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3472/1

Matematik Tambahan Kertas 12 jam Ogos 2012

BAHAGIAN PENGURUSANSEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN

KEMENTERIAN PELAJARAN MALAYSIA

PENTAKSIRAN DIAGNOSTIK AKADEMIK SBP 2012 PERCUBAAN SIJIL PELAJARAN MALAYSIA

ADDITIONAL MATHEMATICS

Paper 1

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MARKING SCHEME

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PERATURAN PEMARKAHAN- KERTAS 1

No. Solution and Mark Scheme Sub Marks

Total Marks

1(a) 25 1 2

(b) 4 1

2(a) 2 1 3

(b) 4 2

B1:2x

OR3y

2 [use k 1(2) y].x 3 y 2

3(a) x 3 2 3

B1:5

5

f (x) x 3

(b) 1 1

4 m 1

B2: (6)2 4(2 m)(3) 0

B1: 2 mx 2 6x 3 0

3 3

5 2 p 1

B2 : ( p 2)( p 1) 0 or -2 -1

B1: p 2 3 p 2 0

3 3

6(a) x 1 1 3

(b) 1 1

(c) (1, 4) 1

7 x 2

B2: 2(2x 3) 1 5 6x or 2x 3 5 6x

2

2(2 x3) 1

B1: 3 2 35 36 x

3 3

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6

25

or

B1 log5

4B

oreq

uiva

lent

1

log5

9

(

1

(*

B1

B2

2

3

4 3 3 3

4 3 1 2 3

1

6253

n or6

25n3

m

n 3

m n 3

B1:

log5

m(f

or c

hang

e ba

se)

20) =

– k3(

20

20

- 7 +

7)) =

1

] OR

h –

(-

1)

]

d

h

11 [b

oth

k =

11

d =

20

- 7 +

3(

1 2

3

1)

9(h

1 or

h =

2 a

nd k

=

B2:

h =

2 o

r

B1:

- 7 +

3 O

R

r23

3 10

S

75h

55 10

[2(3

h

B1:

a

3h

8 10 (b)

11

4

12

(a)

(b)

p 3

q

B1:y pq

1 p or

x x

5

pq 3

2

1

3

13 y 3

x 5

or 4y 6x 5 or equivalent2 4

B3 : y 1 3 x

3

2 2

B2 : m 3

or 3

,12 2 2

B1 : m2

OR S(3,0) and T(0, 2)1

3

4 4

14 x2 y2 4x 4 y 92 0

B2 : (x 2)2 ( y 2)2 (6 (2))2 (4 2)2

B1 : PS = PQ OR (6 (2))2 (4 2)2

3 3

15(a) 5i 12 j

1 513 12

1 3

(b)

5 i

12 j or

13 135i 12 j

13or 2

B1 : | OR |=13

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16 k 3

4

B2: 4k 3 0

B1: (4k 3)i (4k 2) 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° 3 3

B2 : 2x 300 ,1500 , 3900 , 5100

or sin 2x = 1

2

B1 : 2(2sin x cos x) 1

19(a) 80 32x 1 3

(b)x 2.5 or x

5 2

2B1 : 80 32x 0

20 y 4x 5

or equivalent2

B2 : y 3 4x 1 or equivalent

2

B1 :dy

4 ordy

(1) 3dx dx

3 3

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21(a) 10

3

B2 : hx3

5

71 2 2

3 3 f (x) 7B1: hdx 2

dx 21 1

1 4(b)

3

22(a) 3

1.648 or 1.6475

B2:163

( 51

)2 or equivalent20 20

51 2(0)2 3(1)2 2(2)2 8(3)2 5(4)2

B1 : x or 2 20 20

1 4

(b) 3

23(a) 252 1 3

(b) 66 2B1 : 4C 6C or 60 OR 4C 6C or 6

3 2 4 1

24(a) 110

B1 : 3

1

2

5 4 3

1920B1 : 1

3

1

1 5 4 3

2 4

(b)2

25(a) 0.1741 1 4

(b) 49.69 3

B2 : 0.938 k 45

5B1 : z = 0.938

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3472/2

Matematik Tambahan Kertas 2Ogos 20122 ½ jam

BAHAGIAN PENGURUSANSEKOLAH 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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No Solution and Mark Scheme Sub Marks

