⇒ △PQR ||| △QTR
| Sides in proportion
△PQR ||| QTR and △PQR ||| △QTR
⇒ △PQR ||| △PTQ ||| QTR
7.3
9.1
9.2
Answers
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2013/05/31 11:19:12 AM
2.3
No
^ S = 2R
^
PO
| ∠ at centre = 2 × ∠ on circumference
PS does not subtend equal angles at O and R.
3.3
In △CDG and △EAD:
^ =E
^ = 90° | Given and proved in 3.2
1)
C
2
2
^ = x | Proved in 3.1
^
2)
D4 = A
2
^ =D
3)
G
| ∠sum △
2
⇒ △CDG ||| △EAD | AAA
CD ___
DG
___
| △CDG ||| △EAD
EA = AD
3.4
CD ___
DG
___
=
TOPIC 10: REVISION TEST
1.1
PT
PV
2
____
= ___ = __
TW
VR
| TV || WR
3
1.2.1
QZ ____
QW ___
10
___
=
=
ZV
WT
1.2.2
2
__
| Common vertex Q, same height
1.2.3
4
Area △TQV = __
Area △PQV
5
| WZ || TV
EC = AE
CD2 = EC.EG
3.5
5
| Common vertex V, same height
4
__
Area △TQV ___________
8
5 Area PQV
4 2
__________
=
= __ __ = ___
Area △PQR
2.1
2.2
2.3
^ =x
E
6
^ =x
E
3
^ =x
A
2
Area △PQR
( )
5 5
In △FEA and △CED:
^ =D
^ =x
1)
A
| Proved
2
^
^
2)
E2 = E5
| Vertically opposite ∠s
^
^=C
3)
F
| ∠sum of △
1
⇒ △FEA ||| △CED | AAA
CE
ED
2.5
3.1
3.2
^ =F
^
C
1
FA || BD
^ = AB
^C
A
1
^ =x
C
1
^ =x
D
4
^ =x
A
2
D1 = x
^ B = 90°
FD
^
E2 = 90°
^ = 90°
E
1
AE = EC
| Proved in 2.2
| Alternate ∠s equal
| Corresponding ∠s, FA || BC
| ∠s on chord AD
| Alternate ∠s, FG || AC
| tan GD, chord DC
| tan FD, chord DA or alternate
∠s, FG || AC
| Radius ⊥ tangent
| Co-interior ∠s, FG || AC
| Straight ∠
| OE ⊥ AC
| Equal chords subtend equal
angles
| Proved in 3.2
| ∠ on diameter BD
^ D = 90°and
| In △BAD, BA
AE ⊥ BD
^ = 180° − x + y
A
(
)
quadrilateral ACDE
| Opposite ∠s cyclic
4.1.2
^ =x+y
F
4
quadrilateral ABFE
| Opposite ∠s cyclic
4.1.3
^ =x+y
F
2
| Vertically opposite ∠s
4.2
| △FEA ||| △CED
^ =F
^ | tan BA, chord AE
A
3
^
^
C1 = F
| Proved in 2.2
^
^
C1 = A
3
BAEC is a cyclic quadrilateral.
| Exterior ∠ = interior opposite ∠
| From 3.3
4.1.1
FE.ED = CE.EA
2.4
^ D = 90°
BA
AE2 = DE.EB
144 = 6.EB
EB = 24 units
25
| tan BD, chord EC
| Vertically opposite ∠s
| tan HE, chord FE
FE
EA
___
= ___
AD
CD.AD = AE.DG
CD = AD
TW = 3x
3
EA
4.3
^ = x | Exterior ∠ of △EFD
E
2
^B = x
CD
| Given
^
^
CDB = E2 = x
CD is a tangent to circle EFD
| Converse tan/chord theorem
In △CDF and △CED:
^ is common.
1)
C
2
2)
3)
4.4
^F = E
^ =x
CD
2
| Proved
^ = CD
^E = x + y
F
2
| Proved
⇒ △CDF ||| △CED
| AAA
CD ___
CF
___
=
| △CDF ||| △CED
CE
CD
CD2 = CE.CF
5.1
T3 = 90°
^ = 90°
R
1
| PT ⊥ MN
| ∠ on diameter MN
TSRN is a cyclic quadrilateral
| Exterior ∠ = interior opposite ∠
Answers
9780636143319_plt_mat_g12_lb_eng_zaf.indb 377
377
2013/05/31 11:19:12 AM
5.2
In △MTP and △MTQ:
1)
MT is common.
^ =T
^ = 90° | PT ⊥ MN, straight ∠
2)
T
2
1
3)
PT = QT
| OT ⊥ PQ
⇒ △MTP ≡ △MTQ | SAS
^
^ =Q
P
| △MTP≡△MTQ
1
2
5.3
In △PSM and △RSQ:
^ =R
^
1)
P
| ∠s on chord MQ
1
2
^
^
2)
S4 = S2
| Vertically opposite ∠s
^
^S=Q
3)
PM
| ∠s on chord PR
1
⇒ △PSM ||| △RSQ | AAA
5.4
PS
SM
___
= ___
RS
SQ
| △PSM ||| △RSQ
5.10
P
1
4 5
| ∠ on diameter MN
^N = 90° and
| In △MPN, MP
PT ⊥ MN
MP
MN
___
= ____
| △MPN ||| △MTP
MP
MP2 = MT.MN
5.6
5.7
^ =N
^
P
| △MPN ||| △MTP
1
1
MP is a tangent to circle PTN.
| converse tan/chord theorem
In △MTS and △MRN:
^ is common.
1)
M
2
2)
3)
^ =R
^ = 90° | Exterior ∠ cyclic
T
1
1
quadrilateral TSRN
^S = N
^
4
2
MS
MT
___
= ____
MR
MN
MR.MS = MT.MN
MP2 = MT.MN
⇒ MP2 = MR.MS
5.9
PT2 = MT.TN
PT ⊥ MN
PT = TQ
PT.TQ = MT.TN
36
21
O
72
2
1
2
R
1
Q
5.10.1 PT = 36units
________
_____
5.10.2 MR = √ 752 − 212 = √5 184 = 72 units
| Pythagoras’ Theorem in △MNR
5.10.3 TS: SQ = 7: 25
5.10.4 MS: SR = 225: 351 = 25: 39
5.10.5 No, sides are not proportional.
5.11
9
___
5.12
7
___
6.1
| Exterior ∠ cyclic
quadrilateral TSRN
| △MTS ||| △MRN
N
2
48
T
⇒ △MTS ||| △MRN | AAA
5.8
1
45
^ N = 90°
MP
△MPN ||| △MTP
MT
2
3
27 2 4
1
14 3
M 3 S1 2
PS.SQ = RS.SM
5.5
Learners should fill in values as they find them.
They must constantly check back to previous
solutions.
6.2.1
64
25
^ =x
E
2
^
E3 = x
^ =x
B
2
^ =x
C
1
^ =x
D
2
| Alternate ∠s, AB || FE
| EF bisects A^ED
| Corresponding ∠
| Corresponding ∠s, EF || CD
| Alternate ∠s, BA || CD
AF
BE
___
= ___
| BA || EF
| Isosceles △ECD
FC
ED
EC = ED
| Proved in 5.5
AF
BE
___
= ___
^N = 90° and
| In △MPN, MP
^ =D
^ =x
A
| Proved in 6.1
1
2
ABCD is a cyclic quadrilateral
| BC subtends equal angles at A and D
FC
6.2.2
EC
| Proved in 5.2
MT.TN
⇒ PT = ______
TQ
378
Answers
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2013/05/31 11:19:12 AM
6.2.3
In △EBC and △EAD:
^ = AE
^D
1)
E
| Vertically opposite ∠s
5
^
^
2)
B1 = A2
| ∠s on chord CD
3)
EC = ED
| Isoceles △ECD
⇒ △EBC ≡ △EAD | AAA
TOPIC 11: EXERCISE 1
1
Mean = 21,25 (using mid-class values on a
calculator
Median = 19
Draw an ogive to determine the median
∴ Mean > Median, so the data is skewed to the
right. The range is fairly big (55), so the data is
distributed over a wide range, but 50% of the
data falls between 15 and 27, so the interquartile
range is small (12). As we can see from the boxand-whisker plot, the data is mostly bunched on
the left, so we can instinctively see that the data
is skewed to the right.
