sulit 3472/2 kertas 2 2017 bahagian pengurusan sekolah ... · additional mathematics [p2] trial spm...

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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

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Page 1: SULIT 3472/2 Kertas 2 2017 BAHAGIAN PENGURUSAN SEKOLAH ... · ADDITIONAL MATHEMATICS [P2] TRIAL SPM 2017 5 Number Solution and Marking Scheme Sub Marks Full Marks 6 (a) x x x x x

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

Page 2: SULIT 3472/2 Kertas 2 2017 BAHAGIAN PENGURUSAN SEKOLAH ... · ADDITIONAL MATHEMATICS [P2] TRIAL SPM 2017 5 Number Solution and Marking Scheme Sub Marks Full Marks 6 (a) x x x x x

ADDITIONAL MATHEMATICS [P2]

TRIAL SPM 2017

2

Number Solution and Marking Scheme Sub

Marks

Full

Marks

1 (a)

(b)

(c)

01141 k

3k

4

3opm or

3

4PQm

13

4

4

3

PQOP mm

5TP

53422 yx

06822 yxyx

K1

N1

P1

N1

P1

K1

N1

7

2

33 2 or

2

2 5(3 2 ) 9 (3 2 ) or

2 5 2 59 or 9

33 2

2

3 5 2 1 0 or 3 1 2 0

5 1,

3 2

1, 2

3

yy x x

x x x x

yx x y

x x y y

x x

y y

P1

K1

K1

N1

N1

5

Page 3: SULIT 3472/2 Kertas 2 2017 BAHAGIAN PENGURUSAN SEKOLAH ... · ADDITIONAL MATHEMATICS [P2] TRIAL SPM 2017 5 Number Solution and Marking Scheme Sub Marks Full Marks 6 (a) x x x x x

ADDITIONAL MATHEMATICS [P2]

TRIAL SPM 2017

3

Number Solution and Marking Scheme Sub

Marks Full

Marks

3 (a)

75.0,......25.20,27,3636 roror

K1

52.175.03652.11

n

n orT K1

75.0log

36

52.1log

136

52.1log75.0log1

norn

K1

12n N1

(b) 75.01

36

K1

75.01

36248

K1

336 cm

N1 7

4 (a)

(i)

(ii)

2

51505100 ..

125.5 (accept without working)

110

(20)275.5225.5(30)(10)175.5(40)125.575.5(10)

180.05

K1

N1

K1

N1

(b)

200.5 or 60 or 30

5030

605825200

..

238

P1

K1

N1

7

Page 4: SULIT 3472/2 Kertas 2 2017 BAHAGIAN PENGURUSAN SEKOLAH ... · ADDITIONAL MATHEMATICS [P2] TRIAL SPM 2017 5 Number Solution and Marking Scheme Sub Marks Full Marks 6 (a) x x x x x

ADDITIONAL MATHEMATICS [P2]

TRIAL SPM 2017

4

Number Solution and Marking Scheme Sub Marks Full

Marks

5 (a)

050125

2 2 xx

(x – 25)(x – 5) = 0

x = 25 , x = 5

Width = 25 – 5 = 20 unit

K1

K1

N1

(b) 12

5

4 x

dx

dy OR 10015

5

2 2 xy

x = 15 OR 40155

2 2 xy

y = – 40 or max depth = 40

K1

N1

N1

6

Page 5: SULIT 3472/2 Kertas 2 2017 BAHAGIAN PENGURUSAN SEKOLAH ... · ADDITIONAL MATHEMATICS [P2] TRIAL SPM 2017 5 Number Solution and Marking Scheme Sub Marks Full Marks 6 (a) x x x x x

ADDITIONAL MATHEMATICS [P2]

TRIAL SPM 2017

5

Number Solution and Marking Scheme Sub

Marks Full

Marks

6 (a)

x

x

x

xx

x

cos

sin

2cos

2sincos

sin

K1

= xxxx

xx

x

x

2cossin2sincos

cos2cos

cos

sin

=

)1cos2(sincossin2cos

cos2cos

cos

sin2

xxxxx

xx

x

x

=

)sincossin2cossin2

cos2cos

cos

sin22 xxxxx

xx

x

x

= cos 2x N1

6 (b)

