Download learner notes | sessions 10 - 15 - Sci

1
SECONDARY SCHOOL IMPROVEMENT
PROGRAMME (SSIP) 2016
GRADE 12
SUBJECT:
MATHEMATICS
LEARNER NOTES
(Page 1 of 55)
© Gauteng Department of Education
2
TABLE OF CONTENTS
SESSION
TOPIC
PAGE
10
Revision of Trigonometry (Grade 11 and12)
3 - 13
11
Revision of Calculus
13 - 21
12
Revision of Analytical Geometry
21 - 28
13
Revision of Algebra
29 - 33
14
Revision of Functions
33 - 42
15
Revision of Gr 11 Euclidean Geometry
43 - 55
© Gauteng Department of Education
3
SESSION NO:
10
(CONSOLIDATION)
TOPIC:
REVISION OF TRIGONOMETRY (GRADE 11 AND 12)
SECTION A: TYPICAL EXAM QUESTIONS
QUESTION 1:
(6 MARKS)
(5 MINUTES)
(DoE papers pre-2014)
This question revises compound and double angles. Please revise reduction
formulae and special angles before attempting this question.
Simplify without using a calculator:
(a)
cos(50  x)cos(20  x)  sin(50  x)sin(20  x)
(3)
(b)
(1  2 sin 75)( 2 sin 75  1)
(3)
QUESTION 2:
(30 MARKS)
(20 MINUTES)
(DoE papers pre-2014)
Revise the use of a calculator when solving a trigonometric equation. The graphical
method is used in the solutions (see Section B). However, you may also use the
reference angle method (see Section B). Revise identities before doing this question.
Determine the general solution for the following equations:
(a)
2  sin   2,5
(4)
(b)
cos 2  0,5
(4)
(c)
2 tan
(d)
3sin   2cos   0
(4)
(e)
sin 2   3cos 2   5sin 
(7)
(f)
cos3x  sin x
(7)

1
2
(4)
QUESTION 3:
(8 MARKS)
(5 MINUTES)
(DoE papers pre-2014)
The equations to be solved are slightly more complicated. Specific solutions must be
given in question (a). Reduction formulae, negative angles and identities feature in
question (b).
© Gauteng Department of Education
4
(a)
Calculate the value of x, correct to one decimal digit if x  90 ; 360 and
cos2x  tan 23
(b)
(3)
Determine the general solution to the equation:
cos( x  360)  cos(90  x)  0
(5)
QUESTION 4:
(20 MARKS)
(10 MINUTES)
(DoE papers pre-2014)
Before attempting this question, ensure that you revise sketching the graphs of
trigonometric functions. The four main shifts need to be mastered before this
question can be done (see Section B). Solving trigonometric equations involving
double angles is dealt with in (a).
Given: f ( x)  1  sin x and g ( x)  cos2 x
(a)
(b)
(c)
Calculate the points of intersection of the graphs of the two functions for
x  180 ; 360
(7)
Draw sketch graphs of f and g for x  180 ; 360 on the same set of
axes.
(4)
For which values of x will f ( x)  g ( x) for x  180 ; 360
(3)
Here are some additional level 4 challenges:
(d)
For which values of x will f ( x).g ( x)  0 for x  180 ; 360
(3)
(e)
For which values of x will f ( x).g ( x)  0 for x  180 ; 360
(3)
QUESTION 5:
(14 MARKS)
(10 MINUTES)
(DoE papers pre-2014)
Before attempting this question, ensure that learners revise sketching the graphs of
trigonometric functions. The four main shifts need to be mastered before this
question can be done (see Section B). Solving trigonometric equations involving
compound angles is dealt with in (b).
(a)
Draw neat sketch graphs of f ( x)  cos  x  30 and g ( x)  sin  x  30 on
the same set of axes for x   120;210 . Indicate the intercepts with the
axes as well as the coordinates of the endpoints of the graphs.
(8)
(b)
Hence solve f ( x)  g ( x) algebraically.
© Gauteng Department of Education
(6)
5
QUESTION 6:
(10 MARKS)
(5 MINUTES)
(DoE papers pre-2014)
This question focuses on three-dimensional trigonometry using the sine, cosine and
area rules. First make sure that you know when to use the rules in given triangles. It
might be useful to construct a model of the diagram using cardboard.
A, B and L are points in the same horizontal plane, HL is a vertical pole of length 3
metres, AL = 5,2 m, the angle AL̂B  113 ° and the angle of elevation of H from B is
40.
H
3m
5,2m
mm
L
113
40
B
A
(a)
Calculate the length of LB.
(2)
(b)
Hence, or otherwise, calculate the length of AB.
(4)
(c)
Determine the area of ABL.
(4)
QUESTION 7:
Given:
(a)
(9 MARKS)
(5 MINUTES)
(Gauteng prelim 2015)
cos 2 x
cos x  sin x
Show that
cos 2 x
 cos x  sin x
cos x  sin x
© Gauteng Department of Education
(2)
6
 cos 2 x  1
  can be simplified to cos 2 x  sin 2 x
 cos x  sin x  2
(b)
Show that cos x 
(c)
Hence determine the general solution of cos x 
 cos 2 x  1