2
Data A: Range = 30 – 10 = 20
IQR = 21 – 15 = 6
Median = 17
Data B: Range = 23 – 1 = 22
IQR = 21 – 15 = 6
Median = 18
EC
BC
EB
___
= ___ = ____
EA
ED
AD
In △DEF and △DBA
^ is common
1)
D
1
^B
^
2)
F2 = DA
| Corresponding ∠s, EF || BA
^
^
3)
E3 = B2
| Corresponding ∠s, EF || BA
⇒ △DEF ||| △DBA | AAA
DE
DF
EF
___
___
___
| △DEF ||| △DBA
DB = DA = BA
DF
EF
___
= ___
BC
BA
AB ___
___
⇒ EF = BC
FD
7.1.1
| BC = AD
^ = 90°
B
| ∠ on diameter CD
1
^ = 90°
O
| AO ⊥ CD
1
BCOE is a cyclic quadrilateral
| Opposite ∠s supplementary
7.1.2
In △AOC and △DBC
^ is common.
1)
C
^ =B
^ = 90° | Proved in 7.1.1
2)
O
1
1
^
^
3)
A=D
| ∠sum △
⇒ △AOC ||| △DBC | AAA
7.1.3
AC
OC
___
= ___ ⇒ DC.OC = BC.AC
7.1.4
DC = 2OC
2OC.OC = BC.AD
⇒ 2OC2 = BC.AD
7.2.1
In △ACO and △ADO:
1)
CO = OD
| Radii
^
^
2)
O1 = O2
| AO ⊥ CD
3)
AO is common
⇒ △ACO ≡ △ADO | SAS
^ O = DA
^O
CA
| △ACO ≡ △ADO
^D
AO bisects CA
7.2.2
DC
BC
Data A values are generally higher than those
of B. Both the maximum and minimum values
of A are higher than those of B, although the
median of B is higher than A. Both sets of
values have the same interquartile range, and
both have their middle 50% of data values
falling between 15 and 21. Data A has at least
one high value of 30, resulting in a longer tail
on the right, while data B has at least one very
low value of 1, resulting in a long tail to the left.
This means that data A is skewed to the right
and data B is skewed to the left.
| △AOC ||| △DBC
| DC diameter, OC radius
3.1.1
96,2 kg
3.1.2
Standard deviation = 11,27
DO ___
AB
If AD || BO, then ____
DC = AC
| Proportional intercepts
1
AB = __
2 AC
| DO radius, DC diameter
Answers
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379
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67,67%
–
–
–
–
–
–
–
–
–
–
3.1.3
70
3.2
75
80
85
90
95
100
1992 Springboks Median = 93 kg (mean = 96,2
kg)
2010 SA Schools Median = 100 kg
(mean = 97,66 kg)
The SA schools players are heavier than the
1992 Springboks.
SA schools have a bigger range of weights.
The data is skewed to the left.
1992 Springbok’s data is skewed to the right.
TOPIC 11: EXERCISE 2
105
110
115
120
D
1.1
r = –0,896
This shows that there is a very strong negative
correlation between x and y.
1.2
A = 52,6 and B = – 0,77
Regression line: y = – 0,769x + 52,876
1.3
When x = 10: the regression line predicts
y = 44,9
This is interpolation.
E
1.1
r = –1
This shows that there is a perfect negative
correlation between x and y.
r = –0,2378
This shows that there is a weak negative
correlation between x and y.
1.2
1.2
A = 200 and B = –2
Regression line: y = –2x + 200
A = 67,71 and B = – 0,2186
Regression line: y = – 0,2231x + 67,65
1.3
1.3
When x = 150: the regression line predicts
y = –100
This is extrapolation.
When x = 100: the regression line predicts
y = –45,85
This is extrapolation.
F
A
1.1
1.1
r = 0,985
This shows that there is a very strong positive
correlation between x and y.
r = 0,509
This shows that there is a moderately strong
positive correlation between x and y.
1.2
1.2
A = –7,69 and B = 3,47
Regression line: y = 3,47x – 7,69
A = 33,2 and B = 1,4186
Regression line: y = 1,4186x + 33,2
1.3
1.3
When x = 67: the regression line predicts
y = 224,8
This is interpolation.
When x = 28: the regression line predicts
y = 72,9208
This is interpolation.
2.1
r = 0,36
This shows that there is a fairly weak positive
correlation between x and y.
A = 37,487 and B = 0,3256
Regression line: y = 0,3256x + 37,487
1.3
When x = 6: the regression line predicts
y = 39,4406
This is extrapolation.
60 40 20 -
2.2
-
-
1.2
380
80 -
-
C
1.1
100 -
-
1.1
Test 1
B
0
50
Test 1
100
r = 0,967
Answers
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Regression line shown in 2.1. Yes a linear model
is very appropriate.
y = 0,8685x + 8,482
2.6
74,488%. This would be an accurate prediction
as the correlation coefficient for this set of
bivariate data is close to 1, and interpolation
was involved in this prediction.
40
3.2.2
A = 66,8 and B = 0,8278
3.3
y = 0,827x x + 66,8
At age 26 months, predicted height
= 88,3228 cm
At birth (x = 0), so predicted length = 66,8 cm
3.5
This is not accurate because it involves
extrapolation.
160
170
180
190
200
2010 SA Schools Players: Height versus weight
y
120 –-110 –-100 –-90 –-80 –-70 –-60 –
150
160
170
180
190
200
x
210
0
-
y = 10,983x + 88,655
-
90 80 70 60 50 40 30 20 10 -
-
5.1
-
≈ 190 cm
-
4.3.2
-
≈ 107 kg
-
60
4.3.1
-
-
40
20
Age (months)
Intuitively an exponential trend where the
rate of increase is decreasing, seems more
appropriate when considering all the points
plotted. When comparing these points to the
regression line, we see that the plotted points
are first below, then above, and then below the
residual line. This trend means that the residual
values would be positive in the middle and
negative at the low and high ends. This trend
confirms our intuitive suggestion that a linear
1
2
3
4
5
6
Hours spent watching TV
5.2
y = −10,983x + 88,655
5.3
r = – 0,95
5.4
There is a strong negative correlation
Answers
9780636143319_plt_mat_g12_lb_eng_zaf.indb 381
x
210
Height (cm)
-
-
Height/length (cm)
3.4
3.7
120 –-110 –-100 –-90 –-80 –-70 –-60 –
–
–
–
–
–
–
–
-
20
Age (months)
r = 0,997
120 100 80 60 40 20 0
1992 Springboks: Height versus weight
y
Height (cm)
3.2.1
3.6
4.2
Test result
-
The 1992 Springboks, because the points are
closer together and would lie closer to the line
of best fit.
150
-
0
4.1
–
–
–
–
–
–
–
120 100 80 60 40 20 -
-
3.1
Height/length (cm)
2.5
trend is not appropriate. This is why Rabia’s
prediction of her birth length was not accurate
when she applied the regression line.
Weight (kg)
2.4
The data values follow a linear trend very
closely and there is a very high positive
correlation between these two test values. This
means that learners who did well in the first test
tended to do well in the second test also, and
those who did poorly in the first test also did
poorly in the second test.
Weight (kg)
2.3
381
2013/05/31 11:19:13 AM
TOPIC 11: REVISION TEST
2.1
B
skewed right
1.1.1
Test A: Mean = 31,267
Test B: Mean = 42,333
2.2
C
symmetrical
2.3
D
skewed left
1.1.2
Test A: Standard deviation = 11,67
Test B: Standard deviation = 5,95
2.4
A
symmetrical
1.2
Test A:
3.1
3.2
Mean = 159,67
–-–--–--–--–--–--–--–--–-30
35
40
45
Five-number summary:
Min = 5; Q1 = 24; Q2 = 35; Q3 = 40; Max = 46
–-–--–--–--–--–--–--–--–---
Test B:
5
10
15
20
25
30
35
40
45
Five-number summary:
Min = 24; Q1 = 40; Q2 = 43; Q3 = 47; Max = 49
120
-
Test B
Sales (thousands
of Rands)
-
20
40
60
1.6
382
y = 0,2695x + 33,905
x
400 300 200 100 -
-
Test A
We saw from the value of r that there is a
moderately strong positive correlation between
the data values. Considering the scatter plot,
it appears that a regression line would fit
reasonably well, especially if we omit the first
data value. Also, the points are randomly above
and below the regression line, which suggests
again that a linear trend is appropriate.