Shape

Amplitude or 2 cycles or modulus

All correct

2

xy

Sketch graph

6 number of solution

P1

P1

P1

N1

K1

N1

8

Use x

x

cos

sin

or sin 2x = 2sin x cos x

or cos 2x = 2cos2

x – 1

Page 6: SULIT 3472/2 Kertas 2 2017 BAHAGIAN PENGURUSAN SEKOLAH ... · ADDITIONAL MATHEMATICS [P2] TRIAL SPM 2017 5 Number Solution and Marking Scheme Sub Marks Full Marks 6 (a) x x x x x

ADDITIONAL MATHEMATICS [P2]

TRIAL SPM 2017

6

Number Solution and Marking Scheme Sub

Marks

Full

Marks

7

(a)

(b)

9

12tan//

9

12tan PORPOQ

1.8548 rad // 1.855 rad

SQR = 9(*1.8548) // 9(*1.855) or 13.53sin9

QM

= 16.6932 // 16.695

Chord QR = 2.7*2

= 14.4

Perimeter = SQR + * Chord QR

= 31.09 → 31.098

K1

N1

K1

N1

N1

K1

N1

(c)

Area of sector OQR or Area of triangle POQ

= 8548.192

1 2 = 912

2

1

= 75.12 cm2 = 54 cm

2

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

2(54) − 75.12

32.88 cm2

K1

K1

N1

10

Page 7: SULIT 3472/2 Kertas 2 2017 BAHAGIAN PENGURUSAN SEKOLAH ... · ADDITIONAL MATHEMATICS [P2] TRIAL SPM 2017 5 Number Solution and Marking Scheme Sub Marks Full Marks 6 (a) x x x x x

ADDITIONAL MATHEMATICS [P2]

TRIAL SPM 2017

7

Number Solution and Marking Scheme Sub

Marks Full

Marks

8

Refer to graph

9 (a)

(i)

(ii)

(b)

OQAOAQ or APOAOP

baAQ4

3

baOP4

1

4

3

OPmOS

=

bam

4

1

4

3

banaOS

4

3

nm 14

3 or nm

4

3

4

1

443 nm and nm 3 shown

Solve simultaneous linear equations

13

12m

13

4n

K1

N1

N1

N1

N1

K1

N1

K1

N1

N1

10

Page 8: SULIT 3472/2 Kertas 2 2017 BAHAGIAN PENGURUSAN SEKOLAH ... · ADDITIONAL MATHEMATICS [P2] TRIAL SPM 2017 5 Number Solution and Marking Scheme Sub Marks Full Marks 6 (a) x x x x x

ADDITIONAL MATHEMATICS [P2]

TRIAL SPM 2017

8

Number

Solution and Marking Scheme Sub

Marks Full

Marks

10 (a)

52 xdx

dy

2

5052 xorx

2

55

2

52

y

4

25y

K1

N1

(b)

A1 = 54

25

A2 = 2

5

3

23 xx

21 AA

42.10//12

510//

4

125

K1

K1

K1

N1

(c) 5

0

22 5 dxxxV or dxxxx 5

0

234 2510

3

25

4

10

5

345 xxx

= 17.104//6

1104//

6

625

K1

K1

N1

Page 9: SULIT 3472/2 Kertas 2 2017 BAHAGIAN PENGURUSAN SEKOLAH ... · ADDITIONAL MATHEMATICS [P2] TRIAL SPM 2017 5 Number Solution and Marking Scheme Sub Marks Full Marks 6 (a) x x x x x

ADDITIONAL MATHEMATICS [P2]

TRIAL SPM 2017

9

Number Solution and Marking Scheme Sub Marks Full

Marks

11 (a) i)

37

7

10 )45.0()55.0()7( CxP

= 0.1665

K1

N1

(ii)

100

0

10 )55.0()45.0(C or 91

1

10 )55.0()45.0(C

)1()0(1 XPXP equivalent

= 0.9767

K1

K1

N1

(b) i)

(ii)

25.0

56.36.3

0.4364

Seen - 0.524 or 0.524

524.025.0

56.3

k

k = 3.429

K1

N1

P1

K1

N1

10

Page 10: SULIT 3472/2 Kertas 2 2017 BAHAGIAN PENGURUSAN SEKOLAH ... · ADDITIONAL MATHEMATICS [P2] TRIAL SPM 2017 5 Number Solution and Marking Scheme Sub Marks Full Marks 6 (a) x x x x x

ADDITIONAL MATHEMATICS [P2]

