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SULIT 3472/2(PP)

3472/2 (PP) 2008 TRIAL SPM Hak Cipta Jabatan Pelajaran Negeri Terengganu [Lihat sebelah SULIT

3

Trial SPM 2008

MARK SCHEME ADDITIONAL MATHEMATICS 2

SECTION A [40 MARKS]

No. MARK SCHEME MARKS

1

x = 2y + 2 or y =2

2

x P1

2 2 55

2 2

y y

y y

or

25

22

2

x

x

x x

= 5 K1

y2 + 2y – 4 = 0 or x2 – 20 = 0

y = 22 2 4(1)( 4)

2(1)

or x = 20 K1

y = –3·24 , y = 1·24 N1

x = 4·47, x = – 4·47 N1 (do not accept x = 4·48 or 4·48)

5

SULIT 3472/2(PP)


4


2

(a) 38dy

hdh

and 2dh

dx P1 OR y =

2

4

(2 5)x

P1

Use dy dy dh

dx dh dx 38(2 5) (2)

dyx

dx

K1

3

82

h

K1 = 3

16

(2 5)x N1

3

16

h

= 3

16

(2 5)x N1

(b) 4 2dx

dt K1 Use

dh dh dx

dt dx dt

12 unit sdx

dt

N1

(c) 16( 0 02)y K1 Usedy

y xdx

16dy

dx or 0 02x P1

= 0∙32 N1

8

3(a) P1 for ∑ fx = 378 + 35·5m

2(5 5) 5(15 5) 6(25 5) (35 5) 3(45.5)26

2 5 6 3

m

m

K1

m = 4 N1

(b) P1 for ∑fx2 = 16145

21614526

20 K1

= 11·46 N1

6

SULIT 3472/2(PP)


5


4(a) 2 cos2 x + (tan2 x sec2 x)

OR tan2 x + 2 cos2 x (1 + tan2 x) K1

2 cos2 x 1 N1

(b)

P1 for shape of cosine

P1 for amplitude (max & min)

P1 for shifted

K1 for straight line (gradient or yintercept)

3 cos2 x 1 = 3x

or y = 3

x

P1

No. of solutions = 4 N1

8

23

2O

y

2

-4

1

2

-3

20 x

SULIT 3472/2(PP)


6


5

(a) Perimeter for 1st rectangle = 2(30) + 2(20) = 100 cm ------T1

2nd rectangle = 2(15) + 2(10) = 50 cm ------T2

3rd rectangle = 2 (7·5) + 2(5) = 25 cm ----- T3

P1 for 100, 50, 25, ….

a = 100 and r = 50

100or

25

50or

1

2 K1

T7 = 100

61

2

= 1·563 N1

(b) Area : 600, 150, 37·5,…….. P1 Tn < 20

600

11

4

n

< 20 K1 for using arn – 1 < 20

(n – 1) log 0·25 < log 0·03333 K1

0·6021n + 0·6021 < 1·477

0·6021n > 2·079

n > 3·453

n = 4 N1

7

SULIT 3472/2(PP)


7


6(a) (i) AC

=

1

2a b

N1

(ii) EF

=1 1

2 2b AB

= 1 1

( )2 2

b a b

K1

= 1

2a

N1

(iii)4 1 1

3 2 2AC a b b

K1

= 2 1

3 3a b

N1

(b)

1

23 1

2 3

a bAE

DB a b

=

1(2 )

21

(2 )3

a b

a b

K1

= 3

2

3

2AE DB

or equivalent N1

7

SULIT 3472/2(PP)


8

SULIT 3472/2(PP)


9


8(a) 7a + 3b = 36 K1 from area of triangle

115 2 4 8 3 5 29 5

2b a a b

59b = 295 or equivalent K1 solving simultaneous eqn.