200
The IQR is fairly small compared to the range
(which is 75 cm), showing that 50% of the data
falls within a small range of 15 cm. There is
one high and one low value which have caused
the range to be large. These could possibly be
outliers. As the mean is almost equal to the
median, we can say that the data is close to
symmetrical.
20 0
Q1 Q2 Q3
180
3.4
40 -
-
160
median = 160 cm, IQR = 168 – 153 = 15 cm
4.1
60 -
1.5
140
3.3
r = 0,53 which show that there is a moderately
strong positive correlation between the two test
results.
-
1.4
Test A has a wider range than test B (41 for
test A and 25 for test B). Test A also has a
larger interquartile range (16 versus 7 for test
B). This shows that the results for test A were
more widely spread. In fact, for test B, 75% of
the results were above 40. Both test values are
skewed to the left because each had one or two
low results that caused the left whisker to be
long.
-
1.3
0
-
25
-
20
-
15
-
10
-
5
-
y
90° –----80° –-----70° –-----60° –-----50° –-----40° –-----30° –------20° –-----10° –-----–-
20
40
50
10
30
Advertising (thousands of Rands)
4.2
r = 0,96 which shows that there is a very strong
positive correlation between the advertising
expenditure and the sales figures. As they spend
more money on advertising, so their sales
figures go up.
4.3
This set of bivariate data has a high correlation
coefficient (close to 1), and therefore has a
strong positive correlation.
Answers
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4.4
The plotted points lie randomly above and
below the regression line, and therefore the
residual values are randomly positive and
negative, so a linear trend is appropriate.
4.5
y = 5,349x + 104 604,36
4.6
y = 5,349(35 900) + 104 604,36 = R296 633,46
4.7
This is interpolation, and therefore reasonably
accurate.
5.1
6.4
r = – 0,55,which shows that there is a
moderately strong negative correlation between
these variables. As the number of days that the
drug is administered passes, there are fewer mice
that are still affected by the virus. However,
this changes after 20 days, with the number
of mice being affected increasing as more days
pass. Thus the correlation coefficient from then
becomes positive. The correlation coefficient for
the first four bivariate data values would show a
strong negative correlation.
7.1
Outlier at (5; 50)
0,03 -
7.2.1
y = – 0,1447x + 78,378
0,02 -
7.2.2
y = – 0,5155x + 95,759
7.3.1
r = – 0,2287
7.3.2
r = – 0, 868
7.4
When the outlier is included, there is a weak
negative correlation, whereas when the
outlier is excluded, there is a strong negative
correlation. In both cases they follow a linear
trend.
8
r = 0,0007. As this value is very close to 0, it
shows that there is no correlation between these
data values. So the height of the learners in this
group did not affect their mathematics results.
0,05 0,04 -
0
-
5
10
Days passed since spillage
-
-
-
0,01 -
-
Percentage oil
0,06 -
15
5.2
This shows a declining exponential trend.
5.3
Regression line: y = − 0,004x + 0,0434
Residuals will be negative for the middle
values and positive at the high and low ends.
This confirms that a linear approach is not
appropriate in this situation.
r = – 0,96, which shows that there is a very
strong negative correlation between these two
sets of data. As the number of days since the
spillage increases, so the percentage of oil still in
the river decreases.
6.1
2 __
1
__
=
200 -
1.2
0
100 -
1.3
1
__
-
1.4
3
__
2.1
4
1
__
P(queen) = ___
52 = 4
2.2
26 __
1
P(red card) = ___
52 = 2
2.3
3
12 ___
P(picture card) = ___
52 = 13
2.4
2
1
___
P(black king) = ___
52 = 26
2.5
P(red or 7) = P(red) + P(7) − P(red and 7)
-
10
-
0
-
-
Variable B
TOPIC 12: EXERCISE 1
1.1
300 -
-
5.4
20
30
Variable A
40
6.2
Parabolic trend. Perhaps the virus becomes
immune to the drug after 20 days
6.3
y = 4,7086x + 160,73
Residuals will be negative for the middle values
and positive on the high and low ends. While
this cannot tell us that the trend is parabolic
and not exponential, it does tell us that a linear
approach is not appropriate. We can see from
the plotted points that there is an increase after
the decrease, which is why we would decide
that a parabolic trend is appropriate.
2.6
3.1
4
2
4
4
1
4
2
7
= __ + ___ − ___ = ___
2
52
52
13
48 ___
12
P(not Jack) = ___
52 = 13
P(A ∩ C) = 0 ∴ A and C are mutually exclusive,
and P(A) + P(C) = 1 ∴ A and C are exhaustive.
Answers
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383
2013/05/31 11:19:14 AM
3.2
3.3
P(B) + P(C) = 1,05 > 1 ∴ B and C are not
mutually exclusive, and therefore they are not
complimentary.
6.6
Independent ∴ P(A ∩ B) = P(A) × P(B) = 0,75 ×
0,8 = 0,6
7
4
Male
Female
Married
75
81
156
Unmarried
15
9
24
90
90
180
6.7
n(C ’ ∪ B ’) = 29
6.8
n(A ’ | B) = 9
5
Green
shirt
Black
trousers
Blue
shirt
White
shirt
Green
shirt
Blue
trousers
Blue
shirt
Striped tie
Black tie
Blue tie
Striped tie
Black tie
Blue tie
Striped tie
Black tie
Blue tie
Striped tie
Black tie
Blue tie
Striped tie
Black tie
Blue tie
Striped tie
Black tie
Blue tie
5.1
1
P(Black Trousers, White Shirt, Striped Tie) = ___
5.2
6
1
P(Striped Tie) = ___
= __
5.3
4
2
P(no Blue) = ___
= __
5.4
P(Trousers and Tie same colour) = P(both Black
6
1
__
or both Blue) = ___
18 = 3
18
18
18
6
13
17
5
9
3
1
Rain
0,2
6.2
31
P(B ∪ C)= ___
60
6.3
43
P(C) ’= ___
60
6.4
n(A ∪ B ∪ C) ’ = 13
6.5
5
P(B | A) = ___
21
384
0,6
0,8
Not
rain
Lose
Win
0,7
0,3
Lose
1.1
2
P(Rain and Win) = 0,2 × 0,4 = ___
25
1.2
P(Not Rain) = 0,8
1.3
16
P(Win) = 0,2 × 0,4 + 0,8 × 0,7 = ___
25
1.4
P(Lose | Not Rain) = 0,3
2.1
P(Knife, Fork and Spoon)
(
(
) (
) (
)
10 ___
8
12 ___
) ( ___
30 × 29 × 28 )
8
10 ___
12
___
) ( ___
30 × 29 × 28 )
8
8
10
12 10
12
= ___ × ___ × ___ + ___ × ___ × ___ +
30 29 28
30 29 28
10
8
8
12
12 10
+ ___ × ___ × ___ + ___ × ___ × ___ +
30 29 28
30 29 28
12 × 10 × 8
48
= 6 ____________ = ____
30 × 29 × 28
203
2.2
B
5
1
___
P(A ∩ B) = ___
60 = 12
Win
0,4
2.3
2.4
(
P(three the same)
(
6.1
2x + 20
x
TOPIC 12: EXERCISE 2
A
16
650
x = 62
∴ there are 372 learners who play neither soccer
nor rugby.
9
60
C
Rugby
6x
12
White
shirt
Soccer
x + 10
75
5
P(married male) = ____
= ___
180
17
P(C | A ’) = ___
39
) (
) (
10
10
9
8
8
12
11
7
= ___ × ___ × ___ + ___ × ___ × ___ + ___ × ___
30
29
28
30
29
28
30
29
6
___
×
28
99
_____
=
1 015
10
12 ___
11 ___
11
____
P(three knives) = ___
30 × 29 × 28 = 203
8
7
2
10 ___
___
___
P(F, S, S) =___
×
29 × 28 = 87
30
)
(
)
2.5
P(K, K,no fork)
3.1.1
10
8
99
12
11
12
11
= ___ × ___ × ___ + ___ × ___ × ___ = _____
30
29
28
30
29
28
1 015
10
1
___
P(run and gym) = ____
100 = 10
90
9
___
P(Cycling) ’ = ____
100 = 10
(
3.1.2
) (
)
Answers
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3.1.3
10 __
2
P(gym | run) = ___
25 = 5
3
11 × 10 = 110
3.2
25
40
1
____
___
P(run) × P(gym) = ____
100 × 100 = 10
= P(run and gym)
Therefore running and going to the gym are
independent events.
