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1
SECONDARY SCHOOL IMPROVEMENT
PROGRAMME (SSIP) 2015
GRADE 12
SUBJECT:
MATHEMATICS
TEACHER NOTES
(PAGE 1 OF 42)
© Gauteng Department of Education
2
TABLE OF CONTENTS
SESSION
TOPIC
PAGE
5
Trigonometry
3 - 15
6
Calculus
15 - 31
7
Analytical Geometry
31 - 42
© Gauteng Department of Education
3
SESSION NO:
5
TOPIC: TRIGONOMETRY
LESSON OVERVIEW
1.
2.
3.
Introduction session – 5 minutes
Typical exam questions:
Question 1
5 minutes
Question 2: 5 minutes
Question 3: 5 minutes
Question 4: 10 minutes
Question 5: 10 minutes
Question 6: 10 minutes
Question 7: 10 minutes
Discussion of solutions: 60 minutes
Teacher suggestions:
Trigonometry is a topic where learners can score a lot of marks. Unfortunately, many
learners struggle with this topic and perform poorly in examinations. It is therefore so
important to spend a lot of time on the basics. Make sure that your learners revise
the basic principles of Trigonometry.
Make sure that your learners master the concepts from Grade 10 which include:
(a) Trigonometric ratios in right-angles triangles (determining lengths of sides and
the size of angles)
(b) The quadrants in which the trigonometric ratios are positive and negative.
(c) Evaluating expressions using a calculator (rounding off is important)
(d) Solving basic trigonometric equations using a calculator.
Important concepts to master in Grade 11 include:
(a) Problems involving Pythagoras.
(b) Reduction formulae.
Make sure that your learners understand that:
(1)
sin(180  )   sin  but sin 2 (180  )  ( sin )2  sin 
(2)
sin(180  )   sin  but sin180  0
(3)
but cos(90  )   sin 
cos(90  )  sin 
(4)
cos2 225   cos2 45
cos 225  cos(180  45)
2
(c)
2
2

2
2 1
 ( cos 45)   
  
4 2
 2 
2
Negative angles.
Make sure that your learners understand that:
© Gauteng Department of Education
4
(a)
sin()   sin 
cos()  cos 
tan()   tan 
sin(  90)  sin  (90  )   sin(90  )   cos 
Angles greater than 360 .
Make sure that your learners understand that whenever the angle is greater
than 360 , keep subtracting 360 from the angle until you get an angle in the
interval 0 ;360 .
For example, tan 765  tan(765  2(360))  tan 45  1
General solutions of trigonometric equations.
You can use either the reference angle approach (see Section B) or the
following method:
If sin   a and  1  a  1
If cos   a and  1  a  1
(b)
(d)
(e)
(f)
then   sin 1 (a)  k .360
(k  Z)
or   180  sin 1 (a)  k .360
If tan   a and a  R
(k  Z)
then    cos1 (a)  k .360
(k  Z)
then   tan 1 (a)  k .180 (k  Z)
When teaching the sine, cosine and area rules, ensure that your learners know
when to use the rules. Focus on basic examples before doing the more
complicated types involving more than one triangle.
Emphasise the following:
The sine rule is used when you are given:
(1)
Two sides and a non-included angle
(2)
More than one angle and a side
The cosine rule is used when you are given:
(1)
Two sides and the included angle
(2)
Three sides
The area rule is used when you are given:
(1)
Two sides and the included angle
It is so important for learners to be able to integrate compound and double angles
into the Grade 11 concepts which include Pythagoras problems, identities,
trigonometric equations, trigonometric graphs and the sine, cosine and area rules.
Once you are confident that your learners have mastered the basics, then discuss
the typical examination questions and then let them do the homework questions.
Section B contains a summary of all relevant theory.
SECTION A: TYPICAL EXAM QUESTIONS
Question 1
This question involves reduction formulae, negative angles, identities and compound
and double angles. Learners need to revise their basic Grade 11 work as well as the
Grade 12 work done on Trigonometry in Term 1. See Section B.
© Gauteng Department of Education
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Simplify the following without using a calculator:
(a)
(b)
tan(60)cos(156)cos 294
sin 492
2
cos 375  cos 2 (75)
sin(50)sin 230  sin 40 cos310
(7)
(7)
Question 2
This question involves compound angles and special angles. Emphasise the
meaning of the word “hence”.
(a)
Show that cos  60    cos  60      3 sin 
(5)
(b)
Hence, evaluate cos105  cos15 without using a calculator.
(5)
Question 3
This question involves compound angles and identities.
Rewrite cos3 in terms of cos .
(6)
Question 4
This question involves double angles and identities.
Prove that:
(a)
(b)
(c)
1  cos 2 x  sin x
 tan x
sin 2 x  cos x
sin   sin 2
 tan 
1  cos   cos 2
cos 2A
cos A  sin A

1  sin 2A cos A  sin A
(6)
(6)
(5)
Question 5
This question involves diagrams using Pythagoras and compound angles.
It is known that 13sin   5  0 and tan   
[90 ; 270] .
3
where  [90 ; 270] and
4
Determine, without using a calculator, the values of the following:
(a)
cos
cos(  )
(b)
Question 6
© Gauteng Department of Education
(5)
(6)
6
This question involves diagrams using Pythagoras and compound angles.
If sin18  t determine the following in terms of t.
cos18
(a)
(b)
sin 78
(4)
(5)
Question 7
Make sure that learners know the sine, cosine and area rules and when to use them.
Thandi is standing at point P on the horizontal ground and observes two poles, AC and
BD, of different heights. P, C and D are in the same horizontal plane. From P the
angles of inclination to the top of the poles A and B are 23° and 18° respectively.
Thandi is 18 m from the base of pole AC. The height of pole BD is 7 m.
B
A
7m
D
18°
C
18 m
23°
42°
P
Calculate, correct to TWO decimal places:
(a)
The distance from Thandi to the top of pole BD.
(b)
The distance from Thandi to the top of pole AC.
(c)
The distance between the tops of the poles, that is the length of AB, if
ˆ  42
APB
SECTION B: NOTES ON CONTENT
Summary of all Trigonometric Theory
© Gauteng Department of Education
(2)
(2)
(3)
7
sin  
sin  
y
r
cos  
x
r
tan  
y
x
sin  
90   90  
cos  
cos  
tan  
tan  
( x ; y)
180  

