marking scheme add math zone c kuching 2008

8/14/2019 Marking Scheme Add Math Zone C Kuching 2008

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Marking Scheme of Additional Mathematic Zone C Common Paper

Paper 1

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

(b) Many to many

2

2. 8 1

8 2

m

18 (8)

2m

4m 3

3.

33

)(1ax

xf

Either3

2=b ora = 2 is correct.

Both3

2=b and a = 2 are correct

3

4. 3 + k= 1

k= 4

3k= p

p = 124

5. 2 4 3 0 + >x x( 1)( 3) 0 >x x

1, 3< >x x3

6. (a) 4x =

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

f(x) = (x 4) 2 4

3

7. 6log (2 1)6 6

log 6 log 11t

6 6log 2 1 log 11t

2 1 11t

5t

4

1


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8.2 34(3 )(3 ) 7(3 )

3

nn n +

1(36 7 )(3 )

3

n +

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

9. 1(2 n 1) = 512 and n = 10

10

10

1(2 1)1023

2 1s

= =

2

10.2 0.42 0.0042 0.000042 ...

h

k= + + + + or 0.42, 0.01a r= =

0.42

2 1 0.01

h

k= + or2.4

1 0.01=

h

k80

, 80, 3333

hh k

k= = =

3

11. (a) 1 22(1) 3 5 2(2) 3 7T or T = + = = + =

2 17 5 2d T T= = =

(b)10 2

S S =10 2

(2(5) (10 1)(2)) (2(5) (2 1)(2))2 2

+ + 128

*Alternative: use 3 9T = as first term and find 8S4

12. 1

2m =

1 12

2y x=

1 1 4

2

x

y x

=

2

1 4

xy

x=

4

13. 2PR =3RQ

2222 )5()3(3)]1([)2(2 +=+ yxyx5x2 38x + 5y2 98y + 286 = 0

3

14.(a)

6

2

(b) 3 5i j

2

2


3/15

15. 35 12r a b

35h

23k 3

16. (a) SQ SP PQ 5 4SQ x y

(b)1

4TR TQ QR SQ PS uur uuur uuur uuur uuur

1 5(5 4 ) 4

4 4TR x y y x y uur

% % 4

17. 3(2sin cos ) 5cos 0 x x x

cos (6sin 5) 0x x

5cos 0 sin6

x or x

90 ,270 ,236.44 ,303.56x 4

18. Length of arc AB = 12 cm or 6 6 6( ) 24

2 radian

2x 180 = 114 35'

3

19. 3'( ) 20( 3) f x x 3'(2) 20(2 3) 20f

2

20. 1

2

xp

,

2(1 )

1 32

xy

,

(1 )3

2

dy x

dx

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

3 182

dy

dx

-118(4) 72 units s

dy

dt

4

21. =

==

3 32

432

44)34(

k k

xx

dxx

[2(3)2 3(3)] [2k2 3k] = 4

(2k 5)(x + 1) = 0, k =2

5( k > 0)

3

3


4/15

22. 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)

4

= 0.3696

4

4


5/15

Paper 2

1. 8 2y x= or equivalent(Expressx ory as the subject)

2 22 (8 2 ) (8 2 ) 18 0 x x x x+ =(use substitution method to eliminatex ory)

24 20 23 0 + =x x

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

2(4)x

=

(use formula correctly to solve quadratic equation)

1.793 3.207 x or =

4.414 1.586 y or =

2. (a)

n

41652= (use the formula,

xx

n

)

n = 8

49 = 22

)52(8

y

(use the formula,2

22 x x

n

)

2 28 [ 49 (52) ] 22024y

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

New standard deviation = 21)49(3

3. (a) 21

3(1) 5(1) 8a S

2

2 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 )

1062T

(c) 23 5 1062n n

18n

5


6/15

4. (a)

: The shape of cos.

: The amplitude is 3: One period for the range 0 x 2: Modulus of the graph.

(b): find the equation 1

2

xy

.

