SULIT

3472/2

Additional Mathematics

Kertas 2

Ogos

2016

BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH

DAN SEKOLAH KECEMERLANGAN

PENTAKSIRAN DIAGNOSTIK AKADEMIK SBP 2016

PERCUBAAN SIJIL PELAJARAN MALAYSIA

ADDITIONAL MATHEMATICS

Kertas 2

PERATURAN PEMARKAHAN

Peraturan pemarkahan ini mengandungi 12 halaman bercetak

ADDITIONAL MATHEMATICS [P2]

TRIAL SPM 2016

2

Number Solution and Marking Scheme Sub

Marks

Full

Marks

1 (a)

1st method

2

2

4

pxpSee

p

pq

2

44 2See

p

pq

pxp)x(f

422

K1

K1

N1

2nd

method

px

2

4See

qpp

pp

f

24

222

p

pq

pxp)x(f

422

K1

K1

N1

3rd

method

2

2

4

pxpSee

qp

p

2

2

4See

p

pq

pxp)x(f

422

K1

K1

N1

(b)

10

2

1

24

42

q

p

p

pqor

p

K1

N1

N1

6

ADDITIONAL MATHEMATICS [P2]

TRIAL SPM 2016

3

Number Solution and Marking Scheme Sub

Marks Full

Marks

2 2442 yx or 2223yxyx K1

yx 212 or 2

12 xy

P1

22*2*

3212212 yyyy or K1

2

2

2*

2

123

2

12

xx

xx

02 yy or 0128 xx K1

,0y 2y

N1

,12x 8x

N1

Area / Luas = 48 cm2 N1 7

3 (a) L0 = 35.5 or F = 16 or fm = 10 P1

510

16255.35

K1

40 N1

(b)

Mid point : 28, 33, 38, 43, 48, 53, 58

3914610511

)58(1)53(4)48(6)43()38(10)33(5)28(11

x

x

x = 8

Bil pekerja yang berpindah : 2 orang

P1

K1

N1

N1

ADDITIONAL MATHEMATICS [P2]

TRIAL SPM 2016

4

Number Solution and Marking Scheme Sub

Marks Full

Marks

4 (a)

2 , 23

2

, 2

3

2

3

2

or 2 ,

9

8,

3

4 or r =

3

2

T5 =

15

3

22

= 81

32

K1

N1

(b)

n = 26

S26 =

3

21

3

212

26

= 5.9998 // 6

P1

K1

N1

(c) (c) S =

3

21

2

= 6

K1

N1

7

5 (a) i) APCACP or ABCACP

K1

xy 46 N1

(ii) CBCR

3

1

xy 126

3

1

xy 42

N1

(b) ADCACD

yxyxyh 366*)46( K1

Compare

yyh 96 or xxh 64 K1

2

3h N1 6

ADDITIONAL MATHEMATICS [P2]

TRIAL SPM 2016

5

Number Solution and Marking Scheme Sub

Marks Full

Marks

6 (a) 2coscot or 1 cos 2 2sin

sin

2coscot dan 1 cos 2 2sin

sin

= sin 2 ( RHS)

P1

N1

(b)

1sin 2

2 or 210 º or 330 º

0 0 0105 ,165 ,285 ,345o

P1

N1

(c)

y

1

2

3

2 2

P1

P1

P1

P1

8

Shape

of sin graph

2 cycles for 20 x

Modulus

Shifted

1

0

3

2

1

0

ADDITIONAL MATHEMATICS [P2]

TRIAL SPM 2016

6

Number Solution and Marking Scheme Sub

Marks Full

Marks

7 (a)i) orXPorXPorXP )12()11()10(

K1

012

12

12111

11

12210

10

12 )25.0()75.0()25.0()75.0()25.0()75.0( CorCorC

orXPXPXP )12()11()10(

K1

012

12

12111

11

12210

10

12 )25.0()75.0()25.0()75.0()25.0()75.0( CCC

= 0.3907

N1

(ii)

75.025.012

25.22

K1

N1

(b) i)

See 40

520515

0.45026

K1

N1

(ii)

See 800

480

See 0.253 or – 0.253

88.509k

P1

K1

N1

10

ADDITIONAL MATHEMATICS [P2]

TRIAL SPM 2016

7

Number Solution and Marking Scheme Sub

Marks

Full

Marks

8 (a)

212 6

2x x

(2,4)A

K1

N1

10

(b) i)

