NATIONAL
SENIOR CERTIFICATE
GRADE 12
MATHEMATICS P2
FEBRUARY/MARCH 2012
MEMORANDUM
MARKS: 150
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QUESTION 1
1.1
Mean
3 102100
n
∑x
1
1
1.2
102100
n
9
= R11 344, 44
=
Standard deviation
n
∑ (x
1
3 answer
(2)
33 answer
− x)2
1
1.3
= R4 460,97
n
Value of one standard deviation above mean
= R11 344,44 + R4 460,97
= R15 805,41
Only one person earned a commission of more than R 15 805,41.
Therefore only 1 person received a rating of good.
(2)
3 adding mean
and std. dev.
3 deduction
(2)
[6]
QUESTION 2
2.1
3 at least
four points
correct
3 all points
correct
(2)
2.2
Exponential
(The increase in growth is showing a virtual doubling for each year).
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3exponential
(1)
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2.3
YEAR
N
(Number in
millions)
Log N
(correct to
1 decimal
place)
3
NSC – Memorandum
DBE/Feb. – Mar. 2012
1995
1996
1997
1998
1999
2000
2001
8
17
34
67
135
281
552
6,9
7,2
7,5
7,8
8,1
8,4
8,7
3 at least
four values
correct
3 all values
correct
(2)
OR (if only log of values in table taken in account)
YEAR
N
(Number in
millions)
Log N
(correct to
1 decimal
place)
1995
1996
1997
1998
1999
2000
2001
8
17
34
67
135
281
552
0,9
1,2
1,5
1,8
2,1
2,4
2,7
3 at least
four values
correct
3 all values
correct
(2)
2.4
3 at least 4
points
correctly
plotted
3 all points
correct
(2)
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OR (if only log of values in table taken in account)
3 at least 4
points
correctly
plotted
3 all points
correct
(2)
2.5
The graph representing log N is a straight line. That is,
log N = mx + c
mx + c
N = 10
Therefore exponential graph.
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3linear
3 reason (2)
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QUESTION 3
3.1
3.2
40
Time, t, in minutes
0 ≤ t<5
5 ≤ t <10
10 ≤ t < 15
15 ≤ t < 20
20 ≤ t < 25
3.3
3 40
(1)
3 for intervals in
table
3 for first three
correct
frequencies
3 for last two
correct
frequencies (3)
Frequency
3
5
10
15
7
Frequency
15
3first three bars
correct
3last two bars
correct
3no gaps between
bars
10
5
0
5
10
15 20
Time intervals
25
(3)
[7]
QUESTION 4
a=7
b = 15
c = 17
d = 23
e = 34
f = 37
g = 42
3 each correct
answer
(7)
OR
g = 42 ; a = 7 ; d = 23 ; f = 37 ; b = 15
42 + 7 + 23 + 37 + 15 + 3c
= 25
7
= 51
3c
c
= 17
e
= 34
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3g
3a
3d
3f
3b
3c
3e
(7)
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QUESTION 5
5.1
5.2
5.3
5.4
y 2 − y1
x 2 − x1
−2−4
=
5 −1
6
3
= −
= −
4
2
mAD =
AD =
3 for substitution
3 for answer
(2)
( x2 − x1 ) 2 + ( y 2 − y1 ) 2
=
(5 − 1) 2 + (−2 − 4) 2
=
16+ 36
=
52
3 for substitution