4.1
33 + 32 = 36
4.2
3! + (3 × 2) = 12
5.1
6! = 720
5.2
66= 46656
6.1
6!
_______
= 120
6.2
63 = 216
4.1.1
58
29
___
P(female) = ____
122 = 61
4.1.2
36
72
___
P(no) = ____
122 = 61
(6 − 3)!
4.1.3
49
P(yes | male) = ____
122
4.1.4
1
P(yes | female) = ___
58
7
13!
______________
= 129 729 600
4.2
64
50
800
____
_____
P(male) × P(yes) = ____
122 × 122 = 3721 and
8
P6
______________
= 25 740
49
P(male and yes) = ____
9
3 × 4 × 2 × 2 = 48
These are not equal values, so the events are not
independent.
10
8! = 40 320
11
25 = 32
There is one female who enjoys fishing, so the
events are not mutually exclusive and therefore
not complimentary. Complimentary events
must be mutually exclusive and exhaustive.
12.1
105 = 100 000
12.2
10P5 = 30 240
12.3
9 × 104= 90 000
12.4
9 × 9P4 = 27 216
12.5
1 × 104 = 10 000
12.6
1 × 8P3 × 1 = 336
12.7
1 × 1 × 93 = 729
12.8
1 × 1 × 8P3 = 336
12.9
10 × 9P3 × 1 = 5040
13
122
4.3
5.1
3 __
1
P(out first round) = __
9=3
5.2
P(winner)
(
) (
) (
) (
) (
)
3
3
4
2
4 __
2 __
4 __
2 __
1
__
__
= 2 __ × __ × __
7 + 9×8×7 + 9×8×7
9
8
3
2
4
2 __
4 __
1
2 __
1 __
4
__
__
+ __ × __ × __
7 + 9×8×7 + 9×8×7
9
8
5
= ___
21
5.3
(
(
5.4
)
P(winner | past first round)
) (
) (
) (
) (
)
3
3
2
2 __
2 __
1
__
__
= 2 __ × __
7 + 8×7 + 8×7
8
3
4
4 __
1
1 __
4
__
__
+ __ × __
7 + 8×7 + 8×7
8
5
= __
7
(
)
) (9
) (9
) (9
)
3
5
4
4
2
2
4
2
1
= __ × __ + __ × __ + __ × __ + __ × __ = ___
8
8
8
2! × 3! × 2! × 2!
12.10 10 × 103 × 1 = 10 000
P(getting to third round)
(9
2! × 3! × 2! × 2!
8
12
13
58 = 390 625
14
5P3 + 5P4 = 180
15.1
64 + 54 = 1 921
15.2
64 + 5P4 = 1 416
15.3
6P4 + 54 = 985
6.1
n(A ∩ B) = 0,1 × 200 = 20
16.1
10! = 3 628 800
6.2
P(A) = 0,2 + 0,1 = 0,3
16.2
4! × 2! × 4! × 3! = 6 912
6.3
0,1
1
P(B | A) = ___ = __
16.3
7! × 4! = 120 960
6.4
P(A ∪ B ’) = 0,6
17.1
4! = 24
17.2
1 × 3! = 6
17.3
3! × 1 = 6
17.4
1 × 1 × 2! = 2
17.5
2 × 3! = 12
17.6
3 × 3! = 18
0,3
3
TOPIC 12: EXERCISE 3
1.1
99 = 387 420 489
1.2
9! = 362 880
2.1
3×2=6
2.2
3 × 2 × 1 × 2 = 12
Answers
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2013/05/31 11:19:14 AM
TOPIC 12: EXERCISE 4
5.2.1
840
7
___
Probability = _____
7 920 = 66
1.1
263 ___
1
___
=
5.2.2
120 ___
5
Probability = ____
864 = 36
1.2
13 800
1
___
Probability = _______
358 800 = 26
5.2.3
24
1
__
Probability = ____
168 = 7
2.1
120 __
1
Probability = ____
720 = 6
5.2.4
30
1
___
Probability = ____
840 = 28
2.2
2 × 5! __
1
Probability = ______
720 = 3
6.1
2 × 3 × 4 = 24
2.3
Ace and two together = 5! × 2 = 240
240 __
1
∴ probability = ____
720 = 3
6.2
8
1
__
Probability = ___
24 = 3
7.1
7 × 5 × 3 = 90
3.1
Total number of cars = 4 × 2 × 2 = 16
7.2
3.2
8
1
__
Probability = ___
16 = 2
30 __
1
Probability = ___
90 = 3
8.1
63 = 216
4.1
12 ___
2
Probability = ___
66 = 11
60 ___
10
Probability = ___
66 = 11
1
Probability = _____
5 040
1 800 ___
5
Probability _____
5 040 = 14
120
1
___
Probability = _____
5 040 = 42
129 729 600
1
___
Probability = _____________
1 816 214 400 = 14
9 979 200
1
____
Probability = _____________
1 816 214 400 = 182
259 459 200
1
__
Probability = _____________
1 816 214 400 = 7
13! ___
1
Probability = ___
=
14
14!
12! ____
1
Probability = ___
=
182
14!
2 × 13! __
1
Probability = _______
=7
14!
8.2
36
1
__
Probability = ____
216 = 6
9.1
5P3 × 5P1 = 300
9.2
24
2
___
Probability = ____
300 = 25
10.1
9! = 362 880
10.2
80 640
2
__
Probability = _______
362 880 = 9
10.3
5 760
1
___
Probability = ______
80 640 = 14
11.1
1
P(walks) = __
3
11.2
P(on time | bus) = 0,95
11.3
3
1
__
P(bus | late) = ___________
= ________
1
1 =3
___
___
P(bus)
11.4
1
P(walking ∩ on time) = __
3 × 0,8 = 15
4.2
5.1
5.2
5.3
6.1
6.2
6.3
7.1
7.2
7.3
264
26
1
__
1.4
6
5
__
6
2 __
1
__
6=3
1 __
1 __
1
__
6+6=3
2.1
255 = 9 765 625
2.2
25P5 = 6 375 600
1.2
1.3
15 + 30
4
___
12.1
x = 0,55
12.2
P(A ’ ∩ B)= 1 – (0,05+0,55+0,1) = 0,3
12.3
P(A | B) = 0,55
12.4
P(A ∩ B) = 0,55; P(A) × P(B) = 0,6 × 0,85
= 0,51 ≠ 0,55. A and B are not independent.
13.1
9
P(A ∩ B) = P(A) × P(B) = 0,75 × 0,15 = ___
80
13.2
P(A ∩ B) = 0
13.3
3
80
P(A | B) = _______
= ____ = __
4
0,15
P(B)
13.4
P(A) + P(B) = 0,9 < 1 so A and B are
not exhaustive and ∴ A and B are not
complimentary.
3.1
10P3
____
= 120
3.2
10P3 = 720
4
210 = 1 024
5.1.1
11P4 = 7 920
5.1.2
7P4 + 4! = 864
14.1
10! = 3 628 800
5.1.3
7P1 × 4P3 = 168
14.2
3! × 3! × 5! × 2! = 8 640
5.1.4
4P1 × 7P3 = 840
14.3
8 640
1
_________
= ____
386
3!
2
__
× 0,05
1
4
___
P(walking) × P(on time) = __
3 × 0,9 = 0,3 ≠ 15
∴ not independent events.
