3472/2 ©2011 Mid Year Exam F5 Hak Cipta SBP
1
3472/2
Matematik Tambahan Kertas 2 2 ½ jam Mei 2011
BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN
KEMENTERIAN PELAJARAN MALAYSIA
PEPERIKSAAN PERTENGAHAN TAHUN TINGKATAN 5
2011
ADDITIONAL MATHEMATICS
Paper 2
Skema Pemarkahan ini mengandungi 12 halaman bercetak
MARKING SCHEME
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3472/2 ©2011 Mid Year Exam F5 Hak Cipta SBP
2
PERATURAN PEMARKAHAN – KERTAS 2
No Solution and Mark Scheme Sub Marks
Total Marks
1
3
21 yx
OR
2
13
xy
42
212
2
212
y
yy
y OR 4
2
13
2
1322
xx
xx
0)7)(52( yy OR 0)5)(2( xx
7,2
5 yy (both) OR 5,2 xx (both)
5,2 xx (both) OR 7,2
5 yy (both)
P1
K1
K1
N1
N1
5
2(a)
463)( 2 xxxf
41)1(3
4)2(3
2
2
x
xx
7)1(3 2 x
K1
N1
(b) Maximum point = (1 ,7)
P1
(c)
N1
N1
N1
6
f(x)
Shape (max)
Maximum point
*(1,7)
y-intercept = 4 and any one
point (-1,-5) or (2,4)
(1,7)
4
x
(-1,-5)
(2,4)
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3472/2 ©2011 Mid Year Exam F5 Hak Cipta SBP
3
3(a)
5nT , 3,80 da
5)3)(1(80 n
26n
P1
K1
Height of the arrangement = 2601026 cm N1
(b) )3)(126()80(2
2
2626 S
1105
The total length of circumference of circles = )5(21105
11050 cm
K1
K1
N1
6
4
(a) Mid point of AC = (-1 , 6 )
use midpoint BD = midpoint AC : 62
0
p
p = 6
(b) Area of ADC =
= )]10)(13()12)(3()2)(1[()]10)(3()2)(13()12)(1[(2
1
= 60 unit2
(c) Gradient of BD = 1
2
Equation of BD = (y-6) = 1( 1)
2x
2y + x =11 or equivalent
P1
K1
P1
K1
N1
P1
K1
N1
8 5
(a)
P1
P1
P1
P1
Shape of sine graph
Amplitude = 1
Period of 1 cycle 0 x
Graph shifted up by 1 unit
2
2
1
0
y
x
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3472/2 ©2011 Mid Year Exam F5 Hak Cipta SBP
4
(b)
xy
22
N1
K1
N1
7 6 (b)
4350
50x
or 29 2 133950fx
22 )29(150
133950
211.7
K1
P1
K1
N1
8
Eq. of straight line ;
Draw the straight line :
(either gradient or y-intercept correct)
Number of solution = 3
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3472/2 ©2011 Mid Year Exam F5 Hak Cipta SBP
5
6(a) Skema Histogram
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3472/2 ©2011 Mid Year Exam F5 Hak Cipta SBP
6
7
Linear law graph
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3472/2 ©2011 Mid Year Exam F5 Hak Cipta SBP
7
7 (a)
Note: If table is not shown, award N1 if all the points are plotted correctly.
