Mathematics/P2
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Avusa Education Exemplar 2009
MATHEMATICS EXEMPLAR EXAMINATION
GRADE 12
PAPER 2
MEMORANDUM
QUESTION 1
y
P(5;2)
β
α
S
x
Q(1 ; − 1)
T
W
θ
R(9; − 5)
1.1
1.2
 1 + (9) −1 + ( −5) 
W
;

2
 2

= W(5 ; − 3)
The equation of PW is x = 5
−5 − (−1) −4
1
mQR =
=
=−
9 −1
8
2
1
∴ mPS = −
(PS||QR)
2
1
y − 2 = − ( x − 5)
2
1
5
∴y−2= − x+
2
2
1
9
∴y = − x+
2
2
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midpoint
x=5
(2)
mQR
mPS
correct substitution
into formula for
equation
1
9
y = − x+
2
2
(4)
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mPT = 2
(PT ⊥ QR)
y − 2 = 2( x − 5)
∴ y − 2 = 2 x − 10
∴ y = 2x − 8
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mPT
correct substitution
into formula for
equation
y = 2x − 8
(3)
1.4
mQR = −
1
2
1
y − (−1) = − ( x − 1)
2
1
1
∴ y +1 = − x +
2
2
1
1
∴y = − x−
2
2
1
1
∴− x − = 2 x − 8
2
2
∴− x − 1 = 4 x − 16
correct substitution
into formula for
equation
1
1
y = − x−
2
2
1
1
− x − = 2x − 8
2
2
x=3
T(3; − 2)
(5)
∴−5 x = −15
∴x = 3
∴ y = 2(3) − 8 = −2
∴ T(3; − 2)
1.5
QT 2 = (1 − 3)2 + (−1 − ( −2))2
∴ QT 2 = 4 + 1
∴ QT 2 = 5
∴ QT = 5
TR 2 = (3 − 9) 2 + (−2 − ( −5))2
∴ TR 2 = 36 + 9
∴ TR 2 = 45
correct substitution to
get QT
answer for QT
correct substitution to
get TR
answer for TR
establishing that
1
QT = TR
3
(5)
∴ TR = 45 = 9 × 5 = 3 5
1
∴ TR = 5 = QT
3
1
∴ QT = TR
3
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∴ tan α =
tan α = −1
α = 135°
β = 63, 43494882°
ˆ = 71,56505118°
TPR
∴α = 180° − 45°
θ = 18, 43°
tan α = mPR
1.6
2 − (−5)
5−9
∴ tan α = −1
(5)
∴α = 135°
tan β = mPT
∴ tan β = 2
∴β = 63, 43494882°
ˆ +β = α
Now TPR
ˆ = α −β
∴ TPR
ˆ = 135° − 63, 43494882°
∴ TPR
ˆ = 71,56505118°
∴ TPR
θ + 90° + 71,56505118° = 180°
∴θ = 18, 43°
QUESTION 2
y
.
D
.
.
C
A(0; 2)
.
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x
B( − 1; − 1)
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mAB = mBC
mAB = mBC
working out gradients
k =5
(3)
2−5 k + 4− 2
=
3− 6
2k − 3
k +2
∴1 =
2k − 3
∴ 2k − 3 = k + 2
∴k = 5
x2 + y 2 − 4 x + 6 y + 4 = 0
∴
2.2
( x − 2)2
( y + 3) 2
∴ x 2 − 4 x + y 2 + 6 y = −4
2
2
2
 −4 
6
 −4   6 
∴ x − 4 x +   + y 2 + 6 y +   = −4 +   +  
 2
2
 2  2
2
2
∴ ( x − 2)2 + ( y + 3)2 = −4 + 4 + 9
∴ ( x − 2)2 + ( y + 3)2 = 9
centre = (2; − 3)
new centre after rotation of 90° clockwise: ( − 3; − 2)
new centre after enlargement through origin: (−6; − 4)
original radius: r = 3
new radius after enlargement through origin : r = 3 × 2 = 6
new circle:
r2 = 9
new centre = (−6; − 4)
new radius: r = 6
new circle:
( x + 6)2 + ( y + 4)2 = 36
(6)
( x + 6) 2 + ( y + 4)2 = 36
2.3.1
3 x + 4 y = −7
∴ 4 y = −3 x − 7
3
7
∴y = − x−
4
4
3
∴ mCB = −
4
4
∴ mtangent =
3
4
y − (−1) = ( x − (−1))
3
4
∴ y + 1 = ( x + 1)
3
4
4
∴ y +1 = x +
3
3
4
1
∴y = x+
3
3
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mCB = −
3
4
4
3
substitution into
equation of line
4
1
y = x+
3
3
mtangent =
(4)
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yC = 2
x = −5
C( − 5; 2)
r =5
( x + 5) 2 + ( y − 2) 2 = 25
(5)
C(x ; 2)
Substitute y = 2 into 3 x + 4 y = −7
3 x + 4(2) = −7
∴ 3 x = −15
∴ x = −5
∴ C( − 5; 2)
∴ ( x + 5) 2 + ( y − 2)2 = r 2
Now r = 5
∴ ( x + 5) 2 + ( y − 2)2 = 25
correct substitution
into midpoint formula
xD = −9
yD = 5
D( − 9;5)
(4)
y + yB 
x +x
C( − 5; 2) =  D B ; D