TOPIC 12: REVISION TEST
1.1
P(bus ∩ late)
P(A ∩ B)
3 628 800
9
___
420
Answers
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14.4
2! × 1 × 2! × 5! × 2! × 1 = 960
960
1
__
∴ probability = _____
8 640 = 9
15.1
n(English or Afrikaans or Zulu) ’= 5
15.2
29
P(English and Zulu) = ___
85
29
P(English | Zulu) = ___
45
15 ___
5
P(Zulu | Afrikaans) = ___
39 = 13
3 __
3 ___
123
5 __
4
____
P(Blue) = __
8 × 8 + 8 × 10 = 320
15.3
15.4
16
( )
4.2.3
1 x
y = __
2 −1
5.1
See graph
(2)
f
y=3
3
3
2
3
2
–1
EXAM PRACTICE: PRELIMINARY PAPER 1
1.1.1
x = 5,70 or x = − 0,70
(4)
1.1.2
x = 3 or x = −2
(4)
1.1.3
0<x<3
(5)
1.2
x = 3 or x = 4
y = −1 or y = 1
(7)
4
__
(4)
1.3
y
x = –2
x
g
(4 + 2)
x ∈ ℝ, x ≠ − 2
f − 1(x): x ∈ ℝ, x ≠ 3
(2)
5.3
y=x+5
(2)
5.4
3
1
y = − __
x + __
6.1
f (x) = lim _____________
h
5.2
2
(2)
2
1.4
3
9
___
10
2.1.1
30%; 32%; 34%
(1)
2.1.2
88%
(2)
2.1.3
1 770
(2)
−4(x2 + 2xh + h2) + 5 + 4x2 − 5
= lim ____________________________
h
2.1.4
59%
(1)
−4x2 − 8xh − 4h2 + 5 + 4x2 − 5
= lim ____________________________
h
∑18(__13)n 1
(5)
h(−8x − 4h)
= lim___________
h
2.2.2
27
(2)
= −8x
2.3.1
60 ; 84
(2)
(5)
h→0
−
Tn = 2n2 − 3n + 6
2.3.3
2n2 − 3n + 6 = 176
2n2 − 3n − 170 = 0
n = 10
(3)
3.1
R230,15
(5)
3.2
13,58%
(4)
3.3
115,2 years
(4)
4.1.1
y = − (x − 2)2 + 5
(4)
4.1.2
(5;2)
(2)
4.1.3
(3;5)
(2)
4.2.1
1 x
y = __
(2)
(2)
4.2.2
y = log__1 x
(2)
2
(5)
2
h
h→0
h→0
h→0
h→0
n=1
2.3.2
2
−4(x + h) + 5 − (−4x + 5)
= lim ________________________
10
2.2.1
f (x + h) − f (x)
6.2.1
(4)
18x + 12
3
__
(3)
3
__
6.2.2
20x2 − x − 2
(3)
6.3
3
__
(4)
2
x
7.1
9
(10)
g
–3
3
y
f
–27
(–1;–32)
7.2.1
–1 < x < 3
(2)
7.2.2
x>3
(2)
7.3
(1;−16)
(3)
7.4
y = 12x − 28
(4)
Answers
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8.1
A
h
B
E
x
C
D
x + h = 12
h = 12 − x
1
1
__
Area △ABE = __
2 xh = 2 x(12 − x)
Perimeter BCDE = 2x + 2BC = 24
24 − 2x
BC = _______
= 12 − x
2
Area BCDE = x(12 − x)
1
Area ABCDE = __
2 x(12 − x) + x(12 − x)
1
__
= (12 − x)( x + x)
2
3
= __x(12 − x)
2
(4)
8.2
x=6
(3)
9.1
144
(3)
9.2.1
(4)
Dogs
Cats
5
7
6
0
1
4
1
Fish
6
9.2.2
Six
(1)
9.2.3
8
__
15
(2)
9.3.1
0,675
(4)
9.3.2
0,325
(3)
388
Answers
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EXAM PRACTICE: PRELIMINARY PAPER 2
Cumulative
frequency (cf)
Height in cm
Number of boys
150 ≤ x < 155
7
7
155 ≤ x < 160
10
17
160 ≤ x < 165
15
32
165 ≤ x < 170
12
44
170 ≤ x < 175
9
53
175 ≤ x < 180
5
58
180 ≤ x < 185
2
60
2.1
y = 19,86 1 − 0,099 x
(4)
2.2
−3,899°C
(2)
3.1
2√ 13
(2)
3.2
D (−5;−2)
(2)
3.3
3y − 2x + 4 = 0
(5)
3.4
E does not lie on line.
(4)
3.5
−15
k = ____
2 = −7,5
(3)
___
(2)
y
x
–
–
–
–
–
–
–
–
–
–
–
60 –55 ––
50 45 –40 –35 –30 ––
25 20 ––
15 10 ––
5 0–
140
145
150
155
160
165
170
175
180
185
190
(4)
1.3.1
164
(1)
1.3.2
12
(2)
1.3.3
18,3%
(2)
1.4.1
Box-and-whisker diagram at the top of
the page below ogive
1.4.2
Data is skewed to the right or is positively
skewed
(2)
3.6
t = −1 or t = 5
(4)
3.7
64,65°
(5)
4.1
125
x2 + (y − 5)2 = ____
4
(5)
4.2
45
1
___
y = − __
2x + 4
(4)
4.3
Centre = (1;−3) and r = 2
(4)
5.1.1
a
(2)
(2)
Answers
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5.1.2
1 − 2a2
(2)
5.1.3
√ 1 − a2
(2)
5.2
1
(7)
6.1
Identity proof
(5)
6.2.1
Identity proof
(3)
6.2.2
x = 0° + n.180° or x = 30° + n.180° n ∈ Z
(5)
_____
11.1
^ R = 90° | Angle in semicircle
PV
STR = 90°
| Angle on straight line
∴ exterior angle of a quadrilateral = the interior
opposite angle
(or opposite angles are supplementary)
∴ TAVR is cyclic.
(4)
7.1
y
y = cos(x – 30)
1
–180 –150 –120 –90
–60
–30
30
–1
60
90
7.2
− 60° ≤ x ≤ 120°
(2)
8.1
164,91 m
(2)
8.2
69,73
(3)
8.3
D = 77,72° or 102,23°
(3)
9.1
^ O = 20°
XZ
(7)
9.2
Proof
(6)
9.3.1
^ =x
C
^B = x
EF
| Corresponding angles FB‖ EC
| Alternate angles FE‖ AC
^B = x
FD
^D = x
FB
^B = x
CD
| Tan chord Theorem
| Isosceles △FB=FD
| Alternate angles CE‖ FB
9.3.3
^ C = 180° − x
FE
| Opposite angles of cyclic quadrilateral
^
DBC = 180° − 2x | Angle sum of △
11.2
11.3
x
180
In △PSR and △PTS:
1) P^SR = 90°
| Angle in semicircle
and T1 = 90°
| Angle on straight line
^ is common.
2) P
2
^
3) P^ST = R
| Angle sum of △
1
∴ △PSR ||| △PTS
| AAA
PS
PR
___
= ___
PT
11.4
∴ △PTA ||| △ PVR (AAA)
11.5
10.2.1 90 mm
(3)
2
10.2.2 __
3
(2)
PT
PA
___
___
PV = PR
(3)
| △PTA ||| △ PVR
∴ PT.PR = PA.PV
but PT.PR = PS2 | Proved in 11.2
PS2 = PA. PV
(4)
10.1.2 If all the corresponding sides of two triangles
are proportional, the triangles are similar.
(1)
(3)
In△PTA ||| △ PVR:
^ R = 90° | Proved above
T2 = PV
P1 is common.
^T = R
^
| Angle sum of △
PA
2
(6)
(3)
| △ PSR ||| △ PTS
PS
∴ PS2 = PT.PR
(2)
10.1.1 A line drawn parallel to one side of a triangle
divides the other 2 sides in proportion.
(1)
390
150
y = sin 3x
(7)
9.3.2
120
EXAM PRACTICE A: PAPER 1
1.1.1
x = 0 or x = 5
(2)
1.1.2
x ≤ −4 or x ≥ 3,5
(5)
1.1.3
x=2
(3)
1.1.4
x = 3 only
(5)
Answers
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1.2
b2 − 4ac = − 8
1.3
The roots are non-real.
2
3.1
6.2.2
g ( x ) = log__1 ( −x ), x < 0
(2)
(3)
6.3
h ( x )= log__1 ( x + 2 )
(2)
–2
(5)
6.4
1
m = − __
6
(4)
d=4
a=3
7.1
2
__
(5)
3.2
94,5
(4)
7.2.1
20x − 13
3.3.1
−2 < x < 8
(4)
7.2.2
3x
9
81
27 ___
___
___
g ’ ( x ) = ___
2 + __1 − __3 − __5
(4)
3.3.2
S∞ = 2 500
(3)
7.3.1
f (2) = 0
(2)
3.4
2 641
(6)
7.3.2
4.1.1
7%
(1)
A(0;30)
B ( −3;0 ), C ( 2;0 ) and D ( 5;0 )
4.1.2
3
3
(5)
x2
(4)
1
__
2
2x2
2x2
x2
(
(5)
)
n = 128,135...
