Marking Scheme Kedah Spm 2008 Add Maths Trial P2

2008 SPM TRIAL EXAMINATION Question

2

Marking Scheme Submark

Solution and marking scheme

1.

3 y 2 y  2x  3

x

Full Mark

or P1

Make x or y as the subject

2

 3 y   3 y  2     y  y  10  2   2  or

2

x 2  x  2 x  3   2 x  3  10 K1

Eliminate x or y

9  6 y  y 2  6 y  2 y 2  4 y 2  40  0 or

x 2  2 x 2  3 x  4 x 2  12 x  9  10  0

K1

y = 3.215 , − 3.215 or

Solve quadratic equation N1

x = 3.107 , − 0.107

3 y 2  31  0 31 y2  3 or 2

3x  9 x  1  0 x = 3.107 / 3.108 , − 0.107 / − 0.108 N1

or y = 3.214 / 3.215 , − 3.214 / − 3.215

Answer must correct to 3 decimal places.

3472/2

Additional Mathematics Paper 2

or using formula or completing the square

5

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2008 SPM TRIAL EXAMINATION Question

3

Marking Scheme Submark

Solution and marking scheme

Full Mark

2(a)

16 ,8 , 4 ,...... a  16

r

1 2

P1

  1 7  16 1       2   S7  1 1 2 3 4

K1

Sn 

Use

a( 1  r n ) 1 r

= 31 

N1

3

or 31.75 

(b)

64 ,16 , 4 ,...... a  64

r

1 4

P1

64 1 1 4 1 = 85  3

S 

K1

Use

S 

a 1 r

N1

or 85.33 

3(a)

3

f ( x)  x 2  px  q 2

2

 p  p = x 2  px        q 2 2 2

3472/2

6

K1

p p2  = x   q 2 4 

use x² + bx = ( x + b )² – ( b ) 2



p 3 2

use axis of symmetry 



36  q  5 4

p = －6

2

2

or N1

b 3 2a

q=4 N1

Additional Mathematics Paper 2

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2008 SPM TRIAL EXAMINATION Question

4

Marking Scheme Submark

Solution and marking scheme

Full Mark

Alternative solution 2

x 2  px  q   x  3  5 = x2  6x  9  5 = x2  6x  4

p  6

K1

Comparing coefficient of x or constant term

N1

N1

q=4

3

3(b)

x 2  6 x  4  31  0 x 2  6 x  27  0 K1

( x  9)( x  3)  0 3  x  9

Use f ( x)  31  0 and factorization

N1

2

5

4(a)

cos x 1  sin x sin x cos x = + cos x 1  sin x sin x(1  sin x )  cos 2 x = cos x(1  sin x )

tan x +

sin x  sin 2 x  cos 2 x cos x(1  sin x) sin x  1 = cos x(1  sin x) 1 = cos x = sec x N1

Use tan x 

K1

sin x cos x

=

3472/2

K1

Use identity

sin 2 x  cos2 x  1

Additional Mathematics Paper 2

3

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2008 SPM TRIAL EXAMINATION Question

5

Marking Scheme Submark

Solution and marking scheme

Full Mark

4(b)(i)

y

y

5x 2 

3

y  3sin 2 x

O

 2



3 2

x

2

–3

P1

Negative sine shape correct. Amplitude = 3

[ Maximum = 3 and Minimum = − 3 ]

Two full cycle in 0  x  2

P1

P1 3

4(b)(ii)

3sin 2 x 

5x 2 

or

N1

5x y 2  Draw the straight line

y

Number of solutions is 3 .

5x 2 

K1

N1 3

3472/2

Additional Mathematics Paper 2

9

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2008 SPM TRIAL EXAMINATION Question

6

Marking Scheme

Solution and marking scheme

Submark

Full Mark

7

7

5 (a)

K1

 x  12

 x  240

20

Use x 

x N

N1

The mean X

240  5 + 8 + 10 + 11 + 14 288  25 25

N1

= 11.52

(b)

 x2 20

K1

 122  3

x

2

 3060

N1 Use formula



x N

2

 x2

The standard deviation



3060  52  82  102  112  14 2 2  11.52  25

=

3566 2  11.52  25

=

9.9296

= 3.151

3472/2

K1 For the new x 2 and X



N1

Additional Mathematics Paper 2

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2008 SPM TRIAL EXAMINATION Question

7

Marking Scheme Submark

Solution and marking scheme

6(a)

