jawapan add math sbp

SULIT
3472/2
Additional Mathematics
Kertas 2
Ogos
2017

BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH
DAN SEKOLAH KECEMERLANGAN

PENTAKSIRAN DIAGNOSTIK AKADEMIK SBP 2017
PERCUBAAN SIJIL PELAJARAN MALAYSIA

ADDITIONAL MATHEMATICS
Kertas 2
PERATURAN PEMARKAHAN

Peraturan pemarkahan ini mengandungi 12 halaman bercetak

ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2017

Number

1 (a )

Solution and Marking Scheme

 1  4k  11  0

mop 

N1

3
4

mOP  mPQ 

(c)

or

mPQ 

4
3

P1

3 4

 1
4 3

N1

TP  5

x  42   y  32

P1

5

K1

x 2  y 2  8x  6 y  0

2

N1

y3
2
2 x  5(3  2 x)  9 x(3  2 x) or
y  3  2x

or

2
5

  9
3  2x x

 3x  5  2 x  1  0
x

Full
Marks

K1

k  3

(b)

Sub
Marks

5
1

, x
3
2
1
y , y2
3

x

or
or

7

P1

2
5

 9

y y3
2
 3 y  1 y  2   0

K1

K1
N1
N1
5

2

ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2017
Number

3 (a )

Solution and Marking Scheme

36 or 36, 27 , 20.25,......or r  0.75

Tn  1.52

or

360.75

n 1

n  1 log 0.75  log 1.52 
n  12

(b)

 36 

K1

 1.52 
log

36 

n 1 
log 0.75

36
1  0.75

K1

K1

336 cm

(ii)

(b)

K1

N1

 36 
48  2 

 1  0.75 

4 (a )
(i)

Full
Marks

K1

 1.52

or

Sub
Marks

N1

100.5  150.5
2
125.5 (accept without working)

7

K1
N1

75.5(10)  125.5 (40)  175.5 (10)  225.5(30)  275.5 (20)
110
180.05

N1

200.5 or 60 or

P1

30

 82.5  60 
200.5  
50
 30 

K1

K1
N1

238

3

7

ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2017
Number

5 (a )

Solution and Marking Scheme

Sub Marks

2 2
x 12 x  50  0
5

K1

(x – 25)(x – 5) = 0

K1

Full
Marks

x = 25 , x = 5

Width = 25 – 5 = 20 unit
dy

(b)

dx



4
x  12
5

x = 15
y = – 40

or

N1

OR

y

OR

y



2
x  15 2  100
5

2
x  15 2  40
5



K1

N1
N1

max depth = 40

4

6

ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2017
Number

6 (a )

Sub
Marks

Solution and Marking Scheme

sin x
cos x
sin 2 x sin x

cos 2 x cos x

Use

=

sin x
cos 2 x cos x

cos x cos x sin 2 x  sin x cos 2 x

=

sin x
cos 2 x cos x

cos x cos x 2 sin x cos x  sin x(2 cos 2 x  1)

=

sin x
cos 2 x cos x

2
cos x 2 sin x cos x  2 sin x cos 2 x  sin x)

sin x
cos x

or

sin 2x = 2sin x cos x

or

cos 2x = 2cos2 x – 1

Full
Marks

K1

= cos 2x

N1

Shape
Amplitude or 2 cycles or modulus
All correct

P1
P1
P1

y

N1

6 (b)

x
2

Sketch graph

K1

6 number of solution

N1

5

8

ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2017

Number

7
(a )

(b)

Solution and Marking Scheme

tan POQ 

12
12
// tan POR 
9
9
1.8548 rad // 1.855 rad

SQR = 9(*1.8548) //

9(*1.855)

N1

or

QM

9

 sin 53.13

K1
N1

Chord QR = 2  *7.2
= 14.4

N1

Perimeter = SQR + * Chord QR

K1

= 31.09 → 31.098

Area of sector OQR
1 2
= 9 1.8548
2
= 75.12 cm2

Full
Marks

K1

= 16.6932 // 16.695

(c)