360  
180  
sin  
sin  
cos  
cos  
tan  
tan  
Reduction rules
sin(180  )  sin 
cos(180  )   cos 
tan(180  )   tan 
sin(90  )  cos 
cos(90  )  sin 
sin()   sin 
sin(180  )   sin 
cos(180  )   cos 
tan(180  )  tan 
sin(90  )  cos 
cos(90  )   sin 
cos()  cos 
sin(360  )   sin 
cos(360  )  cos 
tan(360  )   tan 
tan()   tan 
Whenever the angle is greater than 360 , keep subtracting 360 from the angle
until you get an angle in the interval 0 ;360 .
Identities
cos2   sin 2   1
tan  
sin 
cos 
Special angles
Triangle A
2
Triangle B
45
60
2
1
1
30
45
1
3
© Gauteng Department of Education
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From Triangle A we have:
From Triangle B we have:
1
sin 45 

2
1
cos 45 

2
1
tan 45   1
1
2
2
sin 30 
2
2
cos30 
1
2
3
2
1
cos 60 
2
3
tan 60 
 3
1
sin 60 
and
3
2
1
tan 30 
3
and
and
For the angles  0; 90;180;270;360  the diagram below can be used.
90
y
A(0 ;2)
B(1 ; 3)
C( 2 ; 2)
2
60
2
r2
G(  2 ; 0)
D( 3 ; 1)
45
30
180
E(2 ; 0)
F(0 ;  2)
The following identities are important
270
Trigonometry:
for
x
0
360
tackling
Compound angle identities
Double angle identities
sin(A  B)  sin A cos B  cos A sin B
sin(A  B)  sin A cos B  cos A sin B
cos(A  B)  cos A cos B  sin A sin B
cos(A  B)  cos A cos B  sin A sin B
cos 2   sin 2   cos 2
sin 2  2sin  cos 
Grade
cos 2   sin 2 

cos 2  2 cos 2   1

2
1  2sin 
Solving two-dimensional problems using the sine, cosine and area rules

The sine-rule can be used when the following is known in the triangle:
- more than 1 angle and a side
© Gauteng Department of Education
12
9
- 2 sides and an angle (not included)
sin A sin B sin C


a
b
c

The cosine-rule can be used when the following is known of the triangle:
- 3 sides
- 2 sides and an included angle
a2  b2  c2  2bc cos A

The area of any triangle can be found when at least two sides an included
angle are known
Area of ABC 
SECTION C:
1
ab sin C
2
HOMEWORK QUESTIONS
Question 1
Determine the value of the following without using a calculator.
(a)
sin 34 cos10  cos34 sin10
sin12 cos12
(3)
(b)
sin(285)
(5)
(c)
cos 2 15  sin15 cos75
cos 2 15  sin15 cos15 tan15
(6)
Question 2
(a)
(b)
1
cos 2
2
Hence determine the value of sin 75.sin15
Prove that sin(45  ).sin(45  ) 
(5)
(3)
Question 3
Prove that:
sin 4  4sin .cos   8sin 3 .cos 
© Gauteng Department of Education
(4)
10
Question 4
Prove that:
(a)
(b)
(tan x  1)(sin 2 x  2cos2 x)  2(1  2sin x cos x)
cos 2 x
 cos x  sin x
cos x  sin x
(6)
(3)
Question 5
(a)
2(cos   sin )
2
2
Hence prove that sin 2  2sin  45     1
Show that sin(45  ) 
(b)
Question 6
(3)
(6)
p
where p < 0 and  [180 ; 360] , determine, using a diagram, an
5
expression in terms of p for:
(a)
(4)
tan
(b)
(3)
cos 2
If cos  
Question 7
If sin 61  a , determine the value of the following in terms of a:
cos73 cos15  sin 73 sin15
(6)
Question 8
In the diagram below, a foefie-slide platform, G is situated on the opposite side of a
river from points K and I. The angle of elevation from K to G is 51°. The bottom of
the foefie-slide, represented by N, is vertically below G and in the same horizontal
ˆ  38 and KIN
ˆ  48
plane as K and I. KI  15 metres, IKN
© Gauteng Department of Education
11
51
38
48
(a)
Determine GN, the height of the platform, rounded to two decimal places.
(b)
Determine the width of the river, rounded to two decimal places.
(Assume that K,I and N are points on the bank of the river and that the width is
constant).
(3)
SECTION D:
1(a)
SOLUTIONS FOR SECTION A
tan(60) cos(156) cos 294
sin 492
( tan 60)(cos156)( cos 66)

(sin132)

( 3)( cos 24)( sin 24)
(sin 48)
( 3)( cos 24)( sin 24)
2sin 24 cos 24
3

2

(4)






( tan 60)(cos156)
 cos66
sin 48
 3
 sin 24
2sin 24 cos 24
3

2
(7)
© Gauteng Department of Education
12
1(b)
cos 2 375  cos 2 ( 75)
sin(50) sin 230  sin 40 cos 310

cos 2 15  cos 2 75
( sin 50)(  sin 50)  (sin 40)(cos 50)
cos 2 15  sin 2 15
 2
sin 50  (cos 50)(cos 50)

cos 15  sin 15
sin 2 50  cos 2 50
cos 2(15)

1
 cos 30

2






2
cos2 15
cos2 75
sin 2 50
cos2 50
cos30
1
3
2
(7)
3
2
cos  60    cos  60   

2(a)
 cos60.cos   sin 60.sin  
 cos 60.cos   sin 60.sin    cos 60.cos   sin 60cos60
.sin 
   sin 60.sin  
.cos
 3
 3
1
 1
   .cos   
 .sin     .cos   
 .sin 
2
 2
 2 
 2 
  3 sin 
2(b)
1
2
  3 sin 
(5)
cos105  cos15
 cos  60  45   cos  60  45 
 cos  60  45 
 cos  60  45 
  3 sin 45
  3 sin 45
 2
  3 

 2 


3.

