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SULIT 3472/2 3472/2 @ 2010 Hak Cipta JPWP SULIT JABATAN PELAJARAN WILAYAH PERSEKUTUAN KUALA LUMPUR PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2010 SKEMA PEMARKAHAN ADDITIONAL MATHEMATICS PAPER 2

# Jpwpkl Trial 2010 p2 Mark Scheme

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KL SPM 2010 Trial AddMaths P2 Marking Scheme

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SULIT 3472/2

3472/2 @ 2010 Hak Cipta JPWP SULIT

JABATAN PELAJARAN WILAYAH PERSEKUTUAN KUALA LUMPUR

PEPERIKSAAN PERCUBAAN

SIJIL PELAJARAN MALAYSIA 2010

SKEMA PEMARKAHAN

PAPER 2

SULIT 3472/2

[Lihat sebelah

3472/2 @ 2010 Hak Cipta JPWPKL SULIT

1

SECTION A

Question Solution Sub

Mark

Full

Mark

1.

x = 4y – 7

( 4y – 7) 2 – y (4y – 7 ) + y 2 = 9

0404913 2 yy

132

40134)49(49 2 y

196.1,574.2y

x = 3.296 , – 2.216

1

1

1

1

1

[5]

2. (a) 2 2 22 4 30f x x m m m

= 2 22 3 30x m m

(b) (i) 2m = 6

m = 3

(ii)

1

1

1

1

1

1

1

2

5

[7]

3. (a) mtangent = 2

2

6dy h

xdx x

26( 1) 2

1

h

h = 4

24

3( 1)1

k

k = −7

(b) y – (−7) = 2[x – (−1)]

y = 2x − 5

1

1

1

1

1

1

1

5

2

[7]

4

7

xy

0951413

94

7

4

7

2

2

2

xx

xxxx

132

95134)14()14( 2 x

x = 3.295 , −2.218

196.1,574.2y

x

(0, 39) ●

● (8, 7)

y

● (6, 3)

U shape

min point (6, 3)

points: (0, 39) and

(8, 7)

OR

SULIT 2

3472/2 @ 2010 Hak cipta JPWPKL

SULIT

4. (a) List of areas :

2

1xh,

8

1xh,

32

1xh, …

4

1

2

3

1

2 T

T

T

T

This is Geometric Progression and r = 4

1

(b) List of areas of triangles: 20, 5, 4

5, …

a = 20

41

10

41

10

41

4

41

41

])(1[20

1

])(1[20

SorS OR

64

5

4

120

4

5

T

410 SS41

10

41

1

])(1[20

41

4

41

1

])(1[20

41

6

41

645

1

1

105

TtoTS

= 26.6667 – 26.5625

= 0.1042

1

1

1

1

1

1

3

3

[6]

5.

(a)

1

1

1

1

1

1

1

4

3

[7]

ab 48

(ii) AFBABF

ba 510

(b) (i) AEhAG

= )48( abh or

= ahbh 48

(ii) BABFkorBFkABAG

= bkak 5)1010(

1

1

1

1

1

3

(b) 2(1 )x

y

2

Graph of the straight line

No of solutions = 3

0 x

Sine curve

1 cycle in 0 x

Amplitude = 1

shifted upward 1 unit

or y-intercept = 1

2

1

y

OR

SULIT 3472/2

[Lihat sebelah

3472/2 @ 2010 Hak Cipta JPWPKL SULIT

3

8h = 5k or 4h = −10k +10

kh4

5 1010

4

58

kk

2

1h

5

4k

1

1,1

5

[8]

SECTION B

Question Solution Sub

Mark

Full

Mark

7.

Refer to the attachment

8. (a)

41

)0(4)5)(1(,

41

)10(4)0(1R

41

)0(4)5(1

41

)10(4)0(1

yorx

R( 8, –1)

(b) m PQ = 2

1 or gradient of the perpendicular line = −2

y – (–1) = −2 ( x − 8)

y = −2x + 15

(c) Area of OPR= 0150

0800

2

1

= 20 unit2

(d) PR = 22 510

5

4

OR PR = 22 )]5(1[)08(

= 1255

4 or 8.944 cm or 80

2

1

× perpendicular distance × 125

5

4 = 20

perpendicular distance = 5

10 or 4.472 or

20

1

1

1

1

1

1

1

1

1

1

2

3

2

3

[10]

or

SULIT 4

3472/2 @ 2010 Hak cipta JPWPKL

SULIT

9. (a) x2 – 4 = x + 2

x2 – x – 6 = 0

(x – 3)(x + 2) = 0

x = 3 , x = −2

B(3, 5)

(b)

)5)(5(2

1)(1 ABCA

3

2

2

2 4 dxxA or

2

2

2

3 4 dxxA

3

2

3

43

xx

or

2

2

3

43

xx

)2(4

3

2)3(4

3

3 33

or

)2(4

3

2)2(4

3

2 33

Total Area = A1 – A2 + A3

= 3

32

3

7

2

25

= 6

125 or 6

520 unit

2

(c) V = 3

22

2

4x dx

3

4 2

2

8 16 x x dx

35 3

2

816

5 3

x xx

5 3 5 33 8(3) 2 8(2)

16(3) 16(2)5 3 5 3

7.533 or 15

113 or

15

87 unit

3

1

1

1

1

1

1

1

1

1

1

2

5

3

y

Diagram 9

Rajah 9

Q P

y = x2 – 4

A

B(3, 5)

y = x + 2

x

● 2

A3

A2

-2 C

SULIT 3472/2

[Lihat sebelah

3472/2 @ 2010 Hak Cipta JPWPKL SULIT

5

10. (a)