 cos x  sin x  2
QUESTION 8:
(4 MARKS)
(5 MINUTES)
(4)
(3)
(Gauteng prelim 2015)
4x
(a)
Write down an expression for the area of ABC
(1)
(b)
Calculate the value of x for which the area of the triangle will be a
maximum.
(2)
What conclusion can you draw about the type of triangle formed when
the area of the triangle is a maximum?
(1)
(c)
QUESTION 9:
(10 MARKS)
(5 MINUTES)
(a)
Determine the maximum value of the expression
(b)
An equilateral triangle is inscribed in a circle.
(Level 4 challenges)
2cos 2 x  7
(4)
Each side of the triangle measure a units.
a 2
Show that the area of the circle is
3
© Gauteng Department of Education
(6)
7
SECTION B: NOTES ON CONTENT
Summary of all Trigonometric Theory
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 .
© Gauteng Department of Education
8
Identities
cos 2   sin 2   1
tan  
sin 
cos 
The following identities are important for tackling Grade 12 Trigonometry:
Compound angle identities
Double angle identities
cos 2   sin 2   cos 2
sin 2  2sin  cos 
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  2cos 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
- 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 cosA
The area of any triangle can be found when at least two sides an included
angle are known
Area of ABC 
1
ab sin C
2
Trigonometric equations
The following method (called the graphical method) is used in this session to
determine the general solutions of trigonometric equations:
If sin   a and
1  a  1
then   sin 1 (a)  k .360
(k  )
or   180  sin 1 (a)  k .360
(k  )
If cos   a and
1  a  1
then    cos 1 (a)  k .360
If tan   a and
(k  )
a
then   tan 1 (a)  k .180
Note:
© Gauteng Department of Education
(k  )
9
Some schools use a different method which involves ignoring the negative sign when
calculating the acute reference angle and then obtaining two solutions by locating
quadrants. Then the general solution is obtained using these two angles. This
method is acceptable as well.
For example, you solve the following equation using this method:
sin x  0,5
Reference angle (acute)  30
You ignored the negative sign and calculated the reference angle by working out
SHIFT/sin/0.5 on your calculator
Since sin x is negative in quadrant 3 and 4
x  (180  30)  k .360
 x  210  k .360
OR
x  (360  30)  k .360
 x  330  k .360
where k 
Trigonometric Graphs
Amplitude shifts
Graphs of the form y  a sin x, y  a cos x and y  a tan x involve amplitude shifts.
The amplitude of a trigonometric graph is defined to be:
1
distance between the maximum and minimum y  values
2
Vertical shifts
Graphs of the form y  sin x  q, y  cos x  q and y  tan x  q involve vertical
shifts. The basic graphs will not change shape as with amplitude shifts. The basic
graph will move up or down depending on the value of q.
Period shifts
Graphs of the form y  sin kx , y  cos kx, and y  tan kx involve period shifts.
The central principle is that the period of these graphs is determine by taking the
period of the basic graph and dividing by the value of k.
Examples
Period of y  sin x :
Period  360
© Gauteng Department of Education
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360
 180
2
360
Period 
 120
3
180
Period 
 45
4
Period of y  sin 2x :
Period 
Period of y  cos 3 x :
Period of y  tan 4 x :
Mixed shifts
You may be required to sketch graphs involving more than one shift (but not more
than two).
Example
Consider f ( x )  2sin
1
x.
2
This graph involves an amplitude shift as well as a period shift.
The amplitude is 2 and the period is
360
1
2
 720
Horizontal shifts
The graphs of y  sin( x  p), y  cos( x  p), y  tan( x  p) involve shifts to the left
and right.
Examples
The graph of y  sin( x  30) is the graph of y  sin x shifted 30 to the left.
The graph of y  sin( x  30) is the graph of y  sin x shifted 30 to the right.
SECTION C:
QUESTION 1:
HOMEWORK QUESTIONS
(13 MARKS)
(5 MINUTES)
(DoE papers pre-2014)
Simplify without using a calculator:
(a)
cos(140)cos740  sin140 sin(20)
(6)
(b)
cos2 375  cos2 (75)
sin(50)sin 230  sin 40 cos310
(7)
© Gauteng Department of Education
11
QUESTION 2:
(a)
(b)
cos2x  1  0
(3)
(2)
2sin x  cos x  0
(4)
(3)
2sin 2 x  2sin x cos x  0
(7)
(4)
4cos2 x  2sin x cos x  1
(8)
Solve for  if sin    30  0,44 , and   0;720
(6 MARKS)
(b)
(5 MINUTES)
(6)
(DoE papers pre-2014)
On the axes provided on the diagram sheet, draw neat sketch graphs of the
following functions for the interval x   90 ;180 :
g ( x)  sin 2x
(4)
Hence determine graphically the values of x for which:
sin 2 x   tan x
QUESTION 4:
(a)
(DoE papers pre-2014)
(1)
f ( x)   tan x and
(b)
(15 MINUTES)
Determine the general solutions of the following equations:
QUESTION 3:
(a)
(28 MARKS)
(14 MARKS)
(2)
(5 MINUTES)
(DoE papers pre-2014)
Solve for x if cos( x  30)  sin3x where x   60 ;120 .
Sketch the graphs of the following functions on the same set of axes for the
interval x   60 ;120 .
f ( x)  cos( x  30) and g ( x)  sin3x
(c)
(6)
Determine graphically the values of x for which cos( x  30)  sin3x
QUESTION 5:
(a)
(6)
(7 MARKS)
(5 MINUTES)
(2)
(DoE papers pre-2014)
Given: f ( x)  sin x
Sketch the graphs of the following on the same set of axes for the interval
x  0 ;180 :
(1)
g ( x)  f ( x)  3
(2)
© Gauteng Department of Education
12
(2)
h( x)  3 f ( x)  1
(2)
(b)
Determine graphically the value of x for which g ( x)  h( x)
(c)
Explain the transformations involved in the following function:
y  sin(2x  60)
(1)
(2)
QUESTION 6:
(9 MARKS)
(5 MINUTES)
(DoE papers pre-2014)
Two ships, A and B, are 120 km apart. Ship A is at a bearing of 67 from D and 97
km away from D. DN points due north. Ship B is at a bearing of 208 from D.
67
208
(a)
(b)
ˆ ,
Determine the bearing of Ship A from Ship B, i.e, MBA
when BM||DN .
(6)
If Ship A travels due north, and Ship B travels due south, then at some instant
of time, Ship A is due east of Ship B. Calculate the distance between the two
ships at that instant.
(3)
QUESTION 7:
(10 MARKS)
(10 MINUTES)
(Gauteng prelim 2015)
In the following diagram, DE represents a vertical cell phone tower positioned on one
corner of a field. The field is shaped as a cyclic quadrilateral EFGH and E, F, G and
H are all in the same horizontal plane. From H, the angle of elevation to D, the top of
the cell phone tower is α . EH  EF  p units and Ĝ  β .
© Gauteng Department of Education
13
α
β
(a)
Write down DE in terms of α
(b)
Show that:
(1)
(2)
(3)
1
Ĥ 2  β
2
(give reasons)
FH
1
2cos β
2
FH  p 2(1  cosβ)
p
(1)
(3)
(3)
(3)
SESSION NO:
11
(CONSOLIDATION)
TOPIC:
REVISION OF CALCULUS
SECTION A: TYPICAL EXAM QUESTIONS
QUESTION 1:
(21 MARKS)
(15 MINUTES)
(DoE papers pre-2014)
In this question first principles is revised as well as differentiation. Ensure correct
layout and notation. Remember that the gradient of the tangent to a curve at a point
is the derivative of the function at that point.
(a)
Determine f ( x) from first principles if f ( x)  4  x 2
© Gauteng Department of Education
(5)
14
(b)
Determine:
(1)
(2)
(3)
(c)
dy
8
if y  2 x 
dx
x
2
2p  3p  2
Dp 

p2


dy
if xy  5  x3
dx
(4)
(3)
(5)
Determine the value of k for which the gradient of the tangent to
g ( x)  3x2  kx  4 at the point where x  2 is 8.
(4)
QUESTION 2:
(17 MARKS)
(10 MINUTES)
(DoE papers pre-2014)
Curve sketching is revised in this question. Remember that at a turning point the
derivative is zero and at that the point itself can be substituted into the equation of
the function. Simultaneous equations can then be used to determine the value of the
unknowns. Determining the equation of a tangent is also revised.
Consider the function f ( x)  ax3  bx 2
The graph of f ( x)  ax3  bx 2 has a local turning point at (2 ;  4) .
(a)
Determine the value of a and b.
(6)
(b)
Sketch the graph of f on a set of axes.
(7)
(c)
Determine the equation of the tangent to f at x  1
(4)
QUESTION 3:
(12 MARKS)
(10 MINUTES)
(DoE papers pre-2014)
Problems involving area, perimeter, surface area and volume are quite important for
you to master. Spend time discussing these concepts in this question.
A builder wishes to construct a steel window frame in the shape of a rectangle with a
semi-circular part on top. The radius of the semi-circular part is r metres and the width
of the rectangular part is h metres.
© Gauteng Department of Education
15
(a)
(b)
Write down, in terms of h and r:
(1)
the steel perimeter (P) of the frame.
(2)
(2)
the area enclosed by the frame.
(2)
The area enclosed by the frame is to be
4


