NATIONAL
SENIOR CERTIFICATE
GRADE 12
MATHEMATICS P2
FEBRUARY/MARCH 2010
MEMORANDUM
MARKS: 150
This memorandum consists of 14 pages.
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QUESTION 1
1.1
Range = 26 – 4 =22
1.2
Mean
4 + 5 + 8 + 13 + 19 + 22 + 25 + 26 + 23 + 17 + 14 + 7
=
12
183
=
12
= 15,25
Standard deviation = 7,6
(7,59522…..)
1.3
1.4.1
Increase in mean =
The maximum value increases by 1°C and the
minimum value increases by 5°C. This implies that
the range of the range of the data will now decrease.
This will result in the standard deviation getting
smaller. (new SD = 6,27…..)
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9 method
9183
9answer
(3)
99answer
(2)
(3 × 5) + (9 × 1)
12
= 2°C per month.
1.4.2
9maximum and minimum
values
9answer
ANSWER ONLY: Full Marks
(2)
99answer
(2)
9decrease in range
9decrease in standard deviation
(2)
[11]
Mathematics/P2
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DoE/Feb. – March 2010
QUESTION 2
2.1.1
Number of goals scored
Survey of training and goals scored
999plotting the
points
16
14
12
10
A
8
6
4
2
0
0
5
10
15
20
25
Hours of training
2.1.2 A(indicated on the graph)
All 9 point
correct – 3 marks
5 or 7 points
correct – 2 marks
1 or 2 points
correct – 1 mark
0 points correct –
0 marks
(3)
9answer
(1)
2.1.3 8 Goals
99answer
(2)
2.2
Let the mean time for all 560 learners be x.
Then the mean time for the learners living in neighbourhood C is also
x.
(135 × 24) + (225 × 32) + (200 × x)
x=
560
560 x = 3240 + 7200 + 200 x
360 x = 10440
x = 29
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9equal mean
times
9mean × number
9simplification
9answer
(4)
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QUESTION 3
3.1
Time (in
11 ≤ t < 15
minutes)
Frequency
6
Cumulative
6
Frequency
15 ≤ t < 19
19 ≤ t < 23
23 ≤ t < 27
27 ≤ t < 30
9
15
13
28
12
40
8
48
9cumulative
frequency
totals
(1)
3.2
Cumulative Frequency Curve showing the
time taken to complete a task
999plotting
points at upper
limits
6 correct – 3
marks
3 to 5 correct
– 2 marks
1 or 2 correct
– 1 mark
0 correct – 0
marks
60
Cumulative Frequency
50
40
30
20
9curve
10
(4)
0
0
3
6
9
12
15
18
21
24
27
30
33
Time (in minutes)
3.3 Median value at position 24. Reading off the ogive gives Median ≈ 22 minutes
LQ value at position 12. Lower quartile ≈ 18 minutes (from ogive)
UQ value at position 36. Upper quartile ≈ 25,5 minutes (from ogive)
NOTE: Allow margin of error for reading off the graph.
9 median
9 lower
quartile
9 upper
quartile
(3)
3.4
0
10
20
30
40
50
60
3.5 The times are skewed to the right. A small number of people finished this task
very quickly whilst others took more time.
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9 box
9 whiskers
(2)
9 skewed to
the right
(1)
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QUESTION 4
4.1
4.2
4.3
2−0
1
=−
0−4
2
⎛0+ 4 2+0⎞
A: ⎜
;
⎟
2 ⎠
⎝ 2
A (2 ; 1)
9 substitution
m AB .mPQ = −1
9 m AB .mPQ = −1
m AB .( −1 / 2) = − 1 , ∴ m AB = 2
Equation of AB is y = 2x + c
∴ 1 = 2(2) + c
c = –3
Equation of AB is y = 2x – 3 .
