Sarawak ZonC SPM 2008 AM P1P2 Answer · - 2 - 4(3 )(3 ) 7(3 ) 2. 3 3. n n n 1 (36 7 )(3 ) 3 n. 8. 88(3 ) n 1. Since 88 is a multiple of 8 and (3. n – 1) is an integer for all positive

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Marking Scheme of Additional Mathematic Zone C Common Paper

Paper 1

(a) {(1, a), (5, a), (5, u), (9, i)}

1.

(b) Many to many 2

8 18 2

m

18 (8)2

m

2.

4m 3

33)(1 axxf

Either 32

b or a = −2 is correct.

3.

Both 32

b and a = −2 are correct

3

3 + k = 1

k = 4

3k = p

4.

p = 12 4

2 4 3 0 x x ( 1)( 3) 0 x x

5.

1, 3 x x 3 (a) 4x

(b) Either minimum value = 4 or (x – 4)2 is correct

6.

f(x) = (x − 4) 2 − 4 3

6log (2 1)6 6log 6 log 11t

6 6log 2 1 log 11t

2 1 11t

7.

5t 4

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2 34(3 )(3 ) 7(3 )3

nn n

1(36 7 )(3 )3

n

8.

188(3 )n . Since 88 is a multiple of 8 and (3n – 1) is an integer for all positive integer of n(n = 1, 2, 3, ….) , then 4(3n+2) – 7(3n) + 3n – 1 is divisible by 8.

3 1(2 n 1) = 512 and n = 10

9.

10

101(2 1) 1023

2 1s

2

2 0.42 0.0042 0.000042 ...hk or 0.42, 0.01a r

0.4221 0.01

hk

or 2.4

1 0.01

hk

10.

80 , 80, 3333

h h kk

3

1 22(1) 3 5 2(2) 3 7T or T

(a)

2 1 7 5 2d T T

10 2S S = 10 2(2(5) (10 1)(2)) (2(5) (2 1)(2))2 2

11.

(b)

128 *Alternative: use 3 9T as first term and find 8S

4 12

m

1 1 22y x

1 1 42

xy x

12.

21 4

xyx

4 2PR =3RQ

2222 )5()3(3)]1([)2(2 yxyx

13.

5x2 – 38x + 5y2 – 98y + 286 = 0 3

(a) 62

14.

(b) 3 5i j% %

2

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35 12r a b % % %

35h

15.

23k 3

(a) SQ SP PQ uuur uur uuur

5 4SQ x y

uuur

% %

(b) 14

TR TQ QR SQ PS uur uuur uuur uuur uuur

16.

1 5(5 4 ) 44 4

TR x y y x y uur

% % % 4

3(2sin cos ) 5cos 0x x x

cos (6sin 5) 0x x

5cos 0 sin6

x or x

17.

90 ,270 , 236.44 ,303.56x 4 Length of arc AB = 12 cm or 6 6 6( ) 24

2 radian

18.

2 x 180 = 114 35'

3

3'( ) 20( 3)f x x 19.

3'(2) 20(2 3) 20f 2

12

xp ,

2(1 )1 32

xy

,

(1 )32

dy xdx

When p =6, 1 2(6) 11x , (1 ( 11))3 182

dydx

20.

-118(4) 72 units sdydt

4

3 32

432

44)34(k k

xxdxx

[2(3)2 – 3(3)] – [2k2 – 3k] = 4

21.

(2k – 5)(x + 1) = 0, k = 25 ( k > 0)

3

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Either (2 + x) < (12 + 8 + 7 + 3) or (2 + x + 12) > (8 + 7 + 3)

Both (2 + x) < (12 + 8 + 7 + 3) and (2 + x + 12) > (8 + 7 + 3)

22.

4 < x < 28 3 (a) 8 15

3 4C C = 76440

(b) 3 4

3 4P P

23.

=144 4 4 3 3 2 5 4 or or

12 11 12 11 12 11 (any one correct)

114

125

112

123

113

124

(all correct with the summation)

24.

6619

3

(a) 62 6515

z

z = 0.2

(b) 70 65( )15

P z

25.