4 square metres. Show that the perimeter (P) is: P    2  r 
r
2

(c)
(4)
If the steel for the frame costs R10 per metre, calculate the value of r for which
the total cost of the steel will be a minimum.
(4)
QUESTION 4:
(2 MARKS)
(5 MINUTES)
(Level 4 challenge)
Redraw the given diagram of y  f ( x) .
Now draw a possible graph of y  f ( x)
on the same set of axes.
(2)
QUESTION 5:
(2 MARKS)
(5 MINUTES)
(DoE papers pre-2014)
A stone is thrown vertically upwards and its height (in metres) above the ground at
time t (in seconds) is given by:
h(t )  35  5t 2  30t
(a)
Determine its initial height above the ground.
(2)
(b)
Determine the initial speed with which it was thrown.
(2)
(c)
Determine the maximum height above the ground that the stone reached. (5)
(d)
How fast was the stone travelling when it reached a height of 60 metres
above the ground on the way down?
(5)
How fast was the stone travelling when it hit the ground?
(6)
(e)
© Gauteng Department of Education
16
QUESTION 6:
(13 MARKS)
(10 MINUTES)
(Level 4 challenges)
In the diagram alongside, the graph of y  f ( x) is represented where
f ( x)  ax3  bx2  cx represents a cubic function.
y  f ( x)
15
1
5
(a)
Determine the equation of f ( x) and f ( x) where f ( x)  ax3  bx2  cx
(b)
For which values of x will the graph of f ( x)
(c)
(8)
(1)
increase?
(1)
(2)
decrease?
(2)
At which values of x will the graph of f have its turning points?
(2)
QUESTION 7:
(13 MARKS)
(10 MINUTES)
(Level 4 challenges)
3
2
The graph of f ( x)  x  4 x  4 x  k passes through the origin, has a local
maximum at (b ; c) and a local minimum at (a ; 0) .
(a)
Explain why k  0
(b)
Using this value of k , determine the values of:
(c)
(d)
(1)
(1)
a
(4)
(2)
b
(4)
The graph of g with equation g ( x)  mx is the tangent to f at the point
(0 ; 0) . Calculate the value of m.
(2)
Determine the values of p for which x3  4 x2  4 x  2  p will only have one
negative solution.
(2)
© Gauteng Department of Education
17
QUESTION 8:
(10 MARKS)
(10 MINUTES)
(Level 4 challenges)
The profit (P) in rand/day which the owner of a taxi makes when the taxi is driven at
an average speed of x km/hour is given by the formula:
x

P  x 2  10  
12 

(a)
Calculate the average speed at which the owner of the taxi will suffer a loss
in revenue.
(4)
(b)
Calculate the most economical average speed of the taxi and the
corresponding daily profit.
(6)
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:
f ( x)
rate of change, derivative, slope
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
© Gauteng Department of Education
18
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
Rule 2
Rule 3
If f ( x)  axn , then f ( x)  a.nx n1
If f ( x)  ax, then f ( x)  a
If f ( x)  number, then f ( x)  0
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.
point will be a point of
f ( x) , equate it to 0 and
of the turning points and
© Gauteng Department of Education
19
Concavity:
The graph of f is concave up if f ( x)  0
The graph of f is concave down if f ( x)  0
SECTION C:
HOMEWORK QUESTIONS
QUESTION 1:
(a)
(b)
Given:
(13 MARKS)
(10 MINUTES)
(DoE papers pre-2014)
1
f ( x)  2  x 2
2
(1)
Determine the gradient of f at x  2 by using first principles.
(6)
(2)
Determine the equation of the tangent to f at x  2
(3)
Determine
dy
if
dx
y
6
3 6
x

1
6 x6
(4)
(Write your answer with positive exponents)
QUESTION 2:
(4 MARKS)
(5 MINUTES)
(DoE papers pre-2014)
3
2
The function g defined by g ( x)  ax  bx  cx  d has the following properties. Use
this information to draw a neat sketch graph of g without actually solving for a, b, c
and d.
g (0)  32, g (4)  0, g ( 2)  0, g (0)  0, g (4)  0
g ( x)  0 if x  0 or x  4
g ( x)  0 if 0  x  4
(4)
QUESTION 3:
(2 MARKS)
(5 MINUTES)
(DoE papers pre-2014)

The graph of y  g ( x ) is sketched below. This graph represents the derivative graph of
a quadratic function g.
y  g ( x )
1
2
© Gauteng Department of Education
20
(a)
(b)
Determine the values of x for which the graph of g decreases.
Write down the x-coordinate of the turning point of g.
(1)
(1)
QUESTION 4:
(10 MARKS)
(10 MINUTES)
(DoE papers pre-2014)
Consider the diagram below. OCDE is an inscribed rectangle in  ABO.
A(0; 6) and B(8;0) are given points.
A(0;6)
B(8;0)
(a)
Determine the equation of line AB.
(2)
(b)
By using calculus methods, determine the dimensions of rectangle OCDE
so that OCDE has a maximum area.
(5)
(c)
If rectangle OCDE is sketched such that EO = OC (thus forming a square),
determine the length of the square OCDE.
(3)
QUESTION 5:
(2 MARKS)
(5 MINUTES)
(DoE papers pre-2014)
The graph of y  g ( x ) is sketched below. This graph represents the derivative graph of
a quadratic function g.
y  g ( x )
1
2
(a)
Determine the values of x for which the graph of g decreases.
(1)
(b)
Write down the x-coordinate of the turning point of g.
(1)
© Gauteng Department of Education
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QUESTION 6
(2 MARKS)
(5 MINUTES)
(Level 4 challenges)
A is a town 30 km west of town B. Two athletes start walking simultaneously from the
two towns. The athlete who starts from town A, walks due east in the direction of B at
a constant speed of 6 km/h and reaches point P after x hours. The athlete, who starts
at B, walks due north in the direction of another town C at a constant speed of 8
km/h, and reaches point Q after x hours.
C
.
x hours
A
(a)
.
P
30km
x hours
Q
B
Prove that the distance between the athletes after x hours is given by:
PQ  100 x 2  360 x  900
(5)
(b)
How long after they started walking were they nearest to each other?
(3)
(c)
What was this minimum distance between them?
(2)
SESSION NO:
12
(CONSOLIDATION)
TOPIC:
REVISION OF ANALYTICAL GEOMETRY
SECTION A: TYPICAL EXAM QUESTIONS
QUESTION 1:
(18 MARKS)
(20 MINUTES)
(DoE papers pre-2014)
Completing the square to find the coordinates of the centre and radius is revised in
this question. Make sure that you know how to determine the coordinates of the
intercepts of a circle with the axes. The equation of a tangent to a circle is also
revised.
In the figure below, the tangent touches the circle x2  2 x  y 2  4 y  t  0 at
D(  2;  1) . A is the centre of the circle and the circle cuts the y-axis at B and C.
(a)
Determine the coordinates of A, the centre of the circle.
© Gauteng Department of Education
(4)
22
(b)
Express the radius of the circle in terms of t.
(1)
(c)
Show that t  5 .
(3)
(d)
Determine the coordinates of B and C.
(5)
(e)
Write down the length of BC.
(1)
(f)
Determine the equation of the tangent at D.
(4)
C
D(2;  1)
O
A
B
© Gauteng Department of Education
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QUESTION 2:
(23 MARKS)
(20 MINUTES)
(DoE papers pre-2014)
In this question, showing that a point lies on a line is revised. There is a link to
trigonometry and quadrilaterals.
A(  8 ; 2), B(  2 ;  6) and D(0 ; 8) are the vertices of a triangle that lies on the
circumference of a circle with diameter BD and centre M, as shown in the figure
below.
D(0 ; 8)
A(  8 ; 2)
M
O