9 m AB = 2
9 equation of AB
9 y = 2x – 3
9 c = –3
m PQ =
(1)
9 x-coordinate
9 y-coordinate
(2)
OR
m AB .mPQ = −1
(5)
9 m AB .mPQ = −1
m AB .( −1 / 2) = − 1 , ∴ m AB = 2
y − 1 = 2( x − 2)
y − 1 = 2x − 4
y = 2x − 3
9 m AB = 2
9 gradient of AB
9 substitution into
formula
9 equation of AB
(5)
4.4
B is the point (0 ; –3)
BQ = (0 - 4) + (−3 − 0)
2
2
=5
4.5
4.6
(3)
BP = (0 - 0) 2 + (−3 − 2) 2
=5
BP = BQ
∴ΔBPQ is isosceles.
OR
BP = 2 + 3
=5
BP = BQ
∴ΔBPQ is isosceles
If PBQR is a rhombus then A is the midpoint of BR.
Let the coordinates of R be (x ; y)
x+0
=2
2
x=4
∴R(4 ; 5)
and
y −3
=1
2
y=5
OR
RQ || PB so x R = 4
RQ = PB = 5, so y R = 5
∴R(4 ; 5)
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9 coordinates of B
9 substitution
9 answer
9 BP = 5
9 BP = BQ
(2)
9 BP = 5
9 BP = BQ
(2)
9 A is the midpoint of
BR
9 x coordinate
9 y coordinate
(3)
9 RQ || PB
9 x coordinate
9 y coordinate
(3)
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QUESTION 5
C
y
B
45°
P(5 ; 4)
α
β
x
O
A
D
5.1
AB is defined as 5 y − 3 x − 5 = 0 which can be written as y =
3
5
Let α be the inclination of AB.
3
tan α =
5
α = 30.96° .
m AB =
Let β be the inclination of CD
β = 45° + 30,96°
= 75,96°
Gradient of CD = tan 75,96° = 4.
3
x +1
5
9 m AB =
3
5
9 tan α =
3
5
9 α = 30.96°
9 β = 75,96°
9 gradient of CD
(5)
OR
tan β = tan(α + 45°)
tan α + tan 45°
1 − tan α . tan 45°
3
+1
5
=
3
1 − ×1
5
=4
mCD = tan β
=
9 expansion
9 tan 45° = 1
3
9 tan α =
5
9 substitution
9 answer
(5)
mCD = 4
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Equation of CD is y = 4x + c
∴ 4 = 4(5) + c
c = – 16
Equation of CD is y = 4x –16 .
DoE/Feb. – March 2010
9 y- intercept
9 equation of CD
(2)
OR
y − 4 = 4( x − 5)
y − 4 = 4 x − 20
y = 4 x − 16
9 substitution
9 equation of CD
(2)
[7]
QUESTION 6
6.1
x 2 + y 2 + 8 x + 4 y − 38 = 0
x 2 + 8 x + 16 + y 2 + 4 y + 4 = 16 + 4 + 38
( x + 4) 2 + ( y + 2) 2 = 58
Centre is (–4 ; –2) and the radius is
6.2
Centre of second circle is (4 ; 6)
Distance between centres is
6.3
6.4
58
( 4 + 4) 2 + (6 + 2) = 128 = 11,31
9completing the
square (both or one)
9 factor form
9 centre
9 radius
(4)
9centre
9distance
(2)
Sum of radii = 58 + 26 = 12,71
Distance between centres is 11,31.
99 sum of radii
sum of the radii > distance between the centres
9 conclusion
(3)
∴ the circles must overlap and hence the circles must intersect.
Equation of second circle:
(x − 4)2 + ( y − 6) 2 = 26
x 2 − 8 x + 16 + y 2 − 12 y + 36 = 26
x 2 − 8 x + y 2 − 12 y + 26 = 0
9 equation of circle in
form = 0
Let (x ; y) be either of the two points on intersection.
Then
x 2 + y 2 + 8 x + 4 y − 38 = 0
9 statement – two
points of intersection
9 subtracting
and
x 2 + y 2 − 8 x − 12 y + 26 = 0
Subtract
16 y + 16 x − 64 = 0
y = −x + 4
9 simplification
(4)
Both points of intersection lie on this line.