129 months to pay off the loan, final payment
being less than R7 500.
(4)
7.3.3
11
22
___
E ( −1;36 ) and F ___
3 ;−14 27
7.3.4
y = −16x + 32
(3)
8.1
10 − 2x
(2)
4.1.3
R1 020,11
(4)
8.2
V = π​x2 ( 10 − 2x ) = 10π​x2 − 2π​x3
(3)
4.2
R183 189,40
(6)
8.3
116,36 cm3
(4)
5.1
a = −2, b = 12 and c = −9
k = −3, p = −2 and q = 2
9.1
(5)
Contingency table
Boys
Girls
Total
Mathematics
520
300
820
5
4
__
v = __
3 , w = 2 and z = − 3
(12)
5.2.1
x ∈ ( −∞;−2 ) ∪ ( −1;1 )
(3)
Mathematical
Literacy
120
60
180
5.2.2
x ∈ [ 1;3 ]
(2)
Total
640
360
1 000
5.3
x=1
(1)
5.4
9
(4)
9.2
520
P ( Boy taking Maths ) = _____
1 000 = 52% or 0,52
(4)
6.1
1 x
f −1 ( x ) = __
3
( )
(2)
10.1
263 × 104 = 138 240 000
(3)
6.2.1
g ( x ) = − log__1 x = log3 x
(1)
10.2
26 × 25 × 24 × 10 × 9 × 8 × 7 = 78 624 000 (3)
10.3
0,0196
(3)
3
(3)
EXAM PRACTICE A: PAPER 2
1.1
Minimum
Lower quartile
Median
Upper quartile
Maximum
11
41
52
61
79
(4)
1.2
(3)
52
41
61
79
–
–
–
–
–
–
–
–
–
11
0
1.3
10
20
30
40
50
60
70
80
Slightly skewed to the left, slightly negatively skewed.
(1)
Answers
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2.1
Time (seconds)
y
Response time to a stimulus
2-
1-
-
-
-
-
-
-
0
20
40
Age (years)
60
80
(3)
2.2
y = 0,3837 + 0,0262x
(4)
2.3
r = 0,9705916242 ≈ 0,9706
(1)
2.4
very strong, positive correlation
(2)
2.5
y = 2,0867 seconds.
(2)
3.1.1
x=4
(1)
3.1.2
y=5
(2)
3.1.3
y = −2x + 8
(3)
3.1.4
1
y = − __
2x + 2
(3)
3.2.1
AM = 4 units
(1)
3.2.2
MC = MO – CO = 3 units
(2)
3.2.3
AD = 5 units
(2)
3.3
( x − 4 )2 + ( y − 5 )2 = 25
(3)
3.4
θ = 36,9°
(5)
3.5
4
y = − __
3x + 2
(3)
3.6
3x − 4y − 17 = 0
3x − 4y + 33 = 0
(8)
3.7.1
16 units2
(4)
3.7.2
12 units2
(3)
4.1
−2
392
4.2.1
x
Using identities:
cos 343°
= cos
17°
__________
= √_____
1 − sin2 17°
= √ 1 − p2
Using figure:
1
73°
1 – p2
17°
p
_____
cos 343° = cos 17° = √ 1 − p2
(
4.2.2
1__
___
p+
4.3
1
= __
2
4.4
√2
_____
√1 − p2 )
(2)
(4)
(3)
2 cos2 θ − 1 − cos θ
LHS = _________________
2 sin θ cos θ + sin θ
( 2 cos θ + 1 ) ( cos θ − 1 )
= ____________________
sin θ ( 2 cos θ + 1 )
cos θ − 1
= _________
sin θ
= RHS
5.1.1
5.1.2
(5)
n∈ℤ
x = −30° + 360°n
x = 10° + 120°n
(6)
x = −30° or 10° or 130°
(2)
Answers
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5.2
(6)
y
g(x) = sin(x + 60°)
(30°;1)
(0°;–1)
(180°;1)
f(x) = cos(2x)
3
2
–90°
(6)
–60° –45°
45°
90°
120° 135°
(–90°; 12 (
(–90°:–1)
5.3
5.4.1
5.4.2
(90°:–1)
360°
(1)
x ∈ [ −30°;10° ] ∪ [ 130°;180° ]
or
−30° ≤ x ≤ 10° or 130° ≤ x ≤ 180°
6.1
6.2
(2)
x ∈ ( − 60°;− 45° ) ∪ ( 45°;120° ) ∪ ( 135°;180° ]
or − 60° < x < − 45° or 45° < x < 120°
or 135° < x ≤ 180°
7.1
x
180°
(180°;–
3
2
(
^ F = 90° − y − x
DG
(
)
cos ( y − x ) GF
DF = _____________
cos x
h cos x
______
GF = sin x
h cos ( y − x )
DF = ___________
sin x
(3)
➀
(6)
(2)
P
T
90° –x
90° – x
Q
x
R
△PQR ||| △PRT ||| △RQT
| △PQR is a right-angled triangle, with RT ⊥ PQ
PT
| △PQR ||| △PRT
PQ
PR ___
___
=
PR
PR2 = PT.PQ
QR ___
PQ
___
=
| △PQR ||| △RQT
QT
RQ
QR2
= QT.PQ
RT
PT
___
= ___ | △PRT ||| △RQT
QT
RT
RT2 = QT.PT
PR2 + QR2 = PT.PQ + QT.PQ
= PQ ( PT + TQ )
= PQ.PQ
= PQ2
This proves Pythagoras’s Theorem:
The square on the hypotenuse is equal to the sum of the squares on the other two sides.
(8)
Answers
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7.2.1
7.2.2
7.2.3
7.2.5
| Two pairs of adjacent sides
°
| Diagonals of kite cross at 90
| OD ⊥ BD
(4)
^ O = 90°
AB
| Radius OB ⊥ tangent AB
^
ODA = 90
| Radius OD ⊥ tangent AD
ABOD is a cyclic quadrilateral.
| Opposite angles supplementary
(4)
^ =B
^
O
1
1
^ =F
^
B
1
^
^
⇒ O1 = F
7.2.4
| Tangents from common
AB = AD
point A
ADOB is a kite
equal
^ = 90°
G
1
BG = GD
| ∠s on chord AD, ABOD is a cyclic
quadrilateral
| tan AB/chord BD
GB
AG
____
= ___
(6)
2.3.1
−2 < x < 2
(4)
2.3.2
48
S∞ = _______
2
(3)
2.4
n = 12
(5)
3.1
0,072 n×12
5 000 1 + _____
12
(2)
3.2
R8 879,25
(2)
3.3
n = 9,66 years
∴ n = 10
(2)
3.4
R7 111,95
(3)
3.5
n = 10,28
∴ 11 years
(3)
BG
(3)
7.2.6
OD = 15 units
(4)
8.1
4: 2: 3
(4)
8.3
△ADE ||| △ABF
DE: BF = 2: 3
(
)
4.1
y
h
B(–2;4)
A(0;1)
x
| AAA
8
___
(4)
(4)
27
For graph shape
For y-intercept and points on the graph
(2)
4.2
q(x) = y = 2x
(1)
4.3
y = log__1 x
(2)
4.5
y ∈ ℝ, y > 0
(1)
EXAM PRACTICE B: PAPER 1
__
1.1.1
x = 1 ± √5
(4)
1.1.2
x=2
(4)
1.1.3
9
x ≤ − 4 or x ≥ __
2
(5)
1.1.4
27
___
4
(2)
1.1.5
x = −7
(3)
1.2.1
x
__
=4
(3)
1.2.2
16
4
y = __
and x = ___
2.1.1
Pattern 1: 8; 14; 20; 26 + …
Pattern 2: 8; 18; 32; 50 + …
2.1.2
2x − x
| △AGB ||| △BGO
BG
2 = AG.GO
8.4
191,81
(3)
AE || BF
^D = B
^ = 2x
| Corresponding angles equal, BA
4
(6)
GO
2.2
y
5
5
2
4.6
h
y
B(–2;4)
A(0;1)
(3)
x
(1;0)
(4;–2)
(2)
Pattern 1 is linear because the 1st difference is
constant and is equal to 6.
Pattern 2 is quadratic because the 2nd difference
is constant and is equal to 4.