2
2


 x − 1 yD − 1 
∴ C( − 5; 2) =  D
;

2 
 2
x −1
y −1
∴−5 = D
and 2 = D
2
2
∴−10 = xD − 1 and 4 = yD − 1
2.3.3
∴ xD = −9
and
yD = 5
∴ D( − 9;5)
y
QUESTION 3
.
A / ( − 6; 4)
.
.
D(2;3)
A( − 3;2)
.
B( − 2; − 3)
.
.
D / (4;6)
.
x
.
C(3; − 2)
C / (6; − 4)
B / ( − 4; − 6)
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B( − 2; − 3)
B( − 2; − 3)
C(3; − 2)
D(2;3)
C(3; − 2)
D(2;3)
3.2
3.3
ABCD is a square since:
Diagonals are equal in length
Diagonals bisect each other at right angles
square
properties
A / ( − 6; 4)
correct coordinates
indicated
joining points to
form enlarged square
(2)
(2)
B/ ( − 4; − 6)
C / (6; − 4)
D / (4;6)
3.4
(3)
Area ABCD
/ / /
Area A B C D
/
=
1
2
2
=
1
4
1
4
(1)
3.5.1
answer
E(3; 2)
(1)
3.5.2
3.6
Perimeter ABCD 4 × side AB
=
=1
Perimeter EFGH 4 × side EF
(since AB = EF)
1 1 
( x ; y) →  x ; y 
2 2 
1
1 1  1
 x; y →  x;−
2
2
2
2

 
reduction by a factor of
answer
(1)
1
2

y
reflection about x − axis

1  1
1
1

 x ; − y  →  x ; − y − 1 translation of 1 unit downwards
2  2
2
2

reduction
reflection
translation
(3)
1
1

∴ ( x ; y ) →  x ; − y − 1
2
2

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QUESTION 4
y
.
.
A(1 ; 2, 4)
22,5°
22,5°
22,5°
135°
22,5°
x
22,5°
22,5°
22,5°
22,5°
A / (p ; q )
4.1
22,5°× 6 = 135°
22,5°
135°
(2)
4.2
/
x = (1) cos(−135°) − (2, 4) sin( −135°)
∴ x/ = 1
/
y = (2, 4) cos(−135°) + (1) sin(−135°)
y / = −2, 4
∴ A / (1; − 2, 4)
correct substitution
into formula for x /
x/ = 1
correct substitution
into formula for y /
y / = −2, 4
(4)
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QUESTION 5
5.1
(− tan 60°)(cos156°)
− cos 66°
sin 48°
− 3
− sin 24°
2sin 24° cos 24°
3
2
(7)
tan(−60°) cos(−156°) cos 294°
sin 492°
(− tan 60°)(cos156°)(− cos 66°)
=
(sin132°)
=
(− 3)(− cos 24°)(− sin 24°)
(sin 48°)
(− 3)(− cos 24°)(− sin 24°)
2sin 24° cos 24°
3
=
2
=
5.2
cos 2 (180° + x) [ tan(360° − x).cos(90° + x) + sin( x − 90°).cos180°]
= (− cos x)2 [ (− tan x )(− sin x) + (− cos x)(−1) ]
 − sin x 