= 0.3696 4

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Paper 2 1. 8 2y x or equivalent

(Express x or y as the subject)

2 22 (8 2 ) (8 2 ) 18 0x x x x (use substitution method to eliminate x or y)

24 20 23 0 x x

( 20) ( 20) 4(4)(23)

2(4)x

(use formula correctly to solve quadratic equation)

1.793 3.207x or

4.414 1.586y or

(a) n

41652 (use the formula, xxn

)

n = 8

49 = 22

)52(8

y (use the formula,

2 22 x xn

)

2 28[ 49 (52) ] 22024y

2.

(b) New mean = 3(52) + 5 = 161

New standard deviation = 21)49(3

(a) 2

1 3(1) 5(1) 8a S

22 2 1 [3(2) 5(2)] 8 14T S S (find 2T )

14 8 6d

(b) 10 8 9(6)T (use ( 1)nT a n d )

10 62T

3.

(c) 23 5 1062n n 18n

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: The shape of cos.

: The amplitude is 3

: One period for the range 0 x 2

(a)

: Modulus of the graph.

: find the equation 12xy

.

: the straight line of 12xy

in the graph.

(give K1 if the equation is wrong, straight line is drawn correctly according to the wrong equation, but no N1 for answer, even the answer is 4 solutions)

4.

(b)

: 4 solutions (cannot give N1 if equation is wrong or the straight is drawn wrongly)

(a) SQ SP PQ or QR QT TR or PU PQ QU uuur uur uuur uuur uuur uur uuur uuur uuur

(use triangle law)

4 6SQ a b

uuur

% %

4 6QR a b

uuur

% %

1 14 ( 6 )2 2

PU PQ QT or PU a b uuur uuur uuur uuur

% %

4 3PU a b

uuur

% %

5.

(b) 8 6 4 3PR a b or UR a b uuur uuur

% % % %

(Find vector PR or URuuur uuur

)

Show 2PR PU or PU UR uuur uuur uuur uuur

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21(6 4) ; 3 4dy dyx dx x x c

dx dx

2

111 3(2) 4(2) c

21 15; 3 4 15dyc x x

dx

2 3 2

2(3 4 15) ; 2 15y x x dx y x x x c

xxxy 152 23

6.

3x2 – 4x – 15 = 0 ;

35

x or x = 3 (usedxdy = 0)

y = (35

)3 – 2(35

)2 – 15(35

) = 27400

2

2

56( ) 4 143

d ydx

; Maximum value of y = 27400

y = (3)3 – 2(3)2 – 15(3) = – 36

2

2 6(3) 4 14d ydx

; Minimum value of y = 36

xy 4.3 8.8 16.5 27.2 41.0 57.6

2x 1 4 9 16 25 36

correct axes and scale

All point plotted correctly

(a)

(Refer graph on page 8) Line of best fit

7.

(b) abaxxy 2 (express equation in linear form)

a = 936

5.166.57 (Equate a to the gradient)

522.1a

2.78ba (Equate b

a to the xy-intercept)

186.4b

4.9y

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(i) 3

1 2 1 3 5 ( 1)2, ,2 2 2 2

pp

(Midpoint of AC = Midpoint of BM)

52

p

(a)

(ii) 5 ( 1) 721 3 8ADM

72 ( 2)8

y x

8 7 30 0y x

8.

(b)

52( ) 1( )2( 1) 1( ) 21,5 ,

1 2 1 2

yx

52( ) 1( )2( 1) 1( ) 21 51 2 1 2

yxsolve or

5, 10P

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(c) 1 5 3 5 3(1)( ) ( 1)( 1) (3)( ) (5)(5) ( 1)(5) (3)( ) ( 1)(5) (1)( )2 2 2 2 2

= 17 unit 2

(a) 10 POQ = 12 (use s r )

POQ = 1.2 radians

(b) Identify the right angle triangle,

know OPQ = 0.6 radian and radius OP is the hypotenuse

5cos 0.6radradius

(use cos ah

)

(alternative: use sin oh

and follow by Pythagoras theorem)

5

cos 0.6radius

rad , radius = 6.058 cm

9.

