spm trial exam 2010 mark scheme additional …spm trial exam 2010 . mark scheme additional...
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SPM TRIAL EXAM 2010 MARK SCHEME ADDITIONAL MATHEMATICS PAPER 2
SECTION A (40 MARKS)
No. Mark Scheme Total Marks
1
yx 21−=
( ) ( )( ) 521212 22 =−++− yyyy
0377 2 =−− yy
( ) ( ) ( )( )( )72
37477 2 −−−±−−=y
324.0,324.1 −=y
648.1,648.1−=x OR
21 xy −=
52
12
12 2 =
−+
−+
xxxx
0197 2 =−x
( ) ( ) ( )( )
( )72197400 2 −−±−
=x
648.1,648.1−=x
324.0,324.1 −=y
P1
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P1
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5
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2
(a) ( ) ( )
−
−−
−+−−=
−−−=
2124
244
21422
2
2
xx
xxxf
( ) 252 2 +−−= x (b) Max Value = 25 (c) ( )xf 25 (2,25) 21 x -3 2 7 Shape graph Max point ( )xf intercept or point (0,21) d) ( ) ( ) 252 2 −−= xxf
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7
3 a) List of Areas ; xyxyxy
161,
41,
41
2312 =÷=÷ TTTT
This is Geometric Progression and 41
=r
b) 51225
4112800
1
=
×
−n
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262144
141 1
=
−n
91
41
41
=
−n
91 =−n 10=n
(c)
411
12800
−=∞S
= 3217066 cm 2
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7
4 a) 11cos4 2 −−
2cos4 2− ( )1cos22 2−
θ2cos2 b) i) 2 1 π π2 -1 -2
- shape of cos graph - amplitude (max = 2 and min = -2) - 2 periodic/cycle in πθ 20 ≤≤
b) ii) πθ
−= 1y (equation of straight line)
Number of solution = 4 (without any mistake done)
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5 a) Score 0 – 9 10 – 19 20 – 29 30 – 39 40 – 49 Number 3 4 9 9 10
b) ( )
109
73541
5.191
−+=Q
= 21.44
( )10
10
253543
5.393
−+=Q
= 40.75 Interquatile range
44.2175.40 −= = 19.31
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6
6 (a) AQOAOQ +=
( )~~
1 bmamOQ +−=
(b) ( )ORPOnOQPO +=+ ( )
~~31
54 bnanOQ +−=
(c)
(i) mn −=
− 1
54
54 or mn =3
111,
113
== nm
(ii) ~~ 11
3118 baOQ +=
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8
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7
(a)(i) ( )2
2
0
2Area y y dy= −∫
= 23
2
03yy
−
= 243
unit
(ii) ( )1 2
2
0 1
2Area region P y dy y y dy= + −∫ ∫
23
2
1
1 1 12 3
yy = × × + −
= 276
unit
(b) 24 7 13 6 6
Area region Q unit= − =
7 1:6 6
=
= 7 : 1
(c) ( )1
22
0
2Volume y y dyπ= −∫
13 5
4
0
43 5y yyπ
= − +
= 38 15
unitπ
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10
8
(a) Using the correct, uniform scale and axes All points plotted correctly Line of best fit
(b) 10 10 101log log log3
y x p k= +
x 0.000 0.7071 1.000 1.414 1.732
log10 y 1.000 1.330 1.477 1.672 1.826
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(i) 10 log∗ =use c k k = 10.0
(ii) 101.83 1.0 1* 0.47977 log1.73 0 3
use m p−= = =−
27.5p =
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10
9
(a) 126
COD π ∠ =
1 1.0473
radπ= =
(b) (i) 1 20103 3
Arc ABC orπ π π = − =
2 2 120 10 20cos6
Length AC or radπ = −
20 120cos 38.2673 6
Perimeter cmπ π= + =
(ii) ( )21 2 210 sin2 3 3Area of shaded region π π = −
= 61.432cm2
(c) 16
CDE CAD radπ∠ = ∠ = ( alternate segments )
( )21 1102 6Area π =
= 26.183cm2
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10
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10
(a) ( )4, 2T
22
6,42
6=
+=
+ yx
( )2, 2S − (b) ( )2 2 4y x− = −
2 6y x= −
(c) 3 24 3 242 27 7
x y or + += = −
10 38,3 3
U − −
(d) ( ) ( ) ( ) ( )2 2 2 22 2 2 4 2x y x y− + + = − + − 2 23 3 28 20 72 0x y x y+ − − + =
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10
11
(a) (i) ( ) 10 0 1000 (0.6) (0.4)P X C= = or ( ) 10 1 911 (0.6) (0.4)P X C= = [ ]( 2) 1 ( 0) ( 1)P X P X P X≥ = − = + = = 1 ─ 10 0 100 (0.6) (0.4)C ─
10 1 91(0.6) (0.4)C
= 0.9983
(ii) 28005
×
= 320 (b)(i) ( )0.417 1.25P z− ≤ ≤ = 1057.03383.01 −− = 0.556 (ii) ( ) 0.7977P X t> = Z = 0.833−
4.50.8331.2
t −− =
3.5004t =
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10
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No Mark Scheme Sub Marks Total Mark
12a i)
1 (14) (5)sin 212
θ =
36.87 36 52 'orθ = ° ° 180 36.87BAC∠ = ° − ° 143.13 143 8'or= ° °
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3
ii)
2 2 214 5 2(14)(5)cos 143.13BC = + − ° 2 333BC = 18.25BC cm=
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2
iii)
sin sin 143.135 18.25θ °=
9.46 9 28'orθ = ° °
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b i)
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ii)
180 143.13 9.46ACB∠ = ° − ° − ° 27.41= ° ' ' ' 180 27.41A C B∠ = ° − ° 152.59 152 35'or= ° °
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2
14 cm 5 cm
'A
'B 'C
10
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No Mark Scheme Sub Marks Total Mark
13 a)
4.55 1003.50
m = × or 100 1124n× =
130m = RM 4.48n =
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3
b)
*110(70) 130( ) 120( 1) 112(2) 116.5
7 1 2x xx x
+ + + +=
+ + + +
3x =
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2
c i)
See 140
(116.5) 140100
x=
120.17 / 120.2x =
P1
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3
ii)
100 14025x× =
RM 35x =
K1
N1
2
10
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No Mark Scheme Sub Marks Total Mark
15 a)
10 30v ms−= −
N1
1
b)
23 21 30 0t t− + − > ( 5)( 2) 0t t− − < 2 5t< <
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2
c)
6 21a t= − +
5 6(5) 21a = − +
25 9a ms−= −
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3
d)
3 23 21 30
3 2t tS t−= + −
2
3 21 302tS t t= − + −
2
33
21(3)(3) 30(3) 22.52
S = − + − = − or
2
35
21(5)(5) 30(5) 12.52
S = − + − = −
Total distance = 22.5 ( 22.5) ( 12.5)− + − − − 32.5 m=
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K1
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4
10
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Answer for question 14
10
20
70
60
50
80
90
40
30
y
(20,50)
10 20 30 40 50 60 70 0 80 x
(a) I. 70x y+ ≤
II. 2x y≤ III. 2 10y x− ≤
(b) Refer to the graph, 1 graph correct 3 graphs correct Correct area (c) i) 15 40y≤ ≤ ii) k = 10x + 20y max point ( 20,50 ) Max fees = 10(20) + 20(50) = RM 1,200
10
N1
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N1 K1
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N1
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10log y
x 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8
1.0
1.2
1.4
1.1
1.5
1.3
1.6
1.7
1.8
1.9
2.0
X
X
X
X
X
Answer for question 8
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