1
SECONDARY SCHOOL IMPROVEMENT
PROGRAMME (SSIP) 2016
GRADE 12
SUBJECT:
MATHEMATICS
LEARNER SOLUTIONS
(Page 1 of 61)
© Gauteng Department of Education
2
SESSION NO:
10
TOPIC:
REVISION OF TRIGONOMETRY (GRADE 11 AND 12)
SECTION A:
1(a)
1(b)
(CONSOLIDATION)
SOLUTIONS TO HOMEWORK QUESTIONS
cos(140) cos 740  sin140 sin(20)
 cos140 cos 20  (sin 40)( sin 20)
 ( cos 40)(cos 20)  sin 40 sin 20
  cos 40 cos 20  sin 40 sin 20
 (cos 40 cos 20  sin 40 sin 20)
1
  cos 60  
2
2
cos 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)

2(a)(1)
2(a)(2)
cos 2 15  sin 2 15
sin 2 50  (cos 50)(cos 50)
cos 2 15  sin 2 15
 2
sin 50  cos 2 50
cos 2(15)

1
3
 cos 30 
2
cos 2 x  1
 2 x  180  k 360
 x  90  k180 where k 
2sin x  cos x  0
 2sin x   cos x
sin x
1


cos x
2
 tan x  0,5
x  26, 6  k180 where k 





 cos 40
cos 20
sin 40
 sin 20
 cos60
 







1
2
(6)
cos2 15
cos2 75
sin 2 50
cos2 50
cos30
1
3
2
(7)
 2x  180  k 360
 x  90  k180
 k
(3)
sin x
1

cos x
2
 tan x  0,5
 x  26, 6  k180
 k
(4)

© Gauteng Department of Education
3
2(a)(3)
2sin 2 x  2sin x cos x  0
2(a)(4)
 sin 2 x  sin x cos x  0
 sin x(sin x  cos x)  0
 sin x  0
or sin x   cos x
sin x
 x  0  k .360
 1
cos x
or x  180  k .360
tan x  1
where k 
x  45  k .180
2
4 cos x  2sin x cos x  sin 2 x  cos 2 x
2(b)
3cos 2 x  2sin x cos x  sin 2 x  0
(3cos x  sin x)(cos x  sin x)  0
3cos x  sin x or cos x   sin x
sin x
sin x
3
or
1 
cos x
cos x
tan x  3
or
 1  tan x
x  71, 6  k .180 or x  45  k .180
where k 
sin    30   0, 44















sin x(sin x  cos x)  0
sin x  0
tan x  1
x  0  k.360
x  180  k.360
x  45  k.360
k
1  sin 2 x  cos2 x
standard form
factorising
tan x  3
tan x  1
x  71, 2  k .180
x  45  k.180
k
 176,10;303,90;536,10;663,90
3(a)
y
2
1
x
–90
–45
45
(8)
   56,10  k .360
   176,10  k .360
  answers
(6)
  30  26,10388114  k .360
  56,10  k .360 or
  30  206,1038811  k.360
  176,10  k .360 where k 
3
(7)
90
135
–1
–2
–3
© Gauteng Department of Education
180
4
3(b)
4(a)
4(b)
For f
asymptotes
shape
For g
 shape
 range
 0  x  90
 x  180
sin 2 x   tan x
 0  x  90 or x  180
cos( x  30)  sin 3 x
 cos( x  30)  cos(90  3 x)








 x  30  90  3 x  k 360 x  30  360  (90  3 x)
 4 x  120  k 360
 x  30  270  3 x  k 360
 x  30  k 90
2 x  300  k 360
 x  150  k180
x  30 or
x  120 or
x  60
see diagram below
2
f
(4)
(2)
cos(90  3 x)
4x  120  k 360
2x  300  k 360
x  30  k 90
x  150  k180
x  30
x  120
(8)
x  60
f ( x)  cos( x  30) :
 shift of 30 right
 amplitude
 range
g ( x)  sin 3 x :
 period of 120
 amplitude
 intercepts
(6)
y
1
x
–60
–30
30
60
90
120
g
–1
–2
4(c)
cos( x  30)  sin 3x
60  x  120 where x  30
 60  x  120
 x  30
OR x   60 ;30   30
© Gauteng Department of Education
(2)
5
5(a)
2 y
1
x
g
90
180
–1
–2
h
–3
–4
5(b)
 shape
 shape
x  90
5(c)
y  sin(2 x  60)
For g:
For h:
 domain and range
 domain and range
 x  90
(1)
 180 period shift of
 shift of 30 to the right
 y  sin  2( x  30) 
(4)
(2)
The graph of f ( x)  sin x has changed to a
period of 180 and then shifted 30 to the
right.
6(a)
ˆ  360  208  152
NDB
ˆ  28
 MBD
ˆ  208  67  141
BDA
ˆ
sin DBA
sin141

97
120
ˆ  97 sin141
 sin DBA
120
ˆ  0,5087006494
 sin DBA
ˆ  30,58
 DBA
 MB̂D  28
ˆ  141
 BDA
 sine rule
 substitution
ˆ  30,58
 DBA
 answer
ˆ  30,58  28
 MBA
ˆ  58,58
 MBA
© Gauteng Department of Education
(6)
6
6(b)
B̂  30,58
E
A
 definition
 substitution
 answer
(3)
 answer
(1)
D
120km
28 30,58
B
EA
 sin(28  30,58)
120
EA  120sin(28  30,58)
EA  102, 4 km
7(a)
DE
 tan 
p
 DE  p tan 
7(b)(1)
Ê1  180  
opp angles of cyc quad
ˆ  180 sum of angles of 
Eˆ 1  Fˆ1  H
2
ˆ  Fˆ
But H
2
1
angles opp equal sides
 Ê1  180  
ˆ  Fˆ  H
ˆ  180
 E
1
1
2
ˆ  180
 180    2H
2
(3)
ˆ  180
180    2H
2
ˆ 
 2H
2
7(b)(2)
1
 Ĥ 2  
2
p
FH

ˆ
sin H
sin Eˆ 1
2
ˆ
FH . sin H
2
sin(180  )
1
FH . sin 
2
p
sin 
1
FH . sin 
FH
2
p

1
1
1
2sin  cos  2cos 
2
2
2
p
p
FH

ˆ
sin H
sin Eˆ 1
2
1
FH . sin 
2
 p
sin 
1
1
 2sin  cos 
2
2

© Gauteng Department of Education
(3)
7
FH 2  p 2  p 2  2( p )( p ) cos(180  )
 FH2  2 p2  2 p2 ( cos )
 FH 2  2 p 2  2 p 2 ( cos )
 FH2  2 p2 (1  cos )
 answer
7(b)(3)
 FH 2  2 p 2  2 p 2 cos 
(3)
 FH 2  2 p 2 (1  cos )
 FH  2 p 2 (1  cos )
 FH  p 2(1  cos )
SECTION B:
1(a)
1(b)
SOLUTIONS TO TYPICAL EXAMINATION QUESTIONS
cos(50  x) cos(20  x)  sin(50  x) sin(20  x)  cos (50  x)  (20  x)
 cos  (50  x)  (20  x) 
 cos30
3
 cos 30

2
3

2
(1  2 sin 75)( 2 sin 75  1)
 1  2sin 2 75
 cos150
3
 
2
 (1  2 sin 75)(1  2 sin 75)
 1  2sin 2 75
 cos 2(75)
 cos150   cos 30  
2(a)
2  sin   2,5
 sin   2,5  2
 sin   0,5
2(b)
  sin 1 (0,5)  k .360
  30  k .360
where k 
cos 2  0,5
1
2(c)
(3)
(3)
3
2
 sin   0, 5
   30  k.360
   150  k.360
 k
or   180  sin 1 (0,5)  k.360
 2   cos (0,5)  k.360
 2  120  k.360
  60  k.180

2 tan  1
2
(4)
  150  k .360
(k  )




2  120  k.360
  60  k .180
  60  k .180
k
 tan
(4)

 0,5
2
© Gauteng Department of Education
8
 tan

 0,5
2

  tan 1 (0,5)  k .180
2

  27  k .180
2
  54  k .360
2(d)
2(e)
2(f)

 27  k .180
2
   54  k .360
 k
(4)
3sin  2 cos 

cos 
cos 
2
 tan  
3
   34  k .180
 k
(4)

(k  )
3sin   2cos   0
 3sin   2 cos 
3sin  2 cos 


cos 
cos 
 3 tan   2
2
 tan  
3
 2 
  tan 1    k .180
 3 
  34  k .180 (k  )

 1  sin 2 
 2sin 2   5sin   3  0
 sin 2   3  3sin 2   5sin 
 (2sin   1)(sin   3)  0
2sin 2   5sin   3  0
1
or sin   3
 sin  
 2sin 2   5sin   3  0
2
 (2sin   1)(sin   3)  0
   30  k.360
   150  k.360
1
 sin  
or
sin   3
 no solution for sin   3
2
  30  k .360
no solution since  1  sin   1
or   150  k .360
k
In this example, we will make use of the  cos 3 x  cos(90  x)
reduction rule: sin x  cos(90  x) .
 sin 2   3(1  sin 2 )  5sin 
Using this rule, the right side of the equation
cos3x  sin x can also be expressed in terms
of the cosine of an angle as follows:
cos 3 x  cos(90  x)
One general solution of the equation can be  3x  (90  x)  k .360
obtained by equating the angles and adding  x  22,5  k .90
k .360 :
 cos3x  cos 360  (90  x)
 3 x  270  x  k .360
© Gauteng Department of Education
(7)
9
3x  (90  x)  k .360
k
 x  135  k .180
 k
 4 x  90  k .360
 x  22,5  k .90
There is another general solution. If we use
the reduction rule
cos   cos(360  ) , the equation can be
rewritten as follows:
(7)
cos 3x  cos(90  x)
 cos 3x  cos 360  (90  x)
 cos 3x  cos 360  90  x 
3(a)
 cos 3x  cos(270  x)
 3x  270  x  k .360
 2 x  270  k .360
 x  135  k .180
cos 2 x  tan 23
 x  11,5  k .180
 x  168,5  k .180
2 x  tan 1 (tan 23)  k .360
 2 x  23  k .360
 x  11,5  k .180
OR
 x 168,5 ;191,5
(3)





