solaf 1 - skema pemarkahan math

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SULIT 1449/1&2 1449/1&2 Matematik Kertas 1&2 Peraturan Pemarkahan April 2011 JABATAN PELAJARAN PERAK PROGRAM SOLAF SIJIL PELAJARAN MALAYSIA 2011 MATEMATIK SOLAF 1 Jumlah Markah =

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Page 1: Solaf 1 - Skema Pemarkahan Math

SULIT 1449/1&21449/1&2MatematikKertas 1&2PeraturanPemarkahanApril 2011

JABATAN PELAJARAN PERAK

PROGRAM SOLAF

SIJIL PELAJARAN MALAYSIA 2011

MATEMATIK

SOLAF 1

Peraturan pemarkahan ini mengandungi 11 halaman.

[Lihat sebelah

1449/1 SULIT

Jumlah Markah =

Page 2: Solaf 1 - Skema Pemarkahan Math

SKEMA PEMARKAHAN SOLAF SET 1

MATHEMATICS PAPER 1

1449/1 SULIT

Question Answer

1 D

2 C

3 C

4 A

5 C

6 B

7 D

8 A

9 B

10 D

11 B

12 B

13 A

14 C

15 A

16 C

17 A

18 D

19 A

20 C

Question Answer

21 B

22 D

23 B

24 B

25 A

26 D

27 C

28 D

29 A

30 B

31 A

32 D

33 A

34 A

35 C

36 D

37 D

38 B

39 B

40 C

2

Page 3: Solaf 1 - Skema Pemarkahan Math

1449/2 SULIT

SKEMA PEMARKAHAN SOLAF SET 1

MATHEMATICS PAPER 2

Section A[ 52 marks ]

No Marking Scheme Marks

1

Straight dashed line x=9 drawn correctly.

Region shaded correctly.

1

2 3

2. 3 x2+10 x−8=0(3 x−2)( x+4 )=0

x=23

,−4

1

1

1, 1 4

3. 3 m−n=18−3 n=9n=−3m=5

1

1

1

14

3

Page 4: Solaf 1 - Skema Pemarkahan Math

No Marking Scheme Marks

4. (a) ÐEBA

(b) tan ÐEBA =

125

ÐEBA = 67.38 or 67o23’

1

1

1 3

5.(a) False

(b) Converse : If the square of the number is positive, then the number is a negative number. ( False )

(c) PQRS is a square.

1

1 , 1

1 4

6.

(a)

m=7−32−6

m=−1

(b) mAZB = mXY = −1

−2 = (−1)(−3)+c OR y−(−2)=−1 [ x−(−3 ) ]

y=− x−5

(c) x−intercept ; y=0

− x −5 = 0

x = − 5

1

1

1

1

1

1

6

7.

(a)(2 −13 −4 )

−1

= 1(2)(−4 )−(3)(−1 )(−4 1

−3 2 )

=

1−5 (−4 1

−3 2 )=

1−d ( e 1

−3 2 )

d = 5 ; e = −4

1 , 1

4

Page 5: Solaf 1 - Skema Pemarkahan Math

No Marking Scheme Marks

(b)(2 −13 −4 )(x

y )=(119 )

(xy)= 1

−5 (−4 1−3 2 )(11

9 )

(x

y)= (73 )

x = 7

y = 3

1

1

1

1 6

8.

227

×7×7×15

13×22

7×7×7×6

227

×7×7×15 −

13×22

7×7×7×6

2002

1

1

1

1 4

9.

(a)(i) P( J )=4000

16500

=

833

(ii) P(US )=4500

16500

=

311

(b) Number of tourist of Japan =

833

×33000

1

1

1

1

1

1 6

5

Page 6: Solaf 1 - Skema Pemarkahan Math

No Marking Scheme Marks

= 8 000

10.

