Jawapan modul 3 k1 mt jpnk 2015

MODUL 3 [Kertas 1]: MATEMATIK TAMBAHAN JPNK 2015 Muka Surat: 1

JAWAPAN MODUL 3 KERTAS 1

No. Soalan Penyelesaian dan Skima Pemarkahan Sub

Marks Total

Marks

1 (a)

(b)

{2,4}

( )3

xf x =

1 1

2

2 (a)

(b)

4x-5

3

1 1

2

3

2

3 , 4 ( )

2 : 3 4

1: ( 2)

a b both

B a or b

B a x b

= == =

− −

3

4

2

2

1m , 5 ( )

31

B3: m 53

22 : 3 5 2 0 2 2

3 3

1: 3( 2) ( 2) 2 ( 2) and ( )( 2)

n both

or n

nB m m or m and m

B n or HTP m TDP m

= =

= =

+ − = − = − − = −

− + − − = + − = −

4

5

2

1 1( )

3 22 : (2 1)(3 1) 0

B1: 6 x 1 0

f x x

B x x or

x

= − ≤ ≤ −

− + ≤

− − ≤

3

6 (a)

(b)

4

[ ] [ ]

15 5

400

15 52 : 2(2) (15 1)(4) 2(2) (5 1)(4)

2 215 5

B1: [2(2) (15 1)(4)] [2(2) (5 1)(4)]2 2

B

S S or or

+ − − + −

− + − + −

1 3

4

7 9, 3 ; 18, 12 ( )

B3: 9, 3 18, 12

B2 : 27 x y 2127

1: 27 x y 2127

x y x y both

x y or x y

x yand

xx y

B orx

= = = == = = =

+ + = =

+ + = =

4

1/2 -1/3


8 (a)

(b)

0.003,k 0.0003 ( )h both= =

12

0.031: 0.1 atau 0.08

1

p

B rr

=

= +−

1 2

3

9

B2: or equivalent

B1: ( changing to base m)

3

10

2

25

1: 5 2 @ (2 )x xB =

2

11

8

3

3

2 : 2 2

B1: (2 )(2 )

x

x

B =

3

12 B1: 2y

hx kx

= +

B2: 2 25 (1) atau 1 (4)h k h k− = + = + h = 2 or k = -7

B3: h = 2 atau k = -7

B1: h = 2 dan k = -7 (kedua-duanya)

3

13 2 2

2 2 2 2

2 2 2 2

4 4 11 3 2 0

2 : ( ( 2)) (y 3) 3 ( 1) (y 0)

1: ( ( 2)) (y 3) ( 1) (y 0)

x y x y

B x x

B x or x

+ − + − =

− − + − = − + −

− − + − − + −

3


14 (a)

(b)

2

2

4

2

41: 2

2 2

k k

k kB

−

−

2

1

1: 1 2sin2

k

kB θ

=

− =

2 2

4

15

2

33 41', 56 19', 213 41',236 19' (all)

3 : 33 41' , 56 19 '

2 : (2 tan 3)(3tan 2) 0

1: 6(1 tan ) 13tan 0

B

B x x

B x x

° ° ° °° °

− − =+ − =

3

16 2 , 4 ( )

2 : 1, 2 atau ( 2)( 4) 0

1: 3 6 atau 4 (2 )

k k both

B k k

B k kλ λ

= = −− − + == = +

3

17 (a)

(b)

3 5i j+

34

1 1

2

18

2 2

2 2

6.60

1 12 : (7) (OP) (1.125)

2 21 1

1: (7) (OP) (1.125)2 2

B

B or

π

π

=

3

19 (a)

(b)

0.0159

11: 0.1, (Perimeter 2 )

2

drB use r

dtπ

π= × =

*

4.854

1: 5( 0.0159), (30 2 r)newB r r use π= + =

2 2

4


20

49 atau 24.5

2

3: 7B x =

2 : 7 0B x− =

11: ( )(14 )

2B L x x= −

4 2

4

21 (a)

(b)

3t +

3, 4 ( )

1: 1 2 1 3

t t both

B t or t

= =− = − =

1 2

3

22 (a)

(b)

233

1: 18 15 3 11

cm

B or+ ×

1 2

3

23 (a)

(b)

120

B1: 5! or

72

B1: 4! 2! or 48(seen)

2 2

4

24 (a)

(b)

8 8 08

0.05764

1: (0.7) (0.3)B C

8 0 8 8 1 7 8 0 80 1 0

8 1 71

0.9987

1: 1 (p) (q) (p) (q) (0.7) (0.3)

(0.7) (0.3)

B C C or C or

C

− +

2 2

4

x

2

0

X (3,11)

y


25 r = 14.59

B2: = 1.53

B1: 1-0.937 or or 1.53 (seen)

3

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