7a + 3b = 36

8b a = 37

b = 5 or a = 3 N1

R (3, 5) N1

(b) mperpendicular = 8 P1

y – 5 = 8 (x – 3) K1

8x + y = 29 or equivalent N1

(c) 2 2 2 22 ( 5) ( 4) ( 3) ( 5)x y x y

K1 for using distance PN or NR

K1 for using 2PN = NR

3x2 + 3y2 + 46x 22y + 130 = 0 N1

10

SULIT 3472/2(PP)


10


9(a) (i)

dy

dx= kx2 – x = 5 K1

k(1)2 – 1 = 5

k = 6 N1

(ii) y = 36

3

x

2

2

x + c K1 (integrate)

c = 7

2P1

y = 36

3

x

2

2

x

7

2N1

(b) 212 2

3 K1 (volume of cone)

02 2

1

( 3 )x x dx

K1

04 5

3

1

63

4 5

x xx

K1 (integrate)

212 2

3 +

04 5

3

1

63

4 5

x xx

K1 (sum)

= 131

30 or 4·367 or 13·72 N1

10

SULIT 3472/2(PP)


11


10(a) cos =

8

12K1

= 48·19

= 0·8412 rad. N1

(b) Arc length PR = 8(0·8412) or Arc length RST = 850 3 142

180

K1

= 6·7296 or 6·730 = 6·982

RT2 = 82 + 82 2(8)(8) cos 50 or equivalent K1

RT = 6·762

QR = 2 212 8 = 8·944

Perimeter = 4 + 8·944 + 6·730 + 6·982 + 6·762 K1

= 33·42 N1

(c) Area ORQ , A1 = 1

8 8 9442 or Area ORT, A2 = 21

8 sin 502 K1

= 35·77 = 24·51

Area sector OPR, A3 = 218 0 8412

2 = 26·92

or Area sector ORT, A4 = 218 0 8728

2 K1

= 27·93

Area shaded region = (A1 A3) + (A4 A2) K1 = 12·26 or 12·27 N1

10

SULIT 3472/2(PP)


12


11(a) (i) Mean, np = (8)(p) = 6·4 K1

p = 0·8 N1

(ii) Find P(X 3)

= 1 80C (0·8)0(0·2)8 8

1C (0·8)1(0·2)7 82C (0·8)2(0·2)6 K1 for

using nrC (0·8)

r(0·2)

n r

K1 for using 1 – P(X = 0) – P(X = 1) – P(X = 2)

OR for using P(X 3) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)

= 0·9988 or 0·99877 N1

(b) Use 5 3

P z

= 0·0668 or 4 8

P z

= 0·1587 K1

5 3

= 1·5 or

4 8

= 1 K1

Solve simultaneous equation (until 1 variable left) K1

= 0·2 N1 = 5 N1

10

SULIT 3472/2(PP)


13

SECTION C [20 MARKS]


12

(a)(120)(7) (135)(3) (110)(4) (90)(4)

18I

K1

W = 18 P1

I = 113·61 N1

(b) 07 100 113 611000

P K1

07 1136 10P RM N1

= RM 1136 N1

(c) (i) 120

113 61100

K1

= 136·3 N1

(ii)550

100108

K1

= RM 509·26 N1

10

SULIT 3472/2(PP)


14

P

Q

R

324·6 cm

5·3 cm

4·6 cm

Q


13(a) (i)

5 3

sin Q

=

4 6

sin 32

K1

sin Q = 0·6106

PQR = 37·63 or 37 38 N1

(ii) 4·82 = 2·82 + 5·32 – 2(2·8)(5·3) cos R K1

cos R = 0·4343

PRS = 64·26 or 64 16 N1

(iii) Area ∆ PQR = 1

2(4·6)(5·3) sin 110·37

Area ∆ PRS = 1

2(2·8)(5·3) sin 64·26 K1 (sine rule - either one)

N1 for 6·684 or 11·43

Total area = A1 + A2 K1 (adding)

= 18·11 cm2 N1

(b) (i)

(ii) PQR = 142·37 N1

10

N1 for drawing PQR

SULIT 3472/2(PP)


15

SULIT 3472/2(PP)


16


15(a) – 6 ms1 N1

(b)dV

dt = 2t – 1 = 0 K1

t = 1

2N1

v = 61

4 ms1 N1

(c) v = t2 – t – 6 < 0

(t + 2)(t – 3) < 0 K1

0 t < 3 N1

(d) Use |

3

0

vdt | + |

4

3

vdt | K1 for limits

=

33 2

03 2

6t t

t

+

43 2

33 2

6t t

t

K1 for integration

=81

6 +

17

6P1 (substitution)

= 161

3 m N1

OR

s = 3 2

3 26

t tt K1

t 0 1 2 3 4

s 0 37

6

34

3

27

2

32

3

Total distance = 27

2 +

17

6K1

= 161

3 m N1

10

P1 for the table value or route

032

3

27

2

t = 0t = 3

t = 4

END OF MARK SHEME

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