B(  2 ;  6)
(a)
Calculate the coordinates of M.
(2)
(b)
Show that (  8 ; 2) lies on the line y  7 x  58
(2)
(c)
What is the relationship between the line y  7 x  58 and the circle centred
at M?
(5)
(d)
Calculate the lengths of AD and AB.
(4)
(e)
ˆ  90
Prove that DAB
(3)
(f)
Write down the size of angle 
(1)
(g)
A circle, centred at a point Z inside triangle ABD, is drawn to touch sides AB,
BD and DA at N, M and T respectively. Given that BMZN is a kite, calculate
the radius of this circle.
(6)
© Gauteng Department of Education
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QUESTION 3:
(23 MARKS)
(20 MINUTES)
(Level 4 challenges)
(a)
The lines ax  3 y  5 and 2 x  by  3 are parallel. Detemine ab .
(b)
A(1; 5) , B(3 ; 7) , C(8 ; 2) and D(x ; 0) are points in the Cartesian plane.
(4)
(1)
ˆ  90
Show that ABC
(2)
(2)
Show that the area of ABC  10
(5)
(3)
Determine the equation of AB.
(3)
(4)
If area of ABC  area of ABD , calculate the values of x.
(10)
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
or AB  ( xB  xA )2  ( yB  yA )2
The Midpoint Formula
 x  xB yA  yB 
M 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
25
The equation of the line
y  yA  m  x  xA 
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
mradius  mtangent  1
The radius is perpendicular to the tangent:
SECTION C:
QUESTION 1:
HOMEWORK QUESTIONS
(17 MARKS)
(15 MINUTES)
(DoE papers pre-2014)
O is the centre of the circle in the figure on the next page. P( x ; y) and Q (12 ; 5) are
two points on the circle. POQ is a straight line. The point R(t ; – 1) lies on the tangent
to the circle at Q.
Q(12 ; 5)
O
R(t ;  1)
P(x ; y)
(a)
Determine the equation of the circle.
© Gauteng Department of Education
(3)
26
(b)
Determine the equation of the straight line through P and Q.
(2)
(c)
Determine the coordinates of P.
(2)
(d)
Show that the gradient of QR is 
(e)
Determine the equation of the tangent QR in the form y  ...
(3)
(f)
Calculate the value of t.
(2)
(g)
Determine an equation of the circle with centre Q(12 ; 5) .
(3)
QUESTION 2:
(22 MARKS)
12
.
5
(2)
(15 MINUTES)
(DoE papers pre-2014)
Given below is a circle with centre M passing through R(3; 2), T(5; 4) and V on the
y-axis. Q(  2;  2) lies on the line RP. TV is drawn and then extended to meet RP at
P.
T(5;4)
R(  3;2)
Q(  2;  2)
(a)
Determine the equation of the circle with centre M.
(4)
(b)
Show that PR is a tangent to the circle at R.
(3)
(c)
Determine the equations of the lines:
(1)
(2)
PR
TP
(3)
(6)
(d)
If the equations of PR and TP are y  4 x  10 and y  x  1 respectively,
determine the coordinates of P.
(3)
(e)
Calculate the angle of inclination of the line TVP (refer to the angle as θ ).
© Gauteng Department of Education
(3)
27
QUESTION 3:
(a)
(13 MARKS)
(11 MINUTES)
(DoE papers pre-2014)
Determine the centre and radius of the circle with the equation defined by
x2  y 2  8x  4 y  38  0
(4)
A second circle has the equation ( x  4)2  ( y  6)2  26 . Calculate the
distance between the centres of the two circles.
(2)
(c)
Hence, show that the circles intersect each other.
(3)
(d)
Show that the circles intersect along the line y   x  4 .
(4)
(b)
QUESTION 4
(13 MARKS)
(11 MINUTES)
(DoE papers pre-2014)
The straight line through PQ with equation 2 x  y  1 cuts the y-axis at Q. The
straight line through PR with equation 3x  2 y  12 cuts the x-axis at R. The
coordinates of P are (2;3) .

P(2;3)
Determine:
(a)
the coordinates of Q and R.
(2)
(b)
the equation of QR.
(3)
(c)
the coordinates of point S, the midpoint of QR.
(2)
(d)
the equation of PS.
(1)
© Gauteng Department of Education
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(e)
the size of angle  rounded off to the nearest degree.
(6)
(f)
the area of PQR rounded off to two decimal places.
(6)
QUESTION 5:
(9 MARKS)
(10 MINUTES)
(Level 4 challenge)
The coordinates of trapezium ABCD are A(4 ; 3), B( x ; 6), C(4 ; y ) and D(2 ;  1)
AD||BC and BC  2AD . Determine the values of x and y.
(9)
B(x ; 6)
A(  4 ; 3)
C(4 ; y )
D(  2 ;  1)
© Gauteng Department of Education
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SESSION NO:
13 (CONSOLIDATION)
TOPIC:
REVISION OF ALGEBRA
SECTION A: TYPICAL EXAM QUESTIONS
QUESTION 1:
(34 MARKS)
(25 MINUTES)
(DoE papers pre-2014)
This question revises Grade 11 Algebra including algebraic fractions, the quadratic
formula, quadratic inequalities, surd equations, simultaneous equations, the concept
of real and non-real and the nature of roots.
(a)
Solve for x rounded off to two decimal places where necessary:
x2  x  2
1  0
(1)
x 1
(4)
(2) 5x( x  3)  2
(4)
(3) 20  ( x  1)( x  2)
(4)
(4)
(b)
2  7 x  2x  0
(5)
Solve simultaneously for x and y in the following set of equations:
(1)
y  2 x  2 and
2 x2  2  y 2
(2)
(3 x  y ) 2  ( x  5) 2  0
(6)
(4)
4 x
be non-real?
x6
(c)
For which value(s) of x will the expression
(d)
Show that the roots of the equation 5 p 2  x( x  3 p) are non-real if
p  0.
(3)
(4)
QUESTION 2:
(20 MARKS)
(20 MINUTES)
(DoE papers pre-2014)
Simplifying exponential and surd expressions and solving exponential equations is
dealt with in this question. The use of the k-method is ideal for 2(b).
(a)
Simplify the following without the use of a calculator:
(1)
3
81 .
 3
3
4
(4)
© Gauteng Department of Education
30
3
(2)
(b)
16  3 54  3 128
3
(5)
128
Solve for x:
32 x1  4.3x2  81  0
(1)
(5)
1
x  x4  2  0
(2)
(6)
2
x3  4
(3)
QUESTION 3:
(10 MARKS)
(a)
Show that
a2
(b)
Determine the values of p for which the equation 3  1  p will have real
solutions.
(3)
(c)
For which real values of x will
(d)
If r  0 and p  0 and the equation p( x  2) x  rx( x  3)   p has equal
1
can be written as
a
(Level 4 challenges)
a (a  1)
a
( x  3)2 be a real number?
16 p
and hence determine the equal roots.
9
Exponential and surd laws
Laws
a . a  a m n
m
m
2.
3.
(3)
(8)
If P  x 2  5 x  6 and x 2  5 x  6  0 , determine the possible values for P. (5)
SECTION B: NOTES ON CONTENT
1.
(4)
x
roots prove that r 
(e)
(15 MINUTES)
n
a
 a mn
n
a
(a m ) n  a mn  (a n ) m
1.
Definitions
a 1
2.
1a  1 , with aR
3.
xn 
0
1
xn
© Gauteng Department of Education
31
4.
 ab m  ambm
5.
am
a
   m
b
b
4.
1
 xn
n
x
m
5.
n
a
m

