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8/8/2019 SPM Perlis Addmath P2 2010 Ans
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ANSWER TRIAL PAPER 2, 2010SECTION A
1.3
333
xyoryx 1M
Substitutex ory into eqn. 2. 1M
10633 2 yy or
10
3
36
2
xx or
equivalent.
07362 yy or
0122722 xx
)6(2
)7)(6(433 2 y 1M
Or
)2(2
)12)(2(4)9()9( 2 x
y = 0.859, 1.359 1M
x = 5.577 // 5.576,1.077 //1.076 1M
2. (a) 54 kkxy 1M
(b) 82
25 2 kdk
dA1M
082
25 2 k 1M
4
5,
4
5 kk 1M
3
2
225 k
dk
Ad 1M
A minimum,4
5k 1M
3. a) 4 + (251)x = 40 1M
x = 1.50 1M
b) )50.1)(125()4(22
25 1M
= 550 1M
c) )5.1)(1()4(22
nn
1M
nnn 30)5.1)(1()4(22 1M
67.35n 1M
N= 36 1M
4. (a)
sec
tan1M
1
cos
cos
sin 2
1M
(b) (i)
graph cos 1M,Max 2 and min 2 1M
2
11 cycle 1M
(ii) 22
xy 1M,
Draw straight line graph 1MNo. of solution = 3 1M
5. (a)k
k
2425.3410359 1M
2.41242
5.3410359
k
k1M
k= 58 1M
(b) 5934652 fx 1M
2)2.41(300
593465 1M
= 16.76 1M
6. (a) 1128
xy
or 83
2 xy or
equivalent 1M
(b)
31
)8(1)0(3,
31
)0(1)12(3
Either x ory correct 1M
(9, 2) 1M
c) Equation CD or gradient CD 1M
))9((232 xy or
xmCD
9
02
Findx, wheny = 0 or2
3
9
02
x
and solve forx 1M
0,
3
23C 1M
2
2
2
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8/8/2019 SPM Perlis Addmath P2 2010 Ans
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Section B
7. a) BCABAC or ABDADB 1M
yxAC 8 1M
yxDB 46 1M
b) (i) )46( yxkDH 1M
(ii) )8(4 yxmyDH 1M
)8(4 yxmy )46( yxk 1M
km 44 and
km 68 1MSolve equations 1M
19
16k 1M,
19
12m 1M
8. a)
y10log
0.1206 0.301 0.4983 0.699 0.8633 1.061
1M
b) qxpy 101010 logloglog 1M
All points plotted 2M
Line of best fit 1M
c) (i) erceptyp intloglog 1010 1M
p= 0.8320.871 1M(ii) gradientq 10log 1M
q = 0.013 1M
(iii) y = 6.03 1M
9. a) AR = 12 + 9 = 21 1M
b) oSOQ 120 1M
= 2.095 rad 1M
c) arc RQ = 180
6021 or
arc CQ = 095.29 1M
perimeter = 095.29 +
180
6021 +9+3 1M
= 52.85 1M
d) Area of sector 1M
Area of triangle 1M
Area of shaded region =
2222612)(12(
2
1095.2)9(
2
1047.1)21(
2
1
1M
= 83.66 1M
10. a) Solve equations
1322 yxandxy 1M
1,2
1 yy or
2,4
5 xx 1M
Q (2, 1) 1M
Area of trapezium = )1(232
21
1M
yy
dyy 3)1(1
0
2 1M
Area =
01
3
1
4
71M
=12
5// 0.417 1M
(c) V dxy
1
0
22 1
dxyyy
1
0
35
3
2
5
013
)1(2
5
135
867.1//15
28
11. (a)p = 0.75 and q = 0.25 1M
Use nnnC 1010 25.075.0 1M
(i) P(X = 10) = 0.05631 1M
(ii) P(X = 9) + P (X = 10) 1M
= 0.244 1M
b)8
174
xz 1M
(i) 0.6915 1M
(ii) P 75.075.1 z 1MFinding correct area 1M
0.7333 1M
1M
1M
1M
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8/8/2019 SPM Perlis Addmath P2 2010 Ans
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12. (a) 15,0 vt 1M
(b) 015183 2 tt and solve 1Mt=1, t= 5 2M
(c) 186 ta 1M
= 6 1M(d)
Shape minimum 1M
3 coordinates shown 1M
(e) Distance = 503 159 ttt 1M= 425 1M
240RM07 P
13. (a) (i) CD = 10 1M
(ii) oCE
70cos10
1M
CE= 3.42 1M
(b) oAB 70cos)9)(12(2912 222 1M
29.12AB 1M
29.12
70sin
12
sin oABC 1M
9175.0sin ABC
ooABC 57.66//56.66 1M
c) Area = 70sin)9)(12(2
1or
Area = 70sin)42.3)(10(2
11M
AreaABED = 70sin)9)(12(2
1 70sin)42.3)(10(
2
1
1M
= 34.67 1M
14. (a) 1101004.1
54.1x
1151002
y
, y = 2.30 1M
10310018.6
z
, z = 6 1M
(b) (i)
360
)90(103)65(115)45(120)160(110 1M
4.110I 1M
(ii) 4.110100110
09 P
1M
44.12109 P
(c) 8009/10 P 1M
100
08/0909/10 PP =100
4.110801M
= 88.32 1M
15. (a) I : 80 yx 1M
II : yx 2 1M
III : 20 xy 1M
(b)
One straight line drawn 1M
The other two straight lines drawn 1M
Region R 1M
(b) (i) 5010 x 1M(ii) yxp 4030
Finding maximum point (30, 50)
Maximum profit = 30 (30) + 40 (50) 1M= 2900 1M
1M
15
1 5
1M