Total Marks

1x

8 3yOR y

8 2xP12 3

2

y2 3y 8 3y 6 0 OR 8 2x

3x 8 2x 6 0 K1 2 3 3

Replace a, b & c into formula K1

(24) (24)2 4(7)(12) (20) (20)2 4(7)(59)y OR x

2(7) 2(7)

y 0.443, 3.871 OR x 4.664, 1.807 N1x 4.664, 1.807 OR y 0.443, 3.871 N1

5 5

2 (a)

y x2 2x 3k

y (x 1)2 1 3k K11 3k 4 K1

k 1 N1

(b)

(1,4)3

-1 3

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

3

3

6

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3

3(a) 16,8, 4,...... OR r 1

2

1 n

16 1 a(1 rn ) 2 30.5 Use Sn

1 1 r1

2

n 4.416

n 5

64 ,16 , 4 ,......OR r 1

4

S64

1 a1 Use S

4 1 r

= 85 1 or 85.33

3

P1

K1

K1

N1

P1

K1

N1

4 7

(b) 3

4(a)(i) Change 3y x 6 0to y

1 x 2 or m

3 BC

OR mAB 3

y 5 3 x (6) OR any correct method

y 3x 23(ii) Use simultaneous equation to find point B*y 3x 23 and 3y x 6 0 or y

1 x 2

3

B = 15

, 1

2 2

* 15

, 1

2(x) 3(6) , 2( y) 3(5)

2 2 5 5

D = 39

, 25

4 4

13

K1

K1

N1

K1

N1

K1

N1

5 7

(b)2

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5(a)y y

5x 2

3

y 3sin 2x

O 3 2 x

2 2

–3

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

3sin 2x 5x

2 or y 5x

2 N1

Draw the straight line y 5x

2 K1

Number of solutions is 3 N1

4 7

(b)3

6(a) L = 79.5 OR F = 24 OR fm = 4 P1 3

(36) 24 79.5 4 10 K1

4

87 N1

3 8

(b) (i) 5

X (44.5 4) (54.55) (64.5 6) (74.59) (84.5 4)

36

OR 2602K1

36= 72.28 N1

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(ii)(44.5)2 4 (54.5)2 5 (64.5)2 6 (74.5)2 9 (84.5)2 4 (94.5)2

8 K1

197689

(*72.28)2

36 K1 16.34 N1

7 Rujuk Lampiran

8(a)4 2 K1

y x3 dx

x2 c

(4) 2

c , c 2 K1

(1)2

y 2

2 N1

x2

2 2 x2 2 dx

52

2x1 K1 1

2x 5 2(2)1 2(5)1 K1

1 2(2) 1

2(5)

33

or 6.6N1

52 2(i) Volume ( 2)2 dxx2 K15

2 4x3 8x1

4x K1 3 1 5

2 4 ( 2 ) 3 8 ( 2 ) 1

2 4 5 3 8 5 1 3

1

4 2 3

1

4 5 K1 5 5

14.56 N1

3 10

(b)

3

(c )4

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6

9(a) AC 7 x 5 y

OD 7x 3

(7 x 5 y)4

7 x

15 y4 4

3 7 x 5 y h 5 y k 7

x 15

y 4

4 4

21 7k

4 4k 3

15

5h 15

k4 4

h 3

50 1 5 4 t

2t 5

N1

K1

N1

K1

K1

N1

K1

N1

K1

N1

3 10

(b)5

(c)

2

10(a) (i) P X 6 10C 0.36 0.74K1 5 10

= 0.03676 N1

(ii) 10C 0.39 0.71OR 10C 0.310 0.70 K1

9 10

P X 9 10C 0.39 0.71 10C 0.310

K19 10

= 0.0001437 N1

5

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(b) (i) P40 X 48 P 40 45

Z 48 45

3.5 3.5

= 0.7278

(ii) P X m 0.7m 45

0.524 3.5

m 43.166

K1

N1

K1

K1

N1

11(a) OR RQ PR 7cm tan 1

rad 0.7855 rad

4

7(1.571) OR 7(2.3565)

