sbp trial spm 2011 p1 marking scheme
TRANSCRIPT
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Matematik Tambahan Kertas 1 2 jam Ogos 2011
BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN
KEMENTERIAN PELAJARAN MALAYSIA
PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2011
ADDITIONAL MATHEMATICS
Paper 1
Skema Pemarkahan ini mengandungi 6 halaman bercetak
MARKING SCHEME
2
PERATURAN PEMARKAHAN- KERTAS 1
No. Solution and Mark Scheme Sub
Marks
Total
Marks
1(a)
(b)
2
3
B1:
13
m
m
2
1
2
1
3
2(a)
2
3x
1
3
(b)
2
5
B1: k(3) = 2
2
3 a = 6 and b = -11
B2: a = 6 or b = -11
B1: bx
axfg
)2
5()(1
3 3
4
2
1,5 xx
B2:
(2 1)( 5) 0 orx x
B1: 0592 2 xx
3
3
5
1 11p
B2 : ( 1)( 11) 0p p
B1:
2( 1) 4(3)( 1) 0p p
3
3
6(a)
(b)
(c)
m = 1
x = 1
(1, 3)
1
1
1
3
5 1
2
1 11
3
7
2
5x
B2: 1
2( 2) 4( 1) or 2 4 42
x x x x
B1:
12( 2)
24( 1)
15
5
x
x
3 3
8
1
3x
B3: 3 23 9x
B2: 2
9log (3 ) 2x x
B1: 9
9
log 3
log 81
x (for change base)
4 4
9(a)
(b)
h = 1
B1: 5h 1 (2h – 4 ) = 6h + 4 – (5h – 1)
1460
B1: 20
202(16) 19(6)
2s
OR 23 3s s 23 3
2( 2) 22(6) 2( 2) 2(6)2 2
OR 20
2016 130
2s
2
2
4
10
1
3r and a = 4
B3 : 1
3r or a = 4
B2 : 8
3(1 )(1 )3
r r
B1 : 3
1
r
a
or 3
8 ara
4
4
11
42300
B2 : 36
362(300) 35(50)
2S
B1 : 36 or 300 or 300,350,400S a
3
3
4
12
2and 4
3q p
B2: 2
or 43
q p
B1: 21 3 0 3or
6 4 2
pxy x
q q
3 3
13(a)
(b)
t = 1
B1: 4
1
80
02
t
y = 4x + 2
B1: Using m1• m2 = -1 and m2 = 4
2
2
4
14 m = 1 and n = 3
12 :
3
( 4) (4) (2) (6)1: 2 or 3
mB
n
n m n mB
n m n m
3
3
15(a)
(b)
x3 + y6
2x + y2
2 1B1 : (3 6 ) or (6 3 ) or equivalent
3 3x y y x
1
2
3
16(a)
(b)
8 4i j
1 1
(8 4 ) or (2 )80 5
i j i j
B1:22 48 AB = 80
1
2
3
5
17(a)
(b)
r = k =10
B1: 21(0.8) 40
2r
72 cm
)8.0(30Sor 10(0.8) :1 CD ABSB
2
2
4
18
65
33
B1 : 5
3sin B OR
13
5cos A
2
2
19
2
13
B2 : )5(2
1
2
2
x
B1 : 1
2
1
2
)( dxxfdxx OR 2
2x OR
1
2
5)( dxxf
3 3
20 27
B2 : 2 2 3 2 3 3( 1) (9( 1) )(( 1) 4) (( 1) 4) (2( 1))
B1 : 2 2 3 2 3 3(9 )( 4) ( 4) (2 )x x x x x or equivalent
3
3
21(a)
(b)
5
B1 : 450 90 0h
1125
2
1
3
6
22(a)
(b)
p = 9
8
B1 : 2 2 2 2 2
2 23 7 9 5 1(5)
5
1
2
3
23(a)
(b)
20
240
B1 : 5! X 2!
1
2
3
24(a)
(b)
1
2
5
12
B1 : 1 2 3 1
4 3 4 3
1
2
3
25(a)
(b)
0.1515
58.09
B2 : 55
1.033
k
B1 : z = 1.03
1
3
4