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2008 SPM TRIAL EXAMINATION Marking Scheme
3472/2 Additional Mathematics Paper 2 [Lihat sebelah SULIT
2
Question Solution and marking scheme Sub-mark
Full Mark
1.
3
2yx
or
2 3y x
223 3 10
2 2y y y y
or 22 2 3 2 3 10x x x x
2 2 29 6 6 2 4 40 0y y y y y or
2 2 22 3 4 12 9 10 0x x x x x y = 3.215 , − 3.215 or x = 3.107 , − 0.107 x = 3.107 / 3.108 , − 0.107 / − 0.108 or y = 3.214 / 3.215 , − 3.214 / − 3.215 Answer must correct to 3 decimal places.
5
P1 Make x or y as the subject
Eliminate x or y
Solve quadratic equation
23 31 0y
2 313
y
or 23 9 1 0x x
or using formula
or completing the
square
N1
N1
K1
K1
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2008 SPM TRIAL EXAMINATION Marking Scheme
3472/2 Additional Mathematics Paper 2 [Lihat sebelah SULIT
3
Question Solution and marking scheme Sub-mark
Full Mark
2(a)
(b)
16 ,8 ,4 ,......
16a 12
r
7
7
116 12
112
S
= 3314
or 31.75 64 ,16 , 4 ,......
64a 14
r
64
114
S
= 1853
or 85.33
3 3
6
3(a)
2( )f x x px q
= 2 2
2
2 2p px px q
= 2 2
2 4p px q
32p
p = -6
36 54
q q = 4
N1
K1
Use rraS
nn
1)1(
N1
K1
K1
N1
Use 1
aSr
use x² + bx = ( x + 2b )² – 2
2 )( b
or
use axis of symmetry 32ba
P1
P1
N1
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2008 SPM TRIAL EXAMINATION Marking Scheme
3472/2 Additional Mathematics Paper 2 [Lihat sebelah SULIT
4
Question Solution and marking scheme Sub-mark
Full Mark
Alternative solution
22 3 5x px q x
= 2 6 9 5x x = 2 6 4x x
6p q = 4
3
3(b) 2 6 4 31 0x x
2 6 27 0x x
( 9)( 3) 0x x
3 9x
2
5
4(a)
tan x + cos1 sin
xx
= sincos
xx
+ cos1 sin
xx
= 2sin (1 sin ) cos
cos (1 sin )x x x
x x
= 2 2sin sin cos
cos (1 sin )x x x
x x
= sin 1cos (1 sin )
xx x
= 1cos x
= sec x
3
K1
N1
Use ( ) 31 0f x and factorization
K1
K1
N1
Use sintancos
xxx
Use identity 2 2sin cos 1x x
Comparing coefficient of x or constant term
N1
K1
N1
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2008 SPM TRIAL EXAMINATION Marking Scheme
3472/2 Additional Mathematics Paper 2 [Lihat sebelah SULIT
5
Question Solution and marking scheme Sub-mark
Full Mark
4(b)(i) Negative sine shape correct. Amplitude = 3 [ Maximum = 3 and Minimum = − 3 ] Two full cycle in 0 x 2
3
4(b)(ii) 53sin 2 2xx
or 5 2xy
Draw the straight line 5 2xy
Number of solutions is 3 .
