spm addmath2 ans (kedah)

8/14/2019 SPM AddMath2 Ans (Kedah)

1/16

3472/2 Additional Mathematics Paper 2 [Lihat sebelah

SULIT

Question Solution and marking schemeSub-

mark

Full

Mark

1. 3

2

yx

+= or

2y x= 3

2

23 3 102 2

y yy y

+ + + =

0

0

or ( ) ( )22 2 3 2 3 1 x x x x + =

2 2 29 6 6 2 4 40 0 y y y y y+ + + =

or

2 2 22 3 4 12 9 10 x x x x x + + + =

y = 3.215 , 3.215

or

x = 3.107 ,

0.107

x = 3.107 / 3.108 , 0.107 / 0.108

or

y = 3.214 / 3.215 , 3.214 / 3.215

Answer must correct to 3 decimal places. 5

Make x or y asthe subject

P1

Eliminate

x or y

Solve quadratic

equation23 31 0y =

2

31

3y =

or23 9 1 0x x =

or

using formula

or

completing the

square

N1

K1

K1

N1

j2k


2/16


SULIT


mark

Full

Mark

2(a)

(b)

16 ,8 , 4 ,......

16a = 1

2r=

7

7

116 1

2

11

2

S

=

=3

314

or 31.75

64 ,16 , 4 ,......

64a = 1

4r=

64

11

4

S

=

= 1853

or 85.33

3

3 6

3(a) 2( ) f x x px= + + q

=

2 2

2

2 2

p p x px q

+ + +

=

2 2

2 4

p p

x q

+ +

32

p = p =6

365

4q + = q = 4

N1

K1

User

raS

n

n

=

1

)1(K1

N1

K1

N1

P1

P1

N1

1

aS

r =

Use

2

2 )2b

(

b

use x + bx = ( x + ) or

32

b

a =use axis of symmetry

j2k


3/16


SULIT


mark

Full

Mark

Alternative solution

( )

223 5 x px q x+ + =

=2 6 9 5x x +

=2

6 4x x +

6p =

q = 43

3(b)2 6 4 31x x + 0

0

0

2 6 27x x

( 9)( 3)x x +

3 9x 2 5

4(a)

tanx +cos

1 sin

x

x+

=sin

cos

x

x+

cos

1 sin

x

x+

=2sin (1 sin ) cos

cos (1 sin )

x x x

x x

+ +

+

=2 2

sin sin coscos (1 sin ) x x x

x x+ +

+

=sin 1

cos (1 sin )

x

x x

+

+

=1

cosx

= secx3

K1

N1

Use ( ) 31 0f x andfactorization

K1

K1

Usesin

tancos

xx

x=

Use identity2 2

sin cos 1x x+ =

N1

K1

N1

Comparing coefficient of x

or constant term

N1

j2k


4/16


SULIT


mark

Full

Mark

4(b)(i)

Negative sine shape correct.

Amplitude = 3 [ Maximum = 3 and Minimum = 3 ]

Two full cycle in 0 x 2

3

4(b)(ii)

53sin 2 2

xx

=

or

52

xy

=

Draw the straight line

5

2

x

y =

Number of solutions is 3 .

3 9

K1

N1

N1

P1

P1

P1

y

3

3

2

2

3

2 x

52

xy =

O

y 3sin2x=

j2k


5/16


SULIT


mark

Full

Mark

5 (a)

(b)

1220

x=

240x =

The mean

240 5 + 8 + 10 + 11 + 14 288

25 25X

+= =

= 11.52

2

212 3

20

x =

2 3060x =

The standard deviation

( )2 2 2 2 2

23060 5 8 10 11 1411.52

25

+ + + + +=

= ( )23566

11.5225

= 9.9296

= 3.151

7 7

K1

xx N=

Use

N1

N1

K1

Use formula

22x

xN

=

N1

N1

For the new

K1

Xand2

x

j2k


6/16


SULIT


mark

Full

Mark

6(a)