(i)

uuur uuur uuur PR  PO  OR = 6a  15b % %

Full Mark

K1 N1

Use

uuur uuur uuur PR  PO  OR

uuur uuur uuur (ii) OQ  OP  PQ

uuur oruuur uuur OQ  OP  PQ

3 uuur = 6a  OR % 5 N1

= 6a  9b % %

3

(b)

uuur uuur (i) OS  hOQ

= h(6a  9b) % % uuur uuur uuur (ii) OS  OP  PS

N1

uuur = 6a  k PR %

= 6a  k  6a  15b  % % %

N1

h(6a  9b) = 6a  k  6a  15b  % % % % % 6h  6  6k

9h  15k

h  1 k

5 h k 3

K1

5 k  1 k 3 k

3 8

N1

53 5 h   = 38 8

3472/2

Equate coefficient of a or b %and % Eliminate h or k

N1

Additional Mathematics Paper 2

5

8

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2008 SPM TRIAL EXAMINATION

Qn. 7(a)

(ii)

3472/2

Marking Scheme

Submark

Solution and Marking Scheme log y 

log x log y

(b)(i)

8

1 1 log x  log 4k n n

P1

0.18

0.30

0.40

0.60

0.74

0.48

0.54

0.59

0.69

0.76

gradient 1 = n

n=2

N1

N1 N1 N1 N1

K1

Correct axes and scale

N1

All points plotted correctly

N1

Line of best-fit

K1

K1

Full Mark

N1

6

intercept 1 = log 4k n

k = 1.51

Additional Mathematics Paper 2

4

10

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2008 SPM TRIAL EXAMINATION

9

Marking Scheme

log 10 y Graph of log10 y against log10 x 0.9

0.8  0.7



0.6

 

0.5



0.4

39 0.3

0.2

0.1

0 0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

log 10 x 3472/2

Additional Mathematics Paper 2

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2008 SPM TRIAL EXAMINATION

Qn. 8. (a)

10

Submark

Solution and Marking Scheme

Use  ( y 2  y1 ) dx

Full Mark

K1

Solving simultaneous equation

P(– 2, 2)

(b)

Marking Scheme

N1

Q(4, 8)

N1

3

K1

x2  ( x  4  2 ) dx Use correct

K1

4

Integrate  ( y 2  y1 ) dx

K1

limit  into 2

x2 x3  4x  2 6

2

x x3  4x  2 6

18

4

N1

Note : If use area of trapezium and



ydx , give the marks

accordingly.

(c)

Integrate   (

x2 2 ) dx 2

K1

4

K1

5

x  =    20 

limit



into

0

 x5     20  8  5

3472/2

Use correct

N1

Additional Mathematics Paper 2

3

10

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2008 SPM TRIAL EXAMINATION

Qn.

11

Marking Scheme

Submark

Solution and Marking Scheme

Full Mark

9(a) Equation of AD :

Use m = – 2 and find K1 equation of straight line

y – 6 = –2 ( x – 2 )

N1

y = –2x + 10 or equivalent

2 (b)

y = –2x + 10 and x – 2y = 0

K1

D(4, 2)

(c)

N1

2

P1

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

K1

q = – 20

(d)

Solving simultaneous equations

N1

Area OABC  4   OAD

3

K1

Using formula 0 4 2 0 1  OAD   2 0 2 6 0

K1

40

Find area of triangle

N1

3 10

Alternative solution :

B(10, 10)

P1

Using formula Area OABC  3472/2

0 8 10 2 0 1  0 4 10 6 0 2

K1

Find area of parallelogram

Additional Mathematics N1 Paper 2

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2008 SPM TRIAL EXAMINATION

12

Marking Scheme

40 Qn. 10(a)

Submark

Solution and Marking Scheme

6   or equivalent  10  K1  POQ  2  

  sin

1 

Full Mark

K1

 POQ 1.287 rad

N1

3

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

 10 2 10 2 12 2 POQ  cos 1  2(10)(10) 

   

Using (2π – 1.287)

N1

K1 K1

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

(c )

Use formula s = r

N1

1 2

3

Lsector =  (10 ) 2  1.287

Ltriangle =

Use cosine rule

N1

 POQ 1.287 rad

(b)

K1

K1

1  (10 ) 2  sin 1.287 2

Using formula Lsector = 12 r2 K1 Using1 formula LΔ = 2 absin C K1

Lsector - LΔ

2

= 16.35 cm 3472/2

N1

Additional Mathematics Paper 2

4

10 [Lihat sebelah SULIT

2008 SPM TRIAL EXAMINATION

13

Marking Scheme

= 3.9149 cm

Qn. 11 (a)(i)

p=

3 ,p+q=1 5

Use P(X = r) = n Cr prqn–r , p+q=1

3 2 = 5C0( )0( )5 5 5

K1

= 0.01024

b)(i)

