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SULIT3472/2
Additional MathematicsPaper 2Sept 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 13 printed pages
Question Mark Scheme of Paper 2 Trial SPM 2010 Melaka Sub Marks
Total Marks
1.n= -6 – 4m or n = m2 +m-2 or
m2+m+6+4m=2 or 4m+m2+m-2+8=2 or
(m+4)(m+1)=0 or (n-10)(n+2)=0 or use formula/completing the squarem= -4, -1 or n=10, -2n=10, -2 or m= -4, -1
P1
K1K1N1N1
52.(a)
(b)
(c)
m2 = 2y -5= 2(x+1) or equivalenty = 2x +7
Solve simultaneous equations:
E( -2, 3)
C( -3, 1)
K1K1N1
K1
N1
K1N1
73.(a)
(b)
Shape of sin xGraph of sin x shifted up 1 unitMiximum = 4, minimum = -2
1½ cycles in
or equivalent
Sketch the straight line involves x and yNumber of solutions = 4
P1P1P1
P1
N1
K1N1
74.(a) Gradient of tangent = 5
kx2-x = 5 substitute x = 1 to kx2-x = 5
K1K1K1
x
4
1
-2
02
2
3
3
y
(b)
k = 6
or equivalent
N1
K1
K1
N1 75.(a)(i)
(ii)
(b)
L = 20.5 F = 57 fQ3 = 28
= 26.93
=19.4
= 11.30
= 126.02
P1
K1
N1
K1
K1
N1
K1
N18
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
P1K1N1
K1N1
N16
7.(a)
(b)
(c)
(i)
(ii)
log10x 0.15 0.28 0.40 0.51 0.61 0.74log10y 0.52 0.72 0.90 1.08 1.20 1.42
Refer to the appendixOne point correctly plotted with uniform scales6 points correctly plotted with uniform scalesLine of best fit
-5k =1.52
log10p = 0.29
N1N1
K1N1N1
P1
K1N1
K1
N110
8(a)(i)
(ii)
(b)
(c)
or equivalent
½ x 8 x h = 20h = 5
K1
N1
N1
P1P1
K1
N1
N1
K1N1
109(a)
(b)
(c)
(d)
or equivalent
OQ = 6
// 0.2618 rad
Area of sector ROS =
=32.99 or 10.5π
Perimeter of shaded region =
=20.52
K1N1
K1
N1
P1
K1
N1
K1K1N1
1010(a)
(b)
(c)
Use b2 – 4ac = 0(-4)2 – 4(1)k =0k = 4Solve equation : (x – 2)(x – 2) = 0A(2, 2)
=
K1N1K1N1
K1
K1
N1
(d)=
= 8.1π
K1
K1
N110
11(a)(i)
(ii)
(b)(i)
(ii)
= 0.2151 – P(x=0) – P(x=1) or P(x=2)+P(x=3)+P(x+4)+ …………….+P(x=10)
=
= 0.9983
= 0.1587 // 0.15866
P(x >342) = 1 – P(z > 1.6) or other valid methodNumber of cakes = 0.9452 x 1500 = 1417 // 1418
K1N1P1
K1N1
K1
N1
K1K1N1
10
12(a)
(b)
(c)
(d)
Use
x = 135, y = 80, z = 140
m = 6
P2005 = 493.50
= 106.38
K1
N2,1,0
K1K1
N1
K1
N1
K1
N110
13 (a)
(b)
(c)
= 9.22
= 47.80° // 47°48 or 92.2°
QR = 12.44
K1N1
K1
N1P1N1
K1
N1
K1N1
10
14.(a)
(b)
(c)(i)
(ii)
x + y ≤ 40 or equivalent4x + 3y ≥36 or equivalentx ≤ 2y or equivalent
Refer to the appendixDraw correctly at least one straight line involves x and yDraw correctly all the three straight linesRegion R shaded correctly
y = 6min passengers = 11substitute any point in the region R into 40x + 30ySubstitute ( 26, 14) to 40x + 30y and maximum fares = 1460
N1N1N1
K1N1N1
K1N1K1N1
1015(a) (b)
(c)
(d)
18 ms-1
2t -9 = 0
=
Use sketch quadratic graph // number lines // others valid method fort2-9t + 18 < 0
3 < t < 6
P1K1
K1
N1
K1
N1
K1
3 6
Substitute t = 3 or t = 5 into
S3 = 22.5
S5 =
Total distance traveled
=
OR
=
K1
K1
N1
K1K1
K1
N110
7(a)
14(b)
R
x+y=40
x=2y
4x+3y=36