NATIONAL
SENIOR CERTIFICATE
GRADE 12
MATHEMATICS P2
FEBRUARY/MARCH 2011
MEMORANDUM
MARKS: 150
This memorandum consists of 15 pages.
Copyright reserved
Please turn over
Mathematics/P2
2
NSC – Memorandum
DBE/Feb. – Mar. 2011
QUESTION 1
1.1
55 + 55 + 50 + 47 + 42 + 3x
= 48,375
8
249 + 3 x
= 48,375
8
3 x = 138
x = 46
249 + 3x
= 48,375
8
9 3 x = 138
(2)
9
9 max and min
9 median
9 Q1 and Q3
9 whiskers
1.2
42
46,5
QUESTION 2
2.1
Mass
(kg)
60 ≤ x < 70
70 ≤ x < 80
80 ≤ x < 90
90 ≤ x < 100
100 ≤ x < 110
2.2
52,5 54 55 56
Frequency
5
7
7
4
2
99 Frequencies
99 Cumulative
Frequencies
Cumulative
Frequency
5
12
19
23
25
(4)
Cumulative Frequency Curve
99 plotting points
1 mark: 3 – 5 points
correctly
0 marks : 2 or less
points correctly
plotted
30
25
Cumulative Frequency
(4)
[6]
20
9 graph
15
(3)
10
5
0
40
50
60
70
80
90
100
110
120
Mass (kg)
2.3
Mean = 79,28
99 answer
(2)
Copyright reserved
Please turn over
Mathematics/P2
2.4
3
NSC – Memorandum
DBE/Feb. – Mar. 2011
Standard Deviation = 11,02
79,28 – 11,02 = 68,26
79,28+11,02 = 90,3
17 players lie in this interval.
17
= 68%
25
999 sd = 11,02
9 17 players
9 68%
(5)
[14]
QUESTION 3
Question 3.1
3.1
&
3.2
Scatter Plot showing Arm Span vs Height
4 marks:
All points plotted
correctly.
3 marks:
9 – 11 points
correct
2 marks:
6 – 8 points
correct
1 marks:
3 – 5 points
correct
0 marks if less than
3 points plotted
correctly.
(4)
200
195
190
Height (cms)
185
180
175
170
165
160
155
150
150
160
170
180
190
200
Arm Span (cm)
3.3
Yes.
The relationship between arm span and height is a positive, linear one
so we can expect a person with below average arm span to have below
average in height.
Copyright reserved
Question 3.2
99 linear best fit
Line
(2)
9 Yes
9 Reason
(2)
[8]
Please turn over
Mathematics/P2
4
NSC – Memorandum
DBE/Feb. – Mar. 2011
QUESTION 4
4
P
y
2
R
-5
Q
5
x
-2
-4
4.1
Let β be the angle of inclination of PQ.
tan β = m PQ
9 tan β = mPQ
9 tan β = 4
2 − (−2)
− 1 − (−2)
tan β = 4
β = 75,96°
tan β =
9 answer
(3)
4.2
4.3
⎛ −1+ 3 2 + 0 ⎞
;
M⎜
⎟
2 ⎠
⎝ 2
M (1 ; 1)
9 x-value
9 y-value
(2)
9 substitution into
correct formula
9 answer
PQ = (−1 + 2) 2 + (2 + 2) 2
= 17
PR = (−1 − 3) 2 + (2 − 0) 2
=
20
QR = (0 − (−2)) + (3 − (−2))
2
=
4.4
2
29
Perimeter = 29 + 20 + 17
= 13,98 units
= 14 to the nearest whole number
y − 1 = 4( x − 1)
y = 4x − 3
9 answer
9 sum
9 answer
(5)
9m=4
9 substitution of
(1 ; 1)
9 answer
(3)
[13]
Copyright reserved
Please turn over
Mathematics/P2
5
NSC – Memorandum
DBE/Feb. – Mar. 2011
QUESTION 5
5.1.1
x 2 + y 2 − 8x + 6 y
= (2) 2 + (−9) 2 − 8(2) + 6(−9)
= 4 + 81 − 16 − 54
= 15
Hence, the point lies on the circumference of the circle.