= (cos 2 x) 
 (− sin x) + cos x 
 cos x 

2
 sin x

= (cos 2 x) 
+ cos x 
 cos x

 sin 2 x + cos 2 x 
= (cos 2 x) 

cos x


 1 
= (cos 2 x) 
 cos x 
= cos x
5.3
sin 61° =
a
1
sin 2 x + cos 2 x
cos x
1
cos x
cos x
(9)
y
( 1− a ; a)
x 2 + ( a )2 = (1)2
∴ x2 = 1 − a
1
∴ x = 1− a
61°
cos 73° cos15° + sin 73° sin15°
cos 2 x
− tan x
− sin x
− cos x
−1
− sin x
cos x
x
diagram
x = 1− a
cos 58°
2 cos 2 29° − 1
2 sin 2 61° − 1
2a − 1
(6)
= cos(73° − 15°)
= cos 58°
= 2 cos 2 29° − 1
= 2sin 2 61° − 1
= 2( a ) 2 − 1
= 2a − 1
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QUESTION 6
6.1.1
sin(45° + θ).sin(45° − θ)
= [sin 45° cos θ + cos 45° sin θ][sin 45° cos θ − cos 45° sin θ]
 2
 2

2
2
=
cos θ +
sin θ  
cos θ −
sin θ 
2
2
 2
 2

 2
 2

=
(cos θ + sin θ)  
(cos θ − sin θ) 
 2
 2

2
= (cos θ + sin θ)(cos θ − sin θ)
4
1
= (cos 2 θ − sin 2 θ)
2
1
= cos 2θ
2
6.1.2
sin 75°.sin15°
sin 45° = cos 45° =
2
2
(cos 2 θ − sin 2 θ)
1
cos 2θ
2
(5)
45° + 30° ; 45° − 30°
1
cos 60°
2
1
4
(3)
= sin(45° + 30°).sin(45 − 30°)
1
cos 2(30°)
2
1
11 1
= cos 60° =   =
2
22 4
=
6.2
expansion of
sin(45° + θ)
expansion of
sin(45° − θ)
sin 2 x + 2sin x + cos 2 x + cos x = 0
∴ 2sin x cos x + 2sin x + cos 2 x + cos x = 0
∴ 2sin x(cos x + 1) + cos x(cos x + 1) = 0
∴ (cos x + 1)(2sin x + cos x) = 0
∴ cos x = −1
or
2sin x = − cos x
sin x
1
∴
=−
cos x
2
∴ tan x = −0, 5
2sin x cos x
factorising by
grouping
( )( ) = 0
cos x = −1
tan x = −0,5
general solutions
Deduct 1 mark if
k ∈ is not stated.
(7)
x = 0° + k 360°
x = 180° + k 360°
OR
x = 153, 4° + k 360°
x = 333, 4° + k 360°
x = 0° + k180°
OR
x = 153, 4° + k180°
x = 180° + k180°
x = 333, 4° + k180°
OR
x = ±180° + k 360°
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x = −45° + k180°
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QUESTION 7
B
m
12
110°
m
20
C
7m
D
7.1
28m
AC2 = (12m)2 + (20m)2 − 2(12m)(20m) cos110°
∴ AC2 = 708,1696688
7.2
∴ AC = 26, 6m
ˆ
sin BAC
sin110°
=
12m
26, 6m
ˆ = 12 × sin110°
∴ sin BAC
26, 6m
ˆ = 0, 4239214831
∴ sin BAC
A
substitution into cosine
rule
answer
(2)
substitution into sine
or cosine rule
answer
(2)
ˆ = 25°
∴ BAC
OR
ˆ
(12m)2 = (20m)2 + (26, 6m)2 − 2(20m)(26, 6m) cos BAC
ˆ = 963, 56m 2
∴1064 cos BAC
ˆ = 0,9056015038
∴ cos BAC
ˆ = 25°
∴ BAC
7.3
ˆ
(26,6m)2 = (7 m) 2 + (28m)2 − 2(7 m)(28m) cos D
ˆ = 125, 44
∴ 392 cos D
ˆ = 0,32
∴ cos D
7.4
∴ D̂ = 71°
Area ABCD
1
1
= (12m)(20m) sin110° + (7 m)(28m) sin 71°
2
2
= 205, 4m 2
substitution into cosine
rule
ˆ = 0,32
cos D
answer
(3)
1
(12m)(20m) sin110°
2
1
(7 m)(28m) sin 71°
2
answer
(3)
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QUESTION 8
cos( x − 30°) = sin 3 x
8.1
∴ cos( x − 30°) = cos(90° − 3 x)
∴ x − 30° = 90° − 3 x + k 360°