3 52
(2)
⎛ x + x 2 y1 + y 2 ⎞
;
M =⎜ 1
⎟
2 ⎠
⎝ 2
⎛1+ 5 4 − 2 ⎞
M =⎜
;
⎟
2 ⎠
⎝ 2
M = (3 ; 1)
mBC = mAD
Lines are parallel
3
= −
2
y – y1 = m (x – x1)
3
y – 1 = − (x + 3)
2
2 y − 2 = −3 x − 9
3x + 2 y + 7 = 0
3 x-value
3 y-value
(2)
3 value mBC
3 subst (–3 ; 1)
3 equation
(3)
OR
3
y =− x+c
2
3
1 = − (−3) + c
2
7
c=−
2
3
7
y =− x−
2
2
3x + 2 y + 7 = 0
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3 value mBC
3 subst (–3 ; 1)
3 equation
(3)
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5.5.1
3
2
3
tan β = −
2
β = 180° − 56,31°
m AD = −
A(1 ; 4)
α
E
B(–3 ; 1)
β = 123,69
3 tan β = mAD
β
F
3 123,69°
D(5 ; –2)
(2)
5.5.2
− 2 −1 − 3
=
5 − (−3)
8
3
tan α = −
8
α = 180° − 20,56°
α = 159,44°
FEˆ D = 180° − 159,44° = 20,56°
m BD =
3 m BD =
3159,44°
EFˆD = 123,69°
FDˆ E = 180° − (20,56° + 123,69°) = 35,75°
5.6
−3
8
Co-ordinates of centre M (3 ; 1)
Radius of circle:
1
1
1
52
of AD =
(2 13 ) = 13 =
2
2
2
2
2
Equation of the circle is: (x − 3) + ( y − 1) = 13
320,56°
3123,69°
335,75°
(5)
3 value of radius
3 substitution into
equation of circle
centre form (2)
OR
r = (3 − 1) + (1 − 4) = 13
Equation of the circle is:
(x − 3)2 + ( y − 1)2 = 13
2
5.7
M(3 ; 1)
MB =
2
2
B(– 3 ; 1)
(3 + 3)
2
+ (1 − 1)
2
)
MB = 6
Point B lies outside the circle because MB > radius
OR
M(3 ; 1) B(– 3 ; 1)
MB = 3 + 3 = 6
Radius of the circle = 13 < 6
Point B lies outside the circle because MB > radius
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3 value of r2
3 substitution into
equation of circle
centre form (2)
3substitution
3outside
(2)
3substitution
3outside
(2)
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QUESTION 6
6.1
33 coordinates of
centre
3 calculation
3 value
(4)
Coordinates of centre M (−2 ; 1)
(1 + 2)2 + (− 2 − 1)2 = 18 = r 2
Radius = 18 or 3 2
6.2
−3
= −1
3
m MS xm RS = −1
m MS =
3 gradient MS
OR
tangent ⊥ radius
m RS = 1
y – y1 = m (x – x1)
y + 2 = 1(x – 1)
y = x–3
3 gradient RS
3 subst (1 ; –2)
3 equation
(4)
OR
−3
= −1
3
m MS xm RS = −1
3 gradient MS
m MS =
3 gradient RS
m RS = 1
y = x+c
3 subst (1 ; –2)
− 2 = 1+ c
3 equation
c = −3
y = x−3
6.3
(4)
MS 1
=
MP 3
∴ MP = 3MS
3 MP = 3MS
MP 2 = 9MS 2
(a + 2) 2 + (b − 1) 2 = 9(32 + 32 ) = 162
MS ⊥ SR and PS ⊥ SR
b+2 3
=
= −1
a −1 − 3
b + 2 = −a + 1
b = −a − 1
∴ mPS = mMS
(1)
3 equation
3 equal gradients
3 gradient
(2)
3 b = -a - 1
Subst (2) into(1)
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3substitution
(a + 2) 2 + (− a − 1 − 1) 2 = 162
(a + 2) 2 + (a + 2) 2 = 162
2(a + 2) 2 = 162
(a + 2) 2 = 81
a + 2 = 9 or − 9
a = 7 or − 11
b = − a − 1 = −8
P(7 ; − 8)
3a=7
3 b = -8
(8)
OR
MS 1
=
MP 3
∴ MP = 3MS
3 MP = 3MS
MP 2 = 9MS 2
(a + 2) 2 + (b − 1) 2 = 9(32 + 32 ) = 162
MS ⊥ SR and PS ⊥ SR
b+2 3
=
= −1
