spm trial 2012 addmath a perak

7/28/2019 Spm Trial 2012 Addmath a Perak

1/14

1

SPM TRIAL EXAM 2012

Marking Scheme

Additional mathematics Paper I

Number Solution and marking scheme

Sub

Marks

Full

Marks

1 (a) {p, r, s}

(b) {a , b, c, d}(c) Many to One

1

11

3

2(a)

2

7

B1: 2x 5 = 2 atau f(x) =2

5+x

2 2

3 (a)5(b) 42 +x

B2: ( ) 7653 =+ xxf B1 : ( ) 53 + xf

1

3

4

4

p = 7 , q = 10 ( both )

B2: p = 7 or q = 10

B1:33

7 p= or33

10 q= or 01073 2 = xx

or 3(1)2

+p(1)+q = 0, 3(10/3)2

+p(10/3) + q = 0

3 3

5 h =2 andk= 3

B2 : h = 2 atau k= 3

B1 : 1

3

=h

or 922 =++ kh

3 3

6

45 x

B2: ( ) 0)5(4 + xx or

B1 : x2 +x 20 0

3 3

4-5

http://www.chngtuition.blogspot.com


2/14

2

7

pr

12+ or

pr

rp +2

B3:mm 3

3

2

2

log

3log2log

2log +

B2: log m2 + 2logmmm 3

2

3

2

2

log

3log

log

2log+3or

B1 : log 2 + log32

or 2logmmlog

2log3 or or

mlog

3log

4 4

8 158 += px

B2: )32(43 += px

B1: 32 or )32(42 +p

3 3

9 a) 3b) 360

B1: [ ]18

2(3) (18 1)22

+

1

2

3

10 13=a , 4=d (both)

B2: 13=a or 4=d

B1: 52 =+ da or 157 =+ da

3 3

11 (a) r=x

2

(b) 31

B1:2

2

18

1

x

x

=

1

2

3

12a) qpx

x

y+= 2

b) 2=p , 8=q B2 : 2=p or 8=q

B1:42

04

=p or q+= )4)(2(0 or 4 =p(2) + q

1

3

4



3/14

3

13 2,12

B2: 110

)1(3

4

73=

hh or form equation using Pythagoras

theorem

B1:4

73

hor

10

)1(3

h of find the lengths of AB, BC and AC using

distance formula

3 3

14(a)

w

1

(b) 212 ww B1:

21 w or cos2w

1

2

3

15 26.57 ,116.57 , 206.57 , 296.57 B3: 26.57 and 116.57 B2: )2)(tan1tan2( + xx

B1: 02tan3tan2 2 =+ xx

4 4

16(a)13(b)k = 13

B1:

+

++25

112 kor jik )25()112( ++++

1

2

3

17 (a) 4a + 4b(b) ba 42 +

B1: )44(6 baa ++

12

3

18 8

B2: 9.42)3()3()3(3.13.1 =+++++ rrrrrr

B1: r3.1 or )3(3.1 r

3 3

19

5

12

B2:

+0

12

)2(32

2

B1:

2

0

2 1

32

+x

x

3 3



4/14

4

20 122 + aa

B2: ))4()4(()( 22 ++ aa

B1: xx +2

3 3

21 k= 8

B3: (k+1)(k8) = 0

B2:

22 2 22 5 2 56

3 3

k k+ + + + =

B1:3

52 kx

++= or 2222 52 kx ++=

4 4

22 1 2( , )3 3

B2 : x =1

3, y = 2()(32)

B1 : 12x 4 = 0

3 3

23 (a) 1(b)i) 5040

ii) 288

B1: 2 3! 4!