(2)
h –1
(8;–3)
0<x≤8
(4)
2.1.3
Tn = 6n + 2
(2)
5.1
C(–2;1)
(2)
2.1.4
Tn = 2n2 + 4n + 2
(5)
5.2
A(0;3) B(−3;−3)
(5)
394
Answers
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5.3
−3
∴ −2 < x ≤ ___
2 and x ≤ − 6
(3)
9.1
60
5.4
x ∈ ℝ, x ≠ 1
(2)
9.2
12 ___
4
P(hockey) = ___
75 = 25 = 0,16
6.1
f ’(x) = lim ____________
h
f(x + h) − f(x)
h→0
= −2x + 3
6.2.1
(4)
dy
1
1
___
= ___ + ___
dx
2
__
3x3
3x2
6.2.2
dy
___
= 10x + 5
6.3.1
5 = a(2)3 + 1
Rugby
7
12
(4)
dx
2
(2)
6.3.2
14
(3)
6.3.3
y = 6x − 7
(3)
6.3.4
f(−x) = C
f −1(x) = A
−f(x) = B
(3)
f(2) = − (2)3 + 6(2)2 − 9(2) + k = 0
k = 8 − 24 + 18
=2
(2)
7.2
A(0;2)
(2)
7.3
a = −2;b = 3 and c = 2
(4)
7.4
x = 2 E (3,73;0) D(0,27;0)
1: root is real and rational and 2: roots
are real and irrational
(4)
7.5
T(1;−2) P(3;2)
(5)
7.6
1<x<3
(2)
8.1
h = 8 − 2x − πx
8.2
πr2
8.3
Hockey
(3)
4 = 8a
1
a = __
2
7.1
(3)
A = 4x × h + ___
2
π(2x)2
= 4x(8 − 2x − πx) + ______
2
= 32x − 8x2 − 4πx2 + 2πx2
= 32x − 8x2 − 2πx2
1
40
0
9
Squash
4
(6)
(3)
(3)
dA
___
= 32 − 16x − 4πx = 0
dx
x = 4,48 m
(4)
Answers
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9.3
5
8
G
P(GGG) =
7 6
10 9
5
8
3
8
B
P(GGB) =
7 6
10 9
3
8
6
8
G
P(GBG) =
7 3
10 9
6
8
2
8
B
P(GBB) =
7 3
10 9
2
8
G
P(BGG) =
3 7
10 9
6
8
B
P(BGB) =
3 7
10 9
2
8
G
P(BBG) =
3 2
10 9
7
8
B
P(BBB) =
3 2
10 9
1
8
G
6
9
G
7
10
3
9
B
6
8
3
10
G
7
9
2
8
B
7
8
2
9
B
1
8
9.3.1
9.3.2
(4)
7
___
(2)
24
21
___
40
(3)
EXAM PRACTICE B: PAPER 2
1.1
SD = 3,41
(4)
1.2.1
12 cm
(2)
1.2.2
25 cm
(2)
2.1
5
-
-
-
0
-
440 420 400 380 360 340 320 300 280 260 240 220 200 180 160 140 120 100 80 60 40 20 -
-
Number of bacteria (millions)
Age of population versus number of bacteria
10
15
Age (days)
396
Answers
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2.3.2
f
(4)
–75°
(4)
3.1.2
a = 6 or a = −2
(5)
3.2.1
D(−10;0)
C(−10 + 4;0 – 4) = (–6;– 4)
(4)
3.2.2
y = 3x + 14
(4)
3.2.3
(− 4;2)
(2)
3.2.4
8
mAD = __
8=1
90°
135°
-
45°
-
–45°
-
–90°
The original least squares regression line
would not give an accurate prediction for the
number of bacteria on day 16. This would be
extrapolation (it lies outside the given data). (2)
a = −4
180°
x
–1 -
(6)
7.3
−75° < x < 45°
8.1
EB = 55,53 m
8.2
(2)
ED = 117,79 m
(4)
Perimeter = 270,56 m
Area = 2678,99 m2
(4)
9.1
B
E
120°
120°
C
4
mAB = ____
= −1
−4
A
∴ AD ⊥ AB (mAD × mAB = −1)
∴ ABCD is a parallelogram with right angles
and therefore it is a rectangle.
(4)
3.2.5
^ B = 26,56°
AD
(5)
4.1
M(−2;−3)
(5)
4.2
A(− 6;0) and B(2;0)
(4)
4.3
4
y = __
3x + 8
5.1
sin(B − A) =
6.1
–135°
-
–180°
-
The growth of the bacteria slows down and
the new points do not lie on the original least
squares regression line
(2)
3.1.1
5.2
g
-
2.3.1
y
-
y = 24,29 + 35,21x
r = 0,99
Therefore a very strong positive correlation
between age and number of bacteria
7.2
-
2.2
(3)
-
See scatter plott
-
2.1
radius = 5
40°
30°
30°
120°
70°
O
60°
T
9.1.1
D
^
AOC = 120°
(2)
(4)
9.1.2
^ C = 60°
AD
(2)
(6)
9.1.3
^ C = 120°
AB
(2)
9.1.4
^ D = 70°
CA
(2)
9.1.5
^ C = 120°
AE
(2)
9.2
Theorem bookwork
(5)
9.3.1
a)
(cos x + sin x)(cos x + sin x)2
9.3.1
b)
cos x − sin x
= _________________________
sin2 x + 2 sin x cos x + cos2x
9.3.2
___
__
− √155 − 2√5
______________
18
1__
tan 30° = ___
√
(5)
3
cos 2x
cos x − sin x
_____________
= ___________
(cos x + sin x)3
2
1 + sin 2x
2
cos x − sin x
LHS = _____________
3
(cos x + sin x)
(cos x − sin x)(cos x + sin x)
= ________________________
cos x − sin x
= ___________ = RHS
1 + sin 2x
(5)
6.2
x = 135°
(5)
7.1
x = 45° + n.120°, n ∈ Z
or
x = −135° + n.360°, n ∈ Z
(5)
^ = 20°
O
2
(4)
^ = 90°
D
1
(2)
^ = 90°
O
| Given
1
^ C = 90° | Angle in semicircle
BD
The exterior angle of quadrilateral COED is
equal to the interior opposite angle
COED is a cyclic quadrilateral.
Answers
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(3)
397
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9.3.3
10.1
^ = 90°
^ =O
D
| Proved
1
1
The angles in the same segment of quadrilateral
ADOB are equal.
∴ ADOB is a cyclic quadrilateral.