∴ y = –x + 4 is the equation of the common chord.
OR
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Check that the line y = – x + 4 cuts the two circles at the same points:
(x − 4)2 + (− x − 2) 2 = 26
9 substitution
x 2 − 8 x + 16 + x 2 + 4 x + 4 = 26
2x 2 − 4x − 6 = 0
x 2 − 2x − 3 = 0
( x − 3)( x + 1) = 0
x = 3 or x = – 1
9 answer
x 2 + y 2 + 8 x + 4 y − 38 = 0
9 substitution
x 2 + (4 − x) 2 + 8 x + 4(4 − x) − 38 = 0
x 2 + 16 − 8 x + x 2 + 8 x + 16 − 4 x − 38 = 0
2x 2 − 4x − 6 = 0
9 answer
x 2 − 2x − 3 = 0
x = 3 or x = −1
(4)
[13]
QUESTION 7
7.1.1
7.1.2
7.2.1
9 answer
P / (5 ; − 2)
(1)
9 x coordinate
9 y coordinate
/
P (5 ; 2)
K → K ′′ : (14 ; 4) → (2 ; 2)
U → U ′′ : (18 ; 6) → (3 ; 9)
H → H ′′ : (16 ; 8) → (4 ; 8)
L → L ′′ : (18 ; 10) → (5 ; 9)
E → E ′′ : (14 ; 12) → (6 ; 7)
So “halve” and ‘interchange” or ‘interchange” and “halve”.
Reflection across y = x followed by contraction by
1
2
OR
Contraction by
7.2.2
H/=
OR
7.2.3
9 reflected
9 the line y = x
9 enlarged
9 scale factor of
1
followed by reflection across y = x.
2
1
2
(4)
9 (8 ; 4)
1
(16 ; 8) = (8 ; 4)
2
H / (8 ; 16)
9 (8 ; 16)
(2)
2
⎛2⎞
Area KUHLE : Area K // U // H // L// E // = ⎜ ⎟ = 4 : 1
⎝1⎠
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(2)
9 9answer
(2)
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QUESTION 8
8.1
For anti-clockwise rotation:
x / = x cosθ − y sin θ
= 3 cos120° − 2 sin 120°
= 3(− cos 60°) − 2 sin 60°
9 simplification
9 substitution
⎛ 1 ⎞ ⎛ 3 ⎞⎟
= 3⎜ − ⎟ − 2⎜⎜
⎝ 2 ⎠ ⎝ 2 ⎟⎠
−3− 2 3
2
y / = x sin θ + y cosθ
= 3 sin 120° + 2 cos120°
= 3 sin 60° + 2(− cos 60°)
=
⎛ 3⎞ ⎛ 1⎞
⎟ + 2⎜ − ⎟
= 3⎜⎜
⎟ ⎝ 2⎠
2
⎝
⎠
3 3−2
2
⎛−3−2 3 3 3 −2⎞
⎟
P / ⎜⎜
;
⎟
2
2
⎝
⎠
=
8.2
⎛ 3⎞
⎛ 1⎞
⎟
− 2 = x⎜ − ⎟ − y⎜⎜
⎟
2
⎝ 2⎠
⎠
⎝
− 4 = −x − 3y
9 formula
…… equation 1
9 answer
9 simplification
9 answer
(6)
9 − 4 = −x − 3y
⎛ 3⎞
1
⎟ + y⎛⎜ − ⎞⎟
0 = x⎜⎜
⎟
⎝ 2⎠
⎝ 2 ⎠
0 = 3x + y
y = − 3x
…… equation 2
Substitute equation 2 into equation 1
− 4 = − x − 3 − 3x
− 4 = − x + 3x
− 4 = 2x
x = −2
(
y=2 3
(
Q −2;2 3
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)
9 y = − 3x
)
9 x-coordinate
9 y-coordinate
(4)
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QUESTION 9
9.1.1
sin θ = −
3
4
and cos θ = −
5
5
sin θ + cos θ = −
9.1.2
tan 2θ =
7
5
sin 2θ
2 sin θ cos θ
=
cos 2θ cos 2 θ − sin 2 θ
⎛ 3 ⎞⎛ 4 ⎞
2⎜ − ⎟⎜ − ⎟
5 ⎠⎝ 5 ⎠
= ⎝
16 9
−
25 25
24
=
7
θ
–4
−3
5
9 correct quadrant
and values.