(5)
2 x  360  tan 1 (tan 23)  k .360
 2 x  337  k .360
 x  168,5  k .180
 x  168,5 ;191,5
3(b)
cos( x  360)  cos(90  x)  0
 cos x  ( sin x)  0
 cos x  sin x  0
 sin x   cos x
 tan x  1
x  135  k .180
x  315  k .180
where k 
cos x
 sin x
tan x  1
x  135  k.180
x  315  k.180
Deduct 1 mark if k 
© Gauteng Department of Education
is omitted
10
4(a)
1  sin x  cos 2 x





1  sin x  1  2sin 2 x
 sin x  2sin 2 x  0
 sin x(1  2sin x)  0
 sin x  0 or
sin x  
1  2sin 2 x
sin x(1  2sin x)  0
two equations
general solutions
answers
(7)
1
2
For sin x  0 :
x  0  k 360
or
x  180  k 360
or
x  180  k 360
1
For sin x   :
2
x  30  k 360
or
x  210  k 360
 x  180 ; 210 ; 330 ; 360
4(b)
2 y
1
x
0
180
225
270
315
360
–1
–2
4(b)
see diagram
For


For


f ( x)  1  sin x
max and min values
shape
g ( x)  cos2x
amplitude
intercepts
© Gauteng Department of Education
(4)
11
4(c)
f ( x)  g ( x)
180  x  210 or 330  x  360
4(d)
180  x  225
x  270
315  x  360
180  x  225
270  x  315
4(e)
180  x  210
330  x  360
inequality signs correct






180  x  225
x  270
315  x  360
(3)


180  x  225
270  x  315
(2)
(3)
5.
1.5 y
1
f
g
0,9
0.5
x
–120
–90
–60
–30
30
60
90
120
150
180
210
–0.5
–1
–1.5
5(a)
For f :  shape
 x-intercepts: 60; 120
 y-intercept:
 endpoints:
5(b)
3
 0,9
2
(120 ;  0,9)
(210 ;  1)
For g :  shape
 x-intercepts: 30; 150
 y-intercept:
 endpoints:
1
 0,5
2
(120 ;  1)
(210 ;  0,9)
cos  x  30  sin  x  30
 cos  90  ( x  30) 
cos  x  30  cos  90  ( x  30) 
 cos  x  30  cos  60  x 
cos  x  30  cos  60  x 
Quad 1
 x  30   60  x   k 360
2x  90  k 360
x  45  k180
 x  45
Quad 2

(8)
 x  30   60  x   k 360
 x  45
  x  30  360   60  x   k 360
 no solution
(6)
© Gauteng Department of Education
12
 x  30  360   60  x   k 360
 x  30  300  x  k 360
6(a)
6(b)
no solution
3
 tan 40
LB
3
 LB 
tan 40
 LB  3,58 m
AB2  AL2  BL2  2.AL.BL.cos113
 AB2  (5, 2)2  (3,58) 2  2(5, 2)(3,58) cos113
 AB  54, 40410138 m
 AB  7,38 m
1
ˆ
Area of ABL  AL.BL.sin ALB
2
1
 (5, 2)(3,58)sin113
2
 8.568059176
2
6(c)
 definition
 answer
(3)
 cosine rule
 substitution
 answer
(4)
 area rule
 substitution
 answer
(4)
 cos 2 x  sin 2 x
 (cos x  sin x)(cos x  sin x)
(2)
1
2
1
2
 cos x  cos x sin x 
2
2
 2cos x  1  2sin x cos x
 cos 2 x  sin 2 x
(4)
2
 8,57 m
7(a)
7(b)
cos 2 x
cos x  sin x
cos 2 x  sin 2 x

cos x  sin x
(cos x  sin x)(cos x  sin x)

(cos x  sin x)
 cos x  sin x
 cos 2 x  1
cos x 

 cos x  sin x  2
1
 cos x(cos x  sin x) 
2
1
 cos 2 x  cos x sin x 
2
2
 2cos x  2cos x sin x  1
 cos x(cos x  sin x) 
 2cos 2 x  1  2sin x cos x
 cos 2 x  sin 2 x
© Gauteng Department of Education
13
7(c)
8(a)
8(b)
 cos 2 x  1
cos x 

 cos x  sin x  2
 cos 2 x  sin 2 x
cos 2 x sin 2 x


cos 2 x cos 2 x
1  tan 2 x
 2 x  45  k .180
 x  22,5  k .90
k
 1  tan 2x
 2 x  45  k.180
 x  22,5  k.90
(3)
1
(m)(n)sin 4 x
2
The graph of y  sin 4x has a maximum
 answer
(1)
 sin 4 x  1
 x  22,5
(2)
 answer
(1)
 2cos 2 x  1  8
 cos 2 x  8
 max value of cos2x
 answer
(4)
Area 
value of 1.
8(c)
9(a)
 sin 4 x  1
 4 x  90
 x  22,5
Since 4x  90 , the triangle is right-angled.
2cos 2 x  7
 2cos 2 x  1  8
 cos 2 x  8
The maximum value of cos2x is 1 since
the range of y  cos 2 x is [1;1]
Therefore the maximum value of
cos 2 x  8 is 1  8  9  3
9(b)
cos30 
a
2
r
3 a


2
2r
 3r  a
a
r 
3

a
2r
a
2 30
 30
 a2
r
 cos30 
a
2
r
3 a

2
2r
a
 r
3

© Gauteng Department of Education

a
2r
14
2
a 2 a 2
 a 



Area of circle   

3
3
 3
SESSION:
11
TOPIC:
REVISION OF CALCULUS
SECTION A:
1(a)(1)
 a 
 

 3
2
(6)
(CONSOLIDATION)
SOLUTIONS TO HOMEWORK QUESTIONS
f ( x  h)  f ( x )
h 0
h
1
1 

2  ( x  h) 2   2  x 2 
2
2 

 f ( x)  lim
h 0
h
1
1
2  ( x 2  2 xh  h 2 ) 2  2  x 2
2
2
 f ( x)  lim
h 0
h
1 2
1 2
1
2  x  xh  h  2  x 2
2
2
2
 f ( x)  lim
h 0
h
1
 xh  h 2
2
 f ( x)  lim
h 0
h
1 

h  x  h 
2 
 f ( x)  lim 
h 0
h
1 

 f ( x)  lim   x  h 
h 0 
2 
 f ( x)   x
 f (2)  (2)  2
f ( x)  lim
1
 2  ( x  h) 2
2
1
 2  x 2
2
1
 xh  h 2
2

h
1 

  x  h 
2 

 f ( x)   x
 f (2)  (2)  2
© Gauteng Department of Education
(6)
15
1(a)(2)
 f (2)  0
1
f (2)  2  (2) 2
2
 f (2)  0
(2 ; 0) lies on the tangent
mt  2
 y  0  2( x  (2))
 y  2x  8
(3)
 y  0  2( x  (2))
 y  2( x  4)
 y  2x  8
1(b)
1
 6 x 2  x 6
6
3
 12x
 x7
12 1
  3 7
x
x
6 1 6
 x
x2 6
1
 y  6 x 2  x 6
6
dy
  12 x 3  x 7
dx
dy
12 1
  3  7
dx
x
x
y




2.
y
.
g
(0 ; 32)
.
( 2 ; 0)
(4 ; 0)
y-intercept
x-intercepts
turning points
shape
x
g (0)  32
y-intercept is (0 ; 32)
g (4)  0
x-intercept is (4 ; 0)
g ( 2)  0
x-intercept is ( 2 ; 0)
g (0)  0
Turning point at x  0
g (4)  0
Turning point at x  4
g ( x)  0 if x  0 or x  4
g increases for x  0 or x  4
© Gauteng Department of Education
(4)
(4)
16
g ( x)  0 if 0  x  4
g decreases for 0  x  4
3(b)
4(a)
4(b)
g decreases if g ( x)  0
 x  1
x  1
 x  1
(1)
 x  1
(1)
6
y  x6
8
3
y  x6
4
 gradient
 y-intercept
(2)
 3

 A( x)  x   x  6 
 4

3
 A( x)   x 2  6 x
4
3
  x60
2
 x4
 y3
(5)
Area of OCDE  xy
5(b)
 3

 A( x)  x   x  6 
 4

3
 A( x)   x 2  6 x  And A / ( x)  0
4
3
 x  6  0
2
3 x  12  0
3
 x  4 and  y   (4)  6  3
4
3
 x6 x
4
3x  24  4 x
7 x  24
24
x 
7
g decreases if g ( x)  0
 x  1
x  1
6(a)
PQ 2  (30  6 x) 2  (8 x) 2
4(c)
5(a)
3
  x6 x
4
 3x  24  4x
24
 x
7
 PQ 2  900  360 x  36 x 2  64 x 2
x hours
A
6x
Q
8x
.
P
30km
 x  1
(1)
 PQ 2
 answer
.
 PQ  100 x 2  360 x  900
(1)
 30  6x
 8x
 PQ 2  (30  6 x) 2  (8 x) 2
C
 PQ 2  100 x 2  360 x  900
(3)
 x  1
x hours
3(a)
30  6x
B
© Gauteng Department of Education
(5)
17
6(b)
We will first minimize the expression
100 x2  360 x  900
Let f ( x)  100 x 2  360 x  900
 f ( x)  200 x  360
 0  200 x  360
200 x  360
360
x 
200
9
 x  hours
5
Minimum distance:
6(c)
2
9
9
PQ  100    360    900  24 km
5
5
SECTION B:
1(a)
 f ( x)  200 x  360
 0  200x  360
 x
9
5
 substitution
 answer
(3)
(2)
SOLUTIONS TO TYPICAL EXAMINATION QUESTIONS
f ( x  h)  f ( x )
h 0
h
4  ( x  h)2  (4  x 2 )

 f ( x)  lim
h 0
h
2
4  ( x  2 xh  h 2 )  4  x 2
 f ( x)  lim
h 0
h
2
4  x  2 xh  h 2  4  x 2
 f ( x)  lim
h 0
h
2
2 xh  h
 f ( x)  lim
h 0
h
 f ( x)  lim(2 x  h)
f ( x)  lim