OR
x 2 + y 2 − 8 x + 6 y = 15
9 substitution
9 answer
(2)
9 substitution
9 answer
( x − 4) 2 + ( y + 3) 2 = 15 + 16 + 9
(2)
( x − 4) + ( y + 3) = 40
2
2
( x − 4) 2 + ( y + 3) 2
= (2 − 4) 2 + (−9 + 3) 2
5.1.2
= 22 + 62
= 40
∴The point lies on the circumference of the circle.
x 2 + y 2 − 8 x + 6 y = 15
( x − 4) 2 + ( y + 3) 2 = 15 + 16 + 9
( x − 4) 2 + ( y + 3) 2 = 40
Circle centre (4 ; − 3)
− 3 − (−9)
mrad =
4−2
mrad = 3
99 ( x − 4) 2 + ( y + 3) 2 = 40
9 centre
9 gradient of radius
1
3
1
y + 9 = − (x − 2)
3
1
25
y =− x−
3
3
m tan = −
9 gradient of tangent
9substitution
9 answer
(7)
5.2
A(6 ; 4)
B
10
(3 ; – 1)
9 radius = 10
Copyright reserved
Please turn over
Mathematics/P2
6
NSC – Memorandum
Radius AB = 10
Distance from A to centre of circle is
=
(6 − 3)2 + (4 + 1)2
= 9 + 25
= 34
AB 2 = 34 − 10
AB 2 = 24
AB = 24
DBE/Feb. – Mar. 2011
9 subs into distance
formula
9 34
9 AB 2 = 34 − 10
9 answer
(5)
AB = 2 6
AB = 4,90
OR
r 2 = 10
r = 10
Radius ⊥ tangent
By Pythagoras
AB 2 = (6 − 3) 2 + (4 + 1) 2 − 10
= 24
AB = 4,90
Copyright reserved
9 r = 10
99
AB2 = (6 − 3) 2 + (4 + 1) 2 − 10
9 AB = 4,90
(5)
[14]
Please turn over
Mathematics/P2
7
NSC – Memorandum
DBE/Feb. – Mar. 2011
QUESTION 6
y
C
B
x
A
6.1
9x=0
9 + ( y + 2) 2 = 25
( y + 2) 2 = 16
y + 2 = ±4
y = 2 or y = – 6
B(0 ; 2)
OR
x=0
(0) 2 − 6(0) + y 2 + 4 y = 12
9 factors
9 answers
9 answer for B
(4)
9x=0
y 2 + 4 y − 12 = 0
( y + 6)( y − 2) = 0
y = – 6 or y = 2
B(0 ; 2)
9 factors
9 answers
9 answer for B
(4)
6.2
99 answer
C(6 ; 2)
(2)
6.3
2
2
3⎞ ⎛
3⎞
⎛
⎛ 3⎞
⎜ x − 3× ⎟ + ⎜ y + 2 × ⎟ = ⎜5× ⎟
2⎠ ⎝
2⎠
⎝
⎝ 2⎠
2
9⎞
⎛
⎛ 15 ⎞
2
⎜ x − ⎟ + ( y + 3) = ⎜ ⎟
2⎠
⎝
⎝2⎠
2
9 each part ×
2
3
2
2
9⎞
⎛
2
⎜ x − ⎟ + ( y + 3) = 56,25
2⎠
⎝
6.4.1
AB = (12 − 3) + (10 − (−2))
2
9 answer
(2)
2
= 9 2 + 12 2
= 15
6.4.2
The radii are 5 and 10.
rA + rB = 5 + 10
9 substitution
9 answer
(2)
9 addition of radii
= 15
= AB
The circles will only intersect at one point.
Copyright reserved
9 answer
(2)
[12]
Please turn over
Mathematics/P2
8
NSC – Memorandum
DBE/Feb. – Mar. 2011
QUESTION 7
− 3 = x cos150° − 2 sin 150°
− 3 = − x.
9 expansion
9 substitution
3
1
− 2.