3

2

(b) length of chord PQ = 2 x 3 sin 3

= 5.196 cm

(c) the perimeter of the shaded region = SPRQ + SPQ

= 2

196.5 ( ) + 3(

3

2)

= 598.4 or 14.45 cm

(d) 2

12

2

196.5

( ) or

2

1(3) 2 (

3

2)

2

1(3) 2

sin3

2 or

3cos3

3sin)3)(2(

2

1

= Area semicircle PRQ – [Area sector POQ – Area POQ]

=2

1(

2

196.5) 2 ( ) – [

2

1(3) 2 (

3

2) −

2

1(3) 2

sin3

2]

= 10.6023 – 5.5277

= 5.075 cm 2

1

1

1

1,1

1

1

1

1

1

1

2

3

4

[10]

11. (a)(i) P(x = 2) = 10 2 8

2(0.2) (0.8)C

= 0.3020

(ii) 2 1 0 1P x P x P x

91

1

10100

0

10 )8.0()2.0()8.0()2.0(1 CC

= 0.6242

(b) P ( x ≤ 20 ) = P ( z ≤10

3020)

= P ( z ≤ – 1 )

= 0.1587

P ( x ≥ m) = 2

1587.01

P ( z ≥ 10

30m) = 0.4207

10

30m = 0.2

m = 32

1

1

1

1

1

1

1

1

1

1

5

5

[10]

SULIT 6

3472/2 @ 2010 Hak cipta JPWPKL

SULIT

SECTION C

Question Solution Sub

Mark

Full

Mark

12. (a) v = 5t (t − 4) = 5t2 – 20t

a = 10t − 20 = 0

t = 2

v = 5(2)2 – 20(2)

= − 20 ms-1

(b) s = 25 20 t t dt

= 3

2510

3

tt

32

3 2

510 0

3

5 30 0

tt

t t

5t2(t – 6) = 0

t = 6

(c) 5t2 – 20t = 0

5t (t – 4) = 0

t = 4

3 32 2

4 5

5(4) 5(5)10(4) 10(5)

3 3t ts or s

= 160

3 =

125

3

Total distance travelled = 160 160 125

3 3 3

= 65

1

1

1

1

1

1

1

1

1

1

3

3

4

[10]

OR

5t2 – 20t = 0

5t (t – 4) = 0

t = 4

Total distance travelled

=

45

2 2

40

4 5

3 2 3 2

0 4

45

2 2

40

5 20 (5 20 )

5 20 5 20

3 2 3 2

5(64) 10(16) 0

3

5 5(64)(125) 10(25) 10(16)

3 3

5 20 (5 20 )

65

OR

OR

+

t t dt t t dt

t t t t

or

t t dt t t dt

1

1

1

1

SULIT 3472/2

[Lihat sebelah

3472/2 @ 2010 Hak Cipta JPWPKL SULIT

7

13. (a)

1cos 60

2 oPSR or PSR

2 2 26.5 5.2 2 6.5 5.2 cosPR PSR

2 2 2 16.5 5.2 2 6.5 5.2

2PR

PR = 5.957 cm

(b) '329/04.29

50sin

4.9

sin

oo

o

PQR

PQR

PR

0 0 0 0180 50 29.04 100.96 PRQ

096.100sin957.54.92

1)( PQRAc

060sin2.55.6

2

1PRSA

PRSPQRPQRS AAA

= 42.12 cm 2

1

1

1

1

1

1

1

1

1

1

3

3

4

[10]

14. Refer to the attachment.

15.

(a) (i) y = 110

100100

= 110

04

04

148(ii) 125 100

118.40

P

P RM

(b) 15

)2(110)4(105)3(125)6(130 I

I = 119.67

(c) 67.119100684

07 P

54.81807 RMP

(d) 67.119100

12004/09 I

= 143.6

1

1

1

1,1

1

1

1

1

1

3

3

2

2

[10]

SULIT 8

3472/2 @ 2010 Hak cipta JPWPKL

SULIT

Question 7

1.2

1.0

0.8

0.4

0.2

1.4

0.5 0.4 0.3 0.2 0.1 0.6 0.7

0.6

1.6

1.8

Correct axes and uniform scale .........K1

Plot all the points correctly …………N1

Line of best fit ………N1

5

x10log 0.18 0.30 0.48 0.60 0.70 0.78

y10log 1.60 1.40 1.12 0.88 0.80 0.60

10

0

x

x

x

x

x

x

c = 1.88 ............... K1

6.13.08.0

4.16.0

m ............... K1

xpky 101010 logloglog ............. P1

6.1

p

pm

86.75

88.1log10

k

kc

............................. 1

0.8

y10log

x10log

……......… N1

……… N1

N1

N1

5

1.9

SULIT 3472/2

[Lihat sebelah

3472/2 @ 2010 Hak Cipta JPWPKL SULIT

9

Question 14

x 0 100 400 300 200 700 600 500 800

y

400

100

350

300

250

200

150

450

500

50

R

5x + 7y = 3500

x + 8y = 800

x = 4y

500

(a) 5x + 7y 3500 or equivalent

x + 8y ≥ 800 or equivalent

x 4y or equivalent

(b) Draw correctly any straight line

All three straight lines are correct

(c) (i) x = 518 [accept integer 510 – 520]

(ii) Use 3x + 2y where (x, y)

for point in the shaded region

Maximum point = (518,130) or

130129,520510 yx ,

x and y = integers

Maximum profit = RM1780 to RM1820

N1

N1

N1

3

K1

K1

N1

3 N1

K1

N1

N1

4

10

Question 14

x