4  x2  2xh  h2
4  x2
2xh  h2
(2 x  h)
2x
h 0
 f ( x)  2 x
© Gauteng Department of Education
(5)
18
1(b)(1)
8
x
y2 x
y 
1
2x 2
 8x
 y

1
 8 x 1
1
2
 x
 8x1
1
8

 2
x x
1

dy
 x 2  8 x 2
dx
dy
1
8
 
 2
dx
x x

1(b)(2)
1
2x 2
 (2 p  1)( p  2)
 2 p 1
2
 2 p2  3 p  2 
Dp 

p2


 (2 p  1)( p  2) 
 Dp 

p2


(4)
(3)
 D p  2 p  1
2
1(b)(3)
3
3
xy  5  x 2
 xy 
y 
3
x2
3
x2
x
 x2
1
5

5
x
1
 y  x 2  5 x 1
dy 1  1
  x 2  5 x 2
dx 2
dy
1
5
1
5
  1  2
 2
dx 2 x 2 x
2 x x
1(c)
2(a)
g ( x)  3x 2  kx  4
 g ( x)  6 x  k
 6(2)  k  8
 k  4
(2 ;  4) is a turning point
f ( x)  3ax 2  2bx
 f (2)  3a (2) 2  2b(2)
 0  12a  4b
 0  3a  b
 b  3a
(2 ;  4) is a point on the graph
 y  x 2  5x1
1 1
 x 2
2
 5x 2
1
5

 2
2 x x
 g ( x)  6 x  k
 Substituting x  2
 6(2)  k  8
 k  4
(5)
(4)
 f ( x)  3ax 2  2bx
 0  12a  4b




4  a(2)3  b(2)2
1  2a  b
a 1
b  3
© Gauteng Department of Education
(6)
19
2(b)
4  a (2)3  b(2) 2
 4  8a  4b
 1  2 a  b
 1  2a  (3a )
 1   a
a 1
 b  3
f ( x)  x 3  3 x 2
y-intercept: (0 ; 0)
x-intercepts:
0  x  3x
3
(0 ; 0)
(3 ; 0)
2
 0  x 2 ( x  3)
 x  0 or x  3






y-intercept
x-intercepts
0  3x 2  6 x
x  0 or x  2
(0 ; 0) (2 ;  4)
shape
(7)
(2 ;  4)
Turning points:
f ( x)  3 x 2  6 x
 0  3x 2  6 x
2(c)
f (0)  0 f (2)  4
 0  x2  2 x
(0 ; 0)
(2 ;  4)
 0  x( x  2)
 x  0 or x  2
f (1)  3(1)2  6(1)  9
f (1)  (1)3  3(1)2  2  m
(1;  2) lies on the graph
y  (2)  2( x  (1))
 y  2  2( x  1)
 y  2  2 x  2
 y  2 x  4
3(a)(1)
3(a)(2)




f (1)  9
f (1)  2
y  (2)  2( x  (1))
y  2 x  4
1
P  2h  2r   2r
2
 P  2h  2r  r
 2h  2r
 r
1
A  2rh  r 2
2
 2rh
1
 r 2
2
© Gauteng Department of Education
(4)
(2)
(2)
20
3(b)
1
4  2rh  r 2
2
 8  4rh  r 2
 8  r 2  4rh
8  r 2

h
4r
P  2h  2r  r
 8  r 2 
 P  2 
  2r  r
 4r 
1
 4  2rh  r 2
2
2
8  r
h

4r
 8  r 2 
 P  2
 4r   2r  r


4


 P    2 r 
(4)
r
2

8  r 2
P 
 2r  r
2r
4 r
 P    2r  r
r 2
4 r
 P    2r
r 2
4 

P     2 r
r 2

4


P    2 r 
r
2

3(c)
40


C  10   2  r 
r
2

 C  5r  20r  40r 1
 C(r )  5  20  40r 2
 C(r )  5  20 
 0  5  20 
40
r2
40


 C  10   2  r 
r
2

 C  5r  20r  40r 1
40
 0  5  20  2
r
 r  1, 06m
40
r2
40
 5  20
r2
40

 r2
5  20
40

r
5  20
 r  1, 06m

© Gauteng Department of Education
(4)
21
  graph of f 
4.
(2)
f 1
5(a)
h(0)  35  5(0)2  30(0)  35 metres
5(b)
h(t )  35  5t  30t
 h(t )  10t  30
 h(0)  10(0)  30
 h(0)  30
5(c)
2
h(t )  35  5t 2  30t
 h(t )  10t  30
 0  10t  30
10t  30
t  3
 substitution
 answer
 h(t )  10t  30
 h(0)  30
(2)
(2)
 0  10t  30
 t 3
 y  3x2  12 x  15
 h(3)  35  5(3)2  30(3)
 answer
(5)
 h(3)  35  5(3) 2  30(3)
 h(3)  80 metres
5(d)
60  35  5t 2  30t
 5t 2  30t  25  0
 t 2  6t  5  0
 (t  5)(t  1)  0
 t  5 or t  1
For t  5
h(5)  10(5)  30  20
 20 m / s





60  35  5t 2  30t
t 2  6t  5  0
(t  5)(t  1)  0
h(5)  10(5)  30  20
20 m / s
(5)
© Gauteng Department of Education
22
5(e)
h(t )  35  5t 2  30t






 0  35  5t 2  30t
 5t 2  30t  35  0
 t 2  6t  7  0
 (t  7)(t  1)  0
 t  7 or t  1
But t  1
t  7
h(t )  10t  30
 h(7)  10(7)  30
 h(7)  40
40 metres per second
6(a)
0  35  5t 2  30t
t 2  6t  7  0
(t  7)(t  1)  0
t  7 or t  1
h(t )  10t  30
h(7)  40
y  a ( x  1)( x  5)
Substitute the point (0 ; 15)
15  a(0  1)(0  5)
 y  a ( x  1)( x  5)
 a  3
 y  3x2  12 x  15
15  5a
 a  3
y  3( x  1)( x  5)





 y  3( x 2  4 x  5)
 y  3 x  12 x  15
2
(6)
f ( x)  3ax2  2bx  c
a  1
b6
c  15
f ( x)   x3  6 x 2  15 x
(8)
 f ( x)  3 x 2  12 x  15
Now f ( x)  ax3  bx 2  cx
 f ( x)  3ax 2  2bx  c
Equating coefficients:
3a  3
2b  12
c  15
6(b)(1)
 a  1
b  6
3
2
 f ( x)   x  6 x  15 x
The graph will increase for 1  x  5
6(b)(2)
The graph will decrease for x  1 or x  5
6(c)
7(a)
x  1 or x  5
The y-intercept is at the origin meaning that k  0
 1  x  5
 x  1
 x 5
 x  1
 x 5
 answer
© Gauteng Department of Education
(1)
(2)
(2)
(1)
23
7(b)(1)
 0  x3  4 x 2  4 x
 0  x( x  2)( x  2)
 x  0 or x  2
 answer
f ( x)  x3  4 x 2  4 x
 0  x3  4 x 2  4 x
 0  x( x 2  4 x  4)
 0  x( x  2)( x  2)
 x  0 or x  2
(4)
The graph cuts the x-axis at (2 ; 0)
7(b)(2)
a  2
f ( x)  3x 2  8 x  4
7(c)
 0  3x 2  8 x  4
 0  (3 x  2)( x  2)
2
x 
or x  2
3
2
b 
3
3
x  4 x2  4 x  2  p
 f ( x)  3x 2  8x  4
 0  (3x  2)( x  2)
 x
 x3  4 x 2  4 x  p  2
2
or
3
x2
 answer
(4)
 p2
 answer
(2)
p20
p2
8(a)
x

x 2 10    0
12 

Now x 2  0 for all real values of x
x
10   0
12
120  x  0
x

0
12 

 x2  0
x
0
 10 
12
 x 2  10 
 answer
 x  120
 x  120 km/h
© Gauteng Department of Education
(4)
24
x

P  x 2 10  
12 

8(b)
 P( x)  10 x 2 
1 2
x
4
 0  x(80  x)
 x  0 or x  80
1 3
x
12
1
 P( x)  20 x  x 2
4
1
 0  20 x  x 2
4
 0  80 x  x 2
 0  x(80  x)
 x  0 or x  80
For x  80
80 

P(80)  (80) 2 10  
12 

 P(80)  R21 333,33
 0  20 x 
 P( x)  10 x 2 
 substitution
 answer
SESSION:
12
TOPIC:
REVISION OF ANALYTICAL GEOMETRY
SECTION A:
1(a)
1(c)
(6)
(CONSOLIDATION)
SOLUTIONS TO HOMEWORK QUESTIONS
r 2  OQ 2
 r 2  (5)2  (12)2
 r 2  (5)2  (12)2
 r 2  169
 x 2  y 2  169
 r 2  169
1(b)
1 3
x
12
 x 2  y 2  169
50
mPQ 
12  0)
5
 mPQ 
12
5
y  x
12
P(  12 ;  5)
 mPQ 
(3)
5
12
5
x
12
(2)
answer
(2)
 y
© Gauteng Department of Education
25
1(d)
mPQ  mQR  1
mPQ 
 mPQ 
5
12
5
12
 mQR  
12
5
y  y1  m( x  x1 )
 mQR  
1(e)
 m
12
( x  12)
5
 5 y  25  12( x  12)
 5 y  12 x  144  25
 5 y  12 x  169
12 x  5 y  169  0
12
169
y   x
5
5
Substitute the point R(t ;1) :
 y 5  
1(f)
12
169
t  
5
5
 t  14,5
 x  12 2   y  52  OQ2
2
2
OQ 2  12  0    5  0   169
2
2
  x  12    y  5   169
  x  12    y  5  OQ2
Centre M(1; 3) and r 2  (5  1) 2  (4  3) 2  17
 Equation of circle:
 M(1; 3)
 Substitution to find r
 r 2  17
 ( x  1)2  ( y  3)2  17
 grad PR  4
1
 grad TR 
4
  grad PR  grad TR  1
2(c)(1)
2  2
 4 and
2  (3)
42
1
grad TR 