2

0

3

1 26

x

xA

442

12 A

21 AA

33.13//3

40

K1

K1

K1

N1

(ii)

4

22 4y dy

42

24y y

2 24 4(4) 2 4(2)

4

P1

K1

K1

N1

9 (a)

x 3.16 4.47 5.48 6.32 7.07 7.75

N1

(b)

Refer to graph

Correct axes, uniform scale and one point correctly plotted

All points correctly plotted

Line of best fit

K1

K1

N1

(c) i)

ii)

bxay 5

Use 5a = gradient

a = 0.196

Use b = y – intercept

b = 2.1

x = 36

P1

K1

N1

K1

N1

N1

10

ADDITIONAL MATHEMATICS [P2]

TRIAL SPM 2016

8

Number Solution and Marking Scheme Sub

Marks Full

Marks

10 (a)

i)

2m

)4(29 xy

172 xy

K1

N1

(ii)

2

6172

xx

)1,8(B

K1

N1

(iii) 0

5

2)8(3

x or 3

5

2)1(3

y

)9,12( D

K1

N1

(iv)

area BOD = 1901

81208

2

1

= 72122

1

= 30 unit2

K1

N1

(b)

PBPA

2222 )1()8()9()4( yxyx

42 yx

K1

N1

10

ADDITIONAL MATHEMATICS [P2]

TRIAL SPM 2016

9

Number Solution and Marking Scheme Sub Marks Full

Marks

11 (a)

∠COD =

20

16cos 1

COD 280360

θ = 0.7529 rad

K1

K1

N1

(b) i)

s = (5) (0.7529)

Perimeter = (5) (0.7529) + 10+6+12+20

= 55.529 // 55.53

K1

K1

N1

ii)

Find area of sector OAB or sector OAM or sector OCE

)7529.0()10(2

1 2 or )6436.0()10(2

1 2 or )6436.0()16(2

1 2

Find area of AOB or OCD

13.43sin)10(2

1 2 or )12)(16(2

1

Use the right combination to find area of shaded region

Area of shaded region = 49.23 (accept 49.23 – 49.26)

K1

K1

K1

N1

10

ADDITIONAL MATHEMATICS [P2]

TRIAL SPM 2016

10

Number Solution and Marking Scheme Sub Marks Full

Marks

12 (a)

(b)

056 ta

6

5t

ctt

v 52

6 2

c )0(5)0(312 2

12c

126

55

6

53

2

v

08.14/12

169v

01253 2 tt

0)3)(43( tt

30 t

K1

K1

K1

N1

K1

N1

(c)

ttt

s 122

5

3

3 23

)3(122

)3(5)3(

23

3 s or )4(122

)4(5)4(

23

4 s

Jumlah jarak = )3(122

)3(5)3(

23 + )3(12

2

)3(5)3(

23 -

)4(122

)4(5)4(

23

= 39 m

K1

K1

K1

N1

ADDITIONAL MATHEMATICS [P2]

TRIAL SPM 2016

11

Number Solution and Marking Scheme Sub Marks Full

Marks

OR

3

0

2 )1253( dttt

4

3)12523( dttt

Jumlah jarak = 3

0

2 )1253( dttt + 4

3)12523( dttt

= 39 m

K1

K1

K1

N1

10

13 (a)

5.9

2.43sin

4.12

sin

ACB

32.63ACB

K1

N1

(b)

See 73.48º

48.73cos)4.12)(5.9(24.125.9 222AC

cmAC 30.13

P1

K1

N1

(c)

ADC cos)4.5)(9.9(24.59.93.13 222

71.117ADC

71.117sin4.59.92

1

= 23.66 cm2

K1

K1

N1

(d) i)

ii)

116.68º

P1

N1

10

A ’ C ’

B ’

ADDITIONAL MATHEMATICS [P2]

TRIAL SPM 2016

12

Number Solution and Marking Scheme Sub

Marks Full

Marks

14 Rujuk Lampiran 10

15 (a) 9.90x RM 7.50y RM

115z

N1

N1

N1

(b)

120( ) 150(45) 140(17) 115(20)k 120( ) 150(45) 140(17) 115(20)

135.9045 17 20

k

k

k = 18

K1

K1

N1

(c)

2016 100 135.9400

P

543.60RM

K1

N1

(d)

135.9 118

100I

160.36

K1

N1

10

END OF MARKING SCHEME