a −1 − 3
b + 2 = −a + 1
b = −a − 1
(1)
∴ mPS = mMS
3 equation
3 equal gradients
3 gradient
(2)
3 b = -a - 1
Subst (2) into(1)
a 2 + 4a + 4 + a 2 + 4a + 4 = 162
2a 2 + 8a − 154 = 0
a 2 + 4a − 77 = 0
(a + 11)(a − 7) = 0
a = 7 or − 11
But a > 0
∴a = 7
b = −a − 1 = −8
P(7 ; − 8)
3substitution
3a=7
3 b = -8
(8)
OR
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P(a ; b)
MSP is a straight line
mPM = − 1
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3 MSP a straight
line
3 m PM = −1
(MS ⊥ SR)
b −1
= −1
a+2
b − 1 = −a − 2
b = −a − 1.......(1)
3
3 equation 1
PS = 2MS = 2 9 + 9 = 2 18
3 equation 2
PS 2 = 4(18) = 72
(a − 1) 2 + (b + 2) 2 = 72.......(2)
(a − 1) + (− a − 1 + 2) = 72
2(a − 1) = 72
2a − 4a − 70 = 0
(a − 1) 2 = 36
a 2 − 2a − 35 = 0
(a − 7)(a + 5) = 0
a = 7 or a =/ −5
b = −7 − 1 = −8
P (7 ; − 8)
a − 1 = 6 or − 6
a = 7 or − 5
2
b −1
a+2
2
2
2
OR
3 substitution of
equation 1 into
equation 2
33 coordinates
a=7
b = −8
(8)
P(7 ; -8)
OR
33 diagram
M(−2; 1)
33 (−2; −8)
3
3
(−2; −2)
S(1 ;−2)
3
(−2; −2)
3
(1; −8)
33 P(7 ; −8)
(8)
6
(−2; −8)
3
6
(1; −8)
P(a ;b)
33 division of line
segment into
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given ratio
33substitution
3 equation
P(a ; b)
xS − xM
y − yM 1
= S
=
x P − xM
yP − yM 3
3
1
−3
=
=
b −1 a + 2 3
− 9 = b −1
b = −8
9=a+2
3equation
3coordinates
(8)
[16]
a=7
P (7 ;−8)
QUESTION 7
7.1
y
5
4
K
3
2
N
1
x
-5
-4
-3
-2
N/
1
2
M
3
4
5
-1
M/
-2
L
K/
-1
-3
For correct
coordinates and
label of each
image:
3 K′
3 L′
3M′
3 N′
L/
-4
-5
(4)
7.2.1
Transformation is not rigid, because the area is not preserved under
enlargement.
7.2.2
N // (−2 ; − 2)
7.3
(x ; y) → (–y ; x) → (–2y ; 2x)
3 not rigid
3 size not
preserved
33 coordinates of
N //
(2)
3 –y
3x
3 –2y
3 2x
(4)
(2)
7.4
Area of KLMN : area of K // L// M // N // = 1 : 4
33 answer
7.5
If the point that is furthest away from the origin is sent into the circle,
the whole quadrilateral is sent into the circle. K is furthest away.
3 K – furthest
3 KO = 18
3 answer
KO = 3 2 + 3 2 = 18
1
p.KO = 1 , p =
18
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(2)
(3)
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QUESTION 8
8.
xQ = x cos θ + y sin θ
xQ = −2 cos135° + (−3) sin 135°
xQ =
2
−
3
=
−1
2
2
2
y Q = y cos θ − x sin θ
or
− 2
or - 0,71
2
y Q = −3 cos135° − (−2) sin 135°
yQ =
3
2
+
2
2
=
5
2
=
3 subst −2 and −3
into correct formula
for yQ
3 for y coordinate (in
any format)
(5)
5 2
= 3,54
2
⎛ −1 5 ⎞
Q⎜
;
⎟
⎝ 2 2⎠
OR
xQ = x cos θ − y sin θ
xQ = −2 cos(−135°) − (−3) sin( −135°)
xQ =
2
−
3
=
−1
2
2
2
y Q = y cos θ + x sin θ
or
− 2
or - 0,71
2
y Q = −3 cos(−135°) + (−2) sin(−135°)
yQ =
3
2
+
2
2
=
5
5 2
=
= 3,54
2
2
⎛ −1 5 ⎞
Q⎜
;
⎟
⎝ 2 2⎠
OR