5  (3) 4
 grad PR  grad TR  1 and  PR  TR
grad PR 
 PR is a tangent
y  (2)  4( x  (2))
 y  2  4( x  2)
 y  4 x  10
(3)
 1 
( x  1)2  ( y  3) 2  17
2(b)
12
5
12
( x  12)
5
12
169
 y  x
5
5
12
169
t  
5
5
12t  174
174
t 
12
 t  14,5
2(a)
(2)
 y 5  
1 
1(g)
12
5
2
(2)
2
 OQ2  169

 x 122   y  52  169
 Substitute (2 ;  2)
 Gradient  4
 y  intercept of  10
© Gauteng Department of Education
(3)
(4)
(3)
(3)
26
2(c)(2)
(0  1) 2  ( y  3) 2  17
1  ( y  3) 2  17
2(d)
2(e)
( y  3) 2  16
y  3  4
 y  7 (N/A) or y  1 and  V(0 ;  1)
4  (1)
 grad TP 
1
50
Equation of TP: y  x  1
4 x  10  x  1
5 x  9
9
4
4
4
 x    1 and  y  1  1  2
5
5
5
5
4
 4
 P  1 ;  2 
5
 5
V(0 ;  1)
T(5 ; 4)
4  (1)
tanθ 
1
50
 tanθ  1
 θ  45
Alternatively:
(since equation of TVP is y  x  1 )
mTVP  1
 tanθ  1
 θ  45
3(a)
x 2  y 2  8 x  4 y  38  0
 x 2  8 x  y 2  4 y  38
 x 2  8 x  16  y 2  4 y  4  38  16  4
 ( x  4) 2  ( y  2) 2  58
Centre  (4 ;  2)
3(b)
3(c)
 Substitute x  0
 ( y  3)2  16
 y  7 or y  1
 V(0 ;  1)
 grad TP  1
 y  x 1
 4x 10  x 1
9
 x
5
 y  2
4
5
(3)
1 if coordinates of P
was not stated
4  (1)
1
 mTVP 
50
 tanθ  1
 θ  45
OR
 mTVP  1
 tanθ  1
 θ  45
(3)
 completing the
square
 factor form
 centre
 radius
(4)
radius  58
Centre of second circle: (4 ; 6)
 (4 ; 6)
Distance  (4  4)2  (2  6)2  128

sum of radii  58  26  12, 71
 sum of radii
 conclusion
Distance between centres  128  11,31
 sum of radii  distance between centres
 circles intersect
(6)
128
© Gauteng Department of Education
(2)
(3)
27


4(a)
4(b)
4(c)
4(d)
P(2;3)

Coordinates of Q:
2(0)  y  1
 Q(0;  1)
 R(4;0)
 y  1
Q(0;  1)
Coordinates of R:
3x  2(0)  12
 3x  12
x  4
R(4;0)
Q(0;  1)
R(4;0)
0  (1) 1
mQR 

40
4
1
 y  x 1
4
 0  4 1  0 
S
;

2 
 2
1

 S 2 ;  
2

Equation of PS:
x2
 mQR
 y -intercept:  1
1
 y  x 1
4
(2)
(4)
 formula
 correct coordinates
(2)
 answer
(1)
© Gauteng Department of Education
28
4(e)
tan   mPQ
tan   mPQ
3  (1)
20
 tan   2
  63
 tan  
or
Now 2 x  y  1
 y  2 x  1
 y  2x 1
 tan   2
  63
tan   mPR
tan   mPR
30
24
3
 tan   
2
  180  56
  124
Now 3x  2 y  12
 2 y  3 x  12
3
y   x6
2
3
 tan   
2
  124
 tan  
or
 Using formula for
inclination
 tan   2
   63
3
 tan   
2
   124
   61
(6)
 
  124  63
  61
4(f)
 correct substitution
for PQ2
PQ 2  (2  0) 2  (3  ( 1)) 2
 PQ 2  4  16
 PQ  20
 correct substitution
for PR 2
 PR  13
 Area rule
 7,05
 PQ 2  20
 PQ  20
PR 2  (2  4)2  (3  0)2
 PR 2  4  9
 PR  13
2
 PR  13
Area PQR 
1
2

20
 13  sin 61
 Area PQR  7, 05 units 2
© Gauteng Department of Education
(6)
29
SECTION B:
1(a)
SOLUTIONS TO TYPICAL EXAMINATION QUESTIONS
x2  2 x  y 2  4 y  t  0
 x2  2 x  1  y 2  4 y  4  t  1  4
 completing the square
 A(1;  2)
(4)
(1)
 ( x  1) 2  ( y  2)2  t  5
Centre  A(1;  2)
1(b)
r  t 5
answer
1(c)
A(1;  2) and D(2;  1)
 AD2  (1  (2))2  (2  (1))2
 AD2  (1  (2)) 2  (2  (1)) 2
 AD2  10
 t 5
 AD2  9  1
1(d)
 AD2  10
 t  5  10
t  5
x2  2 x  y 2  4 y  5  0
For intercepts with the y  axis: let x  0
 (0)2  2(0)  y 2  4 y  5  0
1(e)
1(f)
 y2  4 y  5  0
 ( y  5)( y  1)  0
 y  5 or y  1
 B(0;  5) C(0;1)
BC  6 units
A(1;  2) and D( 2;  1)
2  (1) 1
mAD 

1  (2)
3
 mtangent  3
 y  (1)  3( x  (2))
 y  1  3( x  2)
 y  1  3x  6
 y  3x  5
2(a)
2(b)
 0  (2) 8  (6) 
M
;

2
2 

 M(1;1)
y  7 x  58
 y  7(8)  58  2





(3)
y2  4 y  5  0
( y  5)( y  1)  0
y  5 or y  1
B(0;  5)
C(0;1)
answer
(5)
(1)
1
3
 mtangent  3
 mAD 
 y  (1)  3( x  (2))
 y  3x  5
(4)
 x-coordinate
 y-coordinate
(2)
 substitution
 answer
(2)
© Gauteng Department of Education
30
2(c)
mline  7
2 1
1

8  (1)
7
1
mline  mAM  7    1
7
The line is a tangent to the circle
mAM 
2(d)
AD2  (0  8) 2  (8  2) 2
 AD  (0  8) 2  (8  2) 2
 AD  36  64
 AD  10
AB2  (8  2)2  (2  6) 2
 relationship
 mline  7
1
 mAM  
7
 product
(5)
 substitution
 answer
 substitution
 answer
(4)
 gradient of AD
 gradient of AB
 product
(3)
 answer
(1)
 AB  (8  2) 2  (2  6) 2
2(e)
 AB  36  64
 AB  10
8  (2) 6 3
mAD 
 
0  (8) 8 4
2  (6)
8
4
mAB 


8  (2) 6
3
4 3
mAB .mAD     1
3 4
ˆ  90
 DAB
2(f)
  45
2(g)
BM 2  (1  (2)) 2  (1  (6)) 2
 BM 2  1  49
 BM 2  50
Z
3(a)
ax  3 y  5
 3 y  ax  5
a
5
y 
x
3
3
22, 5
 BM  50
ZM
 sin 22,5
BM
ZM

 sin 22,5
50
 ZM  50 sin 22,5
 ZM  2,93
 distance formula
 length of BM
 tan perpendicular to rad
M(  1;1)  definition
 22,5
 answer
B(  2 ;  6)
2 x  by  3
 by  2 x  3
2
3
y 
x
b
b
a
5
x
3
3
2
3
x
 y
b
b
 y
© Gauteng Department of Education
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31
 a 2

3
b
  ab  6
 ab  6
 a 2

3
b
 ab  6


(4)
B(3 ; 7)
A(1; 5)
E(a ; a  4)
C(8 ; 2)
D(x ; 0)
3(b)(1)
 mAB
 mBC
 mAB  mBC  1
See diagram above.
75
mAB 
1
3 1
27
mBC 
 1
83
 mAB  mBC  (1)(1)  1
(3)
 AB  BC
ˆ  90
 ABC
3(b)(2)
Area ABC 
1
(AB)(BC)
2
AB2  (7  5)2  (3  1) 2
 AB2  (7  5) 2  (3  1) 2
 AB2  8
 BC  50  5 2
1
 (2 2)(5 2)
2
 AB  8  2 2
 BC2  (8  3) 2  (2  7) 2
 AB  8  2 2
BC  (8  3)  (2  7)
2
2
2
 BC2  50
 BC  50  5 2
1
 Area ABC  (2 2)(5 2)  10
2
© Gauteng Department of Education
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3(b)(3)
3(b)(4)
75
1
3 1
 y  7  1( x  3)
 y  x4
75
1
3 1
y  7  1( x  3)
 y 7  x3
y  x4
Area ABD  10
 mAB 
mAB 
1
 (AB)(ED)  10
2
1
 (2 2)(ED)  10
2
10
 ED 
2
Let E be the point ( a ; b)
Since E lies on AB produced, the point will be
E(a ; a  4)
ED 2  ( x  a) 2  (0  ( a  4)) 2
 ED 2  ( x  a) 2  ( a  4) 2
2
 10 
2
2

  ( x  a )  ( a  4)
2


(3)
1
(AB)(ED)  10
2
10
 ED 
2
 E(a ; a  4)

 50  ( x  a ) 2  ( a  4) 2
 x  2a  4
1
 (AB)(ED)  10
2
 50  (2a  4  a) 2  (a  4) 2
 0  a2  8a  9
 a  9 or a  1
 x  14 or x  6
 50  ( x  a ) 2  ( a  4) 2
mBE  mED  1
 (1)  mED  1
 mED  1
0  ( a  4)
 1
xa
a  4   x  a

 x  2a  4
 50  (2a  4  a) 2  (  a  4) 2
 50  (a  4) 2  ( a  4) 2
 50  a 2  8a  16  a 2  8a  16
 50  2a 2  16a  32
 0  2a 2  16a  18
 0  a 2  8a  9
 0  (a  9)(a  1)
 a  9 or a  1
 x  2(9)  4  14 or x  2(1)  4  6
© Gauteng Department of Education
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4.
1  3
y6

