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 3OR 2 [use (2) ].

3 2

x yk y

x y

2

3(a)

(b)

3x

B1: 3

5

)(

5

xxf

1

2

1

3

4 1m

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)

1x

1

(1, 4)

1

1

1

3

7 2x

B2: xx 652

1)32(2 or xx 6532

B1: xx

652

1)32(2

333

3 3

or -2 -1

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Muat turun (percum

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uat turun (percuma) skem

a lain di

3

8

3625 n or 1

3625n

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

3r

3

10

B1 :

1

22

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 w

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uat turun (percuma) skem

a lain di : Muat turun (percum

a) skema lain di

4

12

(a)

(b)

3p

q

11:

yB pq p

x x

or 3pq

5

2

1

3

13 3 5

2 4y x or 4 6 5y x or equivalent

B3 : 3 3

12 2

y x

B2 : 2

3

2m or

3,1

2

B1 : 1

2

3m 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 1or or

1213 13 13 13

i ji j

B1 : |OR |=13

1

2

3

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Muat turun (percum

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uat turun (percuma) skem

a lain di

5

16

4

3k

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 02 30 ,150 ,390 ,510x or sin 2x = 1

2

B1 : 2(2sin cos ) 1x x

3 3

19(a)

(b)

80 32x

5.2x or 2

5x

B1 : 03280 x

1

2

3

20

2

54 xy or equivalent

B2 : 142

3 xy or equivalent

B1 : 4 or ( 1) 3dy dy

dx dx

3

3

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Muat turun (percum

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uat turun (percuma) skem

a lain di : Muat turun (percum

a) skema lain di

6

21(a)

(b)

10

3

B2 : 3

1

5 7

2 2hx

B1:

3 3

1 1

( ) 7

2 2

f xhdx 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 20x

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

15 4 3

2

2

4

25(a)

(b)

0.1741

49.69

B2 : 45

0.9385

k

B1 : z = 0.938

1

3

4

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Muat turun (percum

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uat turun (percuma) skem

a lain di : Muat turun (percum

a) skema lain di

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

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uat turun (percuma) skem

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a) skema lain di

2

No Solution and Mark Scheme Sub

Marks

Total

Marks

1

8 3

2

yx

OR

8 2

3

xy

P1

2 8 33 6 0

2

yy y

OR

28 2 8 2

3 6 03 3

x xx

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

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Muat turun (percum

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uat turun (percuma) skem

a lain di

3

3(a) 16 ,8 ,4 ,...... OR

1

2r

P1

116 1

230.5

11

2

n

K1

4.416n K1

5n N1

4

7

(b) 64 ,16 ,4 ,...... OR

1

4r P1

64

11

4

S

K1

= 1

853 or 85.33 N1

3

4(a) (i) Change

13 6 0 2

3y x to y x or

1

3BCm

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 23

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

raS

n

n

1

)1(

Use 1

aS

r

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uat turun (percuma) skem

a lain di : Muat turun (percum

a) skema lain di

Muat turun (percum

a) skema lain di : M

uat turun (percuma) skem

a lain di

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

53sin 2 2

xx

or

52

xy

N1

Draw the straight line 5

2x

y

K1

Number of solutions is 3 N1

3

6(a)

L = 79.5 OR F = 24 OR fm = 4 P1

3(36) 24

479.5 104

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)

36X

2602

36

= 72.28

5

y

3

–3

2

2

3 2 x

52

xy

O

3sin2y x

K1

N1

OR

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Muat turun (percum

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uat turun (percuma) skem

a lain di

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 2y dx c

x x

2

2(4) , 2

( 1)c c

2

22y

x

2

2

5

21

5

1 1

22

22

1

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

1 1

dxx

xx

3

3

10

(c )

33or 6.6

5

2

2

2

5

23 1

5

22 3 13 1

5 5

2( ) Volume ( 2)

