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### Transcript of Skema Matematik k1 & k2 Mibf Spm 2012 Set1

MAJLIS PENGETUA SEKOLAH MENENGAH CAWANGAN KLANG

PERATURAN PERMARKAHAN MODUL ITEM BERFOKUS (1) DAERAH KLANG SIJIL PELAJARAN MALAYSIA 2012

______________________________________________________ MATHEMATICS 1449 KERTAS 1 dan KERTAS 2 ______________________________________________________

Jumlah Markah =

Kertas 1 + Kertas 2 100% 140

Skema jawapan ini mengandungi 14 halaman bercetak.

2

MODUL ITEM BERFOKUS (1) DAERAH KLANG SIJIL PELAJARAN MALAYSIA 2012 MATHEMATICS 1449 SIFIR MARKAH1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= 1% 1% 2% 3% 4% 4% 5% 6% 6% 7% 8% 9% 9% 10% 11% 11% 12% 13% 14% 14% 15% 16% 16% 17% 18% 19% 19% 20% 21% 21% 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= 22% 23% 24% 24% 25% 26% 26% 27% 28% 29% 29% 30% 31% 31% 32% 33% 34% 34% 35% 36% 36% 37% 38% 39% 39% 40% 41% 41% 42% 43% 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= 44% 44% 45% 46% 46% 47% 48% 49% 49% 50% 51% 51% 52% 53% 54% 54% 55% 56% 56% 57% 58% 59% 59% 60% 61% 61% 62% 63% 64% 64% 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= 65% 66% 66% 67% 68% 69% 69% 70% 71% 71% 72% 73% 74% 74% 75% 76% 76% 77% 78% 79% 79% 80% 81% 81% 82% 83% 84% 84% 85% 86% 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= .= 86% 87% 88% 89% 89% 90% 91% 91% 92% 93% 94% 94% 95% 96% 96% 97% 98% 99% 99% 100%

MATHEMATICS (1449/1) PAPER 1 No Soalan 1 2 3 4 5 6 7 8 9 10 Jawapan D B C D B A B B C A No Soalan 11 12 13 14 15 16 17 18 19 20 Jawapan A A D D C C C A B A No Soalan 21 22 23 24 25 26 27 28 29 30 Jawapan C A A D C A B D C A No Soalan 31 32 33 34 35 36 37 38 39 40 Jawapan D C C B B C B D D BSulit

1449 @ Hak Cipta MPSM Cawangan Klang

3 MODUL ITEM BERFOKUS (1) PPD KLANG 2012 MATHEMATICS PAPER 2 (1449/2) Section A [ 52 marks ] No 1 (a) (i) P Marking Scheme R Q 1 mark MarksSubmarks Total Marks

(ii)

R P Q

2 marks

(b) Draw the straight dotted line x = 2 Region shaded correctly y 6

1 mark 2 marks 6

4 x+ y =2 2

y=x+1

4 2

2

0

2

4

x 1 mark 1 mark 2 marks 1 mark 1 mark 2 marks 4

3

16m + 4n = 12 or n = 4m or equivalent 3 2 9m = 18 or equivalent m = 2 dan n = 5 4 6x2 x 5 = 0(6 x + 5)( x 1) = 0

4

5 x = ,x =1 6

4

(a)

E

F1449 @ Hak Cipta MPSM Cawangan Klang Sulit

4 No H Marking Scheme MarksSubmarks Total Marks

GE

D

4 cm

A 3 cm B Mark angle HBE (b)tan HBE = 8 5

C

8 cm

1 mark

1 mark 1 mark 1 mark 1 mark 1 mark 1 mark 1 mark 1 mark 1 mark 1 mark or y = 8 or c = 8 or V(0, 8) 1 mark 1 mark 5 5

HBE = tan 1

8 5 = 58 or 57.99O

3

5

(a) (i) Non-statement or not a statement Statement (b) If 8 x = 2 then x = If x = then 3 x = 2 83

(c) All rectangles have four sides. 6 (a) 3(0) + 4x = 24 x=6 (b) mPV = mOU = 3y 4(0) = 24 4 y = x+8 3 4 3

7

14 14 8 1 22 7 2 14 2 7

1 mark 1 mark

1449 @ Hak Cipta MPSM Cawangan Klang

Sulit

5 No Marking Scheme MarksSubmarks Total Marks

1 22 7 2 14 14 14 8 + 2 72646

1 mark

4

1 mark

8

(a)

4 2 1 3( 4 ) ( 2)( 7 ) 7 3 1 4 2 = 2 7 3 x 4 2 1 1 = y 3( 4 ) ( 2 )( 7 ) 1 3 8 or equivalent x =6 y = 17 2

1 mark 1 mark

(b)

1 mark 1 mark 1 mark 5

Note : 1. Accept only matrix method.

x 6 2. = 1 7 y 29 (a)

award 1 mark

45 22 2 27 360 7 21 3 14 or 297 14 or 21.21 1 99 2 1 22 2 9 4 7

1 mark 1 mark

(b)

45 22 27 2 360 7

or

or

1 mark 1 mark 1 mark 5

45 22 1 22 1 27 2 92 9 9 360 7 4 7 2 1 729 182 or or 182.25 4 4

10

(a) (i)

145 560

or

29 112

1 mark

195 + 220 29 145 or (ii) 1 or 1 560 112 560 1449 @ Hak Cipta MPSM Cawangan Klang

1 markSulit

6 No 415 560 Marking Scheme 83 112 MarksSubmarks Total Marks

or

1 mark

(b) (i) {1, 2, 3, 4, 7, 10, 12, 15} (ii) { 2, 4, 10, 12} 4 1 = 8 2

1 mark 1 mark 1 mark 6

11

(a) (b)