2  (4) 4  x
 y  2x  2
mAD  mBC

1  3
y6

2  (4) 4  x
4 y  6


2 4 x
2(4  x)  y  6
8  2 x  y  6
 y  2x  2

 (4  x) 2  ( y  6) 2
 4 (2  (4))2  (1  3)2 
 (4  x) 2  ( y  6) 2  80
BC2  4AD 2
 (4  x) 2  ( y  6) 2  4 ( 2  ( 4)) 2  ( 1  3) 2 




x2  8x  0
x( x  8)  0
x  0 or x  8
y  2 or y  14
 (4  x) 2  ( y  6) 2  80
 (4  x) 2  (2 x  2  6) 2  80
16  8 x  x 2  4 x 2  32 x  64  80
 5 x 2  40 x  0
 x2  8x  0
 x( x  8)  0
 x  0 or x  8
 y  2(0)  2  2
or y  2(8)  2  14
© Gauteng Department of Education
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34
SESSION:
13
TOPIC:
REVISION OF ALGEBRA
SECTION A:
1(a)(1)
(CONSOLIDATION)
SOLUTIONS TO HOMEWORK QUESTIONS
 standard form = 0
 factorisation
 both answers
 excluding x  2
x2  4
 3x
x2
 x 2  4  3 x( x  2)
(4)
 x  4  3x  6 x
2
2
2 x 2  6 x  4  0
 x 2  3x  2  0
 ( x  2)( x  1)  0
 x  2 or x  1
But x  2
 x  1 is the solution
OR
 factorisation
 simplification
 correct answer
 excluding x  2
x2  4
 3x
x2
( x  2)( x  2)

 3x
( x  2)
 x  2  3x provided x  2
2 x  2
 x  1
1(a)(2) ( x  3)( x  2)  8
 simplification
 standard form
 substitution into formula
 correct answers
(4)
 x2  5x  6  8
 x2  5x  2  0
x 
(5)  (5) 2  4(1)(2)
2(1)
5  33
2
 x  5,37 or x  0,37
x 
© Gauteng Department of Education
35
1(a)(3)
1(a)(4)
7 x 2  18 x  9  0
 (7 x  3)( x  3)  0
3
3  x 
7
 factorisation
 endpoints
 inequality notation
3
5  x 1 x  0
5  x  x2  2x  1
1(b)
3
7

 5  x  x 1
 0  x 2  3x  4
 0  ( x  4)( x  1)
 x  4 or x  1
Solution is x  4
2x  y  7
 y  2 x  7
 y  2x  7
x 2  xy  21  y 2
 x 2  x(2 x  7)  21  (2 x  7) 2
 x 2  2 x 2  7 x  21  (4 x 2  28 x  49)
(4)
5  x  x 1
 0  x 2  3x  4
 0  ( x  4)( x  1)
 x  4 or x  1
 selecting x  4
(5)
 y  2x  7
 substitution
 multiplication
 standard form
 factorisation
 both x-values
 both y-values
(7)
 3x 2  7 x  21  4 x 2  28 x  49
 7 x 2  35 x  28  0
 x2  5x  4  0
 ( x  4)( x  1)  0
 x  4 or x  1
 y  1 or y  5
1(c)
108  18
 36  3  9  2
1(d)
 6 3  3 2  6b  3a
x(4 x  3)   p
 4 x 2  3x  p  0
 b 2  4ac  (3) 2  4(4)( p)
 9  16 p  0
16 p  9
9
p
16
 6 3
3 2
 6b  3a
(3)
 4 x 2  3x  p  0
 (3)2  4(4)( p)
 9  16 p  0
9
 p
16
© Gauteng Department of Education
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1(e)
  (k ) 2  4(k 2  1)(1)
 (k ) 2  4( k 2  1)(1)
  k 2  4k 2  4
   3k 2  4
 (3k 2  4)
  3k 2  4
 (3k 2  4)  0
  (3k 2  4)  0
2(a)
LHS 
9 x 1  6.32 x
 3
4 x 1



(32 ) x 1  6.32 x
4 x 1
2x
18 x 20




2(c)(1)
10 2 x
7 8 x
10 2 x

3
1
2
1
2
 3  3  RHS
1
 32  3
(5)
 3 2x10
 24 2x10
 8
(5)
1
 3x (3  )
3
x
3

3

 x 1
(3)
 7  2 2x
10
3 2 x10
10 2 x10  14 2 x10
3 2 x10
24 2 x10
3 2 x10
8
3x 1  3x 1  8
 3x.31  3x.31  8
1
 3x (3  )  8
3
8
 3x ( )  8
3
x
3  3
x 1
2(c)(2)
20
18 x 20
10
1
2
3
 10 2x10
 14 2x10
10 2 x 20  7 8 x 20
20
2 x 1
3 2
 32 x (32  6)
3
 1
32
2 x 1
3 .3
2(b)
 32 x2
 12 
3 
 
2 x2
3
 6.32 x
3 2
32 x (32  6)
(4)
2
3
6 x  54
2
3
 x 9
1
1
 x 3  3 or x 3  3
 x  27 or  27
© Gauteng Department of Education
(3)
37
2
x3  9
1
 ( x 3 )2  9
1
1
 x 3  3 or x 3  3
 x  27 or x  27
2(d)
4 x. 6 y  4812
 2 .2 .3
 (22 ) x . (2 . 3) y  (2 4. 3)12
 2 .3
 22 x. 2 y. 3 y  2 48. 312





 22 x  y. 3 y  248. 312
 2 x  y  48 and y  12
 2 x  12  48
 2 x  36
 x  18
 x  y  18  12  30
3(a)
2x
y
48
12
y
22 x  y
2 x  y  48
y  12
x  18
x  y  30
(7)
424  816  1612  648
 (22 )24  (23 )16  (24 )12  (26 )8
 (22 )24  (23 )16  (24 )12  (26 )8
 4 . 248
 248  248  248  248
 22. 248
 4.2
48
 250
(4)
 22. 248
3(b)
 250
b  1  2 n
1
b  1  n
2
1
b  1 
[a  1  2n ]
a 1
(a  1)  1
b 
a 1
a
b 
a 1
1
2n
1
 b  1
a 1
(a  1)  1
 b
a 1
a
 b
a 1
 b  1
© Gauteng Department of Education
(4)
38
3(c)
x  3 2 2  3 2 2
 x2 

 
2  2  (3  2
3 2 2
 x2  3  2
2
2
3 2 2


3 2 2 

3 2 2
2)(3  2 2)  3  2 2
 x2  6  2 1
x2  2 x  2  0
x 
3(f)
(2)  (2) 2  4(1)(2)
2(1)
x y 3
x  y 3  0
( x  y  1)( x  y  3)  0
 x  y  1  0 [divide both sides by x  y  3 ]
 x  y  1
1
k
2
x 1
1  kx 2  k
 0  kx 2  k  1
 0  kx 2  0 x  k  1
  (0)2  4(k )(k  1)
3 2 2


2
 2 3 2 2


3 2 2


3 2 2

2
(6)
2  12
2
 x  1 3
 x
2  12 2  2 3
x 

 1 3
2
2
 x2  2x  2  0 for all
x  1  3 or x  1  3
3(e)
2

 x2  6  2 1
 x2  4
 x2
 x2  4
x  2
3(d)


 x  1 3
 x  1 3
(4)
 x  y 3  0
 x  y 1  0
 x  y  1
(4)







1  kx2  k
0  kx2  k 1
  4k 2  4k
4k 2  4k  0
k (k  1)  0
k 0
k  1
  4k 2  4k
For real roots,   0
 4k 2  4k  0
k2  k  0
 k (k  1)  0
 k  0 or k  1
© Gauteng Department of Education
(7)
39
SECTION B:
1(a)(1)
SOLUTIONS TO TYPICAL EXAMINATION QUESTIONS
 x2  2 x  3  0
 ( x  3)( x  1)  0
 x  3 or x  1
 selecting x  3
x2  x  2
1  0
x 1
 x2  x  2  x  1  0
 x  2x  3  0
 ( x  3)( x  1)  0
 x  3 or x  1
But x  1
 x  3 is the only solution
Alternatively:
x2  x  2
1  0
x 1
( x  2)( x  1)

1  0
( x  1)
 ( x  2)  1  0 (provided x  1)
 x  3
2
1(a)(2)
( x  2)( x  1)
1  0
( x  1)
 ( x  2)  1  0
 provided x  1
 x  3

5 x( x  3)  2
 standard form
 substitution into
formula
 correct answers
 5 x 2  15 x  2  0
x 
(4)
(15)  (15) 2  4(5)(2)
2(5)
(4)
15  265
10
 x  3,13 or x  0,13
x 
1(a)(3)
20  ( x  1)( x  2)
 20  x 2  3 x  2
 20  x 2  3 x  2  0

 x  3x  18  0
6
2






3
 x  3 x  18  0
 ( x  6)( x  3)  0
6  x  3
2
© Gauteng Department of Education
 x 2  3x  18  0
x 2  3x  18  0
( x  6)( x  3)  0
6  x  3
(4)
40
1(a)(4)
2  7x  2x  0

