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SULIT 3472/2 SKEMA MATEMATIK TAMBAHAN PEPERIKSAAAN PERCUBAAN SPM JPNS 2008 BAHAGIAN A Soalan Penyelesaian dan Peraturan Markah Markah Sub Markah penuh 1 2 4 m n or m = 4(2 – n) P1 2 12 42 4 m m or (8 – 4n) 2 – 12 = 4n K1 4 5 m m or (4n – 13)(n – 1) = 0 K1 4, 5 m or 13 1, 4 n N1 13 1, 4 n or 4, 5 m N1 5 5 2(a) 2(b) m =3 P1 hx 2 –9=3 K1 h(1) – 9 = 3 K1 h = 12 N1 dx x ) 9 12 ( 2 K1 c x x y 9 3 12 3 N1 c ) 1 ( 9 3 ) 1 ( 12 2 3 K1 y = 4x 3 – 9x + 7 N1 4 4 8 http://mathsmozac.blogspot.com sah@mozac2008

# skema SPM Trial 2008 - Trial Paper Collection · PDF file · 2008-11-06SKEMA MATEMATIK TAMBAHAN PEPERIKSAAAN PERCUBAAN SPM JPNS 2008 ... Straight line graph P1 3 x y ... s P – s

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SULIT 3472/2

SKEMA MATEMATIK TAMBAHANPEPERIKSAAAN PERCUBAAN SPM JPNS 2008

BAHAGIAN A

Soalan Penyelesaian dan Peraturan Markah MarkahSub

Markahpenuh

1 24

mn or m = 4(2 – n) P1

2 12 4 24

mm

or (8 – 4n)2 – 12 = 4n K1

4 5m m or (4n – 13)(n – 1) = 0 K1

4, 5m or13

1,4

n N1

131,

4n or 4, 5m N1 5 5

2(a)

2(b)

m = 3 P1

hx2 – 9 = 3 K1

h(1) – 9 = 3 K1

h = 12 N1

dxx )912( 2 K1

cxx

y 93

12 3

N1

c )1(93

)1(122

3

K1

y = 4x3 – 9x + 7 N1

4

4 8

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sah@mozac2008

SULIT 3472/2

Soalan Penyelesaian dan Peraturan Markah MarkahSub

MarkahPenuh

3(a)

(b)(i)

(ii)

3x, x2

3, x

4

3P1

1

2r P1

6

72

13

xT K1

3

64

xN1

2

11

3

xS K1

= 6x N1

4

2 6

4(a)

4(b)

Score Cumulativefrequency

1 1y 2

10 414 64x 1028 12

53 + y + 4xFor C.F. P1

x – 5 = 7 K1

x = 12 N1

1(1) + m(2) + 10(4) + 14(6) + 4x(10) + 28(12) P1

912

101

yK1

y = 7 N1

4

2 6

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sah@mozac2008

SULIT 3472/2

Soalan Penyelesaian dan Peraturan Markah MarkahSub

MarkahPenuh

5 Shape of y = cos x P1

Amplitude 4 P1

N

6(a)

6(b)

6(c)

Q

m

sah@mozac2008

2 periods P1

4

Modulus P1

Straight line graph P1

3x

y

K1

umber of solutions = 4 N1

5

2 7

6

2

5

3R K1

1

1N1

1

1

45

2h

mK1

2 + m = – (–1) K1

m = – 1 N1

222 )1()1(416 K1

m2 + 16 = 16(2) K1

4m N1

2

3

3 8

O π 2π

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SULIT 3472/2

BAHAGIAN B

Soalan Penyelesaian dan Peraturan Markah MarkahSub

MarkahPenuh

7(a)

7(b)

N1

* If table is not given, see from the points plotted in the graph

Plotx

yagainst x (correct axes and uniform scale) K1

All points plotted correctly N1

The line of best fit (go through at least 3 points) N1

(i) abaxx

y P1

Find gradient K1

a = gradient =05

138.21

K1

a = 4.076 (accept 4.08) N1

(ii) ab = intercept of the axisx

y= 1 K1

b = 0.245 N1

x 1 2 3 4 5

x

y5.00 9.20 13.17 17.30 21.38

4

6 10

8(a)x

dx

dy2 K1

2

12 m K1

12

13 xy K1

2

5

2

1 xy or equivalent N1 4

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sah@mozac2008

SULIT 3472/2

Soalan Penyelesaian dan Peraturan Markah MarkahSub

MarkahPenuh

8(b)

8(c)

Area = dyy 4

3

2

1

)4( or 1

0

2 )4( x or 3 K1

=

4

3

2

3

2

31

4

yor 3

34

1

0

3

xx K1

=3

2unit2 N1

Volume = 4

3

)4( dyy K1

=

2

3)3(4()

2

4)4(4(

22

K1

= 2

1unit3 N1

3

3 10

9(a)