m
an
Rules for exponential and surd equations
(a)
1
even
number
x
(b)
Whenever you have equations of the form x n  number , always rewrite the
 negative number
m
m
1
expression x n in the form ( x n ) m .
A  negative number
(c)
Quadratic formula and nature of roots
x
b  b 2  4ac
2a
b 2  4ac
Roots
0
0
Non-real
Real
0
Real, rational and unequal
Real, irrational and unequal
Real, rational and equal
  0 and   perfect square
  0 and   a perfect square
SECTION C:
HOMEWORK QUESTIONS
QUESTION 1:
(41 MARKS)
(30 MINUTES)
(DoE papers pre-2014)
(a) Solve for x rounded off to two decimal places where necessary:
(1)
x2  4
 3x
x2
(4)
(2)
( x  3)( x  2)  8
(4)
(3)
7 x 2  18 x  9  0
(4)
(4)
5  x 1 x  0
(5)
© Gauteng Department of Education
32
(b)
Solve for x and y simultaneously:
2x  y  7
(1)
(2)
(c)
x2  xy  21  y 2
(7)
( x 2  8 x  16)  ( y 2  2 y  1)  0
(6)
2  a and
If
3  b , express the following in terms of a and b:
108  18
(d)
(e)
(3)
For which values of p will the equation x(4 x  3)   p have equal
roots?
Show that the roots of ( k  1) x  kx  1  0 are non-real for all real
2
2
values of k.
QUESTION 2:
(a)
(4)
(23 MARKS)
9 x 1  6.32 x
Show that:
 3
4 x 1
(15 MINUTES)
(DoE papers pre-2014)
 3 for all values of x.
(5)
10 2 x 20  7 8 x 20
(b)
Simplify:
(c)
Solve for x:
(5)
18 x 20
3x 1  3x 1  8
(1)
(3)
2
3
6 x  54
(2)
(d)
(4)
(3)
If 4 . 6  48 , then find the value of x  y
x
y
12
QUESTION 3:
(28 MARKS)
(20 MINUTES)
(4)
, then find the value of b in terms of a
(4)
Without using a calculator, show that:
(b)
If a  1  2 and b  1  2
n
(Level 4 challenges)
4  8  16  648  250
(a)
n
(7)
24
16
12
© Gauteng Department of Education
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If x  3  2 2  3  2 2 , then determine the value of x without using a
calculator.
(6)
(d)
Solve for x x 2  2 x  2  0 leaving answers in surd form.
(4)
(e)
If ( x  y  1)( x  y  3)  0 and x  y  3 , find the value of x  y
(3)
(f)
For which real values of k will
(c)
1
 k have real roots?
x 1
2
SESSION NO:
14
TOPIC:
REVISION OF FUNCTIONS
(7)
(CONSOLIDATION)
SECTION A: TYPICAL EXAM QUESTIONS
QUESTION 1:
(14 MARKS)
(10 MINUTES)
(DoE papers pre-2014)
In this question, the focus is on revising quadratic functions. Revise the two different
forms of a quadratic function and how to determine the coordinates of the turning
point for each form.
(a)
(b)
Consider the function
f ( x)  ( x  1)2  4
(1)
Draw a neat sketch graph indicating the coordinates of the intercepts
with the axes, the coordinates of the turning point and the equation of
the axis of symmetry.
(6)
(2)
Write down the range.
(2)
(3)
Determine the values of x for which the graph increases.
(1)
Consider the function g ( x)   x2  4 x  3
(1)
Determine the maximum value of g.
(2)
Determine the coordinates of the turning point of the graph of
g ( x  1)
(3)
(2)
© Gauteng Department of Education
34
QUESTION 2:
(13 MARKS)
(10 MINUTES)
(DoE papers pre-2014)
In this question, you are required to determine the equation of a quadratic and
hyperbolic function. This is extremely important for Paper 1.
In the diagram below, the graphs of the following functions are represented:
a
q
x p
f ( x)  a( x  p)2  q and g ( x) 
The turning point of f is (1; 4) and the asymptotes of g intersect at the turning
point of f. Both graphs cut the y-axis at 3.
(1 ; 4)
3
(a)
Determine the equation of f.
(4)
(b)
Determine the equation of g.
(4)
(c)
Determine the coordinates of the x-intercept of g.
(3)
(d)
For which values of x will g ( x)  0 ?
(2)
© Gauteng Department of Education
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QUESTION 3:
(19 MARKS)
(15 MINUTES)
(DoE Nov 2015)
This is quite a challenging question with some level 4 type questions.
Given:
h( x)  2x  3 for 2  x  4 and the x-intercept of the graph is Q.
2
4
P( x ; y)
3
(a)
Determine the coordinates of Q.
(2)
(b)
Write down the domain of h 1
(3)
(c)
Sketch the graph of h 1 , clearly indicating the intercepts and endpoints.
(3)
(d)
For which value(s) of x will h( x)  h1( x) ?
(3)
(e)
P(x ; y) is the point on the graph of h that is closest to the origin. Calculate the
distance OP.
(5)
(f)
Given: h( x)  f ( x) where f is a function defined for 2  x  4
(1)
Explain why f has a local minimum.
(2)
Write down the value of the maximum gradient of the tangent to the
graph of f.
(1)
© Gauteng Department of Education
(2)
36
QUESTION 4:
(4 MARKS)
The function defined as y 
The domain is x 
(5 MINUTES)
(DoE Nov 2015 level 4)
a
 q has the following properties:
x p
where x  2
y  x  6 is an axis of symmetry
The function is increasing for all x 
where x  2
Draw a neat sketch graph showing asymptotes, if any.
(4)
QUESTION 5:
(6 MARKS)
(5 MINUTES)
(Gauteng prelim 2015)
The straight line y  mx  32 is a tangent to the graph of y  2 x2  4 x  30 .
Calculate the possible values of m.
(6)
QUESTION 6:
(7 MARKS)
(10 MINUTES)
(Gauteng prelim 2015)
2
The sketch below shows the graphs of f ( x)  2 x  5x  3 and g ( x)  ax  q
The angle of inclination of graph g is 135 in the direction of the positive x-axis.
P is the tangent point of f and g such that g is a tangent to the graph of f at P.
135
(a)
Calculate the coordinates of the point of contact between the two graphs.
(4)
(b)
Hence, or otherwise, determine the equation of g
(2)
(c)
Determine the values of d for which the line y   x  d will not intersect the
graph of f.
(1)
© Gauteng Department of Education
37
QUESTION 7:
(12 MARKS)
(10 MINUTES)
(Exemplar 2014)
In this question, the focus is on exponential functions. It is important for you to
understand the concept of reflection about the axes as well as translating graphs
horizontally and vertically (see (c) and (d)).
In the diagram below (not drawn to scale), the graph of f ( x)  2a x .
The graph of f passes through the point (1; 4) and cuts the y-axis at A.
( 1 ; 4)
1
and hence write down the equation of f.
2
(a)
Show that a 
(b)
Determine the coordinates of A.
(c)
Show that the equation of g, the reflection of f about the y-axis, can
(2)
be written as g ( x)  2 x 1
(d)
(3)
(3)
Draw a neat sketch graph of y  f ( x  1)  2 indicating the intercepts with the
axes as well as the equation of the asymptote.
(4)
QUESTION 8:
(10 MARKS)
(5 MINUTES)
(DoE papers pre-2014)
This question revises exponential and logarithmic functions.