(a)

12×14×14

OR

90360

×227

×21×21

90360

×227

×21×21 −

12×14×14

248.5

(b)

60360

×2×227

×14 OR

90360

×2×227

×21

14 +

60360

×2×227

×14 + 7 +

90360

×2×227

×21 + 21

89.67

1

1

1

1

1

1

6

11.(a) Gradient of RS = - 2

Equation of PQ,

5=−2 (3 )+c

y = - 2x + 11

(b) y – intercept = 11

(c) 0=−2x+11

x-intercept =

112

1

1

1

1

1

1 6

6

Page 7: Solaf 1 - Skema Pemarkahan Math

No Marking Scheme Marks

Section B [ 48 marks ]

12.(a) y = 13 , 8

(b)

(c) (i) 5 .3≤ y≤5 .7

(ii) x = −2.8, 0, 2.8

(d) Identify equation of y = −2x + 4

Straight line y = −2x + 4 correctly drawn

1, 1

1

1

1

1

x

y

-3 -2 -1 1 2 3

-4

-2

0

2

4

6

8

10

12

14

7

Page 8: Solaf 1 - Skema Pemarkahan Math

No Marking Scheme Marks

x = 2.4, 0.2, −2.5

Note : Awarded 1 mark for any two correct values.2 12

13.(a) (i) (-3,5)

(ii) (1,5), (-3,3)

(b) (i) M is an enlargement with scale factor 2 at the centre (3,0)

(ii) N is a rotation of 180⁰ about the centre (2,4)

(c) (i)

12

(ii) Area of JKLM = ( 12

)2 x 112

= 28 m2

1

1 , 1

3

3

1

1

1 12

14. (a)Class Interval Midpoint Frequency

15 - 19 17 3

20 - 24 22 5

25 - 29 27 7

30 - 34 32 9

35 - 39 37 7

40 - 44 42 6

45 - 49 47 3

(b) Mean =

3(17 )+5 (22)+7(27 )+9 (32 )+7(37 )+6( 42)+3( 47)3+5+7+9+7+6+3

= 32.25

3

2

1

8

Page 9: Solaf 1 - Skema Pemarkahan Math

No Marking Scheme Marks

(c)

524742373227221712

9

8

7

6

5

4

3

2

1

0

Payment (RM)

frequency

(d) Number of drivers who paid more than RM34 = 16

5

1

12

15. (a) 47

(b)

Mass (kg) Frequency MidpointUpper

boundary

20 – 29 0 24.5 29.530 – 39 12 34.5 39.5

1

1

9

Page 10: Solaf 1 - Skema Pemarkahan Math

No Marking Scheme Marks

40 – 49 15 44.5 49.550 – 59 9 54.5 59.560 – 69 8 64.5 69.570 – 79 6 74.5 79.5

(c) (i) 40 – 49

(ii) Mean =12(34 .5 )+15(44 . 5 )+9(54 . 5)+8 (64 .5 )+6(74 .5 )50

= 50.7

(d)

Note: Axes drawn in the correct directions with uniform scales. Upper boundaries/ midpoints/ class intervals correctly used for x-axis

All rectangular with equal width drawn correctly.

1

1

1

2

1

1

1

2 12

16. (a) y = 6 , –8

(b) Axes drawn in the correct directions with uniform scales.

1, 1

1

10

Page 11: Solaf 1 - Skema Pemarkahan Math

No Marking Scheme Marks

All 8 points and *2 points correctly plotted or curve passes through

all the points. ( 8 or 9 points correctly plotted award 1 mark).

2 smooth and continuous curve without any straight line passes

through all 10 correct points using the given scales.

(c) (i) y = –3.4

(ii) x = –1.2

(d) Identify equation of y=−3 x+1

Straight line y=−3 x+1 correctly drawn

x = −1.8, 2.2

1

1, 1

1

1

1

1

1, 1

12

11

Page 12: Solaf 1 - Skema Pemarkahan Math

Disediakan oleh:Panel Penggubal SOLAF1 JPNTahun 2011

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