3
9
K1
N1
N1
P1
P1
P1
y
3
–3
2
2
3 2 x
5 2xy
O
3sin 2y x
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2008 SPM TRIAL EXAMINATION Marking Scheme
3472/2 Additional Mathematics Paper 2 [Lihat sebelah SULIT
6
Question Solution and marking scheme Sub-mark
Full Mark
5 (a)
(b)
1220
x 240x
The mean
240 5 + 8 + 10 + 11 + 14 28825 25
X
= 11.52
2212 3
20x
2 3060x
The standard deviation
2 2 2 2 2
23060 5 8 10 11 14 11.5225
= 23566 11.5225
= 9.9296 = 3.151
7
7
K1
N1
Use formula 2
2xx
N
For the new 2x and X
K1
K1
N1
N1
Use x
xN
N1
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2008 SPM TRIAL EXAMINATION Marking Scheme
3472/2 Additional Mathematics Paper 2 [Lihat sebelah SULIT
7
Question Solution and marking scheme Sub-mark
Full Mark
6(a)
(b)
(i) PR PO OR
uuur uuur uuur
= 6 15a b
% %
(ii) OQ OP PQ
uuur uuur uuur
= 365
a ORuuur
%
= 6 9a b
% %
(i) OS hOQ
uuur uuur
= (6 9 )h a b
% %
(ii) OS OP PS
uuur uuur uuur
= 6a kPR
uuur
%
= 6 6 15a k a b
% % %
(6 9 )h a b% %
= 6 6 15a k a b % % %
6 6 6h k 9 15h k
1h k 53
h k
5 13
k k
38
k
5 33 8
h
= 58
3
5
8
N1
N1
K1
Use PR PO OR uuur uuur uuur
or OQ OP PQ uuur uuur uuur
N1
N1
Equate coefficient of a
% or b
%
and Eliminate h or k
K1
N1
N1
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2008 SPM TRIAL EXAMINATION Marking Scheme
3472/2 Additional Mathematics Paper 2 [Lihat sebelah SULIT
8
Qn. Solution and Marking Scheme Sub- mark
Full Mark
7(a)
(b)(i)
(ii)
kn
xn
y 4log1log1log
log x 0.18 0.30 0.40 0.60 0.74 log y 0.48 0.54 0.59 0.69 0.76
6
4
10
N1 N1
P1
Correct axes and scale
All points plotted correctly
Line of best-fit
N1
K1 intercept
= kn
4log1
n = 2 k = 1.51
K1
N1
N1
K1
N1
gradient
= n1
N1 N1
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2008 SPM TRIAL EXAMINATION Marking Scheme
3472/2 Additional Mathematics Paper 2 [Lihat sebelah SULIT
9
0.6
0.7
Graph of log10 y against log10 x
0.8
0.5
0.4
0.1
0.2
0.3
log 10 x
log 10 y
0.9
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
0.39
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2008 SPM TRIAL EXAMINATION Marking Scheme
3472/2 Additional Mathematics Paper 2 [Lihat sebelah SULIT
10
Qn. Solution and Marking Scheme Sub- mark
Full Mark
8. (a)
(b)
(c)
Solving simultaneous equation P(– 2, 2) Q(4, 8)
Use dxyy )( 12
dxxx )2
4(2
Integrate dxyy )( 12
64
2
32 xxx
18
Note : If use area of trapezium and ydx , give the marks
accordingly.
Integrate dxx 2
2)
2(
=
20
5x
58
3
4
3
10
K1
N1 N1
N1
K1
K1
K1
Use correct
limit
4
2into
64
2
32 xxx
N1
K1
K1
Use correct
limit 4
0
into
20
5x
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2008 SPM TRIAL EXAMINATION Marking Scheme
3472/2 Additional Mathematics Paper 2 [Lihat sebelah SULIT
11
Qn. Solution and Marking Scheme Sub- mark
Full Mark
9(a)
(b)
(c)
(d)
Equation of AD : y – 6 = –2 ( x – 2 ) y = –2x + 10 or equivalent
y = –2x + 10 and
x – 2y = 0 D(4, 2)
p = 3 or C(8, 4)
Substitute (8, 4) into y = 3x + q
q = – 20
OADOABCArea 4 Using formula
00
62
24
00
21OAD
40 Alternative solution : B(10, 10) Using formula
00
62
1010
48
00
21OABCArea
2
2
3
3
10
K1
N1
Use m = – 2 and find equation of straight line
Solving simultaneous equations
N1
K1
K1
P1
N1
N1
Find area of triangle
K1
K1
N1
Find area of parallelogram
K1
P1
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2008 SPM TRIAL EXAMINATION Marking Scheme
3472/2 Additional Mathematics Paper 2 [Lihat sebelah SULIT
12
40
Qn. Solution and Marking Scheme Sub- mark
Full Mark
10(a)
(b)
(c )
106sin 1 or equivalent
2POQ
radPOQ 287.1 Alternative solution : 122 = 102 + 102 – 2(10)(10) cos POQ
)10)(10(2121010cos
2221POQ
radPOQ 287.1