TRIAL SPM 2017

10

Number Solution and Marking Scheme Sub Marks Full

Marks

12 (a)

(b)

(c)

24306 2 tt

014 tt

4t and 1t

3012 t

2

5t

subtitute t into v

242

530

2

56

2

v

1ms2

113 v or 1ms

2

113 moves to the left

t < 2

5

1241151223

1 S or

2

524

2

515

2

52

23

2

5S

Total distance = 2

51 *2 SS

= 24.5

K1

K1

N1

K1

K1

N1

N1

K1

K1

N1

10

Page 11: SULIT 3472/2 Kertas 2 2017 BAHAGIAN PENGURUSAN SEKOLAH ... · ADDITIONAL MATHEMATICS [P2] TRIAL SPM 2017 5 Number Solution and Marking Scheme Sub Marks Full Marks 6 (a) x x x x x

ADDITIONAL MATHEMATICS [P2]

TRIAL SPM 2017

11

Number Solution and Marking Scheme Sub

Marks Full

Marks

13 (a) zor

yor

x 100

15

18100100

20175100

70

x = 40, y = 20, z = 120

K1

N2, 1, 0

(b)

100

461202410010125121508175 )(*)()()()(I

123.7

K1

N1

(c)

7.123*

10088006

P

RM 711.40

K1

N1

(d)

Seen orI 25.16615/18 165 or 137.5 or 110 or 132

100

4613224110105.13712165825.166 I

97133.

N1

K1

N1

10

Page 12: SULIT 3472/2 Kertas 2 2017 BAHAGIAN PENGURUSAN SEKOLAH ... · ADDITIONAL MATHEMATICS [P2] TRIAL SPM 2017 5 Number Solution and Marking Scheme Sub Marks Full Marks 6 (a) x x x x x

ADDITIONAL MATHEMATICS [P2]

TRIAL SPM 2017

12

Number Solution and Marking Scheme Sub Marks Full

Marks

14 Refer to graph

15 (a)

(i)

sin sin 22.18

7.71 3.46

122.73

ADC

ADC

K1

N1

(ii)

2 2 27.71 4.98 4.07 2(4.98)(4.07)cos

116.49

ABC

ABC

K1

N1

(iii)

Area of triangle ADC

49.116sin)07.498.42

1

9.070 cm2 // 9.07 cm

2

07.971.72

1 BE

BE = 2.353 cm

K1

N1

K1

N1

(b)

2 22.35 2.06

=3.125

BD

cm

P1

N1 10

END OF MARKING SCHEME

Page 13: SULIT 3472/2 Kertas 2 2017 BAHAGIAN PENGURUSAN SEKOLAH ... · ADDITIONAL MATHEMATICS [P2] TRIAL SPM 2017 5 Number Solution and Marking Scheme Sub Marks Full Marks 6 (a) x x x x x

x 3 5 6 9 10 12

y10log 0.37 0.22 0.15 0.09- 0.16- 0.31-

0 2 4 6 8 10 12 14 x

0.3-

y10log

x

x

x

x

x

x

(b)

Correct axes, uniform scale and K1

one point correctly plotted

All points correctly plotted K1

Line of best fit N1

(c) qxpy 101010 logloglog P1

(i) Use p10log = y – intercept K1

p = 3.98 N1

(ii) Use q10log = gradient K1

q = 1.19 N1

(iii) y = 3.31 N1

QUESTION 8

0.4-

0.2-

0.1-

0.1

0.2

0.3

0.4

0.5

0.6

N1

10

Page 14: SULIT 3472/2 Kertas 2 2017 BAHAGIAN PENGURUSAN SEKOLAH ... · ADDITIONAL MATHEMATICS [P2] TRIAL SPM 2017 5 Number Solution and Marking Scheme Sub Marks Full Marks 6 (a) x x x x x

45

( 4,38)

(a) I. y - x ≥ 10 N1

II . x ≥ 𝑦

10. or 10x ≥ y N1

III. 3000 x + 1800y ≤81 000 N1

Note: Don’t accept x : y = 1: 10

(b) Correct axes with uniform scale and 1 graph correct K1

All graphs correct N1

Correct shaded region N1

(c) (i) 33 (From point (3,30)) N1

(ii) k = 10y + 25x , let k = 250 and

max point ( 13, 23) K1

maximum allowance = 10 (23) +25 (13) K1

= RM 555 N1

R

(13, 23)

QUESTION 14