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3472/2 ©2011 Mid Year Exam F5 Hak Cipta SBP

1

3472/2

Matematik Tambahan Kertas 2 2 ½ jam Mei 2011

BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN

KEMENTERIAN PELAJARAN MALAYSIA

PEPERIKSAAN PERTENGAHAN TAHUN TINGKATAN 5

2011

ADDITIONAL MATHEMATICS

Paper 2

Skema Pemarkahan ini mengandungi 12 halaman bercetak

MARKING SCHEME

www.banksoalanspm.com


2

PERATURAN PEMARKAHAN – KERTAS 2

No Solution and Mark Scheme Sub Marks

Total Marks

1

3

21 yx

OR

2

13

xy

42

212

2

212

y

yy

y OR 4

2

13

2

1322

xx

xx

0)7)(52( yy OR 0)5)(2( xx

7,2

5 yy (both) OR 5,2 xx (both)

5,2 xx (both) OR 7,2

5 yy (both)

P1

K1

K1

N1

N1

5

2(a)

463)( 2 xxxf

41)1(3

4)2(3

2

2

x

xx

7)1(3 2 x

K1

N1

(b) Maximum point = (1 ,7)

P1

(c)

N1

N1

N1

6

f(x)

Shape (max)

Maximum point

*(1,7)

y-intercept = 4 and any one

point (-1,-5) or (2,4)

(1,7)

4

x

(-1,-5)

(2,4)



3

3(a)

5nT , 3,80 da

5)3)(1(80 n

26n

P1

K1

Height of the arrangement = 2601026 cm N1

(b) )3)(126()80(2

2

2626 S

1105

The total length of circumference of circles = )5(21105

11050 cm

K1

K1

N1

6

4

(a) Mid point of AC = (-1 , 6 )

use midpoint BD = midpoint AC : 62

0

p

p = 6

(b) Area of ADC =

= )]10)(13()12)(3()2)(1[()]10)(3()2)(13()12)(1[(2

1

= 60 unit2

(c) Gradient of BD = 1

2

Equation of BD = (y-6) = 1( 1)

2x

2y + x =11 or equivalent

P1

K1

P1

K1

N1

P1

K1

N1

8 5

(a)

P1

P1

P1

P1

Shape of sine graph

Amplitude = 1

Period of 1 cycle 0 x

Graph shifted up by 1 unit

2

2

1

0

y

x



4

(b)

xy

22

N1

K1

N1

7 6 (b)

4350

50x

or 29 2 133950fx

22 )29(150

133950

211.7

K1

P1

K1

N1

8

Eq. of straight line ;

Draw the straight line :

(either gradient or y-intercept correct)

Number of solution = 3



5

6(a) Skema Histogram



6

7

Linear law graph



7

7 (a)

Note: If table is not shown, award N1 if all the points are plotted correctly.

Plot x

y against x

(Correct axes and uniform scales) 6 *points plotted correctly Or at least 5 points correctly plotted if the table is not shown Line of best fit

(b) pkxk

x

y

2

(can be implied)

(i) Use *m = 2

k

k = 1.26 ± 0.05

(ii) Use *c = pk

p = 1.59 ± 0.05

(c) y = 21.6 ± 0.05

x 2 3 4 5 6 7

3.20 3.90 4.50 5.12 5.75 6.40

N1

K1

N1

N1

P1

K1

N1

K1

N1

N1

10



8

8 (a) (OR)(0.9273) = 13.91

OR = 15 cm

(b) OP = 9 cm

Area = 2 21 1(9) (0.9273) (9) sin 0.9273

2 2

= 5.156 cm2

(c) PR = 6 cm

RQ = 15 sin 0.9273 or 2 215 9

Perimeter = arc length PQ + PR + RQ

= 9(0.9273) + 6 + 15 sin 0.9273

= 26.35 cm

K1

N1

P1 K1K1 N1 P1 K1K1 N1

10 9

(a) (i)

(ii)

=

=

(b) (i)

(ii)