The graph of f ( x)  a x 1 passes through the point  0 ;
1
.
2
(a)
Calculate the value of a.
(2)
(b)
Write down the equation of the inverse of f in the form f 1 ( x)  ....
(2)
(c)
Sketch the graph of f and its inverse on the same set of axes.
(5)
(d)
Write down the domain of f 1
(1)
© Gauteng Department of Education
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SECTION B: NOTES ON CONTENT
QUADRATIC FUNCTIONS
(a)
The general equation for the quadratic function is y  a( x  p) 2  q .
(remember that the exponent of x is always squared).
(b)
Draw the “mother graph” y  ax 2 by using a table of selected x-values.
(c)
The value of a tells us if the graph is concave (happy) or convex (sad).
(d)
If a  0 , the shape is concave
If a  0 , the shape is convex.
2
The graph of y  a( x  p)  q is obtained by shifting the graph of y  ax 2 by p
units to the left or right and then q units up or down.
If p  0 , the shift is left
If p  0 , the shift is right
If q  0 , the shift is upwards
If q  0 , the shift is downwards




(e)
For graphs of the form y  ax 2  q , the value of p is 0 meaning that the mother
graph will shift up or down but not left or right.
The equation of the axis of symmetry of the graph y  a( x  p) 2  q is obtained by
putting the expression x  p  0 and solving for x.
The equation of the axis of symmetry for the graph of a parabola of the form
y  ax 2  q is x  0 (the y-axis). The axis of symmetry passes through the x-
(f)
coordinate of turning point of the parabola. The value of q is the y-coordinate of the
turning point.
To determine the coordinates of the turning point of parabolas of the form
y  ax 2  bx  c , express y  ax 2  bx  c in the form y  a( x  p)2  q by
completing the square or use the formula x  
(g)
b
to find the x-coordinate and
2a
hence substitute this x-value into the original equation to find the y-coordinate.
The y-intercept can be determined by letting x  0 in the equation of
y  a( x  p)2  q .
The y-intercept of the graph of y  ax 2  q is the value of q.
(h)
If the parabola cuts the x-axis, the x-intercepts can be determined by letting
(j)
y  0 in the equation of y  a( x  p)2  q and solving for x.
If a  0 the parabola has a minimum value at q.
If a  0 the parabola has a maximum value at q.
If a  0 then the graph of the parabola decreases for all x  p and increases for all
x p.
If a  0 then the graph of the parabola increases for all x  p and decreases for all
x p .
(k)
For any quadratic function of the form y  ax 2  q or y  a( x  p) 2  q :
(i)
Domain:
x    ;  
© Gauteng Department of Education
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Range:
y q ; 
y    ; q
if a  0
if a  0
HYPERBOLIC FUNCTIONS
(a)
(b)
(b)
(c)
(d)
The general equation for the hyperbola is y 
(remember that x is in the denominator).
a
The hyperbola y   q has two asymptotes:
x
Vertical asymptote: x  0 (y-axis)
Horizontal asymptote:
a
The hyperbola y 
 q has two asymptotes:
x p
Vertical asymptote: x  p  0
Horizontal asymptote:
Shape:
a0
a0
(g)
yq
yq
(The dotted lines are the asymptotes)
a
Start by drawing the graph of y  by using a table of selected x-values (preferably
x
two positive and two negative factors of a).
a
a
The graph of y   q is obtained by shifting the graph of y 
q units up or down.
x
x
a
If q  0 , then the graph of y  shifts q units upwards.
x
a
If q  0 , then the graph of y  shifts q units downwards.
x
a
a
 q is obtained by shifting the graph of y 
The graph of y 
p units left or
x p
x
right and then q units up or down.
If p  0 , the shift is left
If q  0 , the shift is upwards
(e)
(f)
a
q
x p
If p  0 , the shift is right
If q  0 , the shift is downwards
The y-intercept can be determined by letting x  0 .
The x-intercept can be determined by letting y  0 .
a
For any hyperbolic function of the form y   q :
x
Domain: x    ;   where x  0
Range: y    ;   where y  q
For any hyperbolic function of the form y 
a
q:
x p
Domain: x    ;   where x  p  0
Range: y    ;   where y  q
© Gauteng Department of Education
40
EXPONENTIAL FUNCTIONS
(a)
The general equation for the exponential function is y  ab x  p  q where b  0 and
(b)
b  1 (remember that the exponent is x).
The exponential graph has one asymptote:
Horizontal asymptote: y  q
(c)
Draw the basic graph y  ab x by using a table of selected x-values
(best to use x 1; 0 ;1 ).
(d)
The graph of y  ab x  p  q is formed by shifting the graph of the function y  ab x
p units horizontally (left or right) and then q units vertically (up or down).
If p  0 , the shift is left
If q  0 , the shift is right
If q  0 , the shift is upwards
If q  0 , the shift is downwards
(e)
Determine the x-intercept of the graph of y  ab x  p  q by letting y  0 .
(f)
(g)
The y-intercept can be determined by letting x  0 in the equation of y  ab x  p  q .
For any exponential function of the form:
y  ab x  p  q [ a  0, b  0 and b  1 ]
x    ;  
Domain:
y q ; 
Range:
Summary of the exponential and logarithmic functions
Increasing functions
Decreasing functions
f ( x)  a x
a 1
yx
f ( x)  a x
0  a 1
f 1 ( x )  log a x
(0 ;1)
a 1
yx
(0 ;1)
(1;0)
(1;0)
f 1 ( x )  log a x
0  a 1
Some important theory:
The expression log a x is defined only if:
0  a  1 or a  1
(a cannot be negative, 0 or 1)
x  0 (x cannot be negative or 0)
© Gauteng Department of Education
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SECTION C:
QUESTION 1:
HOMEWORK QUESTIONS
(10 MARKS)
(5 MINUTES)
(DoE papers pre-2014)
2
x
Functions y  f ( x )  bx and y  g ( x)  a are not sketched according to scale.
9
2
8
(a)
Determine the values of a and b.
(b)
Write down a possible restriction for the domain of f ( x) so that the inverse
of f will be a function.
(1)
(c)
Hence determine the equation of the inverse function of f ( x) .
(3)
(d)
Write down the equation of the inverse of g ( x) in the form g 1( x)  .....
(2)
(e)
If h( x)  g ( x  2)  3 , state the transformation that was applied to g ( x) to
form h( x) and hence or otherwise determine the equation of h( x)
(3)
QUESTION 2:
Given:
(3)
f ( x) 
(23 MARKS)
(15 MINUTES)
(DoE papers pre-2014)
2
x 1
(a)
Write down the equations of the asymptotes.
(2)
(b)
Sketch the graph of f indicating the coordinates of the y-intercept
as well as the asymptotes.
(6)
© Gauteng Department of Education
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(c)
Determine graphically the values of x for which
(d)
State the domain of f.
2
1
x 1
(2)
(2)
QUESTION 3:
(23 MARKS)
(15 MINUTES)
(DoE papers pre-2014)
The straight line f ( x)  4 x  32 passes through the turning point of the parabola
with equation g ( x)  a( x  p) 2  q . The line and parabola intersect at the point
(7 ; 4) .
(7 ; 4)
(a)
Determine the coordinates of D.
(2)
(b)
Determine the equation of g.
(4)
(c)
State the range of g.
(1)
QUESTION 4:
(6 MARKS)
(5 MINUTES)
(DoE Exemplar 2014)
x
An increasing exponential function with equation f ( x)  a . b  q has the following
properties:
Range is y  3
The points (0 ;  2) and (1;  1) lie on the graph
(a)
Determine the equation that defines f
(4)
(b)
Describe the transformation from f ( x) to h( x)  2 . 2 x  1
(2)
© Gauteng Department of Education
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SESSION NO:
15
(CONSOLIDATION)
TOPIC:
REVISION OF GRADE 11 EUCLIDEAN GEOMETRY
SECTION A: TYPICAL EXAM QUESTIONS
QUESTION 1:
(a)
(b)
(10 MARKS)
(10 MINUTES)
(DoE papers pre-2014)
Complete:
(1)
The angle in a semi-circle is …….
(1)
(2)
The perpendicular from the centre of a circle to a chord ……..
(1)
In the diagram below, AOB is a diameter of the circle centre O.
OD||BC. The length of the radius is 10 units.
(1)
What is the size of Ĉ ? State a reason.
(2)
(2)
What is the size of Ê1 ? State a reason.
(2)
(3)
Why is AE  EC ? State a reason.
(1)
(4)
If AC  16 units, calculate the length of ED.
(3)
© Gauteng Department of Education
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QUESTION 2:
(15 MARKS)
(10 MINUTES)
(DoE papers pre-2014)
In the figure below, RDS is a tangent to
the circle centre O at D.