1

12

4

24(a)

15

4

(b) 35

B1 :3 2 2 1 3 1 2 2

1 or5 3 5 3 5 3 5 3

+ +

1

2

3

25(a) 53 (b)

625

144

B1:

32

2

5

5

2

5

3

C

1

2

3



5/14

5



6/14

3472/2 SULIT

1

SPM TRIAL EXAM 2012

Marking Scheme

Additional Mathematics Paper 2SECTION A

Question Important Steps Marks

1 y = 3x 4 1

5x2 4x(3x 4) + (3x 4)2

or 2x= 9

21

8x + 7 = 0

)2(2

)7)(2(4)8()8( 2 =x

1

x = 2.707, 1.293 1y = 3(2.707) 4 , y = 3(1.293) 4

= 4.121 = 0.121 1TOTAL 5

2(a)

Change base of logarithm :4

2

2

log (1 2 )

log 4log (1 2 )

xx

= or

equivalent

1

Use n logx = logx n : 2 log 2 ( 1 2x ) = log2 ( 1 2x ) 12

Solve : (2x + 5 ) = ( 1 2x ) 12

x = 221 , 1

x = 21

1(b)

31 1

TOTAL 6

3(a)

8 1

(b)Use Tn 1= a + (n 1 ) d : 8 + ( 22 1 ) ( 3 )

Use Sn ])([ dnan 122 += : 2 [ 2 ( 8) ( 1) (3) ]n n + 1

Solve : ])()()([ 31822

+ nn = 55 1

n = 10 1

(c)T 11328 1

TOTAL 7

4

(a)x = 3

1

(b)5

5

9105.9

+=median

1



7/14

3472/2 SULIT

2

Median = 10.5 1

(c) All midpoints are correct. 1

11

20

220

20

)22(2)17(4)12(5)7(6)2(3==

++++=x 1

=++++= 3150)22(2)17(4)12(5)7(6)2(3222222fx

or f(x-x )21

= 7302

2

20

220

20

3150

= or

20

7302 = 1

= 36.5 1

TOTAL 8

5(a)

y

xy

2

=

x

Shape1

Max/min1

Oneperiod

1

Completefrom 0 to

2

1(b)

Equation xy

2= 1

Straight line xy

2= 1

2 solutions 1

TOTAL 7

6

(a)

3

2

QSm = 1

)6(2

31 = xy 1

82

3= xy 1

(b) Q(0, 8) 1

[(x 6)2 + (y 1)2 [(6 0)] or 2 + (1+8)2 1]

x2 12x + 36 +y2 1 2y + 1 = 117

x2

+y2

1 12x 2y 80 = 0TOTAL 7

22

32

3

-3

O



8/14

3472/2 SULIT

3

SECTION B

7(a)

x 1 1.5 2 2.5 3 3.5 4log10 0.04y 0.18 0.28 0.40 0.52 0.62 0.76

1

(b)Plot log10 1y againstx [ correct axes and uniform scales ]All 7 points plotted correctly 1Line of best fit 1

(c) (i) 3.72 1

(ii)22101010 hk xy

loglog)(log = 1

Use :log210 hc = 20

210 .

log=

h 1

h = 2.5 1(iii)

Use :log210 km = 240

210 .

log=

k 1

k = 3.0 1TOTAL 10

8(a)

= dxxy 2 1

cx

y +

=2

2 2 1

92 += xy 1

(b)3

2

0( 9)x dx + 1

33

0

93

xx

+

1

)10)(10(2

1

32

0( 9)x dx + or 50

32

0( 9)x dx +

or )10)(10(21

33

0

93x x +

or 50

33

0

93x x +

1

= 32 1

(c) Volume = dyy)9( 1

92

0

92

yy

1



9/14

3472/2 SULIT

4

81

2or equivalent

1

TOTAL 10

9(a)

Angle AOB = 6.5/5 1= 1.3 rad. 1

AnglePOQ = 0.8667 rad. 1(b) MN= 5 sin(0.8667 rad.) = 3.811 cm

or ON = 5 cos( 0.8667 rad.) = 3.2367 cm

1

Length of arcPQ = 6 0.8667 = 5.2002 1

Perimeter = 3.811 + 5.2002 + 1 + (63.2367) 1

= 12.77 cm 1(c) Area of sectorOPQ = 62 1 0.8667

Area of shaded region = 15.60 (3.811)(3.2367) 1

= 9.433 1TOTAL 10

10

(a) (i)RTPRPT +=

)148(2

1xyRT += 1

)148(2

18 xyyPT ++=

= yx 47 +

1

(ii)RS RP PS= +

)14(3

18 xy += 1

yx 83

14= 1

(b) (i) yhxhPM 47 += 1

(ii)RMPRPM +=

)83

14(8 yxky +=

1

= ykxk )88(3

14+ 1

(c)14

73

h k= or hk 488 = 1

2

1=h

1

4

3=k

1

TOTAL 10



10/14

3472/2 SULIT

5

11(a)(i)

