simpor

• View
253

0

Embed Size (px)

Citation preview

• 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

equation23 31 0y =

2

31

3y =

or23 9 1 0x x =

or

using formula

or

completing the

square

N1

K1

K1

N1

j2k

• 8/14/2019 SPM AddMath2 Ans (Kedah)

2/16

3472/2 Additional Mathematics Paper 2 [Lihat sebelah

SULIT

Question Solution and marking schemeSub-

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

• 8/14/2019 SPM AddMath2 Ans (Kedah)

3/16

3472/2 Additional Mathematics Paper 2 [Lihat sebelah

SULIT

Question Solution and marking schemeSub-

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

• 8/14/2019 SPM AddMath2 Ans (Kedah)

4/16

3472/2 Additional Mathematics Paper 2 [Lihat sebelah

SULIT

Question Solution and marking schemeSub-

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

• 8/14/2019 SPM AddMath2 Ans (Kedah)

5/16

3472/2 Additional Mathematics Paper 2 [Lihat sebelah

SULIT

Question Solution and marking schemeSub-

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

• 8/14/2019 SPM AddMath2 Ans (Kedah)

6/16

3472/2 Additional Mathematics Paper 2 [Lihat sebelah

SULIT

Question Solution and marking schemeSub-

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

• 8/14/2019 SPM AddMath2 Ans (Kedah)

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

3472/2 Additional Mathematics Paper 2 [Lihat sebelah

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

=n

1

2k

• 8/14/2019 SPM AddMath2 Ans (Kedah)

8/16

3472/2 Additional Mathematics Paper 2 [Lihat sebelah

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

• 8/14/2019 SPM AddMath2 Ans (Kedah)

9/16

Qn. Solution and Marking SchemeSub-

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

3472/2 Additional Mathematics Paper 2 [Lihat sebelah

SULIT

N1

K1Use correct

limit into

4

2K1

64

2

32x

xx

+

N1

K1

K1

Use correct

limit into2

0

20

5

x

2k

• 8/14/2019 SPM AddMath2 Ans (Kedah)

10/16

Qn. Solution and Marking SchemeSub-

mark

Full

Mark

9(a)

(b)

(c)

(d)

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

Using formula

0

0

6

2

2

4

0

0

2

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

3472/2 Additional Mathematics Paper 2 [Lihat sebelah

SULIT

2k

• 8/14/2019 SPM AddMath2 Ans (Kedah)

11/16

Qn. Solution and Marking SchemeSub-

mark

Full

Mark

10(a)

(b)

(c )

=

10

6sin 1 or equivalent

= 2POQ

Alternative solution :

122

= 102

+ 102

2(10)(10) cos POQ

+=

)10)(10(2

121010cos

2221POQ

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

3472/2 Additional Mathematics Paper 2 [Lihat sebelah

SULIT

2k

• 8/14/2019 SPM AddMath2 Ans (Kedah)

12/16

Qn. Solution and Marking SchemeSub-

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

3472/2 Additional Mathematics Paper 2 [Lihat sebelah

SULIT

K1

N1

K1

N1

Use P(X = r) = n Crprqnr,

p + q = 1

K1

N1

K1

use

Z =

X

K1

P1

N1

2k

• 8/14/2019 SPM AddMath2 Ans (Kedah)

13/16

3472/2 Additional Mathematics Paper 2 [Lihat sebelahSULIT

Qn. Solution and Marking SchemeSub-

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

• 8/14/2019 SPM AddMath2 Ans (Kedah)

14/16

3472/2 Additional Mathematics Paper 2 [Lihat sebelahSULIT

Qn. Solution and Marking SchemeSub-

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

• 8/14/2019 SPM AddMath2 Ans (Kedah)

15/16

3472/2 Additional Mathematics Paper 2 [Lihat sebelahSULIT

Qn. Solution and Marking SchemeSub-

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

(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

• 8/14/2019 SPM AddMath2 Ans (Kedah)

16/16

3472/2 Additional Mathematics Paper 2 [Lihat sebelahSULIT

Qn. Solution and Marking SchemeSub-

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