BC = DC and C D S = 40°
40
(a)
What is the size of B̂1 . State a reason.
(2)
(b)
What is the size of D̂ 2 . State a reason.
(2)
(c)
(d)
What is the size of Ĉ . State a reason.
Calculate the size of Ô 2 State a reason.
(2)
(2)
(e)
Calculate the size of Ô1 . State a reason.
(2)
(f)
Calculate the size of D̂3 State reasons.
(3)
(g)
Calculate the size of  State a reason.
(2)
QUESTION 3:
(9 MARKS)
(5 MINUTES)
(DoE papers pre-2014)
In the diagram, O is the centre of the circle
passing through A, B, C and D.
AB||CD and B̂  20
20
(a)
Calculate the size of Ĉ1 ? State a reason.
(2)
(b)
Calculate the size of Ô1 ? State a reason.
(2)
(c)
(d)
(e)
Calculate the size of D̂ ? State a reason.
Calculate the size of Ê1 ? State a reason.
Why is AOEC a cyclic quadrilateral?
(2)
(2)
(1)
© Gauteng Department of Education
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QUESTION 4:
(9 MARKS)
(5 MINUTES)
(DoE papers pre-2014)
LOM is a diameter of circle LMT. The centre is O. TN is a tangent at T. LN  NP .
MT is a chord. LT is a chord produced to P.
L
.
O
M
1
2
N
1
2
3
4
T
2
1
P
Prove that:
(a)
MNPT is a cyclic quadrilateral
(4)
(b)
NP = NT
(5)
QUESTION 5:
(9 MARKS)
(5 MINUTES)
(DoE papers pre-2014)
In the diagram below, AB is a diameter of the circle ABCD. AE is a tangent to the
circle at A. B̂1  x .
(a)
Prove that AB is a tangent to the circle through A, D and E.
(7)
(b)
ˆ  Eˆ
Prove that C
1
(2)
© Gauteng Department of Education
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QUESTION 6:
(4 MARKS)
(5 MINUTES)
(DoE Nov 2014)
In the diagram, AB and AE are tangents to the circle at B and E respectively. BC is a
diameter of the circle. AC  13, AE  x and BC  x  7 .
Calculate the length of AB.
(4)
x
13
x7
QUESTION 7:
(9 MARKS)
(10 MINUTES)
(Level 4 challenge)
ˆ . T lies on
PR is a common chord of circles PQSR and PXRY. PR bisects XRY
chord PS such that ST  SR  SQ . MPY touches the larger circle at P.
Prove that:
(a)
ˆ Q
ˆ
Q
1
2
(6)
(b)
QP is a tangent to the smaller circle.
(7)
© Gauteng Department of Education
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SECTION B: NOTES ON CONTENT
REVISION OF GRADE 11 THEOREMS
THEOREM 1
The line segment joining the centre of a circle
to the midpoint of a chord is perpendicular to the chord.
If
.
O
AM  MB then OM  AB which means that
ˆ M
ˆ  90
M
1
2
A
||
1 2 ||
M
B
THEOREM 1 CONVERSE
The perpendicular drawn from the centre of a
circle to a chord bisects the chord.
If OM  AB then
.
O
AM  MB
A
THEOREM 2
||
1 2 ||
M
B
The angle which an arc of a circle subtends at the centre of a circle is twice the angle
it subtends at the circumference of the circle.
C
C
x
x
C
.
.
O
1
2x
A
B
A
A
1
.
2x
2x
O
x
B
O
1
B
ˆ  2C
ˆ
For all three diagrams: O
1
C
THEOREM 3
The angle subtended at the circle by a diameter is a A
right angle. We say that the angle in a semi-circle is
90 .
In the diagram Ĉ  90
© Gauteng Department of Education
.
B
48
C
THEOREM 3 CONVERSE
If the angle subtended by a chord at a point on the circle is
90 , then the chord is a diameter.
If Ĉ  90 , then the chord subtending Ĉ is a diameter.
.
A
B
THEOREM 4
D
An arc or line segment of a circle subtends equal
angles at the circumference of the circle. We say
that the angles in the same segment of the circle
are equal.

