Skema SPM SBP Add Maths Paper 2012

1
3472/1
Matematik
Tambahan
Kertas 1
2 jam
Ogos 2012
BAHAGIAN PENGURUSAN
SEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN
KEMENTERIAN PELAJARAN MALAYSIA
PENTAKSIRAN DIAGNOSTIK AKADEMIK SBP 2012
PERCUBAAN SIJIL PELAJARAN MALAYSIA
ADDITIONAL MATHEMATICS
Paper 1
Skema Pemarkahan ini mengandungi 6 halaman bercetak
MARKING SCHEME
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2
PERATURAN PEMARKAHAN- KERTAS 1
No. Solution and Mark Scheme
Sub
Marks
Total
Marks
1(a)
(b)
25
4
1
1
2
2(a) 2 1 3
(b) 4
B1: 12 3
OR 2 [use (2) ].
3 2
x y
k y
x y

 
 
2
3(a)
(b)
3x
B1:
3
5
)(
5


xxf
1
2
1
3
4 1m
B2: 0)3)(2(4)6( 2
 m
B1:   0362 2
 xxm
3 3
5 12  p
B2 : 0)1)(2(  pp
B1: 0232
 pp
3 3
6(a)
(b)
(c)
1x
1
(1,  4)
1
1
1
3
7 2x
B2: xx 65
2
1
)32(2  or xx 6532 
B1:  x
x
652
1
)32(2
333 

3 3
or -2 -1

3
8 3
625 n or
1
3
625n
B3: 625
3
1

n
m
or equivalent
B2: 4log
3
15 
n
m
or equivalent
B1:
125log
log
5
5 m
(for change base)
4 4
9 h = 2 and k = 11 [both]
B2: h = 2 or k = 11
B1: - 7 + 3d = 20 OR - 7 + 3(20 – k) = 20
OR - 7 + 3( h – (- 7)) = 20
3 3
10
(a)
(b)
2
3
r  
3
10
B1 :
1
2
2
1 (* )
3
S 
 
1
2
3
11 75 55h
B2 :
10
[2(3 1) 9( 1)]
2
h h  
B1: 3 1a h  or 1d h 
3 3
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4
12
(a)
(b)
3
p
q

1
1:
y
B pq p
x x
  or 3pq 
5
2
1
3
13 3 5
2 4
y x  or 4 6 5y x  or equivalent
B3 :
3 3
1
2 2
y x
 
   
 
B2 : 2
3
2
m  or
3
,1
2
 
 
 
B1 : 1
2
3
m   OR (3,0) and (0,2)S T
4 4
14 2 2
4 4 92 0x y x y    
B2 : 2 2 2 2
( 2) ( 2) (6 ( 2)) ( 4 2)x y        
B1 : PS = PQ OR
2 2
(6 ( 2)) ( 4 2)    
3 3
15(a)
(b)
5 12i j 
5 12 55 12 1
or or
1213 13 13 13
i j
i j
   
   
 
B1 : |OR |=13
1
2
3

5
16
4
3
k
B2: 034 k
B1: (4 3) (4 2)k i k j  
3 3
17(a)
(b)
0.9506 rad / 0.9505 rad / 0.9507 rad
15.36 or 15.355
B1 : arc OS = 5 (*0.9506) or PS = 3.602
1
2
3
18 x =15°,75°,195°,255°
B2 : 0 0 0 0
2 30 ,150 ,390 ,510x  or sin 2x =
1
2
B1 : 2(2sin cos ) 1x x 
3 3
19(a)
(b)
80 32x
5.2x or
2
5
x
B1 : 03280  x
1
2
3
20
2
5
4  xy or equivalent
B2 :  14
2
3
 xy or equivalent
B1 : 4 or ( 1) 3
dy dy
dx dx
    
3 3

6
21(a)
(b)
10
3
B2 :  
3
1
5 7
2 2
hx  
B1:
3 3
1 1
( ) 7
2 2
f x
hdx dx  
1
3
4
22(a)
(b)
3
1.648 or 1.6475
B2: 2163 51
( )
20 20
 or equivalent
B1 :
2 2 2 2 2
251 2(0) 3(1) 2(2) 8(3) 5(4)
or
20 20
x 
   
 
1
3
4
23(a)
(b)
252
66
B1 : 4 6 4 6
3 2 4 1or 60 OR or 6C C C C 
1
2
3
24(a)
(b)
1
10
B1 :
3 1 2
5 4 3
 
19
20
B1 :
3 1 1
1
5 4 3
 
   
 
2
2
4
25(a)
(b)
0.1741
49.69
B2 :
45
0.938
5
k 

B1 : z = 0.938
1
3
4

1
3472/2
Matematik
Tambahan
Kertas 2
Ogos 2012
2 ½ jam
BAHAGIAN PENGURUSAN
SEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN
KEMENTERIAN PELAJARAN MALAYSIA
PENTAKSIRAN DIAGNOSTIK AKADEMIK SBP 2012
PERCUBAAN SIJIL PELAJARAN MALAYSIA
ADDITIONAL MATHEMATICS
Paper 2
Skema Pemarkahan ini mengandungi 10 halaman bercetak
MARKING SCHEME