(3)
AE
1
___
= __
(2)
AC
6
CG __
1
10.1.2 ___
=
CB
2
AC ___
BC
AB
___
___
10.2
PQ = PR = QR
(2)
(1)
10.3.1 P = N2 = x
Q2 = P = x
(6)
10.3.2 Q1 = Q2 = x
P = N2 = x
N1 = M
QN
QP
____
10.3.3 ____
QM = QN
| Proved
| Proved
| Angles in △
(2)
| △QNP ||| △QMN
∴ QN2 = QP.QM
(2)
10.3.4 QM = 19,31
398
Answers
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Glossary
A
F
amortise a loan to fully pay back a loan or debt, including
interest
factorial (symbol !) multiplying a series of descending
natural numbers
annuity a number of regular payments of a fixed amount
made over a determined time period
function a rule or equation linking x and y so that for every
x value in an equation there is only one corresponding
y value
arithmetic sequence a sequence of numbers with a
common difference between consecutive terms
asymptote a straight line that a function or graph
approaches (gets very close to), but never touches
B
bivariate data two sets of data values that both vary
book value the depreciated value of a vehicle at a point in
time
C
circumcircle of a triangle the circle that passes through all
of the vertices of that triangle
G
general solution a formula that lists all possible solutions
to a trigonometric equation; it takes into account the
period of the trigonometric functions, so the angle can
be positive or negative
geometric sequence a sequence of numbers with a
common ratio between consecutive terms
gradient slope of a line
I
identical exactly the same
common difference the constant amount by which
consecutive terms increase (or decrease)
independent variable a variable that does not depend on
other values
common ratio the constant amount by which consecutive
terms are multiplied
infinite series a series that has no end, and for which it is
impossible to determine the number of terms
common vertex the point at which three or more lines
meet to form two or more angles
interpolation using a regression equation to predict values
within the data range
consecutive terms terms which follow one after the other
L
converge grow smaller and smaller in value
correlation coefficient the measure of association
between two variables
corresponding angles the angles which join sides which
are in proportion
corresponding sides the sides which join equal angles
cyclic quadrilateral a quadrilateral with all four vertices on
a circle
D
linear pattern an arithmetic sequence with a common
difference between consecutive terms
linear trend a pattern that closely follows a straight line
N
negative correlation the measure that indicates that the
two variables move in opposite directions, so as one
increases so the other decreases
nominal interest rate the quoted annual interest rate
dependent variable a variable that is affected by other
values
non-real roots
occur when b2 − 4ac < 0; for example
____
x = 3 ± √− 4
derivative the gradient of a tangent to the function at a
point on the curve
O
diverge grow further apart in value
domain the set of x values of a function
E
effective interest the actual rate of interest that is
obtained
extrapolation using a regression equation to predict values
outside of the data range
optimisation the process you follow to solve practical
problems using calculus
oscillate swing back and forth between large and small
values, and between positive and negative values
P
permutation an arrangement of items where the position
or order of the items is important
positive correlation the measure that indicates that the
two variables increase or decrease together
Glossary
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R
range the set of y values of a function
rational roots occur when the equation factorises
or b2 − 4ac is a perfect square; for example (x − 2)
3
(2x + 3) = 0 ∴ x = 2 or − __
2
real roots occur when b2 − 4ac ≥ 0 and can be rational or
irrational;__for example x2 − 3x + 1 = 0 has irrational roots
3 ± √5
x = _______
2
regression line the ‘line of best fit’ for a set of plotted data
points; also called the line of least squares
residuals the deviations from the line of best fit, that is:
y − ^y
retirement annuity a future value annuity where regular
payments are invested into an account to save for
retirement
S
scatter plot the graphical representation of bivariate data
in the form of points plotted on a Cartesian plane
scrap value the depreciated value of an item once it has
reached the end of its useful life
secant a line which passes through a circle, intersecting the
circle at two points
similar polygons polygons which have the same shape as
each other, but may be different in size
solution a value of the angle which satisfies a given
trigonometric equation
specific solutions solutions that satisfy a given
trigonometric equation in a restricted interval, such
as – 360° ≤ x ≤ 360°
sum to infinity no limit to the number of terms being
added to determine the sum
T
tangent a line which touches a curve at the point of
contact
V
vertex the point at which two straight lines meet to form an
angle
400
Index
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Index
A
analytical geometry 188–197
angles
of depression 120
of elevation 120
annuities
derivation and use of, formulae for 62–70
future value See future value annuities
present value See present value annuities
annuity applications and problem solving 71–78
arithmetic
sequences 4–7, 25–28
series, sum of 12–14
derivations to learn for examination 318
derivative
definition of 150
determining 150
second 162–166
differential calculus 142–183
differentiation
of functions from first principles 150–153
specific rules for 154–157
domain 34
double angle identities 90–108
for tan (α ± β) 96–98
for tan 2α 96–98
B
E
bivariate data 243–252
C
circles
centered at origin 188
centered off origin 188
equations of 188–192
tangent to 193–197
common
difference 4
ratio 8
vertex 221
complementary events 256–260
compound angle identities 92–95
for tan (α ± β) 99–100
for tan 2α 99–100
compound interest 60
calculating time periods 79
consecutive terms 7
contingency tables 261–266
correlation 243–252
coefficient 244
negative 244
positive 244
corresponding
angles 218
sides 218
counting 256–279
principle, application to solve probability problems
275–279
cubic
graphs 167–175
polynomials 135–139
D
data
bivariate 243–252
skewed 240–242
symmetric 240–242, 240–242
dependent variable 146
empirical probability 256
equations
of circles 188–192
exponential 49–50
of tangents to functions 158–161
of tangent to circle 193–197
trigonometric See trigonometric equations
Euclidean geometry 214–236
events 256–260
examination
learning derivations and proofs for 318
preparing for 316
structure of 296–298
writing 317
exponential
equations 49–50
functions 48–50, 48–50
extrapolation 246
F
factorial 268
factor theorem 135–139
finance 60–87
first principles, differentiation of functions from 150–153
function
derivative of 150
f, using limits to define derivative of 146–149
functions 34–36
differentiation of, from first principles 150–153
exponential 48–50
inverse of 37
logarithmic 51–52, 54–55
many-to-one 35
one-to-one 35
fundamental counting principle 267–274
future value annuities 62–66
delayed start of payments 77–78
that end early 73–75
Index
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401
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G
M
general
formula 145
solution 103
geometric
sequences 8–11, 25–28
many-to-one function 35
mean 241
median 240
midpoint theorem 290
modal group 241
mutually exclusive events 256–260
series sum of 15–17
geometry
analytical 188–197
Euclidean 214–236
gradient 144
graphs
cubic 167–175
exponential 48–50
growth and decay 60–87
H
hire purchase agreement 85
horizontal line test 35, 37
I
identities
compound angle 92–95
double angle 96–98, 99–100
proving 101–102
independent
events 256–260
variable 146
inequalities, logarithmic graphs and 57
interpolation 245
interquartile range 240
inverse
functions 37–42
graphs 37–38
investments, analysing 82–85
L
limits 142–145
using to define derivative of a function f, 146–149
linear
patterns 4
trend 243
loan options 85–87
logarithmic
functions 51–52, 54–55
graphs 57
laws 51–52
logarithms 51
N
negative correlation 244
nominal interest rate 62
O
one-to-one functions 35
optimisation 176–183
P
parabola
equations of 145
symmetry line of 145
parallelogram 219
patterns 4–27
permutation 269
polygons, similar 218–220
polynomials 124–131
cubic 135–139
equations, third degree 132–134
positive correlation 244
present value annuities 66–70, 71–73
probability 256–273
problems, application of counting principle to solve
275–279
proofs to learn for examination 318
Proportionality Theorem 221–226
Pythagoras, theorem, and similarity 235–236
Q
quadratic patterns 25–28
R
range 34, 240
rate of change 176–183
rectangle 219
regression lines 245
remainder theorem 135–139
residuals 245
retirement annuities 82–83
rhombus 219
S
scatter plots 243–252
second derivative 162–166
sequences 4–22
arithmetic 4–7, 25–28
geometric 7–9, 8–11
using to calculate time periods 79–81
402
Index
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series 4–27
arithmetic, sum of 12–14
Sigma notation 18–20
similarity
equiangular triangles and 227–230
Pythagoras's Theorem and 235–236
triangles with proportional sides and 231–234
similar polygons 218–220
sinking funds 71–73
skewed data 240–242
solution 103
specific
rules for differentiation 154–157
solutions 103
square 219
statistics 240–252
sum
of arithmetic series 12–14
of terms, formula for 15
symmetric data 240–242
T
tangent 144
tangents
to circle 193–197
to functions, equation of 158–161
third degree polynomial equations 132–134, 138–139
three dimensional problems 125–129
time periods, calculating using logarithms 79–81
tree diagrams 261–263, 261–266
triangles
equiangular 227–230
with proportional sides 231–234
trigonometric equations
with double angles and more than one ratio 105–108
ending with one ratio 103–105
solving using compound and double angle identities
101–102
trigonometry 90–108, 120–129
problems in three dimensions 125–129
problems in two dimensions 120–124
turning point formula 145
two-dimensional problems 120–124
V
Venn diagrams 261–262
vertex 221
vertical line test 35
X
x-intercept formula 145
x-intercept of graph 145
Y
y-intercept of graph 145
Index
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403
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Platinum Mathematics Grade 12
Learner's Book
Maskew Miller Longman (Pty) Ltd
Forest Drive, Pinelands, Cape Town
Offices in Johannesburg, Durban, King William’s Town, Polokwane, Bloemfontein, Mbombela,
Mahikeng and representatives in companies throughout southern and central Africa.
website: www.mml.co.za
© Maskew Miller Longman (Pty) Ltd 2013
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or
transmitted in any form or by any means, electronic, mechanical, photocopying, recording, or
otherwise, without the prior written permission of the copyright holder.
Every effort has been made to trace the copyright holders of material produced in this title. We would
like to apologise for any infringement of copyright so caused, and copyright holders are requested to
contact the publishers in order to rectify the matter.
First published in 2013
Print ISBN 978-0-636-14331-9
ePDF ISBN 978-0-636-15111-6
Acknowledgements:
Edited by Penny Adnams
Proofreading by Lorrainne Bowie
Artwork by Will Alves and Tina Nel
Typesetting by Lizette van Greunen (LVG Maths) and Stronghold Publishing
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