2
2
 6

2
 6
2
cos 3
 cos(2  )
 cos 2.cos   sin 2.sin 
 cos(2  )
 cos 2.cos   sin 2.sin 
 (2 cos 2   1).cos   (2sin  cos ).sin 
 2 cos   cos   2sin .cos 
3
3
2
2
 2 cos3   cos   2(1  cos 2 ) cos 




(5)
2cos2   1
2sin  cos 
1  cos2 
4cos3   3cos 
 2 cos3   cos   2(cos   cos3 )
 2 cos3   cos   2 cos   2 cos3 
 4 cos3   3cos 
© Gauteng Department of Education
(6)
13
4(a)
4(b)
4(c)
5(a)
1  cos 2 x  sin x
sin 2 x  cos x
1  (1  2sin 2 x)  sin x

2sin x cos x  cos x
1  1  2sin 2 x  sin x

2sin x cos x  cos x
2sin 2 x  sin x

2sin x cos x  cos x
sin x(2sin x  1)

cos x(2sin x  1)
sin x

 tan x
cos x
sin   sin 2
1  cos   cos 2
sin   2sin .cos 

1  cos    2 cos 2   1




1  2sin 2 x
2sin x cos x
sin x(2sin x 1)
cos x(2sin x 1)
sin x

cos x
 tan x
(6)
 2sin  cos 
 2cos2  1
 sin  1  2cos  
 cos  1  2cos  

sin  1  2 cos  
cos   2 cos 2 

sin  1  2 cos   sin 

 tan 
cos  1  2 cos   cos 

sin 
cos 
 tan 
(6)
cos2 A  sin 2 A
cos 2A
1  sin 2A
cos 2 A  sin 2 A

1  2sin A cos A
(cos A  sin A)( cos A  sin A)

sin 2 A  cos 2 A  2sin A cos A
(cos A  sin A)( cos A  sin A)

sin 2 A  2sin A cos A  cos 2 A
(cos A  sin A)( cos A  sin A)

(sin A  cos A)(sin A  cos A)
cos A  sin A

cos A  sin A




5
13
y  5 r  13
 sin  
(5)
sin  
x 2  (5) 2  (13) 2
 x 2  144
 x  12
5
13
α
– 12
2sin Acos A
(cos A  sin A)( cos A  sin A)
(sin A  cos A)(sin A  cos A)
cos A  sin A

cos A  sin A




5
13
diagram
Pythagoras
x  12
12
cos   
13
© Gauteng Department of Education
14
 cos   
5(b)
(5)
12
13
3
4
y  3 x  4




tan   
3
r 5
5
β
–4
 cos18 
6(b)
(6)
 12   4   5   3 
    .      . 
 13   5   13   5 
48  15 33


65
65
t
sin18  t 
1
1
sin 78
t
r 5
cos .cos   sin .sin 
 12   4   5   3 
    .      . 
 13   5   13   5 
33

65
cos(  )
 cos .cos   sin .sin 
6(a)
diagram
Pythagoras
x2  r 2  y2
 x 2  12  t 2
18
 x2  1 t 2
x
 x  1 t
1 t2
 1 t2
1
 sin  60  18 
 sin 60.cos18  cos 60.sin18
 3
1
2
 
 . 1  t    . t
2
 2 
 diagram
 Pythagoras
 x  1 t2
1 t2
 1 t2
 cos18 
1
(4)
2
 sin  60  18 
 sin 60.cos18  cos60.sin18
 3
1
 
 and  
2
 2 

3 1  t 2   t
3 1 t2 t

 
2
2
2

1  t 2 and t
3 1  t 2   t
© Gauteng Department of Education
2
(5)
15
7(a)
7(b)
7(c)
7
 sin 18
PB
7
PB 
sin 18
PB  22,65 m (22,65247584...)
Alternatively:
PB
7

sin 90 sin18
 PB  21, 65m
18
 cos 23
PA
18
PA 
cos 23
PA  19,55 m (19,55448679..)
Alternatively:
PA
18

sin 90 sin 67
 PA  19,55m
AB2  (22,65) 2  (19,55) 2  2(22,65)(19,55). cos 42
 237,0847954...
AB  15,40 m
(15,3975581...)
SESSION NO:
 ratio
 answer
(2)
 ratio
 answer
(2)
 use of cosine rule
 237,0847…
 answer
(3)
6
TOPIC: CALCULUS
LESSON OVERVIEW
1.
2.
3.
Introduction session – 5 minutes
Typical exam questions:
Question 1 10 minutes
Question 2: 10 minutes
Question 3: 10 minutes
Question 4: 10 minutes
Question 5: 5 minutes
Question 6: 5 minutes
Question 7: 10 minutes
Discussion of solutions: 60 minutes
© Gauteng Department of Education
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Teacher suggestions:
First principles and differentiation
(a)
When determining the gradient from first principles, learners tend to make the
following mistakes:
f ( x)   x 2
( x  h ) 2  x 2
 f ( x)  lim
h0
h
[ f ( x) is omitted in the second line and  x 2 should be ( x 2 ) ]
Many learners expand the expression ( x  h)2 incorrectly by writing:
( x 2  h2 ) or  x 2  h2 or  x2  2 xh  h2
Sometimes the limit symbol is ignored or written incorrectly:
( x 2  2 xh  h 2 )  x 2
( x 2  2 xh  h 2 )  x 2


f ( x)  lim
f ( x)  lim
h 0
h 0
h
h
( x 2  2 xh  h 2 )  x 2
h
2
 x  2 xh  h 2  x 2
 f ( x) 
h
2 xh  h 2
 f ( x) 
h
h(2 x  h)
 f ( x) 
h
 f ( x)  (2 x  h)
 f ( x) 
(b)
( x 2  2 xh  h 2 )  x 2
h 0
h
2
 x  2 xh  h 2  x 2
lim 
h 0
h
2 xh  h 2
lim 
h
h(2 x  h)
lim 
h
 (2 x  h)
lim 
Learners often make mistakes with the different derivative notations:
The symbol f ( x) is either not used or introduced too early.
3
3
f ( x)  4
f ( x)  4
2x
2x
3 4
3
 f ( x)  x
 f ( x)  x 4
2
2
5
 f ( x )  6 x
 f ( x)  6 x 5
dy
Dy
is sometimes written as
which is incorrect and learners
Dx
dx
may also make the following mistakes:
3
3
y
y  x 4
2x
2
3
dy 3
3
 y   4 x 41
  x 1   x 2
2
dx 2
2
The symbol D x is sometimes handled incorrectly as follows:
The symbol
© Gauteng Department of Education
17
 3 
Dx  4 
 2x 
3

