marking scheme kedah spm 2008 add maths trial p2

2008 SPM TRIAL EXAMINATION Marking Scheme

3472/2 Additional Mathematics Paper 2 [Lihat sebelah SULIT

2

Question Solution and marking scheme Sub-mark

Full Mark

1.

3

2yx

or

2 3y x

223 3 10

2 2y y y y

or 22 2 3 2 3 10x x x x

2 2 29 6 6 2 4 40 0y y y y y or

2 2 22 3 4 12 9 10 0x x x x x y = 3.215 , − 3.215 or x = 3.107 , − 0.107 x = 3.107 / 3.108 , − 0.107 / − 0.108 or y = 3.214 / 3.215 , − 3.214 / − 3.215 Answer must correct to 3 decimal places.

5

P1 Make x or y as the subject

Eliminate x or y

Solve quadratic equation

23 31 0y

2 313

y

or 23 9 1 0x x

or using formula

or completing the

square

N1

N1

K1

K1



3


Full Mark

2(a)

(b)

16 ,8 ,4 ,......

16a 12

r

7

7

116 12

112

S

= 3314

or 31.75 64 ,16 , 4 ,......

64a 14

r

64

114

S

= 1853

or 85.33

3 3

6

3(a)

2( )f x x px q

= 2 2

2

2 2p px px q

= 2 2

2 4p px q

32p

p = －6

36 54

q q = 4

N1

K1

Use rraS

nn

1)1(

N1

K1

K1

N1

Use 1

aSr

use x² + bx = ( x + 2b )² – 2

2 )( b

or

use axis of symmetry 32ba

P1

P1

N1



4


Full Mark

Alternative solution

22 3 5x px q x

= 2 6 9 5x x = 2 6 4x x

6p q = 4

3

3(b) 2 6 4 31 0x x

2 6 27 0x x

( 9)( 3) 0x x

3 9x

2

5

4(a)

tan x + cos1 sin

xx

= sincos

xx

+ cos1 sin

xx

= 2sin (1 sin ) cos

cos (1 sin )x x x

x x

= 2 2sin sin cos

cos (1 sin )x x x

x x

= sin 1cos (1 sin )

xx x

= 1cos x

= sec x

3

K1

N1

Use ( ) 31 0f x and factorization

K1

K1

N1

Use sintancos

xxx

Use identity 2 2sin cos 1x x

Comparing coefficient of x or constant term

N1

K1

N1



5


Full Mark

4(b)(i) Negative sine shape correct. Amplitude = 3 [ Maximum = 3 and Minimum = − 3 ] Two full cycle in 0 x 2

3

4(b)(ii) 53sin 2 2xx

or 5 2xy

Draw the straight line 5 2xy

Number of solutions is 3 .

3

9

K1

N1

N1

P1

P1

P1

y

3

–3

2

2

3 2 x

5 2xy

O

3sin 2y x



6


Full Mark

5 (a)

(b)

1220

x 240x

The mean

240 5 + 8 + 10 + 11 + 14 28825 25

X

= 11.52

2212 3

20x

2 3060x

The standard deviation

2 2 2 2 2

23060 5 8 10 11 14 11.5225

= 23566 11.5225

= 9.9296 = 3.151

7

7

K1

N1

Use formula 2

2xx

N

For the new 2x and X

K1

K1

N1

N1

Use x

xN

N1



7


Full Mark

6(a)

(b)

(i) PR PO OR

uuur uuur uuur

= 6 15a b

% %

(ii) OQ OP PQ

uuur uuur uuur

= 365

a ORuuur

%

= 6 9a b

% %

(i) OS hOQ

uuur uuur

= (6 9 )h a b

% %

(ii) OS OP PS

uuur uuur uuur

= 6a kPR

uuur

%

= 6 6 15a k a b

% % %

(6 9 )h a b% %

= 6 6 15a k a b % % %

6 6 6h k 9 15h k

1h k 53

h k

5 13

k k

38

k

5 33 8

h

= 58

3

5

8

N1

N1

K1

Use PR PO OR uuur uuur uuur

or OQ OP PQ uuur uuur uuur

N1

N1

Equate coefficient of a

% or b

%

and Eliminate h or k

K1

N1

N1



8

Qn. Solution and Marking Scheme Sub- mark

Full Mark

7(a)

(b)(i)

(ii)

kn

xn

y 4log1log1log

log x 0.18 0.30 0.40 0.60 0.74 log y 0.48 0.54 0.59 0.69 0.76

6

4

10

N1 N1

P1

Correct axes and scale

All points plotted correctly

Line of best-fit

N1

K1 intercept

= kn

4log1

n = 2 k = 1.51

K1

N1

N1

K1

N1

gradient

= n1

N1 N1



9

0.6

0.7

Graph of log10 y against log10 x

0.8

0.5

0.4

0.1

0.2

0.3

log 10 x

log 10 y

0.9

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

0.39



10


Full Mark

8. (a)

(b)

(c)

Solving simultaneous equation P(– 2, 2) Q(4, 8)

Use dxyy )( 12

dxxx )2

4(2

Integrate dxyy )( 12

64

2

32 xxx

18

Note : If use area of trapezium and ydx , give the marks

accordingly.