x − 30° = 360° − (90° − 3x )
∴ 4 x = 120° + k 360°
∴ x − 30° = 270° + 3x + k 360°
∴ x = 30° + k 90°
∴−2 x = 300° + k 360°
∴ x = −150° + k180°
x = 30°
or
cos(90° − 3 x)
4 x = 120° + k 360°
−2 x = 300° + k 360°
x = 30° + k 90°
x = −150° + k180°
x = 30°
x = 120°
x = −60°
(8)
x = 120° or
x = −60°
see diagram below
8.2
f ( x) = cos( x − 30°) :
shift of 30° right
amplitude
range
g ( x) = sin 3 x :
period of 120°
amplitude
intercepts with axes
(6)
2
y
1
f
x
–60
–30
30
60
90
120
g
–1
–2
8.3
Points of intersection of the two graphs
correct explanation
(1)
8.4
cos( x − 30°) > sin 3 x
∴−60° < x < 120° where x ≠ 30°
OR x ∈ ( −60° ;30° ) − {30°}
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−60° < x < 120°
x ≠ 30°
(2)
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QUESTION 9
correct second column
(1)
9.1
Number of kilometres
10 < x ≤ 20
20 < x ≤ 30
30 < x ≤ 40
40 < x ≤ 50
50 < x ≤ 60
60 < x ≤ 70
Number of motorists
2
7
4
13
16
8
Cumulative frequency
2
9
13
26
42
50
Cumulative frequency
9.2
52
50
48
46
44
42
40
38
36
34
32
30
28
26
24
22
20
18
16
14
12
10
8
6
4
2
0
.
.
endpoints of class
intervals
cumulative
frequencies
joining points
(3)
.
.
.
. .
5 10 15 20 25 30 35 40 45 50 55 60 65 70 75
Number of kilometres
9.3
median lies in the interval 48 ≤ x ≤ 49
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median in the
allowable interval (1)
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QUESTION 10
plotting of points
10.1
Number of bacteria
(2)
.
80
75
70
65
60
55
50
45
40
35
30
25
20
15
10
5
.
.
.
.
.
.
.
0
1 2
3 4
.
.
5 6 7
.
8 9 10
Time in hours
10.2
quadratic or exponential
answer
(1)
(2)
QUESTION 11
23
20
25
18
22
17
28
24
27
25
11.1
x = 22, 9
answer
11.2
standard deviation = 3,5
answer
(2)
11.3
(x − s; x + s )
= ( 22,9 − 33,5; 22, 9 + 33,5 )
= (19, 4 ; 26, 4)
4 temperatures lie outside the first standard deviation
interval
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(19, 4 ; 26, 4)
4 temperatures
(2)
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QUESTION 12
20
10
10
minimum:
maximum:
median:
Lower quartile:
Upper quartile:
Mean:
20
20
45
x
45
80
52
y
10
80
45
20 + 20
2
51 + 53
52 =
2
10 + 20 + 20 + x + 45 + y + 51 + 53 + 80
= 40
9
x + y + 279
∴
= 40
9
∴ x + y + 279 = 360
∴ x + y = 81
Now 20 < x < 45 and 45 < y < 51
Therefore let x = 34 and y = 47
20 =
51
53
80
min, median and
max
T2 = T3 = 20
T7 = 51 T8 = 53
working with the
mean
value for x and y
nine set of
numbers
Accept variations
for T7 , T8 , x and y
BUT make sure
that the mean of all
nine numbers is 40.
(6)
Therefore the set of nine numbers are:
10; 20; 20; 34; 45; 47; 51; 53; 80
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