2
No Solution and Mark Scheme
Sub
Marks
Total
Marks
1 8 3
2
y
x

 OR
8 2
3
x
y

 P1
2 8 3
3 6 0
2
y
y y
 
   
 
OR
2
8 2 8 2
3 6 0
3 3
x x
x
    
     
   
K1
Replace a, b & c into formula K1
2
( 24) ( 24) 4(7)( 12)
2(7)
y
     
 OR
2
( 20) ( 20) 4(7)( 59)
2(7)
x
     

0.443, 3.871y   OR 4.664, 1.807x   N1
4.664, 1.807x   OR 0.443, 3.871y   N1
5 5
2 (a)
 
kxy
kxxy
31)1(
32
2
2


1
431


k
k
(b)
3
3
6
(1,4)
3
3-1
-Maximum shape P1
-*Maximum point K1
-Another 1 point y-intercept / x-intercept K1
K1
K1
N1

3
3(a)
16 ,8 ,4 ,......   OR
1
2
r 
P1
1
16 1
2
30.5
1
1
2
n


  
  
    

K1
4.416n  K1
5n 
N1
4 7
(b)
64 ,16 ,4 ,......   OR
1
4
r  P1
64
1
1
4
S

 

K1
=
1
85
3
 or 85.33 N1
3
4(a)
(i) Change
1
3 6 0 2
3
y x to y x      or
1
3
BCm  
OR 3ABm  K1
 5 3 ( 6)y x    OR any correct method K1
3 23y x  N1
5 7
(ii) Use simultaneous equation to find point B
* 3 23y x  and
1
3 6 0 2
3
y x or y x      K1
B =
15 1
,
2 2
 
 
 
N1
(b) 15 1 2( ) 3( 6) 2( ) 3(5)
* , ,
2 2 5 5
x y     
    
   
K1
D =
39 25
,
4 4
 
  
 
N1
2
Use
r
ra
S
n
n



1
)1(
Use
1
a
S
r
 

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4
5(a)
(b)
Amplitude = 3 [ Maximum = 3 and Minimum = − 3 ] P1
Sine shape correct P1
Two full cycle in 0  x  2 P1
Negative sine shape correct(reflect) P1
4 7
5
3sin 2 2
x
x

   or
5
2
x
y

  N1
Draw the straight line
5
2
x
y

  K1
Number of solutions is 3 N1
3
6(a) L = 79.5 OR F = 24 OR fm = 4 P1
3
(36) 24
479.5 10
4
 
 
  
 
 
K1
87 N1
3 8
(b) (i)
(44.5 4) (54.5 5) (64.5 6) (74.5 9) (84.5 4) (94.5 8)
36
X
          

2602
36
= 72.28
5
y
3
–3
2

2
3 2 x
5
2
x
y

 
O 
3sin2y x 
K1
N1
OR

5
(ii)
2 2 2 2 2 2
(44.5) 4 (54.5) 5 (64.5) 6 (74.5) 9 (84.5) 4 (94.5) 8          
K1
2197689
(*72.28)
36
  
K1
16.34  N1
7 Rujuk Lampiran
8(a)
(b)
3 2
4 2
y dx c
x x
   
2
2
(4) , 2
( 1)
c c  

2
2
2y
x
 
2
2
5
21
5
1 1
2
2
2
2
1
2( 2) 2( 5)
2( 2) 2( 5)
1 1
dx
x
x
x




 
 
 
  
 
    
        
    

3
3
10
(c )
33
or 6.6
5

 
   
 
2
2
2
5
23 1
5
22 3 13 1
5 5
2
( ) Volume ( 2)
4 8
4
3 1
4 5 8 54( 2) 8( 2)
4 2 4 5
3 1 3 1
14.56
i dx
x
x x
x


 



 

   
 
 
 
   
  
    
         
       


4
K1
N1
K1
K1
N1
K1
K1
K1
N1
K1

6
9(a)
(b)
(c)






yx
yxxOD
yxAC
4
15
4
7
)57(
4
3
7
57
3 7 15
7 5 5
4 4 4
21 7
4 4
3
15 15
5
4 4
3
x y h y k x y
k
k
h k
h
   
            
     


  

5
45
2
1
50


t
t
3
5
2
10
10(a) (i)      
6 410
66 0.3 0.7P X C 
= 0.03676
(ii)        
9 1 10 010 10
9 100.3 0.7 0.3 0.7C OR C
         
9 1 10 010 10
9 109 0.3 0.7 0.3 0.7P X C C  
= 0.0001437
5
5
10
N1
K1
N1
K1
N1
K1
N1
K1
N1
2
3
4
15
5
4
15
3
4
7
4
21
4
15
4
7
557
4
3




















 

h
h
k
k
k
yxkyhyx
K1
K1
N1
K1
N1
2
3
4
15
5
4
15
3
4
7
4
21
4
15
4
7
557
4
3




















 

h
h
k
k
k
yxkyhyx
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7
(b)
  




 



5.3
4548
5.3
4540
4840)( ZPXPi
= 0.7278
 ( ) 0.7
45
0.524
3.5
43.166
ii P X m
m
m
 