Integrate dxx 2

2)

2(

=

20

5x

58

3

4

3

10

K1

N1 N1

N1

K1

K1

K1

Use correct

limit

4

2into

64

2

32 xxx

N1

K1

K1

Use correct

limit 4

0

into

20

5x



11


Full Mark

9(a)

(b)

(c)

(d)

Equation of AD : y – 6 = –2 ( x – 2 ) y = –2x + 10 or equivalent

y = –2x + 10 and

x – 2y = 0 D(4, 2)

p = 3 or C(8, 4)

Substitute (8, 4) into y = 3x + q

q = – 20

OADOABCArea 4 Using formula

00

62

24

00

21OAD

40 Alternative solution : B(10, 10) Using formula

00

62

1010

48

00

21OABCArea

2

2

3

3

10

K1

N1

Use m = – 2 and find equation of straight line

Solving simultaneous equations

N1

K1

K1

P1

N1

N1

Find area of triangle

K1

K1

N1

Find area of parallelogram

K1

P1



12

40


Full Mark

10(a)

(b)

(c )

106sin 1 or equivalent

2POQ

radPOQ 287.1 Alternative solution : 122 = 102 + 102 – 2(10)(10) cos POQ

)10)(10(2121010cos

2221POQ

radPOQ 287.1

Using (2π – 1.287) Major arc PQ = 10 ( 2π – 1.287 ) = 49.96 cm

Lsector = 287.1)10(21 2

Ltriangle = 287.1sin)10(

21 2

= 16.35 cm2

3

3

4

10

Using formula Lsector = 2

1 r2 K1

K1

K1 Lsector - LΔ

N1

Using formula LΔ = 2

1 absin C

Use formula s = r

K1

N1

K1

N1

K1 Use cosine rule

N1

N1

K1

K1



13

= 3.9149 cm


Full Mark

11 (a)(i)

(ii) b)(i) (ii)

p = 53 , p + q = 1

P( X = 0 )

= 5C0(53 )0(

52 )5

= 0.01024 Using P( X ≥ 4 ) = P( X = 4 ) + P( X = 5 )

= 5C4(53 )4(

52 )1 + (

53 )5

= 0.337 P ( 30 ≤ X ≤ 60 )

= P ( 10

3530 ≤ Z ≤ 60 3510 )

Use P( – 0.5 ≤ Z ≤ 2.5 ) = 1 – P( Z 0.5 ) – P( Z 2.5 ) = 1－ 0.30854 – 0.00621 = 0.68525 Number of pupils = P( X 60 ) × 483

3

2

3

K1

N1

K1

N1

Use P(X = r) = n Cr prqn–r , p + q = 1

K1

N1

K1 use

Z =

X

K1

N1

P1



14

3

2

10


Full mark

12. (a) (b) (c) (d)

Subst. t = 0 into dvdt

a= 15 – 6t 15 ms-2

1 175 3/184 4

ms ms

Integrate

Use s = 0

152

/ 17 / 7.52

S4 – S3

2 2 3

K1

N1

K1

N1

K1

Use 0dvdt

and subst. t in v = 15t – 3t2

[t = 52

]

K1

K1

N1

2 3152

s v dt t t

Subst. t = 3 or t = 4 in 2 315

2s t t

K1

N1



15

1152

Note : If use 4

3vdt , give the marks accordingly.

3

10


Full mark

13. (a)

(b) (i)

(ii)

(iii)

Use I = 2007

2005100

PP

Value of m : 25, m, 80, 30 or equivalent 120 25+130m+135 80+139 30

Use i i

i

I WI

W

132.1 = 120×25+130m+135×80+139 30135+m

m =65

100150.00 x132.1

RM 113.55 /I . ( . x . )08 05 132 1 132 1 0 3

3 3 2

N2, 1, 0

K1

x = 48.6 y = 135 z = 80

K1

K1

N1

K1

N1

P1

N1



16

171.73

2

10


Full mark

14. (a) (b) (c)

x + y ≤ 80 or equivalent y ≤ 4x or equivalent

x + 4y ≥ 120 or equivalent

x=30

(16,64)

x + 4y = 120

x + y = 80

y = 4x100

90

80

70

60

50

40

30

20

10

10080604020 9070503010x

y

At least one straight line is drawn correctly from inequalities involving x and y

All the three straight lines are drawn correctly

Region is correctly shaded

(i) minimum = 23

3 3

N1P1 N1

N1

K1

N1

N1

N1

N1



17

(ii) (16,64) Subst. point in the range

in 20x + 40y

RM2880

4

10


Full mark

15. (a) (b) (c)

(d)

120 9.3 6 sin2

BCD

45o48’ / 45.8o

0.74545 4555 Use cosine rule in ΔBCD BD2 = 9.32 + 62 − 2×9.3×6 cos 45°48’ 6.685 Use sine rule in ΔBCD

osin CBD sin '

. .

45 48

9 3 6 685

94o 10’ 4444 qqq11111aaaaaaaaaaaaaaaaaa 14s4 5.555555 5555 z5555555555555555 Sum of area:

4 2

K1

N1

Use area △= ½ ab sin c in △BCD K1

N1

K1

N1

K1

N1

K1

K1

N1

K1

Use area ADB = ½ 6.685 13 sin ADB

Obtain ADB by using 180o – 85o50’ – BAD or equivalent



18

20 cm2 + ΔABD 555 102 58.82 cm2

4

10

marking scheme kedah spm 2008 add maths trial p2

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