 D x  x 4 
2

 D x  6 x 5 
(c)
 3 
Dx  4 
 2x 
3
 x 4
2
  6 x 5
Learners often do not understand the meaning of f (a) and f (a) .
Emphasise the difference:
f (a) is the gradient of the function at x  a whereas f (a) is the value of y
corresponding to the value of x for the function.
Cubic functions and tangents to functions at given points
(a)
Make sure that learners know how to factorise cubic equations before
sketching the graphs of cubic functions.
(b)
Examiners often require learners to write the intercepts with the axes,
stationary points and points of inflection in coordinate form (a ; b) . Make sure
that the learners are aware of this.
(c)
Emphasise the relationship between the graph of a function and the graph of
its derivative is important in that it explains to the learners why the second
derivative is zero at a point of inflection.
(d)
The point of inflection is determined by equating the second derivative to zero
and solving for x. An alternative method is to add up the x-coordinates of the
turning points and divide by 2.
Problems involving maximum and minimum values
Expressions involving area or volume can be seen as graphs of quadratic or cubic
functions. This will assist learners to identify which particular value of x yields a
maximum or minimum value.
© Gauteng Department of Education
18
SECTION A: TYPICAL EXAM QUESTIONS
Question 1
This question involves gradient from first principles and average gradient. Emphasise
correct layout and notation.
(a)
(b)
Given:
f ( x)  2 x 2  1
(1)
Determine f ( x) from first principles.
(2)
Determine the gradient of the graph at x  2 , i.e. f (2)
(3)
Determine f (2) . What does your answer represent?
(4)
Determine the average gradient of f between x  2 and x  4
Use first principles to determine the derivative of f ( x) 
1
x
(5)
(1)
(2)
(4)
(5)
Question 2
In this question, learners need to understand that the gradient of the tangent to a
curve at a point is the same as the gradient of the function at that point.
(a)
Differentiate f by first principles where f ( x)  x 2  2 x .
(b)
Determine the gradient of the tangent to the graph of g ( x)  x at
x  3 by using first principles.
(5)
3
(6)
Question 3
In this question the emphasis must be on correct notation and the use of exponents.
Determine the following and leave your answer with positive exponents:
(a)
D x (2 x  3)( x  4)
(b)
f ( x) if f ( x) 
(2)
1
(3)
4
2 x3
2
(c)
(d)
dy
1 

if y   2 x  
dx
3x 

 1 3

Dx 
x  2 x 2  3x 
 x


(5)

(4)
© Gauteng Department of Education
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Question 4
In this question, learners must know how to determine the equation of a tangent to a
curve at a point and that parallel lines have equal gradients.
(a)
Determine the equation of the tangent to f ( x)  x 2  6 x  5 at x  2 .
(5)
(b)
Find the equation of the tangent to f ( x)  3x  5 x  1 which is parallel
to the line y  7 x  4  0 .
(6)
2
Question 5
The sketching of cubic functions is worth a lot of marks. Emphasis the importance of
indicating coordinates on the graph.
Sketch the graph of f ( x)  2 x3  6 x  4
(15)
Question 6
This question is important in that it deals with not only the gradient at the point but
also that the given point lies on the curve and can be substituted into the original
equation.
(2 ; 9) is a turning point on the graph of f ( x)  ax3  5x 2  4 x  b .
Determine the value of a and b and hence the equation of the cubic function.
(7)
Question 7
This is an important question involving surface area of a cylinder. Show learners how
to determine surface area without the rote learning of a formula.
A drinking glass, in the shape of a cylinder, must hold 200 m of liquid when full.
(a)
Show that the height of the glass, h, can
200
. (2)
 r2
Show that the surface area of the
glass can be expressed as
400
.
(2)
S(r )   r 2 
r
be expressed as h 
(b)
h
r
© Gauteng Department of Education
20
(c)
Hence determine the value of r for which the total surface area of the
glass is a minimum.
(5)
SECTION B: NOTES ON CONTENT
The most important fact in Calculus is that the gradient of the tangent to a curve at a
given point is the gradient of the curve at that point.
Other words for gradient are:
Symbols for gradient are:
rate of change, derivative, slope
f ( x)
Dx
dy
dx
f (a ) is the gradient of f at x  a
f (a ) is the y -value corresponding to x  a
f (a)
mt  m f  f (a )
Average gradient
The average gradient (or average rate of change) of a function
f
between x  a
and x  b and is defined to be the gradient of the line joining the points on the
graph of the function. We say that the average gradient of f over the interval is the
gradient of the line AB.
Gradient of a curve at a point using first principles
The formula to determine the gradient of a function from first principles is given by
the following limit:
f ( x)  lim
h0
f ( x  h)  f ( x )
h
The gradient of a function using the rules of differentiation
You will be required to use the following rules of differentiation to determine the
gradient of a function.
Rule 1 If f ( x)  ax n , then f ( x)  a.nx n1
Rule 2 If f ( x)  ax, then f ( x)  a
Rule 3 If f ( x)  number, then f ( x)  0
© Gauteng Department of Education
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Determining the equation of the tangent to a curve at a point
The gradient of the tangent to a curve at a point is the derivative at that point.
The equation is given by y  y1  m( x  x1 ) where ( x1 ; y1 ) is the point of tangency
and m  f ( x1 )
Rules for sketching the graph of a cubic function
The graph of the form f ( x)  ax3  bx 2  cx  d is called a cubic function.
The main concepts involved with these functions are as follows:
Intercepts with the axes:
For the y-intercept, let x  0 and solve for y
For the x-intercepts, let y  0 and solve for x
(you might have to use the factor theorem here)
Stationary points:
Determine f ( x) , equate it to zero and solve for x.
Then substitute the x-values of the stationary points into the original equation to
obtain the corresponding y-values.
If the function has two stationary points, establish whether they are maximum or
minimum turning points.
Points of inflection:
If the cubic function has only one stationary point, this
inflection that is also a stationary point.
For points of inflection that are not stationary points, find
solve for x. Alternatively, simply add up the x-coordinates
divide by 2 to get the x-coordinate of the point of inflection.
SECTION C:
point will be a point of
f ( x) , equate it to 0 and
of the turning points and
HOMEWORK QUESTIONS
Question 1
(a)
Given:
(1)
(2)
(3)
(4)
(b)
1
f ( x)  1  x 2
4
Determine f ( x) from first principles.
Determine the gradient of the graph at x  4 , i.e. f (4)
Determine f (2) . What does your answer represent?
Determine the average gradient of
x4
f
(6)
(1)
(2)
between x  2 and
(4)
3
Use first principles to determine the derivative of f ( x)  
x
© Gauteng Department of Education
(5)
22
Question 2
(a)
Differentiate f by first principles where f ( x)  2 x .
(4)
(b)
Determine the gradient of the tangent to the graph of g ( x)  2 x3 at
x  2 by using first principles.
(6)
Question 3
Determine the following and leave your answer with positive exponents:
(a)
(b)
(c)
(d)
f ( x) if f ( x)  (4 x  3)2
1 