3
9 sin θ = −
5
4
9 cos θ = −
5
9answer
(4)
sin 2θ
cos 2θ
9 sin 2θ = 2sinθ.cosθ
9 cos 2θ = cos2θ –
sin2θ
9 substitution
9
9 answer
(5)
OR
tan 2θ
2 tan θ
1 − tan 2 θ
⎛3⎞
2⎜ ⎟
4
= ⎝ ⎠2
⎛3⎞
1− ⎜ ⎟
⎝4⎠
24
=
7
cos(360° − x). tan 2 x
sin( x − 180°). cos(90° + x)
=
9.2.1
=
(cos x)(tan 2 x)
(− sin x)(− sin x)
⎛ sin 2 x ⎞⎛ 1 ⎞
⎟⎟⎜ 2 ⎟
= (cos x)⎜⎜
2
cos
x
⎝
⎠⎝ sin x ⎠
1
=
cos x
9.2.2 x = 30°
1
1
2
=
=
cos 30°
3
3
2
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99 expansion
99 substitution
9 answer
(5)
9 cos x
9 −sin x
9 −sin x
9
sin 2 x
cos 2 x
9 answer
(5)
9 x = 30°
9 answer
(2)
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QUESTION 10
10.1.1
sin 48° = sin(36° + 12°)
= sin 36° cos12° + cos 36° sin 12°
9writing 48° in terms of
36° and 12°
9 expansion
9 answer
(3)
= p+q
10.1.2
sin 24° = sin(36° − 12°)
= sin 36° cos12° − cos 36° sin 12°
9writing 24° in terms of
36° and 12°
9 expansion
9 sin 24° = p − q
(3)
= p−q
OR
9writing 24° in terms of
36° and 12°
9 expansion
9 sin 24° = p − q
(3)
sin 24° = sin(36° − 12°)
= sin 36° cos12° − cos 36° sin 12°
= p−q
10.1.3
sin 48° = 2 sin 24° cos 24°
9 cos 48° = 2 cos 2 24° − 1
9 sin 48° = p + q
9 answer
(3)
∴ p + q = 2( p − q ) cos 24°
p+q
∴ cos 24° =
2( p − q )
OR
cos 48° = 2 cos 2 24° − 1
∴ cos 24° =
(
1 + cos 48°
1
1 + 1 − sin 2 48°
=
2
2
=
)
1⎛
2
⎜1 + 1 − ( p + q ) ⎞⎟
⎠
2⎝
9 cos 48° = 2 cos 2 24° − 1
9 sin 24° = p − q
9 answer
(3)
OR
cos 2 24° = 1 − sin 2 24°
cos 2 24° = 1 − ( p − q ) 2
cos 24° = 1 − ( p − q ) 2
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9 cos 2 24° = 1 − sin 2 24°
9 sin 24° = p − q
9 answer
(3)
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sin 2 20° + sin 2 40° + sin 2 80°
940°= 60° 20°
980°= 60° +
20°
= sin 2 20° + (sin(60° − 20°)) 2 + (sin(60° + 20°)) 2
= sin 2 20° + (sin 60° cos 20° − cos 60° sin 20°) + (sin 60° cos 20° + cos 60° sin 20°) 2
2
2
⎛ 3
⎞ ⎛ 3
⎞
1
1
= sin 20° + ⎜⎜
cos 20° + sin 20° ⎟⎟
cos 20° − sin 20° ⎟⎟ + ⎜⎜
2
2
⎝ 2
⎠ ⎝ 2
⎠
2
2
9
9expansions
3
1
3
3
= sin 2 20° + cos 2 20° −
cos 20° sin 20° + sin 2 20° + cos 2 20°
4
4
2
4
1
3
+
cos 20° sin 20° + sin 2 20°
4
2
1
3
= sin 2 20° + cos 2 20° + sin 2 20°
2
2
3
= (sin 2 20° + cos 2 20°)
2
3
=
2
9 substitution
9
simplification
9
factorisation
OR