x ′ = x cos θ − y sin θ
− 2 = x cos135° − y sin 135°
y
−x
−2=
−
2
2
(1)
−3 2 = x− y
(2)
Solving (1) and (2) simultaneously:
− 5 2 = −2 y
−1
5
y=
x=
and
2
2
3 subst −2 and −3
into correct formula
for xQ
3 using −135°
3 x-coordinate (in
any format)
3 subst −2 and −3
into correct formula
for yQ
3 for y-coordinate (in
any format)
(5)
3 subst −2 and 135°
into correct formula
for x′
− 2 2 = −x − y
y ′ = y cos θ + x sin θ
− 3 = y cos135° + x sin 135°
x
−y
−3=
+
2
2
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3 subst −2 and −3
into correct formula
for xQ
3 using 135°
3 x coordinate (in
any format)
3 simplification
3 subst −2 and 135°
into correct formula
for y′
3 y-coordinate
3 x-coordinate
(5)
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OR
Using first principles: Q = (− r cos α ; r sin α )
r
135°
α
θ
r 3
θ
2
Q / = (−2 ; − 3)
tan θ =
3
2
r = 3 2 + 2 2 = 13
θ = 56,31°
∴ α = 135° − 56,31° = 78,69°
Q = (− r cos α ; r sin α )
= (−0,71 ; 3,54)
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3 tan θ =
3
2
3 r = 13
3 θ = 56,31°
3
Q = (− r cos α ; r sin α )
3 answer
(5)
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QUESTION 9
9.1.1
r = 13
cos α =
9.1.2
9.1.3
3 13
12
13
3
(2)
3 180° − (90° + α )
3 90° − α
(2)
TÔR = 180° − (90° + α )
= 90° − α
cos TÔR =
TR
OT
cos(90° − α ) =
3
7,5
OT
cos(90° − α ) =
7,5
cos(90° − α )
7,5
OT =
sin α
7,5
OT =
5
13
OT = 19,5
3
OT =
7,5
OT
7,5
sin α
7,5
OT =
5
13
OT = 19,5
∴ OT =
9.2
cos x. cos x(− tan x)
− cos x
sin x
= cos x.
cos x
= sin x
= RHS
LHS =
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7,5
OQ
7,5
sin α
5
13
319,5
3
(4)
OR
sin(RT̂O) =
12
13
3
sin( RQ̂O) =
7,5
OQ
7,5
sin α
5
3
13
319,5
3
(4)
3 cos x
3 – tan x
sin x
3
cos x
3 answer
(4)
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QUESTION 10
10.1
Period = 120º
3 120°
(1)
10.2
10.3
sin 3x = -1
x = −30° or x = 90°
3 − 30°
3 90°
(2)
Maximum value of f (x) is 1
∴ Maximum value of h(x) is 0
3 max of f(x)
3 answer
(2)
3 − 90° ; 90°
3 (0° ; 3)
3 (180° ; − 3)
10.4
y
g
2
1
f
x
-90
-60
-30
30
60
90
120
150
180
-1
-2
(3)
-3
10.5
sin 3 x
− cos x = 0
3
sin 3 x − 3 cos x = 0
3
sin 3x = 3 cos x
∴ sin 3 x = 3 cos x
There are 2 solutions where graphs f and g are equal
3 answer
(2)
10.6
f(x).g(x) < 0
x∈ (−60º ; 0º) or (60º ; 90º) or (120º ; 180º)
OR
–60°< x < 0° or 60°< x < 90° or 120°< x < 180°
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333 for each
interval
3 correct brackets
or correct
symbols
(4)
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QUESTION 11
11.1.1
11.1.2
sin 61° = p
sin 241° = sin (180° + 61°)
= − sin 61°
= − p
1
p
61˚
1− p
cos 61° = 1 − sin 2 61°
= 1− p
11.1.3
= 2 cos 2 61° − 1
(
3answer
(2)
3 identity
3 answer
(2)
