This document contains the marking scheme for a mathematics additional paper consisting of 25 multiple choice and constructed response questions. The marking scheme provides the full solutions, method marks, and total marks allocated for each question. It also includes relevant formulas, steps, and acceptable equivalent responses for full credit. The marking scheme is 6 pages long and is intended to guide examiners in accurately and consistently scoring student responses on the additional mathematics exam.

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
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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
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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
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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
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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
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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
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