Dx 3 x 
x 

(2)
(3)
D x ( x 2  x ) 2 


2
dy
2x  x  5
if y 
dx
x
(5)
(4)
Question 4
(a)
(b)
Determine the equation of the tangent to the curve y  3x 2  2 x  2 at
x  4 .
The graph of f ( x)  ax 2  bx passes
through the point P(3 ; 6) and cuts the x-axis
y
at (4 ; 0) . Determine the equation of the
tangent to f at P.
.
(5)
P(3; 6)
(6)
4
f
Question 5
Sketch the graph of f ( x)  x  3x  4
Indicate the coordinates of the stationary points, intercepts with the axes and any
points of inflection.
(15)
3
2
© Gauteng Department of Education
x
23
Question 6 (DOE FEB 2009 PAPER 1)
The graph of h( x)   x3  ax 2  bx is shown below. A(1;  3,5) and B(2 ;10) are
the turning points of h. The graph passes through the origin and further cuts the xaxis at C and D.
y
.
B(2 ; 10)
•
C
.
0
D
x
A(–• 1 ; –3,5)
3
and b  6
2
(a)
Show that a 
(b)
Calculate the average gradient between A and B.
(2)
(c)
Determine the equation of the tangent to h at x  2 .
(5)
(d)
Determine the x-value of the point of inflection of h.
(3)
© Gauteng Department of Education
(7)
24
SECTION D:
1(a)(1)
SOLUTIONS FOR SECTION A
f ( x  h)  f ( x )
h0
h
2( x  h) 2  1  (2 x 2  1)
 f ( x)  lim
h 0
h
2( x 2  2 xh  h 2 )  1  2 x 2  1
 f ( x)  lim
h 0
h
2
2 x  4 xh  2h 2  1  2 x 2  1

 f ( x)  lim
h 0
h
h(4 x  2h)
 f ( x)  lim
h 0
h
 f ( x)  lim (4 x  2h)
f ( x)  lim
 2( x  h)2  1
 (2 x2  1)
 2 x2  4 xh  2h2

h(4 x  2h)
h
 4x
(5)
h 0
1(a)(2)
 f ( x)  4 x
f ( x)  4 x
 answer
(1)
 f (2)  4(2)  8
1(a)(3)
f ( x)  2 x 2  1
 f (2)  2(2)2  1
 f (2)  7
The answer represents the y-value corresponding to
x  2
1(a)(4)
f ( x)  2 x 2  1
f (2)  2(2) 2  1
 f (2)  7
1(b)
f (4)  2(4) 2  1
 f (4)  31
(2 ;  7) and (4 ;  31)
31  (7) 24
Average gradient 

 4
4  (2)
6
1
1

 f ( x)  lim x  h x
h0
h
 f (2)  7
 interpretation
(2)
 f (2)  7
 f (4)  31
31  (7)

4  (2)
 4
(4)
1
1

 xh x
h
x ( x  h )
 x ( xh)
© Gauteng Department of Education
25
 f ( x)  lim
h 0
 f ( x)  lim
h 0
 f ( x)  lim
x ( x  h )
x ( xh)

h
x  x h
x( xh)
h
h
x( xh)
h 0
2(a)
h
h
1
 f ( x)  lim

h 0 x ( x  h ) h
1
 f ( x)  lim
h 0 x ( x  h )
1
 f ( x)   2
x
f ( x  h)  f ( x )
f ( x)  lim
h0
h
( x  h) 2  2( x  h)  x 2  2 x
h0
h
2
2
x  2 xh  h  2 x  2h  x 2  2 x
 lim
h0
h
2 xh  h 2  2h
 lim
h0
h
h(2 x  h  2)
 lim
h0
h
 lim(2 x  2  h)
 lim
h
x( xh)
h
1

x ( x  h)
1
 2
x
(5)
 ( x  h)2  2( x  h)
  x2  2 x
2 xh  h 2  2h

h
 (2 x  2  h)
 2x  2
h0
 2x  2
© Gauteng Department of Education
(5)
26
2(b)
( x  h )3  x 3
h0
h
( x  h)( x  h) 2  x 3
 g ( x)  lim
h0
h
( x  h)( x 2  2 xh  h 2 )  x 3
 g ( x)  lim
h0
h
3
2
x  2 x h  xh 2  x 2 h  2 xh 2  h3  x3

 g ( x)  lim
h0
h
2
2
3
3 x h  3 xh  h
 g ( x)  lim
h 0
h
h(3x 2  3xh  h 2 )
 g ( x)  lim
h 0
h
2
 g ( x)  lim (3 x  3 xh  h 2 )
g ( x)  lim