72 72 2(7)(7)(cos135o )

Perimeter

7 7 7(1.571) 7(2.3565) ( 72 72 2(7)(7)(cos135o )

54.4268

1 72

4

1 72 2.3565

1 72 sin135o

2 2

Area 1 72

1 72 2.3565

1 72 sin135o

4 2 2

78.8996

K1

N1

K1 K1

K1

N1

K1

K1

K1

N1

2 10

(b)4

(c)

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No Solution and Mark Scheme Sub Marks

Total Marks

12(a)a = 10 - 5t = 0 Use a = 0 K1t = 2 sv 10t

5 t 2 c Integrate a to K12 find v

30 10(0) 5

(0)2 c2

c = 30v 10t

5 t 2 30

2 Integrate and K1v 10(2)

5 (2)2 30

substitute t = 2

2 N1= 40 ms- 1

v 10t 5

t 2 30 0 Use v > 0 K12

t 2t 6 0 K1

0 t 6 N1

s 5t 2 5

t 3 30t c Integrate v dt K1

6s = 0, t = 0 , c = 0

s 5t 2 5

t 3 30t 6

s 5(6)2 5

(6)3 30(6) or s 5(8)2 5

(8)3 30(8) K16 6

= 180 = 133.33

Total distance = 180 + 180 133.33= 226.67 m N1

OR

4 10

(b) 3

(c) 3

P

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55

1

0t t

2

30

dt

1

0t t

2

30

dtIn

tegr

ate

v2

20

6

K

2K

8 8

I

9

102 2 2 4

6

8

K1

6 8

1

80

46.

67

+

K1

06

2

26.6

7N1

55

100

125

K1

P10

= R

M 4

4N1

110

h

125

4

140

h

3

88 5

h

4

h

3

5* h

= 1

N1

P11

10

0

115

P07

= R

M 2

3N1

See

125P

1I S

=

125

110

K1

110

*1

1

25

4

140

*1

3

1

10

5 K

11

4

4

5=

122

.86N

1

13(a

)

(b)

(c)

(d)

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14 Rujuk Lampiran

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

27 14

BAC 74.64o

BAC (obtuse) 180o 74.64o

105.36o

DCB 105.36o 30o or DC 6 cm

Use cosine rule to find BD.

BD2 62 272 2 627cos135.36

BD 31.55

Use formula correctly to find area of triangle ABC or ACD.ABC 180o 30o 105.36o

44.64o

AC 2 272 142 2 2714cos 44.64

AC 19.67

Area ABC 1

(14)(27) sin 44.64o or2

Area ACD 1

(6)(19.67) sin105.36o

2

Use Area ABCD = sum of two areas

Area ABCD = 189.7 cm2 .

K1

N1

N1

P1

K1

N1

K1

K1

K1

N1

3 10

(b) 3

(c) 4

END OF MARKING SCHEME

10

m x

Pentaksiran Diagnostik Akademik SBP 2012 Paper 2 ADM

No.7(a) 0.1 0.2 0.25 0.4 0.5 0.8 N1

y 62 54 50 38 29 4

y n 1 2m P1

Plot against K1x

(at least one point)

6 points plotted correctly K170

Line of best fit N1

x60

x50 x

ii.

iii.

N1

m 80

n K1

n 2800 N1

1 0.37

x K1

x 2.703 N1

40x

30 x

20

10

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

x

0.8

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N 1

N 1 N 1 N 1

(I.

N

x

y

II

.N

Ans

w

y 2

x

I

(R

efer

to th

e

1 gr

aph

co

rrec

tK

N

Cor

rect

are

a

(c)

max

poi

nt (

i)k

= 10

0x +

80

y M

ax

= 10

0(50

)+ 8

0(12

0)K

= R

M 1

4,

N10

80

y

ii

(50,

120)

x0

23

41

56

78

1 1 1 1 8

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1 6 4 2