Using (2π – 1.287) Major arc PQ = 10 ( 2π – 1.287 ) = 49.96 cm
Lsector = 287.1)10(21 2
Ltriangle = 287.1sin)10(
21 2
= 16.35 cm2
3
3
4
10
Using formula Lsector = 2
1 r2 K1
K1
K1 Lsector - LΔ
N1
Using formula LΔ = 2
1 absin C
Use formula s = r
K1
N1
K1
N1
K1 Use cosine rule
N1
N1
K1
K1
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2008 SPM TRIAL EXAMINATION Marking Scheme
3472/2 Additional Mathematics Paper 2 [Lihat sebelah SULIT
13
= 3.9149 cm
Qn. Solution and Marking Scheme Sub- mark
Full Mark
11 (a)(i)
(ii) b)(i) (ii)
p = 53 , p + q = 1
P( X = 0 )
= 5C0(53 )0(
52 )5
= 0.01024 Using P( X ≥ 4 ) = P( X = 4 ) + P( X = 5 )
= 5C4(53 )4(
52 )1 + (
53 )5
= 0.337 P ( 30 ≤ X ≤ 60 )
= P ( 10
3530 ≤ Z ≤ 60 3510 )
Use P( – 0.5 ≤ Z ≤ 2.5 ) = 1 – P( Z 0.5 ) – P( Z 2.5 ) = 1- 0.30854 – 0.00621 = 0.68525 Number of pupils = P( X 60 ) × 483
3
2
3
K1
N1
K1
N1
Use P(X = r) = n Cr prqn–r , p + q = 1
K1
N1
K1 use
Z =
X
K1
N1
P1
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2008 SPM TRIAL EXAMINATION Marking Scheme
3472/2 Additional Mathematics Paper 2 [Lihat sebelah SULIT
14
3
2
10
Qn. Solution and Marking Scheme Sub- mark
Full mark
12. (a) (b) (c) (d)
Subst. t = 0 into dvdt
a= 15 – 6t 15 ms-2
1 175 3/184 4
ms ms
Integrate
Use s = 0
152
/ 17 / 7.52
S4 – S3
2 2 3
K1
N1
K1
N1
K1
Use 0dvdt
and subst. t in v = 15t – 3t2
[t = 52
]
K1
K1
N1
2 3152
s v dt t t
Subst. t = 3 or t = 4 in 2 315
2s t t
K1
N1
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2008 SPM TRIAL EXAMINATION Marking Scheme
3472/2 Additional Mathematics Paper 2 [Lihat sebelah SULIT
15
1152
Note : If use 4
3vdt , give the marks accordingly.
3
10
Qn. Solution and Marking Scheme Sub- mark
Full mark
13. (a)
(b) (i)
(ii)
(iii)
Use I = 2007
2005100
PP
Value of m : 25, m, 80, 30 or equivalent 120 25+130m+135 80+139 30
Use i i
i
I WI
W
132.1 = 120×25+130m+135×80+139 30135+m
m =65
100150.00 x132.1
RM 113.55 /I . ( . x . )08 05 132 1 132 1 0 3
3 3 2
N2, 1, 0
K1
x = 48.6 y = 135 z = 80
K1
K1
N1
K1
N1
P1
N1
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2008 SPM TRIAL EXAMINATION Marking Scheme
3472/2 Additional Mathematics Paper 2 [Lihat sebelah SULIT
16
171.73
2
10
Qn. Solution and Marking Scheme Sub- mark
Full mark
14. (a) (b) (c)
x + y ≤ 80 or equivalent y ≤ 4x or equivalent
x + 4y ≥ 120 or equivalent
x=30
(16,64)
x + 4y = 120
x + y = 80
y = 4x100
90
80
70
60
50
40
30
20
10
10080604020 9070503010x
y
At least one straight line is drawn correctly from inequalities involving x and y
All the three straight lines are drawn correctly
Region is correctly shaded
(i) minimum = 23
3 3
N1P1 N1
N1
K1
N1
N1
N1
N1
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2008 SPM TRIAL EXAMINATION Marking Scheme
3472/2 Additional Mathematics Paper 2 [Lihat sebelah SULIT
17
(ii) (16,64) Subst. point in the range
in 20x + 40y
RM2880
4
10
Qn. Solution and Marking Scheme Sub- mark
Full mark
15. (a) (b) (c)
(d)
120 9.3 6 sin2
BCD
45o48’ / 45.8o
0.74545 4555 Use cosine rule in ΔBCD BD2 = 9.32 + 62 − 2×9.3×6 cos 45°48’ 6.685 Use sine rule in ΔBCD
osin CBD sin '
. .
45 48
9 3 6 685
94o 10’ 4444 qqq11111aaaaaaaaaaaaaaaaaa 14s4 5.555555 5555 z5555555555555555 Sum of area:
4 2
K1
N1
Use area △= ½ ab sin c in △BCD K1
N1
K1
N1
K1
N1
K1
K1
N1
K1
Use area ADB = ½ 6.685 13 sin ADB
Obtain ADB by using 180o – 85o50’ – BAD or equivalent
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2008 SPM TRIAL EXAMINATION Marking Scheme
3472/2 Additional Mathematics Paper 2 [Lihat sebelah SULIT
18
20 cm2 + ΔABD 555 102 58.82 cm2
4
10
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