cos122° = cos 2(61°)
= 2 1− p
3 − sin 61°
)
2
−1
= 2(1 − p ) − 1
= 2 − 2 p −1
= 1− 2p
11.1.4 cos 73°cos15° + sin 73°.sin15°
= cos(73° − 15°)
= cos 58° = (cos 180° − 122°)
= − (cos 122°)
= −(1 – 2p)
= 2p −1
11.2.1
(cos x + sin x) 2 − (cos x − sin x) 2
LHS =
(cos x − sin x)(cos x + sin x)
cos 2 x + 2 cos x sin x + sin 2 x − (cos 2 x − 2 sin x cos x + sin 2 x)
=
(cos x − sin x)(cos x + sin x)
4 cos x sin x
=
cos 2 x − sin 2 x
2 sin 2 x
=
cos 2 x
= 2 tan x
11.2.2
= RHS
cos x = sin x or cos x = − sin x
11.3.1
x = 45°
x = 135°
sin x = cos 2 x − 1
sin x = 1 − 2 sin 2 x − 1
3 double angle
3 expansion
3 answer
(3)
3 cos(73°-15°)
3 − (cos 122°)
3 answer
(3)
3
(cos x + sin x )
2
− (cos x − sin x )
(cos x − sin x )(cos x + sin x )
3numerator
3 4 cos x sin x
3 cos 2 x − sin 2 x
3 2 sin 2 x
3 cos 2 x
(6)
33 for answer
(2)
3 1 – 2sin2 x
sin x = −2 sin 2 x
(1)
2 sin x + sin x = 0
2
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11.3.2 sin x = cos 2x – 1
2 sin2 x + sin x = 0
sin x (2 sin x + 1) = 0
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3 sin x (2 sin x + 1) = 0
1
2
∴ x = 0° + 180° k; k ∈ Z or x = {210° or 330°} + 360°k; k ∈ Z
sin x = 0 or sin x = –
3 sin x = 0 or sin x = –
x = n.180°
3 0° + 180° k
3 210°
3 330°
3 + 360°k; k ∈ Z
OR
x = n.360° − 30°
x = (2n + 1).180° + 30°, n ∈ Z
11.4
1
2
(6)
tan 1˚ × tan 2˚ × tan 3˚ × tan 4˚ × …….× tan 87˚ × tan 88˚ × tan 89˚
⎛ sin 1° ⎞⎛ sin 2° ⎞ ⎛ sin 45° ⎞ ⎛ sin 88° ⎞⎛ sin 89° ⎞
=⎜
⎟⎜
⎟⎜
⎟
⎟......⎜
⎟......⎜
⎝ cos 1° ⎠⎝ cos 2° ⎠ ⎝ cos 45° ⎠ ⎝ cos 88° ⎠⎝ cos 89° ⎠
⎛ sin 1° ⎞⎛ sin 2° ⎞ ⎛ sin 45° ⎞ ⎛ sin(90° − 2°) ⎞⎛ sin(90° − 1°) ⎞
⎟⎟⎜⎜
⎟⎟
=⎜
⎟......⎜⎜
⎟......⎜
⎟⎜
⎝ cos 1° ⎠⎝ cos 2° ⎠ ⎝ cos 45° ⎠ ⎝ cos(90° − 2° ⎠⎝ cos(90° − 1°) ⎠
⎛ sin 1° ⎞⎛ sin 2° ⎞ ⎛ sin 45° ⎞ ⎛ cos 2° ⎞⎛ cos 1° ⎞
=⎜
⎟
⎟......⎜
⎟⎜
⎟⎜
⎟......⎜
⎝ cos 1° ⎠⎝ cos 2° ⎠ ⎝ cos 45° ⎠ ⎝ sin 2° ⎠⎝ sin 1° ⎠
= tan 45°
=1
3 identity
3 co-ratios
3 simplification
3 for answer
(4)
OR
tan 89° = cot 1° tan 88° = cot 2° ...
∴ product is (tan 1°. cot 1°)(tan 2°. cot 2°)...(tan 44°. cot 44°). tan 45°
= 1 × 1 × 1 × ... × 1 = 1
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3 identity
3 co-ratios
3 simplification
3 for answer
(4)
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QUESTION 12
12
In Δ CBG and ΔCDH:
Pythagoras
CG² = x² + y²
Pythagoras
CH² = x² + y²
In ΔFAE
AE² = x² + x²
= 2x²
= GH²
In Δ CGH
GH² = CG² + CH² - 2 CG.CH. cos GCH
CG 2 + CH 2 − GH 2
cos GĈH =
2CG.CH
2
x + y 2 + x 2 + y 2 − 2x 2
cos GĈH =
2 x2 + y2. x2 + y2
2y2
cos GĈH =
2( x 2 + y 2 )
y2
cos GĈH = 2
x + y2
3 CG²
3 CH²
3 AE²
3 AE² = GH²
3 use of cos rule
3 manipulation of
formula
3 substitution
3
cos GĈH =
2y2
2( x 2 + y 2 )
(8)
[8]
TOTAL: 150
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