2
2
3
x=2
2
4
x=
3
y = x. sin 150° + 2. cos150°
9 answer
9 expansion
⎛
4 1
3⎞
⎟
. + 2.⎜⎜ −
⎟
2
3 2
⎝
⎠
2
= . 3− 3
3
3
=−
3
y=
9 answer
[5]
QUESTION 8
8.1
y
7
N
6
/
999
coordinates
of new
points
(3)
5
M
N
4
/
3
M
2
1
x
-9
-8
-7
-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
7
8
9
-1
-2
-3
-4
P
-5
-6
-7
8.2.1
8.2.2
8.2.3
Copyright reserved
/
99
MN
2
=
/
/
3
M N
area ΔMNP
4
=
/
/ /
9
area ΔM N P
area ΔMNP
⎛4⎞
=⎜ ⎟
//
// //
area ΔM N P
⎝9⎠
P
(2)
99
(2)
n +1
99
(2)
[9]
Please turn over
Mathematics/P2
9
NSC – Memorandum
DBE/Feb. – Mar. 2011
QUESTION 9
9.1
A/ (– 12; – 6)
9.2
x = x cos α − y sin α
− 12 cos α − 6 sin α = −12
− 2 cos α − sin α = −2.........(1)
9 substitution
y / = y cos α + x sin α
9 substitution
9 answer
(1)
/
6 cos α − 12 sin α = −6
cos α = 2 sin α − 1
9 simplification
… (2)
9 substitution
Substitute (2) into (1)
− 2(2 sin α − 1) − sin α = −2
− 4 sin α + 2 − sin α = −2
− 5 sin α = −4
4
sin α =
5
α = 53,13°
9 simplification
9 answer
(6)
OR
y
(– 12 ; 6)
θ
x
α
(– 12 ; –6)
1
2
θ = 26,565°
α = 2(26,565°)
α = 53,13°
tan θ =
1
2
9 θ = 26,565°
99 α = 2(26,565°)
9 answer
99 tan θ =
(6)
[7]
Copyright reserved
Please turn over
Mathematics/P2
10
NSC – Memorandum
DBE/Feb. – Mar. 2011
QUESTION 10
10.1.1
cos 28° = 1 − sin 2 28°
9 1 − sin 2 28°
9answer
= 1− a2
10.1.2
(2)
cos 64°
= cos 2(32°)
= 2 cos 2 32° − 1
9 cos 2(32°)
9 2 cos 2 32° − 1
9answer
= 2b 2 − 1
10.1.3
sin 4°
= sin(32° − 28°)
= sin 32° cos 28° − cos 32° sin 28°
= 1 − b 2 . 1 − a 2 − ab
(3)
9 sin(32° − 28°)
9 expansion
99answer
(4)
OR
sin 4°
= sin(60° − 2 × 28°)
= sin 60° cos(2 × 28°) − cos 60° sin( 2 × 28°)
(
)
3
1
1 − 2a 2 − ( 2a ) 1 − a 2
2
2
3
=
− 3a 2 − a 1 − a 2
2
=
OR
sin 4°
= sin( 2 × 32° − 60°)
= sin( 2 × 32°) cos 60° − cos(2 × 32°). sin 60°
(
)
1
3
= 2.b 1 − b 2 . −
2b 2 − 1
2 2
3
= b 1 − b 2 − 3b 2 +
2
OR
Using sin(A+B) + sin(A – B) = 2.sinA.cosB
With A = 28° and B = 32°
sin 60° + sin( −4°) = 2ab
sin 4° =
3
− 2ab
2
OR
Copyright reserved
Please turn over
Mathematics/P2
11
NSC – Memorandum
DBE/Feb. – Mar. 2011
Using sin(A+B) + sin(A – B) = 2.sinA.cosB
With A = 32° and B = 28°
sin 60° + sin( 4°) = 2 1 − b 2 . 1 − a 2
sin 4° = 2 1 − b 2 . 1 − a 2 −
OR
Using sin 4° = 2 sin 2°. cos 2°
3
2
(
(
(
(
)
)
1
1 − a 2 − 3a
2
1
3 1− b2 − b
and sin 2° = sin(32° − 30°) =
2
1
3 1− a2 + a
and cos 2° = cos(30° − 28°) =
2
1
and cos 2° = cos(32° − 30°) =
3b + 1 − b 2
2
then
1
sin 4° =
3b 1 − a 2 − 3ab + 1 − a 2 . 1 − b 2 − 3a 1 − b 2
2
OR
1
sin 4° = 3 1 − b 2 1 − a 2 + 3a 1 − b 2 − 3b 1 − a 2 − ab
2
and sin 2° = sin(30° − 28°) =
10.2
10.3.1
)
)
{
}
{
}
b 1− a2 − a 1− b2
= cos 32°. 1 − sin 2 28° − sin 28° 1 − cos 2 32°
= cos 32°. cos 28° − sin 28°. sin 32°
= cos(32° + 28°)
= cos 60°
1
=
2
sin 130°. tan 60°
cos 540°. tan 230°.sin 400°
sin 50° × tan 60°
= cos180° × tan 50° × sin 40°
sin 50° × 3
sin 50°
− 1×
× cos 50°
cos 50°
3 cos 50°
=−
cos 50°
= − 3
=
Copyright reserved
9 substitution
9 cos 28°
9 sin 32°
9 compound angle
formula
(4)
9 sin 50°
9 tan 50°
9sin40o
9cos50o
sin 50°
9
cos 50°
9 −1
9answer
(7)
Please turn over
Mathematics/P2
10.3.2
12
NSC – Memorandum
(1 − 2 sin 75°)(1 + 2 sin 75°)
= 1 − 2 sin 2 75°
= cos150°
=
− 3
2
DBE/Feb. – Mar. 2011
9simplification
9 1 − 2 sin 2 75°
9 cos150°
9answer
(4)
OR
sin 75°
= sin( 45° + 30°)
= sin 45°. cos 30° + cos 45°. sin 30°
2 3
2 1
.