(b)

(i) PR PO OR= +uuur uuur uuur

= 6 15a b +% %

(ii) OQ OP PQ= +uuur uuur uuur

=3

65

a O+uuur

%R

r

5%

5%

k

k

= 6 9a b+% %

(i) OS hOQ=uuur uuu

= (6 9 )h a b+% %

(ii) OS OP PS= +uuur uuur uuur

= 6a k PR+uuur

%

= ( )6 6 1a k a b+ +% %

=(6 9 )h a b+% %

( )6 6 1a k a b+ +% %

6 6 6h = 9 15h k=

1h = 5

3h k=

5 13

k k=

3

8k=

5 3

3 8h

=

=

5

8

3

5 8

K1

N1Use

PR PO OR= +uuur uuur uuur

or

OQ OP PQ= +uuur r uuuruuu

N1

N1

N1

N1

N1

K1 Equate coefficient

of a or b % %and

Eliminate

h or k

j2k


7/16

Qn. Solution and Marking SchemeSub-

mark

Full

Mark

7(a)

(b)(i)

(ii)

kn

xn

y 4log1

log1

log +=

log x 0.18 0.30 0.40 0.60 0.74

log y 0.48 0.54 0.59 0.69 0.76

6

410

N1

P1

All points plotted

correctly


SULIT

Correct axes and

scale

Line of best-fit

N1

K1

intercept

= kn

4log1

n = 2 k= 1.51

K1

N1

N1

K1

N1

N1

gradient

=n

1

2k


8/16


SULIT

0.6

0.7

Graph of log10y against log10x

0.8

log 10y

0.5

0.4

0.1

0.2

0.3

0.9

0

0.39

0.4 0.5 0.6 0.7

log 10 x

0.1 0.2 0.8 0.90.3

2k


9/16


mark

Full

Mark

8.

(a)

(b)

(c)

Solving simultaneous equation

P( 2, 2) Q(4, 8)

Use dxyy )( 12

+ dxx

x )2

4(2

Integrate dxyy )( 12

64

2

32 xx

x+

18

Note : If use area of trapezium and ydx , give the marksaccordingly.

Integrate dxx

2

2

)2

(

=

20

5x

5

8

3

4

310

K1

N1 N1

K1


SULIT

N1

K1Use correct

limit into

4

2K1

64

2

32x

xx

+

N1

K1

K1

Use correct

limit into2

0

20

5

x

2k


10/16


mark

Full

Mark

9(a)

(b)

(c)

(d)

Equation ofAD :

y 6 = 2 (x 2 )

y =2x + 10 or equivalent

y =2x + 10

and

x 2y = 0

D(4, 2)

p = 3 or C(8, 4)

Substitute (8, 4) intoy = 3x + q

q = 20

OADOABCArea = 4

Using formula

0

0

6

2

2

4

0

0

2

1= OAD

40

Alternative solution :

B(10, 10)

Using formula

0

0

6

2

10

10

4

8

0

0

2

1=OABCArea

40

2

2

3

3

10

Use m = 2 and find

equation of straight lineK1

N1

Solving

simultaneous

equations

K1

N1

K1

N1

P1

N1

Find area of

triangle

K1

K1

K1

P1

Find area of

parallelogram

N1


SULIT

2k


11/16


mark

Full

Mark

10(a)

(b)

(c )

=

10

6sin 1 or equivalent

= 2POQ

radPOQ 287.1=

Alternative solution :

122

= 102

+ 102

2(10)(10) cos POQ

+=

)10)(10(2

121010cos

2221POQ

radPOQ 287.1=

Using (2 1.287)

Major arc PQ

= 10 ( 2 1.287 )

= 49.96 cm

Lsector= 287.1)10(2

1 2

Ltriangle = 287.1sin)10(2

1 2

= 16.35 cm2

= 3.9149 cm

3

3

4 10

K1

K1

N1

N1

K1 Use cosine

rule

N1

Use formula

s = rK1

N1

K1

Using formula

Lsector= 21 r

2K1

K1

K1Lsector - L

Using formula

L = absin C 21

N1


SULIT

2k


12/16


mark

Full

Mark

11

(a)(i)

