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Matematik Tambahan Kertas 2 Ogos 2012 2 ½ jam

BAHAGIAN PENGURUSANSEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN

KEMENTERIAN PELAJARAN MALAYSIA

PENTAKSIRAN DIAGNOSTIK AKADEMIK SBP 2012PERCUBAAN SIJIL PELAJARAN MALAYSIA

ADDITIONAL MATHEMATICS

Paper 2

Skema Pemarkahan ini mengandungi 10 halaman bercetak

MARKING SCHEME

2

No Solution and Mark SchemeSub

MarksTotal Marks

1 8 32

yx OR P1

2 8 33 6 02

yy y OR28 2 8 23 6 0

3 3x xx K1

Replace a, b & c into formula K1

2( 24) ( 24) 4(7)( 12)2(7)

y OR2( 20) ( 20) 4(7)( 59)

2(7)x

OR N1OR N1

5 5

2 (a)

kxykxxy31)1(

322

2

1431

kk

(b)

3

3

6

(1,4)

3

3 -1

-Maximum shape P1-*Maximum point K1-Another 1 point y-intercept / x-intercept K1

K1K1N1

3

3(a) OR 12

rP1

116 12

30.5112

n

K1

K1

N1

4 7

(b)OR 1

4r P1

64114

S

K1

= 1853

or 85.33 N1

3

4(a)(i) Change or 1

3BCm

OR 3ABm K1

5 3 ( 6)y x OR any correct method K1

N1

5 7

(ii) Use simultaneous equation to find point B

and K1

B = 15 1,2 2

N1

(b)K1

D = 39 25,4 4

N1

2

Use rraS

nn 1

)1(

Use 1

aSr

4

5(a)

(b)

Amplitude = 3 [ Maximum = 3 and Minimum = 3 ] P1Sine shape correct P1Two full cycle in 0 x 2 P1Negative sine shape correct(reflect) P1

4 7

or N1

Draw the straight line K1

Number of solutions is 3 N1

3

6(a) L = 79.5 OR F = 24 OR fm = 4 P13 (36) 24479.5 10

4K1

N1

3 8

(b) (i)(44.5 4) (54.5 5) (64.5 6) (74.5 9) (84.5 4) (94.5 8)

36X

260236

= 72.28

5

y

3

3

2 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

N1

7 Rujuk Lampiran

8(a)

(b)

2

2

25

21

5

1 1

2 2

2 21

2( 2) 2( 5)2( 2) 2( 5)1 1

dxx

x x

3

3

10

(c )2

22

5

23 1

5

22 3 13 1

5 5

2( ) Volume ( 2)

4 8 43 1

4 5 8 54( 2) 8( 2) 4 2 4 53 1 3 1

14.56

i dxx

x x x

4

K1

N1

K1

K1

N1

K1

K1

K1

N1

K1

6

9(a)

(b)

(c)

yx

yxxOD

yxAC

415

47

)57(437

57

3 7 157 5 54 4 421 74 4

3

15 1554 4

3

x y h y k x y

k

k

h k

h

5

452150

t

t

3

5

2

10

10(a) (i) 6 410 6= 0.03676

(ii)

9 1 10 010 109 10

= 0.0001437

5

5

10

N1

K1

N1

K1N1

K1

N1

K1

N1

K1

K1

N1

K1

N1

K1

7

(b)5.34548

5.345404840 )( ZPXPi

= 0.7278

( ) 0.745 0.524

3.543.166

ii P X mm

m

11(a)

(b)

(c)

7tan 1

0.78554

OR RQ PR cm

rad rad

2 2

7(1.571) 7(2.3565)

7 7 2(7)(7)(cos135 )oOR

2 27 7 7(1.571) 7(2.3565) ( 7 7 2(7)(7)(cos135 )

54.4268

o

Perimeter

21 74

2 21 17 2.3565 7 sin1352 2

o

2 2 21 1 17 7 2.3565 7 sin1354 2 2

78.8996

oArea

2

4

4

10

K1

N1

N1

K1

K1

K1

N1

K1K1

K1

N1

K1

K1

K1

N1

8

No Solution and Mark SchemeSub

MarksTotal Marks

12(a)a = 10 - 5t = 0

t = 2 s2

2

c = 302

2

= 40 ms- 1

4 10

(b)2 0

2 6 0t t

0 6t

3

(c)

32

s = 0, t = 0 , c = 0

ttts 30655 32

32 or 32

= 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

30251030

2510

8

6

26

0

2 dtttdttt

13(a)12510055

10P

P10 = RM 44

2 10

(b)

* h = 1

2

(c)115100

2011P

P07 = RM 23

2

(d)

I S =

= 122.86

4

Integrate v

+

K1

K1

N1

N1

K1

N1

K1

K1

N1

See 125 P1K1

K1

N1

10

14 Rujuk Lampiran

15(a) Using sine rule to find BAC .sin sin 30

27 14

oBAC

74.64oBAC (obtuse) 180 74.64

105.36

o o

o

BAC

3 10

(b) 105.36 30 or 6o oDCB DC cm

Use cosine rule to find BD.

22 26 27 2 6 27 cos135.3631.55

BDBD

3

(c) Use formula correctly to find area of triangle ABC or ACD.180 30 105.36

44.64

o o o

o

ABC

22 227 14 2 27 14 cos 44.6419.67

ACAC

o or

o

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

10

20

30

40

50

60

70

80

x

x

0.5

No.7(a)

0.70

x

0.2

x

x

Plot against K1

(at least one point)

6 points plotted correctly K1

Line of best fit N1

0.1 0.2 0.25 0.4 0.5 0.8

y 62 54 50 38 29 4

1 2ny mm x

N1

P1

N1

. 80miin K1

N1

1. 0.37iiix K1

N1

0.8

x

Answer for question 14

(a) I.

II.

III.

(b) Refer to the graph,

1 graph correct3 graphs correct

Correct area

(c) max point ( 50,120 )

i) k = 100x + 80yMax fees = 100(50) + 80(120)

= RM 14, 600

ii)

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