Sub
Marks

or

N1

Area of triangle POQ
1
= 129
2
= 54 cm2

K1

2(Area of triangle *POQ ) − Area of sector *OQR
2(54) − 75.12

K1

32.88 cm2

N1
10

6

ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2017
Number

8

Sub
Marks

OP  OA  AP

K1

Full
Marks

Refer to graph

9 (a )

AQ  AO  OQ

(i)

3
AQ   a  b
4

(ii)

OP 

(b)

Solution and Marking Scheme

or

N1

3
1
a b
4
4

N1

OS  mOP
1 
3
= m  a  b
4 
4

N1

3 

OS  a  n   a  b 
4 

3
m  1 n
4
3m  4n  4

or
and

N1

1
3
m n
4
4
m  3n

K1
shown

Solve simultaneous linear equations
m
n

12
13

N1

K1

N1

4
13

N1

7

10

ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2017
Number

10 (a )

Sub
Marks

Solution and Marking Scheme

dy
dx

 2x  5

K1

2 x  5  0 or

x

5
5
y     5 
2
2
25
y 
4

5
2

2

(b)

N1

 25 
 5
 4 
x3
5 x2
A2 =

3
2

K1

A1  A2

K1

A1 = 

K1

N1

125
5
// 10
// 10.42
4
12

(c)

V 

 x
5

0

2



 5 x dx
2

or

 x 5 10 x 4 25 x 3 




4
3 
 5

=

 x
5

4

0



 10 x3  25 x2 dx

K1

K1

1
625
 // 104  // 104.17 
6
6

N1

8

Full
Marks

ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2017

Number

11 (a ) i)

Solution and Marking Scheme

P ( x  7) 

10

10

C0 (0.45)0 (0.55)10

or

1  P ( X  0)  P ( X 1)

(b) i)

(ii)

10

C1 (0.45)1 (0.55)9

equivalent

K1

K1

= 0.9767

N1

3.6  3.56
0.25

K1

0.4364

N1

Seen - 0.524
k  3.56

0.25

Full
Marks

K1
N1

C7 (0.55)7 (0.45)3

= 0.1665
(ii)

Sub Marks

or 0.524

P1

  0.524

K1
N1

k = 3.429

10

9

ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2017
Number

12 (a )

Solution and Marking Scheme

6t 2  30t  24

and

K1

t 1

N1

12t  30
t

Full
Marks

K1

t  4t  1  0

t4

(b)

Sub Marks

K1

5
2

subtitute  t into v
5
v  6    30
2
1
v  13 ms 1
2


(c)

t<

2



5
   24
2
or

K1
1
13 ms 1 moves to the left
2

5
2

N1

N1

S1  2 1  151  241
3

2

5
5
5
S 5  2    15   24 
2
2
2
2
3

or

Total distance = 2 S1  * S 5

2

K1

K1

2

N1

= 24.5

10

10

ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2017
Number

13 (a )

y
70
18
 100  175 or
 100  100 or
 100  z
x
15
20

x = 40,

(b)

Sub
Marks

Solution and Marking Scheme

I

y = 20,

175( 8 )  150( 12 )  125( 10 )  100( 24 )  *120( 46 )
100

P06 

K1

N1

880  100
*123.7

K1
N1

RM 711.40
(d)

Seen I18 / 15  166.25 or 165 or 137.5 or

I

K1
N2, 1, 0

z = 120

123.7

(c)

Full
Marks

110

or 132

166.25 8  16512  137.510  11024  13246
100

N1
K1

N1

133.97

11

10

ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2017
Number

14

15 (a )
(i)

(ii)

Solution and Marking Scheme

(b)

Full
Marks

Refer to graph

sin ADC sin 22.18

7.71
3.46
ADC  122.73

K1
N1

7.712  4.982  4.072  2(4.98)(4.07) cos ABC
ABC  116.49
Area of triangle ADC
1
4.984.07)sin116.49
2
9.070 cm2 // 9.07 cm2

(iii)

Sub Marks

K1
N1

K1
N1

1
 BE  7.71  9.07
2
BE = 2.353 cm

K1
N1

BD  2.352  2.062

P1
N1

=3.125 cm

END OF MARKING SCHEME

12

10