Post on 04-Mar-2018
SULIT
3472/2
Additional
Mathematics
Paper 2
Sept
2010
PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA
SEKOLAH MENENGAH MALAYSIA (PKPSM) CAWANGAN MELAKA
PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2010
ADDITIONAL MATHEMATICS
Paper 2
MARKING SCHEME
This marking scheme consists of 13 printed pages
Question Mark Scheme of Paper 2 Trial SPM 2010 Melaka Sub
Marks
Total
Marks
1. n= -6 – 4m or n = m
2 +m-2
or
4
6 nm
−−=
m2+m+6+4m=2 or 4m+m
2+m-2+8=2 or 2)
4
6(
2)
4
6( =−
−−+
−−n
nn
(m+4)(m+1)=0 or (n-10)(n+2)=0 or use formula/completing the square
m= -4, -1 or n=10, -2
n=10, -2 or m= -4, -1
P1
K1
K1
N1
N1
5
2.(a)
(b)
(c)
m2 = 2
y -5= 2(x+1) or equivalent
y = 2x +7
Solve simultaneous equations: 22
172 +−=+ xx
E( -2, 3)
)3,2()2
5,
2
)1(( −=
+−+ yx
C( -3, 1)
K1
K1
N1
K1
N1
K1
N1
7
3.(a)
(b)
Shape of sin x
Graph of sin x shifted up 1 unit
Miximum = 4, minimum = -2
1½ cycles in π2
30 ≤≤ x
12
+= xyπ
or equivalent
Sketch the straight line involves x and y
Number of solutions = 4
P1
P1
P1
P1
N1
K1
N1
7
4.(a)
Gradient of tangent = 5
kx2-x = 5
substitute x = 1 to kx2-x = 5
K1
K1
K1
x π x
4 x
1
-2
0 2
π
2
3π
3
y
(b)
k = 6
cxx
y +−=23
6 23
c+−=−2
)1(
3
)1(62
23
2
7
22
23
−−=x
xy or equivalent
N1
K1
K1
N1
7
5.(a)(i)
(ii)
(b)
L = 20.5 F = 57 fQ3 = 28
1028
57)100(4
3
5.203
−
+=Q
= 26.93
100
)5.45(6)5.35(9)5.25(28)5.15(32)5.5(25 ++++=x
=19.4
222222
4.19100
)5.45(6)5.35(9)5.25(28)5.15(32)5.5(25−
++++=σ
= 11.30
222
2
98
20401940
98
204050415 )( −−−
−−=σ
= 126.02
P1
K1
N1
K1
K1
N1
K1
N1
8
6. (a)
(b)
d = 2πr or 2πr, 4πr, 6πr
d = 4πr - 2πr = 6πr - 4πr
2πr = 2πr
T10 = 4 + 9(4) or T10 = 2πr + 9 (2πr )
= 40 = 20 π (r)
Circumference = 2π(40) = 20 π (4)
= 80π = 80 π
P1
K1
N1
K1
N1
N1
6
7.(a)
(b)
(c)
(i)
(ii)
log10x 0.15 0.28 0.40 0.51 0.61 0.74
log10y 0.52 0.72 0.90 1.08 1.20 1.42
Refer to the appendix
One point correctly plotted with uniform scales
6 points correctly plotted with uniform scales
Line of best fit
xkpy 101010 log5loglog −=
-5k =1.52
35.025.0 −↔−=k
log10p = 0.29
N1
N1
K1
N1
N1
P1
K1
N1
K1
N1
0.29.1 ↔=p
10
8(a)(i)
(ii)
(b)
(c)
OBAOAB += or equivalent
yx 36 +−=
xyOD2
1
4
9+=
)2
1
4
9( xymOC +=
)36( yxnBC +−=
)36()2
1
4
9(3 yxnxymy −−+=
5
6=m
10
1=n
½ x 8 x h = 20
h = 5
K1
N1
N1
P1
P1
K1
N1
N1
K1
N1
10
9(a)
(b)
(c)
(d)
1230sin
OQ=° or equivalent
OQ = 6
θπ
62
=