4 84

3 1

4 5 8 54( 2) 8( 2)4 2 4 5

3 1 3 1

14.56

i dxx

x xx

4

K1

N1

K1

K1

N1

K1

K1

K1

N1

K1

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uat turun (percuma) skem

a lain di : Muat turun (percum

a) skema lain di

6

9(a)

(b)

(c)

yx

yxxOD

yxAC

4

15

4

7

)57(4

37

57

3 7 157 5 5

4 4 4

21 7

4 4

3

15 155

4 4

3

x y h y k x y

k

k

h k

h

5

452

150

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

155

4

15

3

4

7

4

21

4

15

4

7557

4

3

h

h

k

kk

yxkyhyx

K1

K1

N1

K1

N1

2

3

4

155

4

15

3

4

7

4

21

4

15

4

7557

4

3

h

h

k

kk

yxkyhyx

K1

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Muat turun (percum

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uat turun (percuma) skem

a lain di : Muat turun (percum

a) skema lain di

7

(b)

5.3

4548

5.3

45404840 )( ZPXPi

= 0.7278

( ) 0.7

450.524

3.5

43.166

ii P X m

m

m

11(a)

(b)

(c)

7

tan 1

0.78554

OR RQ PR cm

rad rad

2 2

7(1.571) 7(2.3565)

7 7 2(7)(7)(cos135 )o

OR

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

54.4268

o

Perimeter

217

4

2 21 17 2.3565 7 sin135

2 2

o

2 2 21 1 17 7 2.3565 7 sin135

4 2 2

78.8996

oArea

2

4

4

10

K1

N1

N1

K1

K1

K1

N1

2

3

4

155

4

15

3

4

7

4

21

4

15

4

7557

4

3

h

h

k

kk

yxkyhyx

K1

K1

K1

N1

K1

K1

K1

N1

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Muat turun (percum

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uat turun (percuma) skem

a lain di : Muat turun (percum

a) skema lain di

8

No Solution and Mark Scheme Sub

Marks

Total

Marks

12(a)

a = 10 - 5t = 0

t = 2 s

cttv 2

2

510

c 2)0(2

5)0(1030

c = 30

302

510 2 ttv

30)2(2

5)2(10 2 v

= 40 ms- 1

4

10

(b)

302

510 2 ttv 0

2 6 0t t

0 6t

3

(c)

cttts 306

55 32

s = 0, t = 0 , c = 0

ttts 306

55 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

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Muat turun (percum

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uat turun (percuma) skem

a lain di

9

67.226

67.46180

302

51030

2

510

8

6

2

6

0

2

dtttdttt

13(a)

12510055

10

P

P10 = RM 44

2

10

(b)

115534

58831404125110

hh

hh

*

h = 1

2

(c)

11510020

11 P

P07 = RM 23

2

(d)

I S = 110125100

88

110 *1 125 4 140 *1 3 110 5

1 4 4 5I

= 122.86

4

Integrate v

6

0

+ 8

6

K1

K1

N1

N1

K1

N1

K1

K1

N1

See 125 P1

K1

K1

N1

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Muat turun (percum

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uat turun (percuma) skem

a lain di w

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uat turun (percuma) skem

a lain di : Muat turun (percum

a) skema lain di

Muat turun (percum

a) skema lain di : M

uat turun (percuma) skem

a lain di

10

14 Rujuk Lampiran

15(a)

Using sine rule to find BAC .

sin sin30

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 cos135.36

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

22 227 14 2 27 14 cos 44.64

19.67

AC

AC

1Area (14)(27)sin 44.64

2

oABC or

1Area (6)(19.67)sin105.36

2

oACD

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

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Muat turun (percum

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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

12

ny m

m x

N1

P1

. 2 70i m

35m

N1

. 80m

iin

K1

2800n

N1

1. 0.37iii

x

K1

2.703x

N1

0.8

x

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Muat turun (percum

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uat turun (percuma) skem

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a) skema lain di

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 70 0 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

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Muat turun (percum

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uat turun (percuma) skem

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uat turun (percuma) skem

a lain di : Muat turun (percum

a) skema lain di