(c) 21 10 1.22

or 21 6.058 2.42

(use 212

A r to find area of sector PQTS or OQRS)

21 10 sin1.22

rad or 21 6.058 sin 2.42

rad

(use Cabsin21 to find the area of tringle PQS or ORS)

60 – 46.6 or 44.04 – 12.39 (find the area of segment QTS or QRS)

= 31.65 – 13.4 (use area of segment QRS – area of segment QTS)

= 18.25 cm2

10. (a) 23 2 1dy x x

dx

Gradient of tangent 23(2) 2(2) 1 9

Gradient of normal 19

15 ( 2) ; 9 479

y x x y

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(i) )1(38 2 yy 0 = )3)(13( yy 3

31

yory

(solve simultaneous equations, not necessary getting the correct answers)

A = ( 8, 3)

(b)

(ii) 3 32 4 2

0 1

64know volume is equal to minus 2 19

y dy y y dy

3

2764

3

0

y

yyy

32

5

353

1

(do integration correctly)

)1

32

51()318

5243(0)27(

2764

(do substitution correctly)

= 30 1514 unit 3

(i) 7 1 6 7 0 71 0.25 .75 .25 .75C or C

(use n r n rrC p q correctly)

7 1 6 7 0 7

1 01 .25 .75 .25 .75C C (knowing ( 2) 1 ( 1) ( 0)p x P x P x )

= 0.5551

(a)

(ii) n(0.25)(0.75) = 375

n = 2000

(i) 53 61 72 61

11 11or (use xz

correctly)

1 – 0.23352* 0.15866* or equivalent

0.60782

11.

(b)

(ii) n(0.60782) = 200

n = 329

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(a) AB2 = 7.92 + 62 − 2×7.9×6 cos 130° (Use cosine rule in ΔABC)

AB = 12.62 cm

sin sin 507.9 6.4

ADC (Use sine rule in ΔADC)

sin 0.9456ADC

108.99 ( is obtuse)ADC ADC

180 50 108.99 21.01DAC

6.4sin 21.01 sin 50

DC

(Use sine rule in ΔADC)

DC = 3 cm

12.

1 (6.4)(3 6)sin108.992

or equivalent

(use area of triangle = Cabsin21 )

= 27.23 cm2

(a) 1120 1004.5P x (Use 1

0100QI

Q )

RM 5.40

(b) 130 x 110 = I x 100 or equivalent

(Use 07/ 06 06/ 04 07/ 04 100I I I )

143

(i) i iI W = 120 3 135(2 ) 130 6 150( ) y y

(Find i iI W correctly and use x = 2y)

120 3 135(2 ) 130 6 150( ) 3 2 6

y yIy y

(divide i iI W by iW correctly)

120 3 135(2 ) 130 6 150( ) 132

3 2 6

y yy y

(Equate I to 132)

x = 4, y = 2

13.

(c)

(ii)

05

33 100 132XQ

(Use 1

0100QI

Q )

RM25

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(a) 4 12dya t

dx (find a using differentiation

4 12 0t (use 0dydx

when v maximum)

2 -1max 2(3) 12(3) 18 msv

(b) 3

2 222 12 63ts t t dt t c

s = 0 when t = 0, thus c = 0. 3

22 63ts t (find s using integration)

3 3

2 22(3) 2(2)6(3) 6(2)3 3A Bs or s (substitute t = 2 or t = 3 is s)

1173A Bs s

(c) Solve

322 6 0

3t t (use s = 0)

9t s

14.

(d) Solve 22 12 0t t (use v = 0)

6t s

x + y ≤ 80 or equivalent

y ≤ 3x or equivalent

(a)

y ≥ 30 or equivalent

(Refer graph on page 13)

At least one straight line is drawn correctly

(from inequalities involving x and y)

All the three straight lines are drawn Correctly

(b)

Region is correctly shaded (i) 45

15.

(c)

(ii) Optimum point (20,60)

Substitute (20,60) in 45x + 60y

RM4500

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