(ii)

b)(i)

(ii)

p =5

3, p + q = 1

P( X = 0 )

=5C0(

5

3)0(

5

2)

5

= 0.01024

Using P( X 4 )= P( X = 4 ) + P( X = 5 )

=5C4(

5

3)4(

5

2)

1+ (

5

3)5

= 0.337

P ( 30 X 60 )

= P (10

3530 Z 60 3510 )

Use P( 0.5 Z 2.5 )

= 1 P( Z 0.5 ) P( Z 2.5 )

= 1 0.30854 0.00621

= 0.68525

Number of pupils = P( X 60 ) 483

3

3

2

3

2 10


SULIT

K1

N1

K1

N1

Use P(X = r) = n Crprqnr,

p + q = 1

K1

N1

K1

use

Z =

X

K1

P1

N1

2k


13/16

3472/2 Additional Mathematics Paper 2 [Lihat sebelahSULIT


mark

Full

mark

12.

(a)

(b)

(c)

(d)

Subst. t= 0 intodv

dt

a= 15 6t15 ms

-2

1 134 4

ms75 /18ms

Integrate

Use s = 0

15

2/

17 / 7.5

2

S4 S3

115

2

Note : If use , give the marks accordingly.4

3vdt

2

2

3

3

10

K1

N1

0dv

dt= K1and subst. t in v = 15t 3t

2Use

5[t= ]

2

N1

K1

N1

K1

K1

K12 315

2dt t t s v = =

Subst. t = 3 or t = 4 in

N1

2 315

2s t t=

2k


14/16



mark

Full

mark

13.

(a)

(b)

(i)

(ii)

(iii)

Use I = 2007

2005

100P

P

Value of m : 25, m, 80, 30 or equivalent

120 25+130m+135 80+139 30

Use i i

i

I WIW

=

132.1 =12025+130m+13580+139 30

135+m

m =65

100150.00x

132.1

RM 113.55

/I . ( . x .132 1 132 1 0 3= + )08 05

171.73

3

3

2

2

10

K1

N2, 1, 0x = 48.6y = 135z = 80

K1

K1

N1

K1

N1

P1

N1

2k


15/16



mark

Full

mark

14.

(a)

(b)

(c)

x + y 80 or equivalent

y 4x or equivalent

x + 4y 120 or equivalent

x=30

(16,64)

x + 4y = 120

x + y = 80

y = 4x

100

90

80

70

60

50

40

30

20

10

10080604020 9070503010

x

y

At least one straight line is drawn correctlyfrom inequalities involving x and y

All the three straight lines are drawn correctly

Region is correctly shaded

(i) minimum = 23

(ii) (16,64)

Subst. point in the range

in 20x + 40y

RM2880

3

3

4 10

N1

N1

N1

K1

N1

N1

N1

K1

N1

N1

2k


16/16



mark

Full

mark

15.

(a)

(b)

(c)

(d)

120 9.3 6 sin

2

BCD=

45o48 / 45.8

o

0.74545 4555Use cosine rule in BCD

BD2

= 9.32

+ 62 29.36 cos 4548

6.685

Use sine rule in BCDo

sin CBD sin '

. .

= 45 489 3 6 685

94

o

10

4444

qqq11111aaaaaaaaaaaaaaaaaa 14s4

5.555555

5555 z5555555555555555

Sum of area:

20 cm2

+ ABD 555 102

58.82 cm2

4

2

4 10

Use area = ab sin c in BCD K1

N1

K1

N1

K1

N1

K1

K1

N1

K1 Obtain ADB by using180o 85o50 BAD or equivalent

Use area ADB = 6.685 13 sinADB

2k

spm addmath2 ans (kedah)

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