- tutor mansor · pdf filep u p = rm 2486.75 k1 n1 n1 k1 n1 n1 k1 n1 k1 n1 10 14 (a) yd3x yt x...

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1
PROGRAM PENINGKATAN PRESTASI AKADEMIK 3472/2
SIJIL PELAJARAN MALAYSIA 2010 Matematik Tambahan Kertas 2
SKEMA JAWAPAN
PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2010
MATEMATIK TAMBAHAN
Kertas 2
PERATURAN PEMARKAHAN
UNTUK KEGUNAAN PEMERIKSA SAHAJA
3472/2 2010 Hak Cipta Jabatan Pelajaran Selangor SULIT

2
Solution Marks Solution Marks
1 yx 34 or 3
4
xy
1)134( yy
or
1)1(3
4
x
x
0153 2 yy or
0732 xx
)3(2
)1)(3(4255 y
542.1,542.4 x
847.1,181.0 y
P1
K1
K1
N1
N1
___
5
2(a)
2
5
2
32 2 xxy
=
2
5
16
9)
4
3(2 2x
=8
49)
4
3(2 2 x
4
3
2
a or 1
88
49
b
2
3a
b = 41
(b) 02)3(2 2 xkx
0)2)(2(4)3( 2 k
0762 kk
0)7)(1( kk
7,1k
K1
K1
N1
N1
K1
K1
N1
___
7
3(a) y2 , y
2 + 2y, y
2 + 4y
yTTyTT 2,2 2312
Since yTTTT 22312 ,then A.P
(b)
i) T7 = 25 + 6(10)
= 85
ii) ]10)1()25(2[2
700 nn
0)14)(10( nn
10n
K1
N1
K1
N1
K1
K1
N1
7
4 (a) LHS = xx 44 sincos
= )sin)(cossin(cos 2222 xxxx
= 1( x2cos )
= x2cos
(b)
Shape of graph cos x
2 cycles
max = 2 and min = 2 within
20 x
2
3xy
Number of solutions = 3
K1
N1
K1
P1
P1
P1
N1
N1 __
8
y
Draw line
2
x 2
0
2
|

3
5 (a) 3 + p + 8 + q + 3 = 24
p + q = 10
(b)(i) p = 9
58
65.443 Q
= 48.25
Modal class = 40 44
7(a)
2
1
3
01
x
x = 1
L( 1, 0)
(b) trapeziumA = 2)1)(31(2
1 unit
2
A1 = 1
0
31
0
2 23
2
y
ydyy
= 23
1
= 3
7unit
2
Area = 3
12
3
7 unit
2
(c) V =
qq
xx
dxx2
2
2
22
)2(
)42(2
22
9 2q
q
0542 qq
( q 5)(q + 1) = 0
q = 5 only
K1
N1
P1
K1
N1
P1
___
6
K1
N1
K1
K1
K1 (limit)
N1
K1
K1
KI
N1
10
6(a) i) OQOT2
1
= y2
1
ii) PSOPOS
= )(3
1OQPOx
= yxx3
1
3
1
= yx3
1
3
2
(b) PR =h PT
PO +OR = h PO( +OT )
yhxhyxkx2
1)
3
1
3
2(
hk
3
23 , hk
2
1
3
1
kk
3
2
3
23
4
3k
2
1h
N1
K1
N1
K1
K1
N1
N1
___
7
8( a)
(b) Refer to the graph
Correct & consistant scale and plot at
least 1 point correctly All 6 points plotted correctly
Line of best fit
(c) i) log10 y = x
2 log 10 a log 10 b
13.7
5.098.0log10
a
a = 1.1918
ii) log10 b = 0.425
b = 0.3758
N1
N1
K1 N1
N1
P1
K1
N1
K1 N1
10
x2 1 2.25 4 5..29 6.25 7.29
log10y 0.5 0.6 0.72 0.83 0.89 0.98

4
9(a) radBAO 621.02
9.1142.3
(b) 10
58.35cosAB
AB = 8.133 cm
BC = 20 8.133 = 11.867
cmSBE 21.6)9.1142.3(5 or
cmSCD 42.12)621.0(20
Perimeter = 10+ 6.21+ 12.42 + 11.867
= 40.497 cm
(c) 22sec 2.124621.0202
1cmA tor
or
22 525.15242.152
1cm
22 83.118.108sin)5(2
1cmAtriangle
Area = 124.2 15.525 11.83
= 96.845 cm2
P1
K1
P1
K1
K1
N1
K1
K1
K1
N1
____
10
10 (a) 2
3,1
2
23
k
k
(b)
)14,9(
24
)3(2,0
4
)3(3
C
yx
(c) Midpoint (3, 6)
4
3
34
1
m
)3(4
36
xy
4
33
4
3
xy
(d) 22 )2()3( yx
22 )2()3( yx =3
4
9x2 + 9y
2 +54x + 36y + 101 = 0
N1
K1
N1
P1
K1
K1
N1
K1
K1
N1
10

5
11(a)
900
)5
4(720
5
4,
5
1
720144
n
n
pq
q
(b) X ~ N ( 28 , 32 )
i) P( 25 X 30 )
6667.013
2830
3
2825
ZP
ZP
= 1 (0.1587 + 0.2525)
= 0.5888 or 0.589
(ii) P(X > m) = 0.1
846.31
282.13
28
1.03
28
m
m
mZP
(accept m = 32)
K1
N1N1
K1
N1
K1
K1
N1
K1
N1
___
10
12 (a)(i)
106cos)10)(15(21015 222AC
AC = 20.1914
(ii)
106sin
1914.20
sin
15
ACB
57.45ACB
(b)i)
43.13457.45180' ABC
(ii)
)]57.45(2180[106' ABC
= 17.14
14.17sin)10)(15(2
1'ABCArea
= 22.1031
K1
N1
K1
N1
P1
K1N1
P1
K1
N1
____
10
A
B
A
C
A

6
13 (a) 1201001500
m
m = 1800
OR
80100900
n
n =1125
(b)
100
3080451501512010130
100
122500
= 122.5
(c)(i) 100
5.122140
2007
2011
I
= 171.5
(c)(ii) 5.1711450
100
P
P = RM 2486.75
K1
N1
N1
K1
N1
N1
K1
N1
K1
N1
10
14 (a) xy 3
50 xy
1000 yx
(b) Refer to graph
Draw at least one line correctly
Draw all the lines correctly
Correct region
(c) (i) RM 720
(ii) (25, 75)
Using x + y = k on any point
in the region
RM 100
N1
N1
N1
K1
N1
N1
N1
N1
K1
N1
10
15 (a) when v = 0,
(2t + 3)(t 2) = 0
t = 2
a = 4 t 1
= 4(2) 1
= 7 m s 2
(b) a = 4t 1 = 0
1
2
ms8
16
6)4
1()
4
1(2
4
1
v
v
st
K1
K1
N1
K1
K1
N1
15 (c)
dttts )62(2
ttt
623
2 23
S2 = 3
28122
3
16
S3 = 2
1418
2
918
d = m6

- tutor mansor · pdf filep u p = rm 2486.75 k1 n1 n1 k1 n1 n1 k1 n1 k1 n1 10 14 (a) yd3x yt x...

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