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3472/1 © 2011 Mid Year Exam F5 Hak Cipta SBP
3472/1
Matematik
Tambahan
Kertas 1
2 jam
Mei 2011
BAHAGIAN PENGURUSAN
SEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN
KEMENTERIAN PELAJARAN MALAYSIA
PEPERIKSAAN PERTENGAHAN TAHUN
TINGKATAN 5
2011
ADDITIONAL MATHEMATICS
Paper 1
Skema Pemarkahan ini mengandungi 6 halaman bercetak
MARKING SCHEME
www.banksoalanspm.com
2
3472/1 © 2011 Mid Year Exam F5 Hak Cipta SBP
PERATURAN PEMARKAHAN- KERTAS 1
No. Solution and Mark Scheme
Sub
Marks
Total
Marks
1(a)
(b)
{2, 4, 6, 10}
One to many
1
1
2
2(a)
g1
(x) =
3
2
x 
1 3
(b) 1
B1 :
2
31
2
3
2
1446)( xxxg 
B2 :
2
7
3 4
2
x  
  
 
B1 : 1 7
( )
2
x
g x 

3 3
4
37.5,372.0 
B2 :
)1(2
)2)(1(4)5(5 2

x
B1 : 0252
 xx
3 3
5
25
2
h 
B2 : ( 10 )2
– 4(2)(h)  0
B1 : 2x2
– 10x + h = 0
3 3
6
x < - , x > 2
B2 : (2x + 11)(x – 2) > 0 or
B1 : 2x2
+ 7x - 22 > 0
3 3
7(a)
(b)
(c)
4
3
B1 : - 8 = - a(0 – 2)2
+ 4
x = 2
1
2
1
4
2
11

2
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3
3472/1 © 2011 Mid Year Exam F5 Hak Cipta SBP
8
2
3 hk 
B3 :
B2 :
B1 :
25log
125
log
5
5 





a
b
4 4
9
n =
B2 : 3462  nn
B1 : 32(n + 3)
= 3- n - 3n + 3
or )1(33
2
1
)3(4
3.3
1
3 

 n
n
3 3
10(a)
(b)
6p
B1 : )13(8)82(13  pppp
nTn 323
B1 : )3)(1(20  nTn
2
2
4
11(a)
(b)
B1 :
12.2 or
5
1
12or
6
61
B1 : or
2
5
4
5
5
4
55 












2
2
4
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4
3472/1 © 2011 Mid Year Exam F5 Hak Cipta SBP
12
x
x
y
52 

B2 : 5
1
2
1







xy
or equivalent
B1 : c
xy







1
2
1
OR c = 5 or 15 = 2(5) + c
3 3
13 a = 10 and b = 100 (both)
B2 : a = 10 OR b = 100 OR
1log2log 1010  aandb (both)
B1 : bxay 101010 logloglog  OR gradient, m = 2
02
15



OR 2log10 b OR 1log10 a
3 3
14 (4,5)
B2 : x = 4 and y = 5
B1 :
13
)2(3)1(10


or
13
)4(3)1(8


or equivalent
3 3
15 049802255 22
 yxyx or equivalent
B2 : 2222
))1(()5(2)4()1(3  yxyx
B1 :
3
2

PB
PA
or equivalent
3 3
16(a)
(b)
5
B1 : 







3
4
or34 jiOR
(b) or
2
1
3
17(a)
(b)
yx 66 
yx 3
B1 : )66(
2
1
yxMK 
1
2
3
www.banksoalanspm.com
5
3472/1 © 2011 Mid Year Exam F5 Hak Cipta SBP
18(a)
(b)
rad or 1.047 rad
37.70 cm2
or 37.69 cm2
B1 : or )094.2()6(
2
1 2
1
2
3
19 60o
, 131.81o
or 131o
49’ , 228.19o
or 228o
11’ , 300o
B3 : ,
B2 :
B1 :
4 4
20
16
1

B2: or equivalent
B1:
3 3
21
(a)
(b)
36
3
4 3
 
x
3
1

036
3
4
:1B 3
 
x
1
2
3
22 n = -3 and p =
B2 : n = -3 or p =
B1 :
)2)(1(
)21( 1

 
n
x n
3 3
www.banksoalanspm.com
6
3472/1 © 2011 Mid Year Exam F5 Hak Cipta SBP
23
(a)
(b)
4
B2 :   5
1
2
4 

 kx
B1 :   

2
1
2
1
5)( kdxdxxf
1
3
4
24 7.43
B2 : 10
25
32-
2
80
40.5median













B1 : lower boundary, L = 40.5 or 32 or 25
3 3
25
3.428
B2 : 2
22222222
2
(7)-
8
1411765553 