 

11(a)
(b)
(c)
7
tan 1
0.7855
4
OR RQ PR cm
rad rad



  

 
2 2
7(1.571) 7(2.3565)
7 7 2(7)(7)(cos135 )o
OR
 
2 2
7 7 7(1.571) 7(2.3565) ( 7 7 2(7)(7)(cos135 )
54.4268
o
Perimeter 
     

21
7
4
 
2 21 1
7 2.3565 7 sin135
2 2
o 
     
 
2 2 21 1 1
7 7 2.3565 7 sin135
4 2 2
78.8996
o
Area 
 
         
 

2
4
4
10
K1
N1
N1
K1
K1
K1
N1
2
3
4
15
5
4
15
3
4
7
4
21
4
15
4
7
557
4
3




















 

h
h
k
k
k
yxkyhyx
K1
K1
K1
N1
K1
K1
K1
N1

8
No Solution and Mark Scheme
Sub
Marks
Total
Marks
12(a)
a = 10 - 5t = 0
t = 2 s
cttv  2
2
5
10
c 2
)0(
2
5
)0(1030
c = 30
30
2
5
10 2
 ttv
30)2(
2
5
)2(10 2
v
= 40 ms- 1
4 10
(b) 30
2
5
10 2
 ttv 0
  2 6 0t t  
0 6t 
3
(c) cttts  30
6
5
5 32
s = 0, t = 0 , c = 0
ttts 30
6
5
5 32

)6(30)6(
6
5
)6(5 32
s or )8(30)8(
6
5
)8(5 32
s
= 180 = 133.33
Total distance = 180 +  33.133180
= 226.67 m
OR
3
Use v > 0
Integrate and
substitute t = 2
Use a = 0
Integrate a to
find v
K1
K1
N1
K1
N1
K1
K1
Integrate v dt K1
K1
N1

9
67.226
67.46180
30
2
5
1030
2
5
10
8
6
2
6
0
2














  dtttdttt
13(a)
125100
55
10

P
P10 = RM 44
2 10
(b)
            115
534
58831404125110



hh
hh
*
h = 1
2
(c)
115100
20
11

P
P07 = RM 23
2
(d)
I S = 110125
100
88

           110 *1 125 4 140 *1 3 110 5
1 4 4 5
I
    

  
= 122.86
4
Integrate v

6
0
+ 
8
6
K1
K1
N1
N1
K1
N1
K1
K1
N1
See 125 P1
K1
K1
N1
www.myschoolchildren.comwww.myschoolchildren.comMuatturun(percuma)skemalaindi:Muatturun(percuma)skemalaindiwww.myschoolchildren.comwww.myschoolchildren.comwww.myschoolchildren.comwww.myschoolchildren.comMuatturun(percuma)skemalaindi:Muatturun(percuma)skemalaindiMuatturun(percuma)skemalaindi:Muatturun(percuma)skemalaindi

10
14 Rujuk Lampiran
15(a) Using sine rule to find BAC .
sin sin30
27 14
o
BAC

74.64o
BAC 
(obtuse) 180 74.64
105.36
o o
o
BAC  

3 10
(b) 105.36 30 or 6o o
DCB DC cm   
Use cosine rule to find BD.
    
22 2
6 27 2 6 27 cos135.36
31.55
BD
BD
   

3
(c) Use formula correctly to find area of triangle ABC or ACD.
180 30 105.36
44.64
o o o
o
ABC   

    
22 2
27 14 2 27 14 cos44.64
19.67
AC
AC
   

1
Area (14)(27)sin 44.64
2
o
ABC  or
1
Area (6)(19.67)sin105.36
2
o
ACD 
Use Area ABCD = sum of two areas
Area ABCD = 189.7 cm2
.
4
END OF MARKING SCHEME
N1
N1
K1
P1
K1
N1
K1
K1
K1
N1

Pentaksiran Diagnostik Akademik SBP 2012 Paper 2 ADM
0.1 0.3 0.4 0.6
1
x
10
20
30
40
50
60
70
80
y
x
x
0.5
No.7(a)
0.7
0
x
0.2
x
x
Plot y against
1
x
K1
(at least one point)
6 points plotted correctly K1
Line of best fit N1
1
x 0.1 0.2 0.25 0.4 0.5 0.8
y 62 54 50 38 29 4
1
2
n
y m
m x
 
   
 
N1
P1
. 2 70i m 
35m  N1
. 80
m
ii
n
   K1
2800n  N1
1
. 0.37iii
x
 K1
2.703x  N1
0.8
x

Answer for question 14
(a) I. 170x y 
II. 80x y 
III. 2 20y x 
(b) Refer to the graph,
1 graph correct
3 graphs correct
Correct area
(c) max point ( 50,120 )
i) k = 100x + 80y
Max fees = 100(50) + 80(120)
= RM 14, 600
ii) 14080  y
10 20 30 40 50 60 700 80
20
40
140
120
100
160
180
80
60
(50,120)
10
N1
N1
N1
N1
N1
N1
K1
N1
K1
N1
y
x

Skema SPM SBP Add Maths Paper 2012

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