.
ˆ B
ˆ
ˆ and Cˆ  D
In the diagram, A

C
A
B
THEOREM 5
D

The opposite angles of a cyclic quadrilateral are
supplementary (add up to 180 )
.
ˆ C
ˆ  180 and B
ˆ D
ˆ  180
In the diagram, A
A
180   C

180  
B
THEOREM 5 CONVERSE
If the opposite angles of a quadrilateral are supplementary, then the quadrilateral is a
cyclic quadrilateral.
D
D


.
C
A
180  
B
A
C
180  
B
© Gauteng Department of Education
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THEOREM 6
D
An exterior angle of a cyclic quadrilateral is equal to
the interior opposite angle.
1
.
If ABCD is a cyclic quadrilateral, then Ĉ1  A .
C
A
B
THEOREM 6 CONVERSE
If an exterior angle of a quadrilateral is equal to the interior opposite angle, then the
quadrilateral is a cyclic quadrilateral.
D
D
1
1
.
C
C
A
A
B
B
THEOREM 7
A tangent to a circle is perpendicular to the
radius at the point of contact.
.
O
If ABC is a tangent to the circle at the point B,
ˆ B
ˆ  90 .
then the radius OB  ABC , i.e. B
1
2
A
.
1 2
B
C
THEOREM 7 CONVERSE
If a line is drawn perpendicular to a radius at the
point where the radius meets the circle, then the line
is a tangent to the circle.
If line ABC  OB and if OB is a radius, then
ABC is a tangent to the circle at B.
.
O
A
© Gauteng Department of Education
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1 2
B
C
50
THEOREM 8
B
1
If two tangents are drawn
from the same point outside
a circle, then they are equal in
length.
||
A
.
O
||
1
C
THEOREM 9
The angle between a tangent to a circle and a chord drawn from the point of contact
in equal to an angle in the alternate segment.
E
E
.
1
.
D
2
A
C
B
D
1 2
B
A
C
ˆ  Eˆ and B
ˆ D
ˆ.
In both diagrams, B
2
1
THEOREM 9 CONVERSE
If a line is drawn through the endpoint of a chord, making with the chord an angle
equal to an angle in the alternate segment, then the line is a tangent to the circle.
E
E
D
1
A
.
D
2
B
C
1
A
2
B
C
ˆ  Eˆ or if B
ˆ D
ˆ , then ABC is a tangent to the circle passing through the points
If B
2
1
B, D and E.
© Gauteng Department of Education
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How to prove that a quadrilateral is cyclic
ABCD will be a cyclic quadrilateral if one of the following conditions is satisfied.
Condition 1
ˆ A
ˆ )  (C
ˆ C
ˆ )  180 or (B
ˆ B
ˆ )  (D
ˆ D
ˆ )  180
(A
1
2
2
3
1
2
1
2
Condition 2
ˆ A
ˆ A
ˆ
C
1
1
2
Condition 3
ˆ or D̂  A
ˆ B
ˆ or B̂  A
ˆ
ˆ or C
D̂1  C
3
2
1
2
1
2
2
How to prove that a line is a tangent to a circle
ABC would be a tangent to the “imaginary” circle drawn through
ˆ  Eˆ
EBD if B
1
PROOFS OF THEOREMS REQUIRED FOR EXAMS
THEOREM 1
The line drawn from the centre of a circle perpendicular to a chord bisects the chord.
Proof
Join OA and OB.
In OAM and OBM :
(a)
(b)
(c)
OA = OB
ˆ M
ˆ  90
M
1
2
OM = OM
OAM  OBM
radii
given
common
RHS
 AM = MB
© Gauteng Department of Education
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THEOREM 2
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).
Proof
Join CO and produce.
For all three diagrams:
ˆ C
ˆ A
ˆ
ext  of OAC
O
1
1
ˆ A
ˆ
OA = OC , radii
But C
1
ˆ  2C
ˆ
O
1
1
ˆ  2C
ˆ
Similarly, in OCB O
2
2
ˆ
ˆ
ˆ
ˆ
O  O  2C  2C
1
2
1
2
ˆ O
ˆ  2(C
ˆ C
ˆ )
O
1
2
1
2
ˆ
ˆ
 AOB  2ACB
For the third diagram:
ˆ O
ˆ  2C
ˆ  2C
ˆ
O
2
1
2
1
ˆ
ˆ
ˆ
ˆ
O  O  2(C  C )
2
1
2
1
ˆ  2ACB
ˆ
 AOB
THEOREM 5
The opposite angles of a cyclic quadrilateral are supplementary (add up to 180 ).
Proof
Join AO and OC.
ˆ  2D
ˆ
O
1
ˆ  2B
ˆ
O
2
 at centre = 2  at circ
 at centre = 2  at circ
ˆ O
ˆ  2D
ˆ  2B
ˆ
O
1
2
ˆ O
ˆ  360  's round a point
and O
1
2
ˆ  B)
ˆ
 360  2(D
ˆ B
ˆ
180  D
ˆ C
ˆ  180
Similarly, by joining BO and DO, it can be proven that A
© Gauteng Department of Education
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THEOREM 9
The angle between a tangent to a circle and a chord drawn from the point of contact
is equal to the angle in the alternate segment.
Proof
Draw diameter BOF and join EF
ˆ B
ˆ  90
B
tan  rad
1
2
 in a semi-circle
Eˆ  Eˆ  90
1
2
ˆ  Eˆ
But B
1
1
ˆ  Eˆ
B
2
arc FD subt =  's
2
ˆ  BED
ˆ
CBD
Draw diameter BOF and join FD
B̂1  90
tan  rad
D̂1  90
ˆ D
ˆ
B
2
 in a semi-circle
arc FE subt =  's
2
ˆ D
ˆ  90 and B
ˆ D
ˆ
Now B
1
1
2
2
ˆ
ˆ
ˆ
ˆ
B  B  D  D
1
2
1
2
ˆ  BDE
ˆ
ABE
SECTION C:
HOMEWORK QUESTIONS
QUESTION 1:
(10 MARKS)
(5 MINUTES)
(DoE papers pre-2014)
In the diagram below, M is the centre of the circle. D, E, F and G are points
on the circle. If F̂1  10 and D̂2  50 , calculate, with reasons, the size of:
(a)
D̂1
D
(2)
1 2
50
(b)
M̂1
(2)
(c)
F̂2
(2)
(d)
Ĝ
(2)
.
M
1 2
E
(e)
Ê1
1
2
(2)
10
2
1
F
© Gauteng Department of Education
G
54
QUESTION 2:
(17 MARKS)
(10 MINUTES)
(DoE papers pre-2014)
In the diagram below, QP is a tangent to a circle with centre O. RS is a diameter of
ˆ and
the circle and RQ is a straight line. T is a point on the circle. PS bisects TPQ
ˆ  22 .
SPQ
Calculate the following, giving reasons:
(a)
P̂2
(2)
(b)
R̂ 2
(2)
(c)
Pˆ3  Pˆ4
(3)
(d)
R̂1
(4)
(e)
Ô1
(3)
(f)
Q̂ 2
(3)
QUESTION 3:
(12 MARKS)
(10 MINUTES)
(DoE papers pre-2014)
ALB is a tangent to circle LMNP. ALB||MP.
Prove that:
(a)
LM = LP
(4)
(b)
ˆ
LN bisects MNP
(4)
(c)
LM is a tangent to circle MNQ
(4)
© Gauteng Department of Education
55
QUESTION 4:
(16 MARKS)
(10 MINUTES)
(DoE papers pre-2014)
EC is a diameter of circle DEC. EC is produced to B. BD is a tangent at D. ED is
produced to A and AB  BE .
Prove that:
(a)
ABCD is a cyclic quadrilateral.
(4)
(b)
ˆ  Eˆ
A
1
(3)
(c)
BD  BA
(5)
(d)
ˆ C
ˆ
C
2
3
(4)
QUESTION 5:
(16 MARKS)
(10 MINUTES)
(Level 4 challenge)
In the diagram, PA is a tangent to the circle at A. AC and AB are chords and PM is
produced to K such that AK  AM . K and M lie on AC and AB respectively. Chord
CB is produced to P.
ˆ
Prove that KP bisects APC
(7)
© Gauteng Department of Education