2

No Solution and Mark Scheme Sub

Marks

Total

Marks

1

24 3 3 2 3y xy x x y

3 2x y

3

2

xy

2

2

4 0

4 3 2 3 2 0

y xy x

y y y y

2

2

4 0

3 34 0

2 2

y xy x

x xx x

2 1 3 0y y 2 9 0x x

1@ 3

2

2@9

y

x

2@9

1@ 3

2

x

y

1

1

1

1

1

5

2(a)

2 2

4 2 3 2

81 27

3 3

8 3 6

6

5

x x

x x

x x

x

1

1

1

7

3

(b)

2 2

2 2

2

4

2

32

3

log 16log 1

log 2

log 216

216

64

4

xx

x

x

x

x

x

1

1

1

1

3(a)

2

3

2220500 1.05

200000

T ar

a

a

2

7

(b)

(ii)

1

1

400000

200000 1.05 400000

1.05 2

1 log1.05 log 2

1 14.21

15.21

16

n

n

n

T

n

n

n

n

2008 415

2009 419.5

2010 423.34

2011 427.57

2012 431.85

2013 436.17

2014 440.53

2993.62

7

7

415 1.01

415 1.01 1

1.01 1

2993.62

a r or

S

5

4

4(a)

(b)

2

sin 2sin cos

1 cos 2cos 1

sin 1 2cos

cos 1 2cos

tan

x x x

x x

x x

x x

x

Proven / Terbukti

Bentuk tan x (di graf) P1

Period (di graf) P1

Straight Line (di graf) P1

1x

y

P1

No Of Solution = 2 N1

2

5

7

2

5

5(a)

(b)

2

3

3

3

y x x

y x x

2

2

3 3 0

1 0

1

1, 2

1, 2

dyx

dx

x

x

x y

x y

2

26

1, 6 min

1, 6 max

d yx

dx

dyx imum

dx

dyx imum

dx

Maximum point is 1,2

0.02x

22, 3 2 3 9dy

xdx

9 0.02

0.18

y

y

4

3

7

6(a)

Find the coordinates of B(6,0) or coordinates C( 0,4)

Find midpoint of BC 3,2

Gradient of BC = 2

3 .

Gradient of the line perpendicular to BC = 2

3

2m

The equation of perpendicular bisector to BC is :

3

2 32

3 5

2 2

2 3 5

y x

y x

or

y x

7

7 N1

N1

K1

N1

6

(b)

Area of triangle ABC

2

14 0 6 4 0 2 2 6 0 4 4

2

124 28

2

26

area

unit

No Solution and Mark Scheme Sub

Marks

Total

Marks

7(a)

(b)

(c)

x

1

1.5

2

2.5

3

3.5

log10 y

1.50

1.22

0.93

0.66

0.34

0.10

Graf = Rujuk Lampiran

Plot 10log y against x and correct axes and uniform scales.

6 points plotted correctly N1

Line of best fit N1

* If table not shown but all points are correctly plotted , award N1 mark.

10 10 10log 3log logy k x p

(Can be implied)

(i) 103log 0.56k

0.65k

K1

K1

P1

N1

K1

N1

N1

K1

7

(ii) 10 log 2.04p

109.65p

(iii) 10 1.38

23.99

loy y

y

8(a)

(b)

10tan

6

59.04

1.031

2 1.031 6

31.518

CD

CD

11.66 1.031

12.02

AB

AB

2 210 6

11.66

OB

OB

11.66 6

5.66

BD AC

Perimeter of shaded region

31.518 5.66 5.66 12.02

54.858

2

5

10

(c )

Area of sector

2111.66 1.031

2

70.09

Or

3

K1

N1

N1

8

Area of right angle triangle

16 10

2

30

Area of shaded region

70.09 30

40.09

9(a)

(i)

(ii)

(b)

Use the triangle law for or PS QR

9

9

OP PS OS

y PS x

PS x y

5 3

3 5

OQ QR OR

x QR y

QR y x

=m 9 ST y x

=n 5 3 RT x y

3

7

10

K1

N1

N1

N1

N1

9

Use the triangle law to find TQ :