( x  h)3  x3
 3x2h  3xh2  h3
 (3x2  3xh  h2 )
 3x 2
 27
(6)
h 0
2
 g ( x)  3 x
3(a)
3(b)
 g (3)  3(3) 2  27
D x  (2 x  3)( x  4) 
 D x  2 x 2  5 x  12 
 4x  5
1
f ( x) 
4
2 x3
1
 f ( x)  3
2x 4
1 3
 f ( x)  x 4
2
1
3  3 1
 f ( x)    x 4
2
4
3 7
 f ( x)   x 4
8
3
 f ( x)   7
8x 4
 2 x2  5x  12
 4x  5
(2)
1  34
x
2
3 7
  x 4
8


3
7
8x 4
(3)
© Gauteng Department of Education
27
3(c)
1 

y  2 x  
3x 

 y  4x 
 squaring
2
1
4
3
1
9
 4 x  x 2  x 2
4 x
1
 2
3x 9 x
 4 
2
3x
1
2
3
2

2
9 x3
4x
1
 x 2
3x 9
4 1 1
 y  4 x  x 2  x 2
3
9
dy
4
1 3 1
  4    x 2   2 x 3
dx
3
2
9
dy
2 3 2
  4  x 2  x 3
dx
3
9
dy
2
2
  4 3  3
dx
3x 2 9 x
 y  4x 
3(d)


 1 3

Dx 
x  2 x 2  3x 
 x

 x3 2 x 2 3x 
 Dx  1  1  1 
x 2 
 x 2 x 2
3
1
 5

 D x  x 2  2 x 2  3x 2 


1
5 3
3 1
 x 2  3x 2  x 2
2
2
1
5 3
3
 x 2  3x 2  1
2
2x 2
4(a)
1
 x2
 D x  x 2  2 x 2  3x 2 


5
3
1
1
5 32
3 1
x  3x 2  x 2
2
2
1
5 32
3
 x  3x 2  1
2
2x 2

(4)
f ( x)  x 2  6 x  5
xT  2
yT  f (2)  (2)2  6(2)  5
 yT  3
mT  f ( x)  2 x  6
 f (2)  2(2)  6  2
(5)





f (2)  2
f ( x)  2 x  6
f (2)  2(2)  6  2
y  (3)  2( x  2)
y  2 x  1
(5)
© Gauteng Department of Education
28
y  yT  mt ( x  xT )
 y  (3)  2( x  2)
 y  3  2 x  4
 y  2 x  1
4(b)






f ( x)  3x 2  5 x  1
 f ( x)  6 x  5
y  7x  4
 6x  5  7
 6 x  12
x  2
f ( x)  6 x  5
6x  5  7
x2
f (2)  3
y  3  7( x  2)
y  7 x  11
(6)
f (2)  3(2) 2  5(2)  1
 f (2)  3
 y  3  7( x  2)
 y  3  7 x  14
 y  7 x  11
5.
y-intercept: (0;  4)
x-intercepts:
 (0;  4)
0  2 x3  6 x  4
 0  x3  3 x  2
 0  ( x  1)( x 2  x  2)
 0  ( x  1)( x  2)( x  1)
 x  1 or x  2
(1;0) (2;0)
Stationary points:
f ( x)  2 x3  6 x  4
(using the factor theorem)




0  2 x3  6 x  4
0  ( x  1)( x2  x  2)
0  ( x  1)( x  2)( x  1)
(1;0) (2;0)
 f ( x)  6 x 2  6
 0  6 x2  6
(At a turning point, f ( x)  0)
 0  x2  1
 x  1
f (1)  8
f (1)  0
Turning points are (1;  8) and ( 1;0)
 f ( x)  6 x 2  6
 0  6 x2  6
 x  1
 (1;  8) and (1;0)
© Gauteng Department of Education
29
Point of inflection:
f ( x)  6 x 2  6
 f ( x)  12 x
 0  12 x
x  0
f (0)  4
Point of inflection at (0;  4)
Alternatively:
The x-coordinate of the point of inflection can be
determined by adding the x-coordinates of the turning
points and then dividing the result by 2.
(1)  (1)
x
0
2
 f ( x)  12 x
 (0;  4)
(1)  (1)
2
 x0

(1;0)
(2;0)
(0;  4)
(1;  8)
The graph is represented above
 intercepts with the
axes
 turning points
 shape
 point of inflection
© Gauteng Department of Education
30
(15)
6.
At the turning point (2 ; 9) , we know that
f (2)  0
f ( x)  ax3  5 x 2  4 x  b
 f ( x)  3ax 2  10 x  4
 f (2)  3a(2)2  10(2)  4
 f (2)  12a  24
 0  12a  24
12a  24
 a  2
We can now substitute a  2 into the original
equation:
y  2 x3  5x 2  4 x  b
In order to get the value of b, substitute the point
(2 ; 9) into this equation:




f ( x)  3ax 2  10 x  4
f (2)  12a  24
a  2
y  2 x3  5x 2  4 x  b
 9  2(2)3  5(2)2  4(2)  b
 b  3
 f ( x)   x 3  5 x 2  4 x  3
(7)
 9  2(2)3  5(2) 2  4(2)  b
 9  16  20  8  b
 9  12  b
b  3
 b  3
The equation of the cubic function is therefore
f ( x)   x 3  5 x 2  4 x  3
7(a)
V   r 2h
200   r 2 h
200
h 2
r
7(b)
Surface Area  2 rh   r 2
200
S(r )   r 2  2 .2 r
r
400
S(r )   r 2 
r
 V   r 2h
 200   r 2 h
(2)
 S  2 rh   r 2
 S(r )   r 2 
200
.2 r
 r2
(2)
© Gauteng Department of Education
31
S(r )   r 2  400r 1
dS
 2 r  400r 2
dr
dS
At minimum :
0
dr
400
2 r  2  0
r
3
 r  200  0
200
r3 

r  3,99 cm
7(c)
SESSION NO:
 S(r )   r 2  400r 1
dS
 2 r  400r 2
dr
400
 2 r  2  0
r
200
 r3 

