skema spm trial 2008 - trial paper collection · pdf file · 2008-11-06skema...
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SULIT 3472/2
SKEMA MATEMATIK TAMBAHANPEPERIKSAAAN PERCUBAAN SPM JPNS 2008
BAHAGIAN A
Soalan Penyelesaian dan Peraturan Markah MarkahSub
Markahpenuh
1 24
mn or m = 4(2 – n) P1
2 12 4 24
mm
or (8 – 4n)2 – 12 = 4n K1
4 5m m or (4n – 13)(n – 1) = 0 K1
4, 5m or13
1,4
n N1
131,
4n or 4, 5m N1 5 5
2(a)
2(b)
m = 3 P1
hx2 – 9 = 3 K1
h(1) – 9 = 3 K1
h = 12 N1
dxx )912( 2 K1
cxx
y 93
12 3
N1
c )1(93
)1(122
3
K1
y = 4x3 – 9x + 7 N1
4
4 8
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SULIT 3472/2
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3(a)
(b)(i)
(ii)
3x, x2
3, x
4
3P1
1
2r P1
6
72
13
xT K1
3
64
xN1
2
11
3
xS K1
= 6x N1
4
2 6
4(a)
4(b)
Score Cumulativefrequency
1 1y 2
10 414 64x 1028 12
53 + y + 4xFor C.F. P1
x – 5 = 7 K1
x = 12 N1
1(1) + m(2) + 10(4) + 14(6) + 4x(10) + 28(12) P1
912
101
yK1
y = 7 N1
4
2 6
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SULIT 3472/2
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5 Shape of y = cos x P1
Amplitude 4 P1
N
6(a)
6(b)
6(c)
Q
m
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2 periods P1
4
Modulus P1
Straight line graph P1
3x
y
K1
umber of solutions = 4 N1
5
2 7
6
2
5
3R K1
1
1N1
1
1
45
2h
mK1
2 + m = – (–1) K1
m = – 1 N1
222 )1()1(416 K1
m2 + 16 = 16(2) K1
4m N1
2
3
3 8
O π 2π
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SULIT 3472/2
BAHAGIAN B
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7(a)
7(b)
N1
* If table is not given, see from the points plotted in the graph
Plotx
yagainst x (correct axes and uniform scale) K1
All points plotted correctly N1
The line of best fit (go through at least 3 points) N1
(i) abaxx
y P1
Find gradient K1
a = gradient =05
138.21
K1
a = 4.076 (accept 4.08) N1
(ii) ab = intercept of the axisx
y= 1 K1
b = 0.245 N1
x 1 2 3 4 5
x
y5.00 9.20 13.17 17.30 21.38
4
6 10
8(a)x
dx
dy2 K1
2
12 m K1
12
13 xy K1
2
5
2
1 xy or equivalent N1 4
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SULIT 3472/2
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8(b)
8(c)
Area = dyy 4
3
2
1
)4( or 1
0
2 )4( x or 3 K1
=
4
3
2
3
2
31
4
yor 3
34
1
0
3
xx K1
=3
2unit2 N1
Volume = 4
3
)4( dyy K1
=
2
3)3(4()
2
4)4(4(
22
K1
= 2
1unit3 N1
3
3 10
9(a)
9(b)
9(c)
9(d)
Midpoint AC = ( 2, –1 ) K1)2(11 xy K1
1 xy N1
5
4ABm K1
5
12
5
4 xy or equavalent N1
124)1(5 xx or 12)1(45 yy K1
Find the value x or y or 7x or 8y K1
(–7, 8 ) N1
42122 yx K1
0114222 yxyx N1
3
2
3
2 10
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10 (a)
10(b)
10(c)
0
0
180
142.360 K1
Arc QRS = 13 (0
0
180
142.360 ) K1
= 13.62 cm N1
Arc QTS = 20 (0
0
180
142.330 ) or 10.47 K1
Perimeter = 20 + 20 + 10.47 K1= 50.47 cm N1
Area =
0
022
180
142.33020
2
113142.3 K1,K1,K1
426.26 cm2 N1
3
3
4 10
11(a)(i)
(ii)
(b)(i)
(ii)
Eiher one )13( xP or )14( xP or )15( xP K115
15
14
14
213
13)75.0(15)25.0()75.0(15)25.0()75.0(15)13( CCCxP
K1= 0.2361 N1
2.10)25.0)(75.0( n K1
n = 554 or 555 N1
)15
5880(80
zPxP or )8.1( zP K1
= 0.0359 / 0.03593 N1
1.0)15
58(
wxP K1
281.115
58
wK1
w = 38.785 N1
3
2
2
3 10
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SULIT 3472/2
BAHAGIAN C
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12(a)
12(b)
12(c)
12(d)
QS2 = 82 + 7.52 – 2(8)(7.5)cos 128o K1
QS = 13.93 N1
o
*
58sin
12
sin
93.13
QPSf.t.* K1
QPS = 79.88o /79o53' (accept 79.87o / 79o52') N1
Area of ∆QRS (A1) =2
1(7.5)(8)sin128o
or Area of ∆QPS (A2) =2
1(13.93)(12)sin 42.12o f.t. QPS K1
see 42.12o/42o7' P1(42.13o/42o8')
A1 + A2 K1= 79. 70 (79.71) N1
ot12.42sin
39.13 /(42.13o)
or1
2
2
4
13 (a)Use I = 100
2
1 P
P
x = 110
y = 33.60
z = 22.75
Note: f.t.= follow thrmentioned
42.12o
S
13.93 58o
Q
htt
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0656122
.))(( t /(56.07) K1
t = 9.343 /(9.344) N1 2 10
K1
N1
N1
N1
ough student’s answer for K marks only when
4
P
t
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SULIT 3472/2
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13(b)
13(c)
13(d)
16
)130(5)125(4)110(7 * I f.t.* K1
= 120 N1
210 ×100
120K1
= 252 N1
100
170120I K1
= 204 N1
2
2
2 10
14(a)
14(b)
14(c)(i)
(ii)
100x + 60y ≤ 6600 or equivalent N1
80x + 120y ≥ 4800 or equivalent N1
y ≥ 2x or equivalent N1
Draw one of the 3 lines correctly f.t. (a) K1
Draw all 3 lines correctly K1
Region R N1
30 ≤ y ≤ 85 N1
50x + 25y = k (any value of k) K1
(30, 60) N1
Maximum profit = RM3000 N1
3
3
4 10
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SULIT 3472/2
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15(a)
15(b)
15(c)
15(d)
h = 8 N1
a = –2t + 8 K1= –2(8) + 8 f.t. h K1= –8 N1
–t(t – 8) = 2t K1t = 6 N1
sP = dttt )( 82
or sQ = dtt 2 see the power of t increased by 1 K1
sP = 23
643
6)(
)(
or sQ = 62 substitute t = 6 f.t.(c) K1
sP – sQ K1= 36 m N1
1
3
2
4 10
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SULIT 3472/2
0
10
20
30
40
50
60
70
80
90
100
y
10 20
y = 2x
5x + 3y = 330
110
R
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(30, 60) ●
30 40 50 60 70 80
2x + 3y = 120
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SULIT 3472/2
0 1
X
X
X
X
2
4
6
8
10
12
14
16
18
20
22x
y
(0, 1)
( 5, 21.4 )
7(a)
( 5, 21.4 )
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2 3
X
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5x
pot.com