 2  7x   2x


2  7x

2
  2 x 
2
 2  7 x  4 x2
 0  4 x2  7 x  2
 0  (4 x  1)( x  2)
1
x 
or x  2
4
 x  2
1(b)(1)
2 x 2  2  (2 x  2) 2
 2 x 2  2  (4 x 2  8 x  4)
 2 x2  2  4 x2  8x  4
 6 x2  8x  2  0
2  7x   2x
 2  7 x  4 x2
 0  (4 x  1)( x  2)
1
or x  2
 x
4
 selecting x  2
(5)
 substitution
 multiplication
 standard form
 factors
 x-values
 y-values
(6)
 3x  4 x  1  0
 (3x  1)( x  1)  0
2
x 
1
3
or
x 1
1
 y  2   2
3
4
y 
or
3
1(b)(2)
1(b)(2)
or
y  2(1)  2
y0
( x 2  8 x  16)  ( y 2  2 y  1)  0
 ( x  4)
 ( x  4)2  ( y  1)2  0
 x  4 and y  1
 ( y  1)
 x  4  y 1  0
 answers
and x  5  0
x  5
 3(5)  y  0
3x  y  0
 y  15
 y  15
1(c)
4 x
will be non-real if:
x6
2
2




3x  y  x  5  0
x 5
3(5)  y  0
y  15
(4)
(4)
 non-real if 4  x  0
 x4
 x6
(3)
© Gauteng Department of Education
41
1(d)
4 x  0
 x  4
x  4
However, the expression will be undefined if x  6 .
Therefore, the expression will be non-real if
x  4 where x  6
5 p 2  x( x  3 p)
 5 p 2  x 2  3 px
 0  x 2  3 px  5 p 2
  (3 p ) 2  4(1)(5 p 2 )




0  x 2  3 px  5 p 2
  (3 p)2  4(1)(5 p 2 )
  11p 2
0
(4)
  9 p 2  20 p 2  11 p 2  0
2(a)(1)
3
81 .
 3
3
4
1
1
  81 3  3 3 
 
 
 34
3
1
3
3
 34

2(b)(1)
 30  1
16  3 54  3 128
3

4
 3 43 




 34. 3
2(a)(2)
 applying surd rule
 same bases
 multiplying exponents
 adding exponents and
obtaining 1
(4)
3
128
8  2  3 27  2  3 64  2
3
64  2
2 2  3 2  43 2
3
3
43 2

3
2
43 2

1
4
 23 2
 33 2
 43 2
3
2
 3
4 2
1

4
32 x  3x  2  32  3x
 3x. 32
 32 x  3x. 32  32  3 x





Let k  3x
 k 2  9k  9  k
 k 2  10k  9  0
 (k  9)(k  1)  0
 k  9 or k  1
 3x  32 or 3 x  30
 x  2 or x  0
© Gauteng Department of Education
(5)
k  3x
k 2  10k  9  0
(k  9)(k  1)  0
k  9 or k  1
x  2 or x  0
(6)
42
2(b)(2)
1
1
1
x2  x4  6  0
 k  x4
 k2  k  6  0
 (k  3)(k  2)  0
1
Let k  x 4
1
1
 k 2  ( x 4 )2  x 2
1
k2  k  6  0
 (k  3)(k  2)  0
 k  3 or k  2
1
1
 x 4  3 or x 4  2
 x  81
(5)
1
 x 4  3 or x 4  2
 x  81
no solution
2(b)(3)
2
1
x3  4
 ( x 3 )2  4
1
1
1
 x 3  2 or x 3  2
 x 8
 x  8
(4)
 ( x 3 )2  4
1
Let k  x 3
k2  4
k2  4  0
 (k  2)(k  2)  0
 k  2 or k  2
1
1
 x 3  2 or x 3  2
1
3(a)
1
 ( x 3 )3  (2)3 or ( x 3 )3  ( 2)3
x  8
or x  8
1
a2
a

a 2  2a  1
a
(a  1) 2
a
a 1

a


a 1
a

a
a

a ( a  1)
a
a 2  2a  1

a
(a  1) 2

a
a 1

a

© Gauteng Department of Education
a 1
a

to get
a
a
answer
(4)
43
3(b)
3(c)
3x  0 for all real values of x
1  p  0
 3 0
 1 p  0
 p  1
 p 1
 p 1
( x  3) 2  0
 ( x  3)  0
 ( x  3) 2  0
 ( x  3)  0
 x3
x
(3)
2
2
The only solution is x  3 since
(3)
( x  3) 2  0 for all real values of x
3(d)
px 2  2 px  rx 2  3rx  p  0
 ( p  r ) x 2  (2 p  3r ) x  p  0
  (2 p  3r )  4( p  r )( p )
2
 standard form
   9r 2  16 pr
 9r 2  16 pr  0
16 p
9
  4 p 2  12 pr  9r 2  4 p 2  4 pr
 r
  9r 2  16 pr
For equal roots,   0
 9r 2  16 pr  0
 substitution of r
 25 px 2  30 px  9 p  0
 (5 x  3)(5 x  3)  0
3
 x
(8)
5
 r (9r  16 p)  0
 9r  16 p  0
[ r  0]
 9r  16 p
16 p
r 
9
16 p  2 

 16 p  
 p 
 x   2 p  3
 x  p  0
9 

 9 

25 p 2 
16 p 
x 2p 
x p 0
9
3 

25 p 2 10 p

x 
x p 0
9
3
 25 px 2  30 px  9 p  0

 25x2  30 x  9  0 [ p  0 ]
 (5 x  3)(5 x  3)  0
3
3
 x  or x  are the equal roots
5
5
3(e)
x2  5x  6  0
 ( x  3)( x  2)  0
2  x  3
P  (2) 2  5(2)  6  20
For x  2
 ( x  3)( x  2)  0
 2 x3
 P  (2) 2  5(2)  6  20
 P  (3) 2  5(3)  6  30
© Gauteng Department of Education
44
 20  x  30
P  (3) 2  5(3)  6  30
For x  3
Possible values of P are:
(5)
20  x  30
SESSION:
14
TOPIC:
REVISION OF FUNCTIONS
SECTION A:
1(a)
(CONSOLIDATION)
SOLUTIONS TO HOMEWORK QUESTIONS
Subst (2 ; 9) into g ( x)
2
1(b)
1(c)
9  a
1
9  2
a
1
a 
3
x  0 or x  0
If f ( x)  2 x 2 , x  0
If f ( x)  2 x , x  0
 x  2 y 2 , y  0 ( f 1 )
 x  2 y 2 , y  0 ( f 1 )
x
, x0
2
 f 1 ( x)  
1(d)
and Subst (2 ;  8) into f ( x)
8  b(2)
2
8  4b
 b  2
2
 f 1 ( x)  
y
1
x 
3
 g ( x)  log 1 x
x
, x0
2
 substituting correct
point into each
function
1
 a
3
 b  2
 for either one
 x  2 y 2
x
x
or 
2
2
 correct sign for root
1
 x 
3
 log 1 x
2(a)
vertical:
horizontal:
(3)
y
(2)
3
Transformation of 2 right and 3 up
1
 h( x )   
3
(1)

3
1(e)
(3)
x2
3
x  1
y0
 2 right
 3 up
x 2
1
   3
(3)
3
 vertical asymptote
 horizontal asymptote (2)
© Gauteng Department of Education
45
2(b)
x  1
(0;2)
(1;1)
 x  1
 y0
 left branch
 coordinates on left
branch
 right branch
 coordinates on right
branch
(6)
 1  x
 x 1
(2)
(3;  1)
(2;  2)
2(c)
x  1
1
(0;2)
(1;1)
y 1
1
(3;  1)
(2;  2)
Therefore
2
 1 for 1  x  1
x 1
2(d)
Domain of f :
x    ;   where x  1
3(a)
Substitute y  12
 x    ;  
 x  1
 12  4 x  32
 D(5 ; 12)
12  4 x  32
x  5
D(5 ;12)
3(b)
y  a( x  5)2  12
 4  a(7  5) 2  12
8  4a
3(c)
Range:
(2)
 y  a( x  5)2  12
 Substituting (7 ; 4)
 a  2
Substitute (7 ; 4)
 a  2
(2)
 g ( x)  2( x  5)2  12
(4)
 y    ;12
(1)
 g ( x)  2( x  5) 2  12
y    ;12
© Gauteng Department of Education
46
4(a)
 y  a . bx  3
[Horizontal asymptote is y  3 ]
Substitute the point (0 ;  2)
2  a . b0  3
2  a .1  3
a  1
Substitute the point (1;  1)
1  (1) . b1  3
b  2
 f ( x)  2 x  3
4(b)
 1 unit left
 4 units up
h( x )  2 . 2 x  1
 h( x )  2
x 1
1
 h( x)  2 x1  3  4
 h( x)  f ( x  1)  4
This is a translation of the graph of f 1 unit left and 4
units up.
SECTION B:
1(a)(1)
SOLUTIONS TO TYPICAL EXAMINATION QUESTIONS
Intercepts with the axes:
x-intercepts: Let y  0
0  ( x  1) 2  4
 0  x2  2 x  1  4
 0  x2  2 x  3
 0  ( x  3)( x  1)
 x  3 or x  1
(3 ; 0) (1; 0)
y-intercept: Let x  0
y  (0  1)2  4
 y  3
(0 ;  3)





(6)
y-intercept
x-intercepts
coordinate of TP
axis of symmetry
shape
© Gauteng Department of Education
(2)
47
Method 1
y  ( x  1) 2  4
Axis of symmetry: x  1
Turning point: (1;  4)
Method 2
y  ( x  1)2  4
 y  x2  2x  1  4
 y  x2  2x  3
x-coordinate of turning point:
b
2
x

 1
2a
2(1)
y-coordinate of turning point:
y  (1)2  2(1)  3
x  1
 y  4
Turning point: (1;  4)
( 3 ; 0)
(1 ; 0)
(0 ;  3)
( 1 ;  4)
y   4 ;  
1(a)(2)
Range:
1(a)(3)
The graph increases for x  1
1(b)(1)
g ( x)   x 2  4 x  3
b
4
xTP  