15 minutes

1 mark 1 mark 1 mark

15 200.75 km min1

or

equivalent 1 mark 1 mark 5

75 (c) =1.5 tt = 50

1449 @ Hak Cipta MPSM Cawangan Klang

Sulit

7 Section B [ 48 marks ] No 12(a) x y (b) Marking Scheme 1.5 13.5 3 18 MarksSubmarks Total Marks

1 mark 1 mark 1 mark

Graph Axes are drawn in the correct direction, uniform scale for x 3. 2.5 7 points and 2 points* plotted accurately Smooth and continuous curve without straight line(s) and passes through all the 9 correct points for x 3. 2.5

2 marks 1 mark

Notes : (1) 7 or 8 points plotted correctly, award 1 mark (2) Other scale being used, subtract 1 mark (i) (c) (ii) 3.45 < x < 3.55 4.25 < y < 4.75 1 mark 1 mark

Note : Do not accept the values of x and y obtained by calculation. (d)y = 4 x 2 + 5 x + 3

0 = 4 x 2 2 x 13 y = 3 x 10

or subtitute 4 x 2 = 2 x + 13 y = 4 x 2 + 5 x + 3 y = (2 x + 13) + 5 x + 3 y = 3 x 10

Identify the equation y = 3x 10 or equivalent The straight line y = 3x 10 drawn correctly The values of x : 2 < x < 2.1 < x< 2.6 2.5 Notes : (1) Award mark(s) if the value(s) of x shown on the graph. (2) Do not accept the value(s) of x obtained by calculation.

1 mark 1 mark 12

1 mark 1 mark

1449 @ Hak Cipta MPSM Cawangan Klang

Sulit

8 x y 2 23 1.5 13.5 1 6 0 3 1 4 1.5 1.5 2 3 3 18 4 41

Graph for question 12 Graf untuk Soalan 12 y 5

2

1

0 5

1

2

3

4

x

10

15

20

x

25

30

35

40

74.5 64.5 54.5

45

1449 @ Hak Cipta MPSM Cawangan Klang

Sulit

9

No 13(a) x y (b)

Marking Scheme 1.5 14.7 4 5.5

MarksSubmarks Total Marks

1 mark 1 mark 1 mark

Graph Axes are drawn in the correct direction, uniform scale for 0.5 x 7. 7 points and 2 points* plotted accurately Smooth and continuous curve without straight line(s) and passes through all the 9 correct points for 0.5 x 7. Notes : (1) 7 or 8 points plotted correctly, award 1 mark (2) Other scale being used, subtract 1 mark (i) 2.25 < x < 2.35 4.5 < y < 5

2 marks 1 mark

1 mark 1 mark

(c)

(ii)

Note : Do not accept the values of x and y obtained by calculation. (d)divide

22 = 4 x 2 + 21x

22 4 x 2 21 x = + x x x

therefore

y = 4 x + 21

Identify the equation y = 4 x + 21 or equivalent The straight line y = 4 x + 21 drawn correctly The values of x : 2.40 < x < 2.50 3.8 < x < 3.9 Notes : (1) Award mark(s) if the value(s) of x shown on the graph. (2) Do not accept the value(s) of x obtained by calculation.

1 mark 1 mark 12 1 mark 1 mark

1449 @ Hak Cipta MPSM Cawangan Klang

Sulit

10

x y

0.5 44

1 22

1.5 14.7

2 11

3 7.3

4 5.5

5 4.4

6 3.7

7 3.1

Graph for question 13 Graf untuk Soalan 13

y 45

40

35

30

25

x

20

15

10

5

74.5 64.5 54.5

0

1

2

3

4

5

6

7

x

1449 @ Hak Cipta MPSM Cawangan Klang

Sulit

11

No 14 (a)

Marking Scheme

MarksSubmarks Total Marks

(i) ( 3) 3, (ii) (4, 3) (i) (a) U = a reflection in the line y = 2 ( or straight line EF or KF or KEF)

2 marks 2 marks 2 marks

(b)

Note : (1) reflection only, award 1 marks. (b) V = Enlargement, centre F, scale factor of 2 (note: accept centre ( 2) 2, get 2 marks get 2 marks get 1 mark 1 mark 1 mark 1 mark 12 3 marks

Note : (1) Enlargement, centre F or ( 2), 2, (2) Enlargement, scale factor 2, (3) Enlargement only, (ii) Area of FKLMJN = 22 x 35 Area of shaded region = (22 x 35) 35 = 105

15 (a)

(i) ( 3, 2) (ii) (0, 2) (i) (a) U = a 180 rotation, at centre E or (2,0) Note : (1) 180 rotation only, get 2 marks. (2) rotation at centre E or (2,0) only, get 2 marks. (3) rotation only, get 1 mark. (b) V = Enlargement, centre (0, 0) or origin, scale factor 3 Note : (1) Enlargement, centre (0, 0) or origin get 2 marks (2) Enlargement, scale factor 3, get 2 marks (3) Enlargement only get 1 mark (ii) Area of shaded region = 32 x 25 = 225

2 marks 2 marks 3 marks

(b)

3 marks

1 mark 1 mark 12

1449 @ Hak Cipta MPSM Cawangan Klang

Sulit

12 MarksSubmarks Total Marks

No 14 (a)

Marking Scheme

(i) ( 3) 3, (ii) (4, 3) (i) (a) U = a reflection in the line y = 2 ( or straight line EF or KF or KEF)

2 marks 2 marks 2 marks

(b)

Note : (1) reflection only, award 1 marks. (b) V = Enlargement, centre F, scale factor of 2 (note: accept centre ( 2) 2, get 2 marks get 2 marks get 1 mark 1 mark 1