 r  3,99 cm

(5)
7
TOPIC: ANALYTICAL GEOMETRY
LESSON OVERVIEW
1.
2.
3.
Introduction session – 5 minutes
Typical exam questions:
Question 1 10 minutes
Question 2: 10 minutes
Question 3: 10 minutes
Question 4: 10 minutes
Discussion of solutions: 30 minutes
Teacher suggestions:
Analytical Geometry is an important topic that carries a lot of marks in the matric final
exam. Make sure that learners know the basic formulae and then practise lots of
examples involving applications of these formulae. The properties of quadrilaterals
are extremely important in Analytical Geometry. Make sure learners know how to
prove that a quadrilateral is a parallelogram, rectangle, square, rhombus or
trapezium by knowing the properties of these quadrilaterals.
Completing the square is an important technique for determining the centre of a circle
and its radius. Ensure that learners know how to do this.
Determining the coordinates of the intercepts of a circle with the axes is important
and learners often struggle with this concept.
Re-writing a linear equation of the form ax  by  c in the form y  ax  q is essential
when finding the gradient of the line. Make sure that your learners know how to do
this.
There are two methods of determining the equation of a straight line joining two
points. The first method is to use the general equation y  mx  c in which m
© Gauteng Department of Education
32
represents the gradient and the value of c can be determined by substituting one of
the points on the line into this equation. The second method is to use the formula
y  y1  m( x  x1) . All you have to do is now substitute m and a point ( x1 ; y1) on the
line into this formula and the equation of the line is easily obtained.
SECTION A: TYPICAL EXAM QUESTIONS
Question 1
Make sure that learners know the properties of a trapezium, the equation of a line
and angle of inclination.
In the diagram, PQRS is a trapezium with vertices P(5;2), Q(1;  1), R(9;  5) and S .
PT is the perpendicular height of PQRS and W is the midpoint of QR. Point S lies on
ˆ  .
the x-axis and PRQ
P(5;2)
S
Q(1 ;  1)
T
W

R(9;  5)
(a)
(b)
(c)
(d)
Determine the equation of PW if W is the midpoint of QR.
Determine the equation of PS.
Determine the equation of PT.
Determine the coordinates of T.
(e)
Show that QT  TR .
(5)
(f)
Calculate the size of  rounded off to two decimal places.
(5)
1
3
© Gauteng Department of Education
(2)
(4)
(3)
(5)
33
Question 2
The properties of parallel and perpendicular lines, the midpoint of a line segment,
collinearity and the distance formula are important in this question.
Consider the following points on a Cartesian plane:
A(1; 2), A(3 ;1), C(3 ; k ) and C(2 ;  3)
Determine the value(s) of k if:
(1; 3) (1; 3) is the midpoint of AC.
(a)
(b)
AB is parallel to CD.
(c)
AB  CD.
(d)
A, B and C are collinear.
(e)
CD  5 2
(3)
(3)
(3)
(3)
(5)
Question 3
Properties of quadrilaterals are important in this question as well as the equation of a
circle and a tangent to a circle.
A(0 ; 5) and B(–8 ; 1) are two points on the circumference of the circle centre M, in a
Cartesian plane. M lies on AB. DA is a tangent to the circle at A. The coordinates of
D are (3 ; –1) and the coordinates of C are (–12 ; –1). Points C and D are joined. K
is the point (0 ; –7). CTD is a straight line.
A(0 ; 5)
M
B(  8;1)
C(  12;  1)
T
D(3; 1)
K(0 ;  7)
(a)
(b)
(c)
Show that the coordinates of M, the midpoint of AB, are (–4 ; 3).
Determine the equation of the tangent AD.
Determine the length of AM.
© Gauteng Department of Education
(1)
(4)
(3)
34
(d)
(e)
Determine the equation of circle centre M in the form
ax2  by 2  cx  dy  e  0
(4)
Quadrilateral ACKD is one of the following:
parallelogram; kite; rhombus; rectangle
Which one is it? Justify your answer.
(4)
Question 4
This question is often asked in an examination. Emphasise the importance of
completing the square to get the centre and radius.
Determine the coordinates of the centre and the length of the radius of the circle:
x2  8x  y 2  6 y  5  0
(7)
SECTION B: NOTES ON CONTENT
If AB is the line segment joining the points A( xA ; yA ) and B( xB ; yB ) , then the
following formulas apply to line segment AB.
The Distance Formula
AB2  ( xB  xA )2  ( yB  yA )2
2
2
or AB  ( xB  xA )  ( yB  yA )
The Midpoint Formula
 x  x y  yB 
M A B ; A
 where M is the midpoint of AB.
2 
 2
The Gradient of a line segment joining two points
Gradient of AB 
yB  yA
xB  xA
Parallel lines
Parallel lines have equal gradients. If AB||CD then mAB  mCD
Perpendicular lines
The product of the gradients of two perpendicular lines is 1 . If AB  CD, then
mAB  mCD  1
© Gauteng Department of Education
35
y  yA  m  x  xA 
The equation of the line
Inclination of a line
tan   mAB
If mAB  0, then  is acute
If mAB  0, then  is obtuse
Collinear points (A, B and C)
mAB  mBC
or mAC  mAB
or mAC  mBC
Circles and tangents to circles
The equation of a circle centre the origin is given by:
x2  y 2  r 2
The equation of a circle centre (a ; b) is given by:
 x  a 2   y  b 2  r 2
The radius is perpendicular to the tangent:
mradius  mtangent  1
© Gauteng Department of Education
36
SECTION C:
HOMEWORK QUESTIONS
Question 1
ABCD is a quadrilateral with vertices A(1;3), B(2;4), C and D(5;  1) .
B(2;4)
A(1;3)
C