2
2a
2(1)
yTP  (2) 2  4(2)  3  1
Turning point is (2 ;1)
 Max value is 1
 answer
(2)
 answer
(1)
 xTP  2
 yTP  1
 Max value is 1
(3)
 completing the square
 Max value is 1
© Gauteng Department of Education
48
Alternatively:
g ( x)   x 2  4 x  3
 g ( x )  ( x 2  4 x )  3
2
2
 2
 4   4  
 g ( x)    x  4 x         3
 2   2  

 g ( x)   ( x  2) 2  4   3
 g ( x)  ( x  2) 2  4  3
 g ( x)  ( x  2) 2  1
 max value is 1
1(b)(2)
The graph of g shifts 1 unit left. Therefore the
turning point of the shifted graph is (1 ; 1)
Alternatively:
g ( x)   x 2  4 x  3
 g ( x  1)  ( x  1) 2  4( x  1)  3
 recognising left shift
 (1 ; 1)
(2)
 algebraic manipulation
 (1 ; 1)
 g ( x  1)  ( x 2  2 x  1)  4 x  4  3
 g ( x  1)   x 2  2 x  1  4 x  1
 g ( x  1)   x 2  2 x
2
xTP  
1
2(1)
yTP  (1) 2  2(1)  1
Turning point is (1;1)
© Gauteng Department of Education
49
2(a)
 y  a( x  1)2  4
y  a ( x  1)2  4
Substitute (0 ; 3)
 3  a(0 1)2  4
 a  1
3  a (0  1)2  4
3  a  4
 a  1
2(b)
2(c)
2(d)
3(a)
 f ( x)  ( x  1)2  4
 f ( x)  ( x  1) 2  4
a
y
4
x 1
Substitute (0 ; 3)
 y 
a
3 
4
0 1
 3  a  4
a  1
1
 g ( x) 
4
x 1
1
0
4
x 1
 0  1  4( x  1)
0  1 4x  4
0  4x  3
4 x  3
3
x 
4
3 
 ; 0
4 
g ( x)  0 for 0  x 
a
4
x 1

a 1

g ( x) 
1
4
x 1
1
4
x 1
3
 x
4
3


  ; 0
4 
(4)
 0
 x0
3
 x
4
 0  2x  3
2 
 Q ; 0
3 
3
4
0  2x  3
2
2 
Q ; 0
3
3 
h(2)  2( 2)  3  7
h(4)  2(4)  3  5
x 
3(b)
(4)
Endpoint coordinates for h are ( 2 ;  7) and (4 ; 5)
 ( 2 ;  7) and (4 ; 5)
 ( 7 ;  2) and (5 ; 4)
 7  x  5
Endpoint coordinates for h 1 are ( 7 ;  2) and
(5 ; 4)
© Gauteng Department of Education
(3)
(2)
(2)
(3)
50
3(c)
Domain of h 1 is 7  x  5
 endpoints
 intercepts
 line
(3)
 x  2y  3
 x  2(2 x  3)  3
 x3
(3)
(5 ; 4)
4
3
2
3
7
5
h 1
2
(7 ;  2)
3(d)
y  2x  3
3(e)
 x  2y  3
(h 1 )
 x  2(2 x  3)  3
 x  4x  6  3
3 x  9
x  3
OP 2  ( x  0) 2  ( y  0) 2
( h)
 OP2  x2  y 2
 OP 2  5 x 2  12 x  9
 0  10x  12
 OP 2  x 2  y 2
 OP 2  x 2  (2 x  3) 2
12 6

10 5
 1,34
 x
 OP  x  4 x  12 x  9
2
2
2
 OP 2  5 x 2  12 x  9
d
 (OP 2 )  10 x  12
dx
 0  10 x  12
12 6
x 

10 5
(5)
2
6
6
 min length of OP  5    12    9  1,8
5
5
2
 min length of OP  1,8  1,34 units
3(f)(1)
Since f ( x)  0 for all 2  x 
axis) and since f ( x)  0 for all
3
2
3
2
(h is below the x-
 f ( x)  0
 x  4 (h is above  f ( x)  0
the x-axis), this means that the graph of f has a local
minimum at x  32
© Gauteng Department of Education
2  x 
3
2
3
2
x4
(2)
51
3(f)(2)
4.
 1,34
The maximum gradient of f occurs at x  4 since
the maximum value of x in the interval 2  x  4 for
the graph of h (which represents the gradient of f is
x4
h(4)  f (4)  2(4)  3  5
Vertical asymptote is x  2
a
y
q
x2
The vertical asymptote is y  q
a
q
x2
a
4
 y
x2
 shape
 asymptotes
(1)
 y
The coordinates of the point of intersection of the
asymptotes is therefore (2 ; q)
This point will lie on the axis of symmetry y  x  6
Substitute the point (2 ; q) into this equation to get
the value of q.
q  2  6
q  4
a
y 
4
x2
Since the graph is increasing a  0
y  x6
(2 ; 4)
y4
x  2
© Gauteng Department of Education
(4)
52
5.
 mx  32  2 x 2  4 x  30
 2 x2  (4  m) x  2  0
mx  32  2 x 2  4 x  30
 2 x 2  4 x  mx  2  0




 2 x 2  (4  m) x  2  0
  (4  m) 2  4(2)(2)
  16  8m  m 2  16
  m2  8m
m2  8m  0
m(m  8)  0
m  0 or m  8
(6)
  m 2  8m
For equal roots (one point of intersection):
0
 m 2  8m  0
 m(m  8)  0
 m  0 or m  8
6(a)
 m  tan135  1
 1  4x  5
 x  1
 P(  1 ; 6)
(4)
P(  1 ; 6)
 y  1x  q
 y  x  5
(2)
 6  1(1)  q
q  5
 y  x  5
The line y   x  5 intersects the parabola at one
 d 5
(1)
Gradient of g is:
m  tan135  1
Gradient of f is:
f ( x)  4 x  5
1  4 x  5
 4 x  4
 x  1
 f (1)  2(1) 2  5(1)  3  6
P(  1; 6)
6(b)
6(c)
y  1x  q
point. It will cut the parabola more than once if the
line shifts down (y-intercept of the line less than 5).
However, if the line shifts up, it will not cut the
parabola (y-intercept of the line greater than 5)
d  5
© Gauteng Department of Education
53
7(a)
 4  2a 1
1
 a
2
y  2a x
Substitute (  1; 4)
 4  2 a 1
2
4 
a
 4a  2
1
a 
2
1
 f ( x)  2  
2
1
 f ( x)  2  
2
7(b)
7(c)
(3)
x
0
0
1
y  2   2
2
(0 ; 2)
1
g ( x)  2  
2
x
1
 2 
2
 (0 ; 2)
x
1
 g ( x)  2  
2
(2)
x
 g ( x )  2(2 x )
 g ( x)  2(2 x )
 g ( x )  21 x
 g ( x)  2x1
(3)
 y-intercept
 x-intercept
 asymptote
 shape
(4)
 g ( x )  2 x 1
7(d)
(0 ; 2)
(1 ; 0)
y  1
© Gauteng Department of Education
54
8(a)
8(b)
y  a x 1
1
  a 01
2
1
  a 1
2
1 1
 
