jawapan matematik kertas 2 set 1 pahang

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1449/2(GMP)
Mathematics
Kertas 2 ( Set 1 )
Peraturan
Pemarkahan
2017

1449/2(GMP)

SKEMA PRAKTIS BESTARI
PROJEK JAWAB UNTUK JAYA (JUJ) 2017

MATHEMATICS
Kertas 2
SET 1
PERATURAN PEMARKAHAN

UNTUK KEGUNAAN GURU MATA PELAJARAN SAHAJA

1449/2(GMP) ©

Peraturan pemarkahan ini mengandungi 13 halaman bercetak
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2

Question

1449/2(GMP)

Solution and Mark Scheme

1(a)

Marks
L

K

M

P1

(b)

K
L
M
P2

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3

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3

Question
2

Solution and Mark Scheme
(3x + 2)2 = ½ (4x + 8)4x
2

1449/2(GMP)
Marks
K1

x － 4x + 4 = 0

K1

(x – 2)(x-2)

KI

Perimeter = 32

N1

3

1449/2(GMP) ©

p + 2q = - 9 ------------ (1)
(1) × 5
5p + 10q = - 45 ------------ (2)
5p - q = - 1 ------------ (3)
(2) - (3),
11q = -44
q=-4
5p + 4 = - 1
5p = - 5

p = -1
∴ p = -1, q = - 4

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4

K1

K1

N1N1

4

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4

Question
4

5

Solution and Mark Scheme
(a) ∠SVR
(b)
9
tan ∠SVR =
11
9
∠SVR = tan−1
11
= 39.29° or 39o 19’

m

4
3
4
4
8   (1)  C or 0   (5)  C
3
3
20
C=
3
4
20
y   x
3
3
DF = 6 unit or DE = 8 unit or EF = 10 unit
Coordinate of point G is (5, 10)

1449/2(GMP)

Marks
P1
K1

N1

3

P1
K1

K1
K1
N1

5

6
= 6 × (10 × 3 × 2)
=

��

=

= 0.52

7

× .

K1

K1

= 40 0.52
= 20.8
= 360 ��
20.8
= 17.31
17 necklaces

K1
N1

4

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5

Question

Solution and Mark Scheme
(a) i. True

ii. False
(b) Implication 1 : If 15x > 45 then x > 3
Implication 2 : If x > 3 then 15 x > 45
(c) ABCD is a trapezium

7

8.

m = 5 , k = -9

9.

−

−

−

= −

240
22
1
22
(2) ( ) (14)or (2) ( ) (7)
360
7
2
7
14 

P1
P1
P1
P1
P1

5

P1

=

=

Marks

P1,P1

=

=

1449/2(GMP)

240
22
1
22
(2) ( ) (14)  (2) ( ) (7)
360
7
2
7

−

K1

N1N1

6

K1

K1
N1

= 94.67
240 22
1 22
( ) (14)(14) or ( ) (7)(7)
360 7
2 7

K1

240 22
1 22
( ) (14)(14)  ( ) (7)(7)
360 7
2 7

K1

= 333.67

N1

1449/2(GMP) ©

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6

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6

1449/2(GMP)

Question

Solution and Mark Scheme

Marks

10

(a) (E,2) (E,0) (E,1) (E,7) (X,2) (X,0) (X,1) (X,7) (A,2) (A,0) (A,1)
(A,7) (M,2) (M,0) (M,1)(M,7)

P2

(b) (i) (E,2) (A,2)

K1

N1

(ii) (E,1) (E,7) (X,1) (X,7) (A,1) (A,7) (M,2) (M,0) (M,1) (M,7)

K1

6

N1

11

(a) 8 – 5 = 3

P1

(b) 50  0

K1

5 0

= 10
(c) 50  50  10
k

k = 10

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N1
K1
N1

5

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7

Question

1449/2(GMP)

Solution and Mark Scheme

12
(a)

Marks

x

-3.5

0.5

y

23.25

- 8.75

P1P1

(b)
Graph
Axes are drawn in the correct direction, uniform scale for
-5  x  3.

P1

7 points and 2 points* plotted accurately

K2

Smooth and continuous curve without straight line(s) and

(c)

passes through all the 9 correct points for -5  x  3.

N1

(i)

6 < y