D(5;  1)
(a)
(b)
(c)
(d)
(e)
(8)
Determine the coordinates of E, the midpoint of BD.
Determine the coordinates of C.
Show that ABCD is a rectangle.
Determine the area of ABCD.
Calculate the size of the angle  rounded off to the nearest degree.
(2)
(5)
(10)
(6)
Question 2
(a)
Determine the numerical value of p if the straight line defined by the equation
px  3 y  6 has an angle of inclination of 135 with respect to the positive
x-axis.
(4)
(b)
Calculate the value of k if the points A(6;5), B(3;2) and C(2k ; k  4) are
collinear.
(3)
Question 3
The equation of a circle with radius 3 2 units is x2  6 x  y 2  2 y  p  0 .
(a)
(b)
Determine the co-ordinates of the centre of the circle.
Find the value of p.
© Gauteng Department of Education
(4)
(2)
37
Question 4
In the diagram below, a circle centre C touches the y-axis at A(0; 2) . A straight line
with equation 3x  4 y  7 cuts the circle at B(  1;  1) and D.
D
C
A(0;2)
B(  1;  1)
(a)
Determine the equation of the tangent to the circle at B.
(b)
(c)
Determine the equation of the circle in the form ( x  a)  ( y  b)  r .
Determine the coordinates of D.
2
© Gauteng Department of Education
(5)
2
2
(4)
(4)
38
SECTION D:
SOLUTIONS FOR SECTION A
Question 1
P(5;2)


S
Q(1 ;  1)
T
W

R(9;  5)
(a)
(b)
 1  (9) 1  (5) 
W
;

2
 2

 W(5 ;  3)
The equation of PW is x  5
 midpoint
 x5
5  (1) 4
1


9 1
8
2
1
 mPS  
(PS||QR)
2
1
y  2   ( x  5)
2
1
5
y2  x
2
2
1
9
y  x
2
2
 mQR
mQR 
(2)
 mPS
 correct substitution
into formula for
equation
1
9
 y  x
2
2
© Gauteng Department of Education
(4)
39
(c)
mPT  2
(PT  QR)
 mPT
 correct substitution
into formula for
equation
 y  2x  8
y  2  2( x  5)
 y  2  2 x  10
 y  2x  8
(3)
(d)
mQR  
 correct substitution
into formula for
equation
1
1
 y  x
2
2
1
1
  x   2x  8
2
2
 x3
 T(3;  2)
1
2
1
y  ( 1)   ( x  1)
2
1
1
 y 1   x 
2
2
1
1
y   x
2
2
1
1
 x   2 x  8
2
2
 x  1  4 x  16
5 x  15
x  3
 y  2(3)  8  2
 T(3;  2)
(e)
(5)
QT 2  (1  3) 2  (1  ( 2)) 2
 QT 2  4  1
 QT 2  5
 QT  5
TR 2  (3  9) 2  ( 2  ( 5)) 2
 TR 2  36  9
 correct substitution to
get QT
 answer for QT
 correct substitution to
get TR
 answer for TR
 establishing that
1
QT  TR
3
(5)
 TR 2  45
 TR  45  9  5  3 5
1
 TR  5  QT
3
1
 QT  TR
3
© Gauteng Department of Education
40
(f)
 tan   1
   135
   63, 43494882
ˆ  71,56505118
 TPR
tan   mPR
2  (5)
59
 tan   1
  180  45
 tan  
   18, 43
(5)
  135
tan   mPT
 tan   2
  63, 43494882
ˆ   
Now TPR
ˆ   
 TPR
ˆ  135  63, 43494882
 TPR
ˆ  71,56505118
 TPR
  90  71,56505118  180
  18, 43
2(a)
2(b)
1  (3) 2  k 
;

2 
 2
2k 
 1;3   1;

2 

2k
3 
2
6  2  k
k  4
 1;3  
mAB  mCD  lines // 
2  1 k  (3)

1 3
3  2
1 k 3


2
5
5  2(k  3)
5  2k  6
 2k  1
1
k  
2

 1  (3) 2  k 
;
  1;3  

2 
 2
2k
 3 
2

k

4

(3)
2  1 k  (3)

1 3
3  2
1
 k  
2
 
(3)
© Gauteng Department of Education
41
2(c)
mAB  mCD  1 lines  
1 k 3

 1
2 5
  k  13
 
1 k 3

 1
2 5
k 3

 1
10
 k  3  10
 k  13
mAB  mBC  lines // 

2(d)
(3)
1
1 k

2 3  (3)

k
4

 
1
1 k

2 3  (3)
1 1 k


2
6
 6  2(1  k )
6  2  2k
8  2k
k  4

2(e)
(3)
2
2
And CD   2  (3)    3  k 
2

 5 2

2
 52  9  6k  k 2
50  25  9  6k  k 2
 0  k 2  6k  16
 0  (k  8)(k  2)
 k  8 or k  2
3(a)
3(b)
3(c)
2
 CD   2  (3)    3  k 
CD  5 2
2

 5 2

2
 52  9  6k  k 2
  0  k 2  6k  16
  0  (k  8)(k  2)
 k  8 or k  2
(5)
 8  0 1  5 
M
;
  (4;3)
2 
 2
53
2 1
mAM 
 
0  (4) 4 2
 mAD  2
y  y1  m( x  x1 )
 y  5  2( x  0)
 y  2 x  5
 (4;3)
AM 2  (0  4) 2  (5  3) 2
 formula
 AM2  (0  4)2  (5  3)2
 AM 2  42  22
 AM  20
2
(1)
 mAM 
1
2
 mAD  2
 y  5  2( x  0)
 y  2 x  5
(4)
 AM  20
(3)
© Gauteng Department of Education
42
3(d)
 x  4 2   y  32  
20

 substituting centre (4;3)
2
 substituting radius
  x  4    y  3  20
2
2

 x  8 x  16  y  6 y  9  20
2
2
 x  4
2
  y  3  20
 x2  y 2  8x  6 y  5  0
 x2  y 2  8x  6 y  5  0
3(e)
20
2
(4)
AT  TK  6 and CD  AK




Therefore ACKD is a kite since diagonal CD bisects
diagonal AK at right angles
AT  TK  6
CD  AK
kite
reason
(4)
4.
x  8x  y  6 y  5  0
2
2
8
  
2
 x  8 x  y  6 y  5
2
2
2
2
2
8
 6 
 8   6 
 x  8 x     y 2  6 y     5      
2
 2
2  2 
2
 x 2  8 x   4  y 2  6 y   3  5   4   3
2
2
 ( x  4) 2  ( y  3) 2  5  16  9
 ( x  4) 2  ( y  3) 2  20
Centre  ( 4;3)
2
2
2
2
2
 6 
  
 2
 ( x  4)2
 ( y  3)2
 20
 centre
 radius
(7)
r 2  20
 r  20
© Gauteng Department of Education