4 9 5 3 5 3

4 9 5 3 5 3

4 5 5 9 3 3

1 5 9 1 5 3 3

11 5 9 45 3 3

7

242 6

7

1

7

ST TQ SQ RT TQ RQ

TQ x m y x TQ x y n x y

x m y x x y n x y

compare

m n m n

m n n n

m n n

m n

n

10

(a)

2dy

xdx

2(1) 2Tm

Equation of tangent

2 2 1

2

y x

y x

3

10

(b) Area of triangle Area under the curve

1

1 22

1

1

2

0

13

0

1

3

11 0

3

4

3

x dx

xx

4

N1

K1

K1

N1

K1

10

Area of shaded region = 24 11

3 3unit

(c)

2

1

22

1

2 2

2

1

2

(2) (1)2 1

2 2

1

2

y dy

yy

unit

3

11

(a)

(i)

(ii)

(b)

(i)

8, 0.8, 0.2n p q

8 0 8 8 1 7 8 2 6

0 1 2

2 0 1 2

0.8 0.2 0.8 0.2 0.8 0.2

0.7969

P X P X P X P X

C C C

250 0.8 0.2

6.325

npq

110, 4

100 120

100 110 120 110

4 4

2.5 2.5

0.9876

P X

P Z

P Z

5

5

10

11

No Solution and Mark Scheme Sub

Marks

Total

Marks

12

(a)

(b)

2

2

32

32

2 18 28

4 18

2 18 28

29 28

3

0, 0, 0

29 28

3

v t t

dva t

dt

s vdt t t dt

ts t t c

s t c

ts t t

2

, 0

18

at A t

a ms

2

1

2 18 28

, 0

28

initial

v t t

v t

v ms

2

2

10

(ii)

100

100 110

4

2.5

0.00621

P X

P Z

P Z

0.00621480

2.98

n AP A

n s

n A

n A

Bilangan helaian yang ditolak ialah 3

12

(c)

2

2

2 18 28 0

9 14 0

2 7 0

2 7

v t t

t t

t t

t

3

(d)

distance s

3 2

3 2

2

2

29 28

3

22 9 2 28 2

3

125

3

t

t

s t t t

s

s

dt

3 2

7

2

27 9 7 28 7

3

116

3

t

t

s

s

The distance is

2 72

1 12 25 16

3 3

67

t ts s s

s

s

3

72

13

13(a)

10

10

116 100250

290

Q

Q

2

10

(b)

10/ 08

10/ 08

115

130 100

149.5

I

I

2006 2008 2010

2006 100 115 10/ 08I

2008 - 100 130

2010 - - 100

2

(c)

I w Iw

110 13 1430

116 25 2900

130 40 5200

x 10 103x

120 12 1440

100 10970 + 103 x

10970 10120

100

103

xI

x

4

14

(d)

08

08

1599120 100

1332.5

Q

Q

2

14a

(i)

2 2 210.5 12.5 2 10.5 12.5 cos80

14.86

QS

QS

2 10

(ii)

0

0

sin sin 35

14.86 9.5

sin 0.8972

63.79

116.21

R

R

QRS

QRS

2

b(i)

1

RQ'

Q

P

15

(ii)

*

2

sin sin80

12.5 14.86

sin 0.8284

55.94

'

1(10.5)(10.5)sin 68.12

2

51.15

Find PQS

Q

Q

PQS

Area of PQQ

cm

0110.5 12.5 sin 80

2

64.63

'

64.63 51.15

13.48

Find area of PQS

Area of PQ S

5

15(a)

(b)

(c)

1.8 2 240x y

3 2 600x y

50y x

Rujuk Lampiran

(i) Jambu batu sebanyak 66 pokok

(ii) Maximum point = (100 ,150)

Maximum total fees = 120 ( 100 ) + 180 (150 )

= 39000

16

Soalan 7(a)

2.0

1.8

1.6

1 0.5 1.5 2 2.5 3

0.2

3.5

0.4

0.6

0.8

1.0

1.2

1.4

X

X

X

X

X

X

log10 y

20

17

Soalan 15(b)

40 20 60 80 100 120 x

20

140

60

80

100

120

140

y

160

160

180

R

(100, 150)

0

40

180