37

7

10 )55.0()45.0()7( CXP == 1

= 0.07460 1

(ii)

P(X= 0, 1, 3)= 10C0(0.45)0(0.55)10 + 10C1(0.45)1(0.55)9 + 10C2(0.45)2(0.55) 18

= 0.09956 1

(b) (i)

6.0 7.2 8.1 7.2( )

1.2 1.2P z

< < OR ( 1 0.75)P z < < 1

= 2266.01587.01 1

= 0.6147 1

(ii) 8.06048

2.12.7 ==

> tzp 1

842.02.1

2.7=

t 1

190.61896.6 ort = 1

TOTAL 10

SECTION C

12(a)

a = 1.4 0.6dv

tdt = 1

= 1.4 0.6(2)= 0.2 1

(b)1.4t0.3t2 1+ 0.5 = 0

(3t+1)(t 5) = 0 or using quadratic formula 1t= 5 1

(c)s = 2(1.4 0.3 0.5)t t dt + = 0.7t

2 0.1t3 1+ 0.5t+ c integrate

At t= 0,s = 0 c = 0 finding c

or +++

5

0

10

5

225.03.04.15.03.04.1 dtttdttt

limits

1

When t= 5,s = 7.5 m, when t= 10,s = 25 mor substitute t=0, 5, 10 in [0.7t2 0.1t3

1+ 0.5t]

Total distance = 7.5 x 2 + 25 or 7.5 + |25 7.5| 1= 40 m 1

TOTAL 10

13(a)

158100

130x = 1

= 121.54 1



11/14

3472/2 SULIT

6

(b)

150100 107.14

140A = 1

121.54100 90.03

135B = 1

120 100 109.09110

C = 1

123100 102.5

120D = 1

(c)

12

)25.102()209.109()503.90()314.107( +++=I 1

= 99.56 1

(d)56.99

100

1202013

=I 1

= 119.47 1

TOTAL 10

14(a) (i)

sin sin 30

12 7

ACB =

1

ACB = 59 1

(ii) 2 2 24 11.47 12cos

2(4)(11.47)AKB

+ = 1

cos AKB = 0.0388AKB = 87.78 or 8747

1

(iii) ABC= 911

AreaABC= (7)(12) sin 91or Area ofAKB = (4)(11.47) sin 87.78

1

Area of quadrilateral= AreaABC+ Area ofAKB= 41.99 + 22.92

1

= 64.91 cm 12

(b)(i)

1

(ii) ACB = 121

1

TOTAL 10

B

AC

12 cm

7 cm

30o



12/14

3472/2 SULIT

7

15(a)(i)

x +y 10 or equivalent 1y x 4 or equivalent 1x 2y or equivalent 1

(b)Draw correctly one straight line from the inequalities 1Draw correctly two more straight line from the inequalities 1RegionR correctly shaded 1

(c)(i)Maximum point ( 3 , 7 ) 1RM [ 10(3) + 25(7) ] = RM 205 1

(ii)Minimum point (2 , 6 ) 1RM [ 10(2) + 25(6) ] = RM 170 1

TOTAL 10



13/14

3472/2 SULIT

8

GRAPH FORQUESTION 7

00.5 1 1.5 2 2.5 3 3.5 4

x

log10 y

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.1

0.2

( 0 , 0.2 )

( 4 , 0.76 )

3.2

0.57



14/14

3472/2 SULIT

9

GRAPH FORQUESTION 15

01

1

x

y

2 3 4 5 6 7 8 9

2

3

4

5

6

7

8

9

10y x = 4

x +y = 10

x = 2y

R

( 3 , 7 )

10x + 25y = 150( 2 , 6 )

y = 6

spm trial 2012 addmath a perak

Documents