=

= or

N1

K1

N1

K1

K1

N1



9

Equate the coefficient of and solve the simultaneous equation

and

(both)

(c) [ t = perpendicular distance]

t = 7 cm

K1

N1

K1

N1

10 10 (a) 2)2(4 xx

3,0 xx

A(0,4) B(3,1)

(b) Area

3

0

2)2()14)(3(2

1dxx

0

3

3

)2(

2

15 3

x

3

)20(

3

)23(

2

15 33

= 4.5

(c) Volume

3

2

24 )1()1)((3

1)2( dxx

32

3

5

)2( 5

x

15

8

K1

N1N1

K1K1

K1

N1

K1

K1 N1

10



10

11 (a) (i) 2

dy

dx

(ii) 2

1normalm

32 xy

(b) )2(3)2(5,2 2 dx

dyx

26

y (26)(-0.02)

= -0.52

(c ) cxx

y 2

3

3

5 23

c

2

)1(3

3

)1(52

23

6

13c

6

133

3

5 3

xx

y

P1

P1 K1

N1

P1

K1

N1

K1

K1

N1

10

12

(a) 5

x 100 = 165 = 115 or 6.9

y 100 = 115 or 4 50

4 00

100 = z

x = 825 / y = 600 / z = 112.5

(b) (i) I = (165 125) (*112.5 100) (115 70) (120 65)

360

[ * follow thro’ from (a)]

W = 360 or 72

= 1326

(ii) 3700

100 132.6*

2009P

[ * follow thro’ from b(i)]

2009

P = RM 279035

K1

N2,1,0

K1

P1

N1

K1

N1



11

(c) 2011

2009

130132.6*

100I [* follow thro’ from b(i)]

OR

4810100

2790 K1

= 172.4

K1

N1

10 13

(a) 2009

2007

160 100100

100 135I

= 118.5

(b) (i) 12.60

1009.40

x

= 134.0 or 134.04 or 134

(ii) 2007118 1009.40

P or

2007

12.60118

134P

2007P RM11.09 = RM11.10

(c) (i) 2005

125.50100

137.50P

= RM91.27

(ii) 118(5) 135( ) 175(3)

137.55 3

k

k

k = 6

K1

N1

K1

N1

K1

K1

K1

N1

K1

N1

10 14

(a) ABC = 45.58◦ ABC = 180◦ - 45.58◦ = 134.42◦ (obtuse)

K1

K1 N1

year 2009 2010 2011

Cost 2790.35 3700 4810



12

14 (b) AC = 72 + 82 – 2(7)(8) cos 134.42◦ = 13.83 cm

(c) ADC = 126.08◦

(d) ACD = 13.92◦

2092.13sin)83.13)(11(2

1 0 = 38.30

K1

N1

K1

N1

P1

K1

N1

10 15

(a) 3.5

55sin

3

sin

C

C = 27.62o

DBC = 180o - 27.62o - 55o = 97.38o

(b) oADE 55 (AE)2

= 92 + 42 – 2(4)(9) cos 55o

AE = 7.463 cm

(c) (i)

(ii) DCB' = 82.62o

(iii) )62.82(2180' ooBBC

= 14.76o

Area = 076.14sin)3.5)(3.5(2

1

= 3.578 cm2

K1

N1

N1

K1

N1

P1

P1

P1

K1

N1

10

C

B’

B 3 cm

5.3 cm 55 D


No. of students

No. 6

95 145 195 245 295 345

10

20

30

40

(a) K1 for the bars with constant width

and proportional height.

K1 for boundary/mid point

N1 all correct

N1 mode = 31 025

Time

11

SULIT 3472/2

3472/2 2011 Mid Year Exam Form 5 Hak Cipta SBP

5

395 445

5

15

25

35

45

50


SULIT 3472/2

3472/2 2011 Mid Year Exam F5 Hak Cipta SBP

6

No. 7

0 1 2 3 4 5 6

1

2

3

4

5

6

7

8

y

x

x 7

SKEMA mid year F5 P2 SBP 2011


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