B1 :
8
56


N
x
or x =7
3 3
www.banksoalanspm.com

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Kertas 1 pep pertengahan tahun sbp 2011

  • 1. 1 3472/1 © 2011 Mid Year Exam F5 Hak Cipta SBP 3472/1 Matematik Tambahan Kertas 1 2 jam Mei 2011 BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN KEMENTERIAN PELAJARAN MALAYSIA PEPERIKSAAN PERTENGAHAN TAHUN TINGKATAN 5 2011 ADDITIONAL MATHEMATICS Paper 1 Skema Pemarkahan ini mengandungi 6 halaman bercetak MARKING SCHEME www.banksoalanspm.com
  • 2. 2 3472/1 © 2011 Mid Year Exam F5 Hak Cipta SBP PERATURAN PEMARKAHAN- KERTAS 1 No. Solution and Mark Scheme Sub Marks Total Marks 1(a) (b) {2, 4, 6, 10} One to many 1 1 2 2(a) g1 (x) = 3 2 x  1 3 (b) 1 B1 : 2 31 2 3 2 1446)( xxxg  B2 : 2 7 3 4 2 x        B1 : 1 7 ( ) 2 x g x   3 3 4 37.5,372.0  B2 : )1(2 )2)(1(4)5(5 2  x B1 : 0252  xx 3 3 5 25 2 h  B2 : ( 10 )2 – 4(2)(h)  0 B1 : 2x2 – 10x + h = 0 3 3 6 x < - , x > 2 B2 : (2x + 11)(x – 2) > 0 or B1 : 2x2 + 7x - 22 > 0 3 3 7(a) (b) (c) 4 3 B1 : - 8 = - a(0 – 2)2 + 4 x = 2 1 2 1 4 2 11  2 www.banksoalanspm.com
  • 3. 3 3472/1 © 2011 Mid Year Exam F5 Hak Cipta SBP 8 2 3 hk  B3 : B2 : B1 : 25log 125 log 5 5       a b 4 4 9 n = B2 : 3462  nn B1 : 32(n + 3) = 3- n - 3n + 3 or )1(33 2 1 )3(4 3.3 1 3    n n 3 3 10(a) (b) 6p B1 : )13(8)82(13  pppp nTn 323 B1 : )3)(1(20  nTn 2 2 4 11(a) (b) B1 : 12.2 or 5 1 12or 6 61 B1 : or 2 5 4 5 5 4 55              2 2 4 www.banksoalanspm.com
  • 4. 4 3472/1 © 2011 Mid Year Exam F5 Hak Cipta SBP 12 x x y 52   B2 : 5 1 2 1        xy or equivalent B1 : c xy        1 2 1 OR c = 5 or 15 = 2(5) + c 3 3 13 a = 10 and b = 100 (both) B2 : a = 10 OR b = 100 OR 1log2log 1010  aandb (both) B1 : bxay 101010 logloglog  OR gradient, m = 2 02 15    OR 2log10 b OR 1log10 a 3 3 14 (4,5) B2 : x = 4 and y = 5 B1 : 13 )2(3)1(10   or 13 )4(3)1(8   or equivalent 3 3 15 049802255 22  yxyx or equivalent B2 : 2222 ))1(()5(2)4()1(3  yxyx B1 : 3 2  PB PA or equivalent 3 3 16(a) (b) 5 B1 :         3 4 or34 jiOR (b) or 2 1 3 17(a) (b) yx 66  yx 3 B1 : )66( 2 1 yxMK  1 2 3 www.banksoalanspm.com
  • 5. 5 3472/1 © 2011 Mid Year Exam F5 Hak Cipta SBP 18(a) (b) rad or 1.047 rad 37.70 cm2 or 37.69 cm2 B1 : or )094.2()6( 2 1 2 1 2 3 19 60o , 131.81o or 131o 49’ , 228.19o or 228o 11’ , 300o B3 : , B2 : B1 : 4 4 20 16 1  B2: or equivalent B1: 3 3 21 (a) (b) 36 3 4 3   x 3 1  036 3 4 :1B 3   x 1 2 3 22 n = -3 and p = B2 : n = -3 or p = B1 : )2)(1( )21( 1    n x n 3 3 www.banksoalanspm.com
  • 6. 6 3472/1 © 2011 Mid Year Exam F5 Hak Cipta SBP 23 (a) (b) 4 B2 :   5 1 2 4    kx B1 :     2 1 2 1 5)( kdxdxxf 1 3 4 24 7.43 B2 : 10 25 32- 2 80 40.5median              B1 : lower boundary, L = 40.5 or 32 or 25 3 3 25 3.428 B2 : 2 22222222 2 (7)- 8 1411765553   B1 : 8 56   N x or x =7 3 3 www.banksoalanspm.com