=∠QOR rad12
πθ = // 0.2618 rad
radradROS 833.1/12
7/1051560180 π°=°−°−°=∠
Area of sector ROS = ROS∠××26
2
1
=32.99 or 10.5π
Perimeter of shaded region = )5.52sin6(2)12
7(6 °+π
=20.52
K1
N1
K1
N1
P1
K1
N1
K1K1
N1
10
10(a)
(b)
(c)
Use b2 – 4ac = 0
(-4)2 – 4(1)k =0
k = 4
Solve equation : (x – 2)(x – 2) = 0
A(2, 2) 3
2
32
32
3
−
xx
=
−
−
3
2
2
)2(3
3
3
2
)3(3 3232 -
K1
N1
K1
N1
K1
K1
N1
(d)
=6
7
3
0
543
54
6
3
9
+−
xxxπ
= 05
3
4
)3(6
3
)3(9 543
−
+−π
= 8.1π
K1
K1
N1
10
11(a)(i)
(ii)
(b)(i)
(ii)
377
10 )4.0()6.0(C
= 0.215
1 – P(x=0) – P(x=1) or P(x=2)+P(x=3)+P(x+4)+ …………….+P(x=10)
= 911
101000
10 )4.0()6.0()4.0()6.0(1 CC −−
= 0.9983
)5
350345(
−<zP
= 0.1587 // 0.15866
P(x >342) = 1 – P(z > 1.6) or other valid method
Number of cakes = 0.9452 x 1500
= 1417 // 1418
K1
N1
P1
K1
N1
K1
N1
K1
K1
N1
10
12(a)
(b)
(c)
(d)
Use 1000
1×=
P
PI
x = 135, y = 80, z = 140
141842
)8(140)4(125150)2(150=
+++
+++
m
m
m = 6
141100350
2005=×
P
P2005 = 493.50
100141
11502005/2006 ××=I
= 106.38
K1
N2,1,0
K1K1
N1
K1
N1
K1
N1
10
13 (a)
(b)
(c)
22 611 −=PQ
= 9.22
8
40sin
22.9
sin=
∠PRQ
= 47.80° // 47°48
PRQQPR ∠°°=∠ --40180 or 92.2°
QR = 12.44
QVR∠−+= cos)10)(11(2101144.12 222 '2972//48.72 °°=∠QVR
4552
48.72sin10112
1 。= °××
K1
N1
K1
N1
P1
N1
K1
N1
K1
N1
10
14.(a)
(b)
(c)(i)
(ii)
x + y ≤ 40 or equivalent
4x + 3y ≥36 or equivalent
x ≤ 2y or equivalent
Refer to the appendix
Draw correctly at least one straight line involves x and y
Draw correctly all the three straight lines
Region R shaded correctly
y = 6
min passengers = 11
substitute any point in the region R into 40x + 30y
Substitute ( 26, 14) to 40x + 30y and maximum fares = 1460
N1
N1
N1
K1
N1
N1
K1
N1
K1
N1
10
15(a)
(b)
(c)
(d)
18 ms-1
2t -9 = 0
2
9=t
18)2
9(9)
2
9(
2max +−=V
= 4
12−
Use sketch quadratic graph // number lines // others valid method for
t2-9t + 18 < 0
3 < t < 6
ttt
s 182
9
3
23
+−=
P1
K1
K1
N1
K1
N1
K1
3 6
Substitute t = 3 or t = 5 into ttt
s 182
9
3
23
+−=
S3 = 22.5
S5 =
6
119
Total distance traveled
= )6
1195.22()05.22( −+−
6
525=
OR 5
3
233
0
23
182
9
318
2
9
3
+−+
+−
ttt
tt
+−−
+−+
−+−= )3(18
2
)3(9
3
3)5(18
2
)5(9
3
50)3(18
2
)3(9
3
3232323
=6
525
K1
K1
N1
K1K1
K1
N1
10
7(a)
14(b)
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
lg x
lg y
(0.15,0.52)
(0.28,0.72)
(0.40,0.90)
(0.51,1.08)
(0.61,1.20)
(0.74,1.42)
R
x+y=40
x=2y
4x+3y=36