Use sin 2 θ =
1 − cos 2θ
2
(7)
LHS
3 1
− {(cos 40° + cos 80°) + cos160°}
2 2
3 1
= − {(cos 60°. cos 40° + sin 60° sin 40° + cos 60°. cos 40° − sin 60° sin 40°) + cos160°}
2 2
3 1
= − {(2 cos 60° cos 20°) − cos 20°}
2 2
⎫
1
3 1 ⎧⎛
⎞
= − ⎨⎜ 2 × cos 20° ⎟ − cos 20°⎬
2
2 2 ⎩⎝
⎠
⎭
=
3
−0
2
3
=
2
=
940°= 60° –
20°
980°= 60° +
20°
9 expansion
of cos 40°
9 expansion
of cos 60°
9
simplification
9
simplification
9 answer for
bracket
(7)
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10.3.1
sin 4 x + sin 2 x cos 2 x
1 + cos x
2
sin x(sin 2 x + cos 2 x)
=
1 + cos x
2
sin x
=
1 + cos x
1 − cos 2 x
=
1 + cos x
(1 − cos x)(1 + cos x)
=
(1 + cos x)
= 1 − cos x
10.3.2 1 + cos x = 0
9 factorisation
9 sin 2 x + cos 2 x = 1
9identity
9 factorisation
(4)
9 1 + cos x = 0
cos x = −1
9180° + k.360°
x = 180° + k .360°; k ∈ Z
Undefined for x = 180° + k .360°; k ∈ Z .
(2)
[22]
QUESTION 11
11.1
1 + sin x = cos 2 x
9 expansion
1 + sin x = 1 − 2 sin 2 x
sin x + 2 sin 2 x = 0
sin x(1 + 2 sin x) = 0
1
sin x = − `
2
x = k.180
or
x = −30° + k .360
x = 210° + k .360
x ∈ {180° ; 210; 330° ; 360°}
sin x = 0
9 factorisation
9 equations
or
k ∈Z
OR
9 x = k .180
9 solution for
1
sin x = − `
2
99answers
(7)
1 + sin x = cos 2 x
1 + sin x = cos 2 x − sin 2 x
9 expansion
1 + sin x = 1 − sin 2 x − sin 2 x
sin x + 2 sin 2 x = 0
9 factorisation
sin x(1 + 2 sin x) = 0
sin x = 0
or
x = k .180
or
1
sin x = − `
2
x = −30° + k .360
x = 210° + k .360
x ∈ {180° ; 210; 330° ; 360°}
9 equations
k ∈Z
9 x = k .180
9 solution for
1
sin x = − `
2
99answers
(7)
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y
1+sinx
9 max and min
values
9 shape
3
2
1
f
45
90
135
180
225
270
x
315
360
cos2x
9 amplitude
9 intercepts
g
(4)
-1
-2
-3
11.3
180° ≤ x ≤ 210° or 330° ≤ x ≤ 360°
999 answer
(3)
[14]
QUESTION 12
12.1
b
BC
=
sin[180° − (α + β ) sinα
BC sin(α + β ) = b sin α
bsinα
BC =
sin(α + β )
9 sine rule
9
ABˆ C = 180° − (α + β )
but BC = DF
bsinα
∴ DF =
sin(α + β )
9 BC = DF
DF
DE
DF
∴ DE =
cosθ
b sin α
∴ DE =
sin(α + β ) cos θ
2000 sin 43°
DE =
sin 79°. cos 27°
= 1559,50 m
9 BC = …
9 manipulation
cos θ =
12.2
9 DE = …
(6)
9substitution
numerator
9substitution
denominator
9answer
TOTAL:
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(3)
[9]
150