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SULIT
3472/2
Additional
Mathematics
Paper 2
Sept
2010
PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA
SEKOLAH MENENGAH MALAYSIA (PKPSM) CAWANGAN MELAKA
DENGAN KERJASAMA
JABATAN PELAJARAN MELAKA
PEPERIKSAAN PERCUBAAN
SIJIL PELAJARAN MALAYSIA 2010
ADDITIONAL MATHEMATICS
Paper 2
MARKING SCHEME
This marking scheme consists of 11 printed pages
MARKING SCHEME FOR ADDITIONAL MATHEMATICS TRIAL SPM 2010
PAPER 2
NO. Solution Mark
Scheme
1
1or 3
1or 3
7
0)1)(73(
02)34()34(
34
2
y
x
xx
xxx
xy OR
1or 3
7
1or 3
0)1)(3(
02)3
4(
3
4
2
x
y
yy
yyy
yx
1
1
1
1
1
[5]
2 (a)
(b)
(i)
6
3
2
0
)2(2
p
p
(ii)
4
17
2
3
)]3([2
34
2
32
xy
xy
m
2
21
)4()2(1
q
q
q OR
2
01
)3()6(1
q
q
q
1
1
1
1
1
1
1
[7]
NO. Solution Mark
Scheme
3(a)
2
32sin
2
3
xx or
2
3
xy
Straight line drawn
5 solutions
1
1
1
1
1
1
[6]
4(a)
(b)
(c)
3
4
042
k
xkx
829
4
)3(2)3(9
42
29
44
3
4
23
23
232
xxy
c
cxxdxxxy
point maximum a ,4,0
43
8
)8,0(
2
2
2
2
dx
ydx
xdx
yd
1
1
1
1
1
1
1
1
[8]
0 2
2
3 2
2
3
2
3
x
y
Graph sinus
Amplitude 1.5 units
Two cycles in 0≤x≤2π
NO. Solution Mark
Scheme
5 (a)
10
5.52)10(16)28(
2
1
5.49
,16,5.49
k
k
k
kfFL m
Uniform scales + height of any bar correct
Use boundaries or midpoints
All heights of 6 bars correct
Mode = 50.5
1
1
1
1
1
1
1
[7]
6(a)
(b)
(c)
50
)5)(15(305 T
hours 9
50)]5)(1()30(2[2
)]5)(1()30(2[2
t
ttt
tt
)]5)(19()30(2[2
9 or )9(50
450 km
1
1
1
1
1
1
1
[7]
NO. Solution Mark
Scheme
7(a)
(b)
(c)
Uniform scale + one of the points plotted correctly
6 Points plotted correctly
Line of best fit
1 2 3 4 5 6 7 8 9
1
2
3
4
5
6
7
8
9
10
11
12
(0.25,0.64)
(1,1.59)
(2.25,3.21)
(4,5.42)
(6.25,8.325)
(9,11.79)
(i)
hkx
kxy
21 2
28.11
k
k = 0.78
(ii) 3.0
2
hk
h = 8.55
x 2 0.25 1.0 2.25 4.0 6.25 9.0
xy 0.64 1.59 3.21 5.42 8.325 11.79
1
1
1
1
1
1
1
1
1
1
[10]
NO. Solution Mark
Scheme
8(a)
(b)
(c)
ODAOAD OR
OCBOBC
xyAD 32
yxBC 52
ykxkOE )55(2
hyxhOE 2)33(
( hk 332 or hk 255 ) and (
2
55332
kk or
hh
22
3355
)
11
9
11
5
k
h
062.849 22
AD
665.3)062.8(11
5
AE
1
1
1
1
1
1
1
1
1
1
[10]
9(a)
(b)
(c)
.107.1
7
14tan
radRQS
RQS
245.28
77)035.2(7
)035.2(7
035.2107.1142.3
Perimeter
S
RQS
)035.2)(7(2
1 2 or )
2
142.3)(14(
2
1 2 or )107.1)(7(
2
1 2
)14)(7(2
1
)2
142.3)(14(
2
1 2− )035.2)(7(
2
1 2− )14)(7(
2
1+ )107.1)(7(
2
1 2
82.22
1
1
1
1
1
1
1
1
1
1
[10]
NO. Solution Mark
Scheme
10(a)
(b)
(c)
1)2(2 kyy
2 ,1)))3(2(32 kk
Limit integral −1, 8
8
1
2
3
)1)(2
3(
)1(
x
)1)(2
3(
)11(
)1)(2
3(
)18( 2
3
2
3
2
8
2
units 9
)3)(6(2
1 or )
2
2(
dx
x
3
3535
3
1
35
units 15
133
13
)1(2
5
13
3
)3(2
5
3
3
2
5
y
yy
1
1
1
1
1
1
1
1
1
1
[10]
NO. Solution Mark
Scheme
11(a)
(b)
(i) mean = (250)(0.75)
175
(ii)
1001.0
)25.0()75.0()8( 088
8CXP
(i)
38292.0
)5.0()5.0(1
6
4043
6
4037
6
zPzP
zP
(ii)
326439or 326438
12500038292.0
n
n
1
1
1
1
1
1
1
1
1
1
[10]
12(a)
(b)
(c)
4
2
1
8 subt. and 165)2()2(2
8or 0)2(4
4
2
q
p
pqqp
pqqp
qpta
s 5
0542
t
tt
mS
ttt
dtttS
3
230
52
4
3
)54(
4
0
23
4
0
2
1
1
1
1
1
1
1
1
1
1
[10]
NO. Solution Mark
Scheme
13(a)
(b)
(c)
3
8.12020
)5(138)1(140)8(120)3(115
4
20)5()1()8(3
y
yyxx
x
yyxx
80.4348RM
8.1201003600
09
09
P
P
115 is seen
74.121
100115
140
1
1
1,1
1
1
1
1
1
1
[10]
NO. Solution Mark
Scheme
14(a)
(b)
(c)
equivalentor 40004050
equivalentor 3
equivalentor 40
yx
xy
yx
10 20 30 40 50 60 70 80 90
10
20
30
40
50
60
70
80
90
100
x
y
Uniform Scale + one of the straight lines drawn correctly
Three straight lines drawn correctly
Shade region R exactly
(i) 36
(ii) (10,30)
50(10) + 40(30) or subt. any point in Region R into 50x+40y
1700
1
1
1
1
1
1
1
1
1
1
[10]
NO. Solution Mark
Scheme
15(a)
(b)
(c)
'3340or 54.40
5.7
32sin
2.9
sin
D
D
BE = 11.25 cm or AE = 13.8 cm
BEC = 72.54 or 72°33’ or 72°32’
BC2 = 9.2
2 + 11.25
2 − 2(9.2)(11.25)cos72.54°
BC = 12.21 cm
BEA = 107.46° or 107°27’ or 107°28’
A1 = Area ∆BEC = 54.72sin)25.11)(2.9(2
1 or
A2 = Area ∆BEA = 46.107sin)25.11)(8.13(2
1
Total area = A1 + A2
= 123.41 cm2
1
1
1
1
1
1
1
1
1
1
[10]