.
+
2 2
2 2
3 1
2 sin 75° =
+ =a
2 2
(1 − 2 sin 75°)(1 + 2 sin 75°)
=
= (1 − a )(1 + a )
= 1− a2
⎛3 1
3 1⎞
. ⎟
= 1 − ⎜⎜ + + 2.
2 2 ⎟⎠
⎝4 4
=−
10.4
3
2
sin x + cos 2 x − cos x = 0
9answer
(4)
2
sin 2 x + (cos 2 x − sin 2 x) − cos x = 0
cos 2 x − cos x = 0
cos x(cos x − 1) = 0
cos x = 0 or cos x = 1
x = ±90° + k .360° or x = 0° + k .360° k ∈ Z
= k .360°
10.5.1
9simplification
9 1 − 2 sin 2 75°
9 cos150°
(i.e. x = 90° + k .180° or x = k .360° ± 90°, k ∈ Z )
x = 0°; 90°; 180°
9 (cos 2 x − sin 2 x)
9 cos 2 x − cos x = 0
9factors
9 cos x = 0 or cos x = 1
9 90° + k .360°
9 k .360°
9 x = −90° + k .360°
(7)
999 each value
(3)
Copyright reserved
Please turn over
Mathematics/P2
10.5.2
13
NSC – Memorandum
sin x
(cos 2 x − sin 2 x).
cos 2 x. tan x
cos x
=
2
2
sin x
sin x
2
2
cos x − sin x
=
cos x. sin x
cos x sin x
−
=
sin x cos x
cos x
− tan x
=
sin x
DBE/Feb. – Mar. 2011
9 (cos 2 x − sin 2 x)
sin x
9
cos x
9 answer
9
cos x sin x
−
sin x cos x
9answer
(5)
[39]
Copyright reserved
Please turn over
Mathematics/P2
14
NSC – Memorandum
DBE/Feb. – Mar. 2011
QUESTION 11
11.1
EC 2 = DE 2 + DC 2 − 2DE.DC cos Ĉ
= (7,5) 2 + (9,4) 2 − 2.(7,5)(9,4) cos 32°
= 25,03521844...
EC = 5,0 metres
11.2
sin DĈE sin 32°
=
7,5
5,0
7,5. sin 32°
sin DĈE =
5,0
= 0,7948788963
DĈE = 52,6°
11.3
Area of ΔDEC
1
= DE.DC sin D̂
2
1
= (7,5)(9,4) sin 32°
2
= 18,7m 2
9substitution into
cosine rule
9 25,03521844...
9answer
(3)
9 sin rule
9 0,7948788963
9answer
(3)
9 substitution
9answer
(2)
OR
Area of ΔDEC
1
= CE.DC sin 52,6°
2
1
= (5,0)(9,4) sin 52,6°
2
= 18,7m 2
11.4
EG
7,5
EG = 7,5. sin 32°
sin 32° =
9ratio
9substitution
= 4,0
EF = (4 + 3,5)
= 7,5 metres
9answer
(3)
[11]
OR
EG = EC.sin 52,6°
= (5,0).sin 52,6°
= 4,0
EF = 4,0 + 3,5
= 7,5
OR
Copyright reserved
Please turn over
Mathematics/P2
15
NSC – Memorandum
DBE/Feb. – Mar. 2011
1
.DC.EG = area Δ DEC
2
1
(9,4)EG = 18,7
2
18,7 × 2
∴ EG =
9,4
= 4,0
EF = 4,0 + 3,5
= 7,5
Copyright reserved
Please turn over
Mathematics/P2
16
NSC – Memorandum
DBE/Feb. – Mar. 2011
QUESTION 12
12.1
Period = 360°
9answer
(1)
12.2
Amplitude =
99answer
1
2
(2)
12.3
9shape
9x intercepts
9 amplitude
y
2
(3)
f
1
g
x
-180
-150
-120
-90
-60
-30
30
60
90
120
150
180
-1
-2
12.4
2 solutions
9answer
(1)
12.5
− 60° ≤ x ≤ 120° or x ∈ [−60° ; 120°]
9 − 60°; 120°
9 notation
12.6
− 90° < x < 30° or x ∈ (−90° ; 30°)
99 − 90°; 30°
9 notation
(2)
(3)
[12]
TOTAL:
Copyright reserved
Please turn over
150