2 a
a  2
y  2 x 1
f
x  2 y 1 f 1
 log 2 x  y  1
 substitution of point
 a2
(2)
 x  2 y 1
 f 1 ( x)  log 2 x  1
(2)
 log 2 x  1  y
 f 1 ( x)  log 2 x  1
8(c)
y
y  2 x 1
y x
y  log 2 x  1
(2 ; 2)
For y  2 x 1 :
 y-intercept
 one other point
For y  log 2 x  1 :
 x-intercept
 one other point
 functions intersecting at
(1 ; 1) and (2 ; 2)
(5)
(1 ; 1)
(0 ; 12 )
x
( 12 ; 0)
8(d)
Domain of f 1 :
x0
 x0
© Gauteng Department of Education
(1)
55
SESSION NO:
15
TOPIC:
REVISION OF GRADE 11 EUCLIDEAN GEOMETRY
SECTION A:
(CONSOLIDATION)
SOLUTIONS TO HOMEWORK QUESTIONS
QUESTION 1
(a)
(b)
ˆ D
ˆ  90
D
1
2
 in a semi-circle
But D̂2  50
given
D̂1  40
ˆ  2D
ˆ
M
ˆ D
ˆ  90
 D
1
2
 D̂1  40
(2)
 at centre  2   at circumference
ˆ  2D
ˆ
 M
1
1
1
1
M̂1  2(40)
 M̂1  80
(2)
 M̂1  80
(c)
(d)
Ê2  50
 's in same segment
ˆ  2D
ˆ
 M
1
1
F̂2  50
 's opp equal sides
( ME  FE , equal radii)
 's in same segment
 F̂2  50
(2)
ˆ  Fˆ  Fˆ
 G
1
2
ˆ  Fˆ  Fˆ
G
1
2
 Ĝ  10  50
(e)
 Ĝ  60
ˆ  Eˆ  180
ˆ D
ˆ G
sum of the  's of a triangle
D
1
2
1
 40  50  60  Eˆ  180
1
 Ê1  30
 Ĝ  60
(2)
ˆ  Eˆ  180
ˆ D
ˆ G
 D
1
2
1
 Ê1  30
(2)
QUESTION 2
(a)
Pˆ2  Pˆ1
given
 Pˆ2  Pˆ1
But P̂1  22
given
 P̂2  22
(2)
 R̂ 2  22
 reason
(2)
 Pˆ2  Pˆ3  Pˆ4  90
 P̂2  22
(b)
R̂ 2  22
tan-chord
(c)
Pˆ2  Pˆ3  Pˆ4  90
 in a semi-circle
But P̂2  22
 Pˆ3  Pˆ4  90  22  68
 P̂3  68
 reason
(3)
© Gauteng Department of Education
56
(d)
Rˆ 1  Rˆ 2  Pˆ1  Pˆ2
But Pˆ  Pˆ  44
1
2
tan-chord
given
ˆ R
ˆ  44
R
1
2
But R̂ 2  22
tan-chord
 R̂1  22  44
ˆ  Rˆ  Pˆ  Pˆ
 R
1
2
1
2
ˆ  Rˆ  44
 R
1
2
 R̂1  22
 reasons
(4)
 R̂1  22
(e)
R̂1  22
proved
 T̂1  22
equal radii,  ’s opp
 Ô1  44
 reasons
(3)
(f)
T̂1  22
equal sides
Ô1  44
ˆ  180
Rˆ  Pˆ  Q
2
2
ext  of triangle
sum of the  's of a triangle
ˆ  180
 22  (90  22)  Q
2
ˆ  180
ˆ  Pˆ  Q
 R
2
2
 Q̂2  46
 reason
(3)
 Q̂2  46
QUESTION 3
(a)
ˆ
Lˆ 3  M
1
ˆL  Pˆ
3
alt  's equal
tan-chord
1
ˆ  Pˆ
M
1
1
(b)
 LM  LP
sides opp equal  's
ˆ  Pˆ
N
1
1
ˆ
Pˆ  M
ML subtends equal  's
ˆ N
ˆ
M
1
2
ˆ
ˆ
N  N
PL subtends equal  's
1
1
(c)
proved
1
ˆ  Pˆ
M
1
1
ˆ
ˆ
N P
1
1
2
proved
ML subtends equal  's
ˆ N
ˆ
M
1
1
 LM is a tangent to circle MNQ.  between line and
chord
ˆ M
ˆ
 L
3
1
ˆ
ˆ
 L P
3
1
 LM  LP
 reasons
(4)
ˆ  Pˆ
 N
1
1
ˆ N
ˆ
 M
1
2
ˆ N
ˆ
 M
1
2
 reasons
(4)
ˆ  Pˆ
 M
1
1
ˆ
ˆ
 N P
1
1
ˆ N
ˆ
 M
1
1
 reasons
(4)
© Gauteng Department of Education
57
QUESTION 4
(a)
D̂3  90
ˆ B
ˆ  90
B
1
 in semi-circle
given
2
ˆ B
ˆ B
ˆ
D
3
1
2
 ABCD is a cyclic quad
(b)
ˆ D
ˆ
A
1
1
ˆ  Eˆ
D
1
ext  of quad equals int opp 
BC subtends equal angles
tan-chord
1
ˆ  Eˆ
A
1
(c)
2
ˆ B
ˆ B
ˆ
 D
3
1
2
 reasons
(4)
ˆ D
ˆ
 A
1
1
ˆ  Eˆ
 D
1
 reasons
(3)
ˆ A
ˆ C
ˆ
 A
1
2
3
ˆ D
ˆ
 C
ˆ A
ˆ C
ˆ
A
1
2
3
ˆ D
ˆ
C
ext   int opp 
ˆ D
ˆ
D
4
2
ˆ
ˆ D
ˆ
 A1  A
2
2
vertically opp  's
ˆ D
ˆ
 D
4
2
ˆ
ˆ
ˆ
 A A  D
 BD  BA
sides opp equal  's
ˆ D
ˆ
C
2
2
ˆ
ˆ
D2  D4
ˆ
D̂  C
AB subtends equal  's
 reasons
(5)
ˆ D
ˆ
 C
2
2
ˆ
ˆ
 D D
3
(d)
 D̂3  90
ˆ B
ˆ  90
 B
tan-chord
4
4
vertically opp  's
3
3
1
2
4
2
2
4
tan-chord
ˆ
 D̂4  C
3
 reasons
(4)
ext  of AMP
ˆ M
ˆ
 Pˆ1  A
1
1
ˆ
ˆ
 M K
ˆ C
ˆ
C
2
3
QUESTION 5
ˆ M
ˆ
Pˆ1  A
1
1
ˆ
ˆ
But M  K
1
 ’s opposite equal sides
1
ˆ K
ˆ
 Pˆ1  A
1
1
ˆ
ˆ
ˆ
But P  C  K
2
1
ˆ C
ˆ
and A
1
 Pˆ  Pˆ
1
2
ext  of AMP
tan-chord
1
1
ˆ K
ˆ
 Pˆ1  A
1
1
ˆ
ˆ
ˆ
 P2  C  K1
ˆ C
ˆ
 A
1
 Pˆ1  Pˆ2
 reasons
(7)
© Gauteng Department of Education
58
SECTION B:
SOLUTIONS TO TYPICAL EXAMINATION QUESTIONS
QUESTION 1
(a)(1)
equal to 90
(a)(2)
bisects the chord
(b)(1)
Ĉ  90
angle in semi-circle
(b)(2)
Ê1  90
corr angles equal
(b)(3)
Perp from centre to chord
(b)(4)
Given
AE  8 units
2
2
2
 OE  10  8
 OE  6 units
But OD  AO  10 units
 ED  4 units
 answer
(1)
 answer
(1)
 answer
 reason
(2)
 answer
 reason
(2)
 answer
(1)
 AE  8 units
 OE  6 units
 ED  4 units
(3)
QUESTION 2
(a)
B̂1  40
tan-chord
(b)
D̂2  40
angles opp equal sides
(c)
Ĉ  100
sum of the  ’s of a 
(d)
Ô2  200
 at centre  2   at circle
(e)
Ô1  160
 ’s round a point
(f)
ˆ  180 sum of the  ’s of a 
ˆ B
ˆ O
D
3
2
1
 B̂1  40
 reason
(2)
 D̂2  40
 reason
(2)
 Ĉ  100
 reason
(2)
 Ô2  200
 reason
(2)
 Ô1  160
 reason
(2)
ˆ  180
ˆ B
ˆ O
 D
3
2
1
ˆ
ˆ
 D B
3
© Gauteng Department of Education
2
59
ˆ B
ˆ  160  180
D
3
2
ˆ B
ˆ  20
D
3
 D̂3  10
(3)
2
ˆ B
ˆ
But D
3
2
ˆ D
ˆ  20
D
3
angles opp equal radii
3
ˆ  20
 2D
3
 D̂3  10
(g)
  80
opp  ’s cyclic quad
or  at centre  2   at circle
 Â  80
 reason
(2)
QUESTION 3
(a)
Ĉ1  20
Alt angles equal
(b)
Ô1  40
 at centre  2   at circle
(c)
D̂  20
(d)
Ê1  40
 at centre  2   at circle
or
angles in same segment
Ext  of triangle
(e)
ˆ  40
Ê1  O
1
 Ĉ1  20
 reason
(2)
 Ô1  40
 reason
(2)
 D̂  20
 reason
(2)
 Ê1  40
 reason
(2)
 answer
(1)
QUESTION 4
(a)
(b)
ˆ N
ˆ  90 given
N
1
2

T̂3  90
 in semi-circle
ˆ N
ˆ  Tˆ
N
1
2
3
 MNPT is a cyclic quad Ext   int opp 
 T̂3  90
ˆ N
ˆ  Tˆ
 N
Tˆ1  Tˆ 4
ˆ
Tˆ  M
vertically opp angles
ˆ  Pˆ
M
1
 Tˆ  Pˆ
ext  of cyclic quad
 NP  NT
sides opp equal  s
4
1
tan chord
ˆ N
ˆ  90
N
1
2
1
2
 reasons
(4)
 Tˆ1  Tˆ 4
ˆ
 Tˆ  M
4
1
ˆ  Pˆ
 M
1
 Tˆ  Pˆ
1
1
 reasons
(5)
© Gauteng Department of Education
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60
QUESTION 5
(a)
ˆ A
ˆ  90
A
1
2
tan  radius
Â2  x
tan-chord
 Â1  x  90
 Â1  90  x
ˆ A
ˆ B
ˆ  Eˆ  180 sum of the  ’s of a 
A
1
2
1
 90  x  Eˆ  180
 Ê  90  x
ˆ  Eˆ
A
1
 AB is a tangent to circle ADE
since  between line and chord equals  in alt
segment.
ˆ A
ˆ
ext  of cyclic quad
C
1
1
ˆ  Eˆ  90  x
proved
A
(b)
1
ˆ  Eˆ
C
1
ˆ A
ˆ  90
 A
1
2
 Â2  x
 Â1  90  x
ˆ A
ˆ B
ˆ  Eˆ  180
 A
1
2
1
 Ê  90  x
ˆ  Eˆ
 A
1
 reasons
(7)
ˆ A
ˆ
 C
1
1
ˆ  Eˆ  90  x
 A
1
(2)
QUESTION 6
AB  x
AC2  AB2  BC2
 (13)2  x 2  (x  7) 2
tangents from the same point
B̂  90 tan  rad
 AB  x
 (13) 2  x 2  (x  7) 2
 0  x2  7 x  60
 0  ( x  12)( x  5)
 x5
(5)
169  x 2  x 2  14 x  49
 0  2 x 2  14 x  120
 0  x 2  7 x  60
 0  ( x  12)( x  5)
 x  12 or x  5
But x  12
x  5
QUESTION 7
(a)
ˆ
Tˆ1  Pˆ2  Q
1
ˆ
ˆ
T̂  Q  Q
1
3
Pˆ2  Rˆ 1
ˆ
R̂  Q
1
3
ˆ
 P̂2  Q
3
2
ext  of 
 s opp equal sides
 s in the same seg
 s opp  sides
ˆ
 Tˆ1  Pˆ2  Q
1
ˆ
ˆ
 T̂  Q  Q
1
3
ˆ
 Pˆ2  R
1
ˆ
 R̂  Q
1
3
ˆ
 P̂2  Q
3
© Gauteng Department of Education
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61
ˆ Q
ˆ
Q
1
2
(b)
ˆ  Pˆ  180
Pˆ1  QPR
5
ˆP  Rˆ
1
2
ˆ R
ˆ
R
2
3
ˆ
ˆ
P  R
1
tan-chord theorem
3
5
1
2
ˆ
 Pˆ1  R
3
ˆ
ˆ  Pˆ  180
 R 3  QPR
5
ˆ
ˆ
ˆ
 R  Y  P  180
given
ˆ  QPR
ˆ  Pˆ  180
R
3
5
ˆ
ˆ
ˆ
Now R  Y  P  180
3
adj  s on a line
ˆ Q
ˆ
 Q
1
2
(6)
ˆ  Pˆ  180
 Pˆ1  QPR
5
ˆ
ˆ
 P R
3
int  s of 
ˆ Y
ˆ
 QPR
QP is a tangent to the circle
 betw line and chord
ˆ Y
ˆ
 QPR
 reasons
(7)
© Gauteng Department of Education
5