Plot x
y against x
(Correct axes and uniform scales) 6 *points plotted correctly Or at least 5 points correctly plotted if the table is not shown Line of best fit
(b) pkxk
x
y
2
(can be implied)
(i) Use *m = 2
k
k = 1.26 ± 0.05
(ii) Use *c = pk
p = 1.59 ± 0.05
(c) y = 21.6 ± 0.05
x 2 3 4 5 6 7
3.20 3.90 4.50 5.12 5.75 6.40
N1
K1
N1
N1
P1
K1
N1
K1
N1
N1
10
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3472/2 ©2011 Mid Year Exam F5 Hak Cipta SBP
8
8 (a) (OR)(0.9273) = 13.91
OR = 15 cm
(b) OP = 9 cm
Area = 2 21 1(9) (0.9273) (9) sin 0.9273
2 2
= 5.156 cm2
(c) PR = 6 cm
RQ = 15 sin 0.9273 or 2 215 9
Perimeter = arc length PQ + PR + RQ
= 9(0.9273) + 6 + 15 sin 0.9273
= 26.35 cm
K1
N1
P1 K1K1 N1 P1 K1K1 N1
10 9
(a) (i)
(ii)
=
=
(b) (i)
(ii)
=
= or
N1
K1
N1
K1
K1
N1
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3472/2 ©2011 Mid Year Exam F5 Hak Cipta SBP
9
Equate the coefficient of and solve the simultaneous equation
and
(both)
(c) [ t = perpendicular distance]
t = 7 cm
K1
N1
K1
N1
10 10 (a) 2)2(4 xx
3,0 xx
A(0,4) B(3,1)
(b) Area
3
0
2)2()14)(3(2
1dxx
0
3
3
)2(
2
15 3
x
3
)20(
3
)23(
2
15 33
= 4.5
(c) Volume
3
2
24 )1()1)((3
1)2( dxx
32
3
5
)2( 5
x
15
8
K1
N1N1
K1K1
K1
N1
K1
K1 N1
10
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3472/2 ©2011 Mid Year Exam F5 Hak Cipta SBP
10
11 (a) (i) 2
dy
dx
(ii) 2
1normalm
32 xy
(b) )2(3)2(5,2 2 dx
dyx
26
y (26)(-0.02)
= -0.52
(c ) cxx
y 2
3
3
5 23
c
2
)1(3
3
)1(52
23
6
13c
6
133
3
5 3
xx
y
P1
P1 K1
N1
P1
K1
N1
K1
K1
N1
10
12
(a) 5
x 100 = 165 = 115 or 6.9
y 100 = 115 or 4 50
4 00
100 = z
x = 825 / y = 600 / z = 112.5
(b) (i) I = (165 125) (*112.5 100) (115 70) (120 65)
360
[ * follow thro’ from (a)]
W = 360 or 72
= 1326
(ii) 3700
100 132.6*
2009P
[ * follow thro’ from b(i)]
2009
P = RM 279035
K1
N2,1,0
K1
P1
N1
K1
N1
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3472/2 ©2011 Mid Year Exam F5 Hak Cipta SBP
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(c) 2011
2009
130132.6*
100I [* follow thro’ from b(i)]
OR
4810100
2790 K1
= 172.4
K1
N1
10 13
(a) 2009
2007
160 100100
100 135I
= 118.5
(b) (i) 12.60
1009.40
x
= 134.0 or 134.04 or 134
(ii) 2007118 1009.40
P or
2007
12.60118
134P
2007P RM11.09 = RM11.10
(c) (i) 2005
125.50100
137.50P
= RM91.27
(ii) 118(5) 135( ) 175(3)
137.55 3
k
k
k = 6
K1
N1
K1
N1
K1
K1
K1
N1
K1
N1
10 14
(a) ABC = 45.58◦ ABC = 180◦ - 45.58◦ = 134.42◦ (obtuse)
K1
K1 N1
year 2009 2010 2011
Cost 2790.35 3700 4810
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3472/2 ©2011 Mid Year Exam F5 Hak Cipta SBP
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14 (b) AC = 72 + 82 – 2(7)(8) cos 134.42◦ = 13.83 cm
(c) ADC = 126.08◦
(d) ACD = 13.92◦
2092.13sin)83.13)(11(2
1 0 = 38.30
K1
N1
K1
N1
P1
K1
N1
10 15
(a) 3.5
55sin
3
sin
C
C = 27.62o
DBC = 180o - 27.62o - 55o = 97.38o
(b) oADE 55 (AE)2
= 92 + 42 – 2(4)(9) cos 55o
AE = 7.463 cm
(c) (i)
(ii) DCB' = 82.62o
(iii) )62.82(2180' ooBBC
= 14.76o
Area = 076.14sin)3.5)(3.5(2
1
= 3.578 cm2
K1
N1
N1
K1
N1
P1
P1
P1
K1
N1
10
C
B’
B 3 cm
5.3 cm 55 D
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No. of students
No. 6
95 145 195 245 295 345
10
20
30
40
(a) K1 for the bars with constant width
and proportional height.
K1 for boundary/mid point
N1 all correct
N1 mode = 31 025
Time
11
SULIT 3472/2
3472/2 2011 Mid Year Exam Form 5 Hak Cipta SBP
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5
15
25
35
45
50
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3472/2 2011 Mid Year Exam F5 Hak Cipta SBP
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y
x
x 7
SKEMA mid year F5 P2 SBP 2011
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