: the straight line of 12

xy

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 solutions

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

5. (a) SQ SP PQ or QR QT TR or PU PQ QU (use triangle law)

4 6SQ a b

4 6QR a b

1 14 ( 6 )

2 2 PU PQ QT or PU a b uuur uuur uuur uuur

% %

4 3 PU a b

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

(Find vector PR or UR)

Show 2 PR PU or PU UR

6


7/15

6. 21

(6 4) ; 3 4dy dy

x dx x x cdx dx

2

111 3(2) 4(2) c

2

115; 3 4 15

dyc x x

dx

2 3 2

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

xxxy 15223

3x2 4x 15 = 0 ;3

5x orx = 3 (use

dx

dy= 0)

y = (3

5 )3 2(

3

5 )2 15(

3

5 ) =

27

400

2

2

56( ) 4 14

3

d y

dx ; Maximum value of y =

27

400

y = (3)3 2(3)2 15(3) = 362

26(3) 4 14

d y

dx ; Minimum value of y = 36

7. (a) xy 4.3 8.8 16.5 27.2 41.0 57.62x 1 4 9 16 25 36

(Refer graph on page 8)

correct axes and scale

All point plotted correctly

Line of best fit

(b)

a

baxxy 2 (express equation in linear form)

a =936

5.166.57

(Equate a to the gradient)

522.1a

2.78b

a (Equate

b

ato thexy-intercept)

186.4b

4.9y

7


8/15

8


9/15

8. (a) (i) 31 2 1 3 5 ( 1)2, ,

2 2 2 2

pp

(Midpoint of AC = Midpoint of BM)

52

p

(ii) 5( 1)

72

1 3 8ADM

72 ( 2)

8y x

8 7 30 0y x

(b)

52( ) 1( )

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

yx

5

2( ) 1( )2( 1) 1( ) 21 5

1 2 1 2

yx

solve or

5, 10P

(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

9


10/15

9. (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

5cos0.6rad

radius (use cos

a

h )

(alternative: use sin oh

and follow by Pythagoras theorem)

5

cos0.6radius

rad , radius = 6.058 cm

(c)

2110 1.2

2or

216.058 2.4

2

(use21

2A 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


11/15

10. (a) 23 2 1dy

x xdx

Gradient of tangent23(2) 2(2) 1 9

Gradient of normal 19

15 ( 2) ; 9 47

9 y x x y

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

3

1 yory

(solve simultaneous equations, not necessary getting the correct

answers)

A = ( 8, 3)

(ii) 3 3

2 4 2

0 1

64know volume is equal to minus 2 1

9 y dy y y dy

327

643

0

y

+ y

yy

3

2

5

353

1

(do integration correctly)

++

)13

2

5

1()318

5

243(0)27(

27

64

(do substitution correctly)

= 30 15

14unit 3

11


12/15

11. (a) (i) 7 1 6 7 0 71 0

.25 .75 .25 .75C or C

(usen r n r

rC 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

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

n = 2000

(b) (i) 53 61 72 61

11 11or

(use

xz

correctly)

1 0.23352* 0.15866* or equivalent

0.60782

(ii) n(0.60782) = 200

n = 329

12


13/15

12. (a) AB2 = 7.92 + 62 27.96 cos 130 (Use cosine rule in ABC)

AB = 12.62 cm

sin sin 50

7.9 6.4

ADC= (Use sine rule in ADC)

sin 0.9456ADC

108.99 ( is obtuse) ADC ADC

180 50 108.99 21.01DAC

6.4

sin21.01 sin50

DC

(Use sine rule in ADC)

DC = 3 cm

1(6.4)(3 6)sin108.99

2 or equivalent

(use area of triangle = Cabsin21

)

= 27.23 cm2

13. (a)1

120 1004.5

P

x (Use1

0 100

Q

I Q )

RM 5.40

(b) 130 x 110 = Ix 100 or equivalent

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

143

(c) (i) i iI W = 120 3 135(2 ) 130 6 150( ) y y (Find i iI W correctly and usex = 2y)

120 3 135(2 ) 130 6 150( )

3 2 6

y yI

y y(divide i iI W by iW correctly)

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

3 2 6

y y

y y(Equate I to 132)

x = 4, y = 2

(ii)

05

33100 132X

Q

(Use 1

0

100Q

I

Q

)

RM25

13


14/15

14. (a)4 12

dya t

dx (find a using differentiation

4 12 0t (use 0dy

dx when v maximum)

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

(b)

32 222 12 6

3

t s t t dt t c

s = 0 when t= 0, thus c = 0.3

22 63

ts t (finds using integration)

3 32 22(3) 2(2)6(3) 6(2)

3 3A B s or s (substitute t= 2 ort= 3 iss)

117

3A Bs s

(c)Solve

322 6 0

3

tt (uses = 0)

9t s(d) Solve 22 12 0t t (use v = 0)

6t s

15. (a) x + y 80 or equivalent

y 3x or equivalent

y 30 or equivalent

(b)

(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

Region is correctly shaded

(c) (i) 45

(ii) Optimum point (20,60)

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

RM4500

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marking scheme add math zone c kuching 2008

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