Full Mark

P1

P( X = 0 )

(ii)

Submark

Solution and Marking Scheme

3

N1

Using P( X ≥ 4 ) = P( X = 4 ) + P( X = 5 ) 3 2 3 = 5C4( )4( )1 + ( )5 5 5 5

K1

= 0.337

N1

2

P ( 30 ≤ X ≤ 60 ) 30  35 60  35 =P( ≤Z≤ ) 10 10

use

K1

Z=

X  

Use P( – 0.5 ≤ Z ≤ 2.5 ) = 1 – P( Z  0.5 ) – P( Z  2.5 )

K1

= 1－ 0.30854 – 0.00621 = 0.68525 (ii) 3472/2

N1

Number of pupils = P( X  60 ) × 483

3

K1

Additional Mathematics Paper 2 N1

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2008 SPM TRIAL EXAMINATION

14

Marking Scheme

2

3

Qn.

Submark

Solution and Marking Scheme

12.

Subst. t = 0 into

(a)

dv dt

10

K1

Full mark

2

a= 15 – 6t N1

15 ms-2

(b) Use

dv  0 and subst. t dt

[t =

in v = 15t – 3t2

K1

5 ] 2

2 75 1 3 ms /18 ms 1 4 4

Integrate s   v dt 

(c) Use s = 0

N1

15 2 3 t t 2

K1

15 1 / 7 / 7.5 2 2

(d)

Subst. t = 3 or t = 4 in 15 s  t 2  t3 2

N1

3

K1

K1

3472/2

K1

S4 – S3

Additional Mathematics Paper 2 N1

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2008 SPM TRIAL EXAMINATION 15

Note : If use

15

Marking Scheme

1 2

4

3 vdt , give the marks accordingly.

3 10

Qn. 13. (a)

Solution and Marking Scheme

Use I =

P2007  100 P2005

x = 48.6 y = 135 z = 80

3

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

P1

I W Use Iˆ  i i

 Wi 120×25+130m+135×80+139  30 132.1 = 135+m

150.00 x

100 132.1

K1

3

N1

(ii)

Full mark

K1

N2, 1, 0

(b) (i)

Submark

m =65

K1

2 N1

(iii)

3472/2

RM 113.55

I 08 / 05  132.1  (132.1x0.3)

K1

Additional Mathematics Paper 2 N1

[Lihat sebelah SULIT

2008 SPM TRIAL EXAMINATION

16

Marking Scheme

171.73

2 10

Qn.

Solution and Marking Scheme

14. (a) (b)

x + y ≤ 80

or equivalent

y ≤ 4x

or equivalent

N1 P1 N1

x + 4y ≥ 120

or equivalent

N1

Submark

Full mark

3

y 100

y = 4x 90

80

x + y = 80 x=30

70

(16,64) 60

50

40

30

20

x + 4y = 12 0 10

x 10

20

30

40

50

60

70

80

90

100

At least one straight line is drawn correctly from inequalities involving x and y K1

All the three straight lines are drawn correctly

Region is correctly shaded (c)

3472/2

(i)

minimum = 23 N1

N1

N1

3

N1

Additional Mathematics Paper 2

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2008 SPM TRIAL EXAMINATION (ii)

17

Marking Scheme

(16,64) Subst. point in the range

K1

in 20x + 40y RM2880 Qn. 15. (a)

N1

Solution and Marking Scheme Use area △= ½ ab sin c in △BCD 1 20   9.3  6  sin BCD 2

4

10

Submark

Full mark

K1

45o48’ / 45.8o N1

0.74545

4555

N1

(c)

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

K1

(b)

K1

Use sine rule in ΔBCD sin CBD sin 45 o 48 '  9.3 6.685 N1

(d)

4

6.685

94o 10’

2

4444 Obtain  ADB by using K1 qqq11111aaaaaaaaaaaaaaaaaa 180o – 85o50’ – 14s4 BAD or equivalent 5.555555 5555

z5555555555555555 K1 Use area  ADB =

½  6.685  13  sin ADB

Sum of area: 3472/2

K1 Additional Mathematics Paper 2

[Lihat sebelah SULIT

2008 SPM TRIAL EXAMINATION

18

20 cm2 + ΔABD 555

102

58.82 cm2

3472/2

Marking Scheme

Additional Mathematics Paper 2

4

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10