9(b)

9(c)

9(d)

Midpoint AC = ( 2, –1 ) K1)2(11 xy K1

1 xy N1

5

4ABm K1

5

12

5

4 xy or equavalent N1

124)1(5 xx or 12)1(45 yy K1

Find the value x or y or 7x or 8y K1

(–7, 8 ) N1

42122 yx K1

0114222 yxyx N1

3

2

3

2 10

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sah@mozac2008

SULIT 3472/2

Soalan Penyelesaian dan Peraturan Markah MarkahSub

MarkahPenuh

10 (a)

10(b)

10(c)

0

0

180

142.360 K1

Arc QRS = 13 (0

0

180

142.360 ) K1

= 13.62 cm N1

Arc QTS = 20 (0

0

180

142.330 ) or 10.47 K1

Perimeter = 20 + 20 + 10.47 K1= 50.47 cm N1

Area =

0

022

180

142.33020

2

113142.3 K1,K1,K1

426.26 cm2 N1

3

3

4 10

11(a)(i)

(ii)

(b)(i)

(ii)

Eiher one )13( xP or )14( xP or )15( xP K115

15

14

14

213

13)75.0(15)25.0()75.0(15)25.0()75.0(15)13( CCCxP

K1= 0.2361 N1

2.10)25.0)(75.0( n K1

n = 554 or 555 N1

)15

5880(80

zPxP or )8.1( zP K1

= 0.0359 / 0.03593 N1

1.0)15

58(

wxP K1

281.115

58

wK1

w = 38.785 N1

3

2

2

3 10

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sah@mozac2008

SULIT 3472/2

BAHAGIAN C

Soalan Penyelesaian dan Peraturan Markah MarkahSub

MarkahPenuh

12(a)

12(b)

12(c)

12(d)

QS2 = 82 + 7.52 – 2(8)(7.5)cos 128o K1

QS = 13.93 N1

o

*

58sin

12

sin

93.13

QPSf.t.* K1

QPS = 79.88o /79o53' (accept 79.87o / 79o52') N1

Area of ∆QRS (A1) =2

1(7.5)(8)sin128o

or Area of ∆QPS (A2) =2

1(13.93)(12)sin 42.12o f.t. QPS K1

see 42.12o/42o7' P1(42.13o/42o8')

A1 + A2 K1= 79. 70 (79.71) N1

ot12.42sin

39.13 /(42.13o)

or1

2

2

4

13 (a)Use I = 100

2

1 P

P

x = 110

y = 33.60

z = 22.75

Note: f.t.= follow thrmentioned

42.12o

S

13.93 58o

Q

htt

sah@mozac2008

0656122

.))(( t /(56.07) K1

t = 9.343 /(9.344) N1 2 10

K1

N1

N1

N1

ough student’s answer for K marks only when

4

P

t

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SULIT 3472/2

Soalan Penyelesaian dan Peraturan Markah MarkahSub

MarkahPenuh

13(b)

13(c)

13(d)

16

)130(5)125(4)110(7 * I f.t.* K1

= 120 N1

210 ×100

120K1

= 252 N1

100

170120I K1

= 204 N1

2

2

2 10

14(a)

14(b)

14(c)(i)

(ii)

100x + 60y ≤ 6600 or equivalent N1

80x + 120y ≥ 4800 or equivalent N1

y ≥ 2x or equivalent N1

Draw one of the 3 lines correctly f.t. (a) K1

Draw all 3 lines correctly K1

Region R N1

30 ≤ y ≤ 85 N1

50x + 25y = k (any value of k) K1

(30, 60) N1

Maximum profit = RM3000 N1

3

3

4 10

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sah@mozac2008

SULIT 3472/2

Soalan Penyelesaian dan Peraturan Markah MarkahSub

MarkahPenuh

15(a)

15(b)

15(c)

15(d)

h = 8 N1

a = –2t + 8 K1= –2(8) + 8 f.t. h K1= –8 N1

–t(t – 8) = 2t K1t = 6 N1

sP = dttt )( 82

or sQ = dtt 2 see the power of t increased by 1 K1

sP = 23

643

6)(

)(

or sQ = 62 substitute t = 6 f.t.(c) K1

sP – sQ K1= 36 m N1

1

3

2

4 10

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sah@mozac2008

SULIT 3472/2

0

10

20

30

40

50

60

70

80

90

100

y

10 20

y = 2x

5x + 3y = 330

110

R

sah@mozac2008

(30, 60) ●

30 40 50 60 70 80

2x + 3y = 120

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SULIT 3472/2

0 1

X

X

X

X

2

4

6

8

10

12

14

16

18

20

22x

y

(0, 1)

( 5, 21.4 )

7(a)

( 5, 21.4 )

sah@mozac2008

2 3

X

http://mathsm

4

ozac.blogs

5x

pot.com