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2007 SPM TRIAL EXAMINATION Marking Scheme

Qn. Solution and Marking SchemeSub-

mark

Full

Mark

7

(a)

(b)(i)

(ii)

(c)

xy = qx2 p

5

3

2 10

JPNKedah 3472/2 [Lihat sebelah]

Additional Mathematics Paper 2 SULIT7

7

xy112135.25375 x2

49162536xy112135.25375 x2

49162536

Correct axes and

scale

t= 4

K1

N1

N1

All points plottedcorrectly

t= 4Line of best-fit

t= 4N1 K1 intercept = p

or

gradient = q

N1

K1

N1

q = 2

p = 3

Use

x2 = 20.25

from graph

or by using

equationy = 9.56

K1

N1

N1

N1 K1

N1

K1

N1y = 9.56

N1N1

N1

N1

P1

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2007 SPM TRIAL EXAMINATION Marking Scheme

JPNKedah 3472/2 [Lihat sebelah]

Additional Mathematics Paper 2 SULIT8

8

60

10 20 30

70

5 15 25

Graph ofxy against x2

x2

xy

80

50

40

10

35

20

30

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2007 SPM TRIAL EXAMINATION Marking Scheme

Qn. Solution and Marking SchemeSub-

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Full

Mark

8.

(a)(i)

(ii)

(b)

Q(3, 0)

Knowing Area = b

a

dxy

Integrate dxx )9(2

=

39

3x

x

3

22

3

26

== BorA

13 : 11

Integrate dxx 22 )9(

421881 xx +

1

0

53

)5

681

+x

xx

5

175

1

5

4 10

JPNKedah 3472/2 [Lihat sebelah]

Additional Mathematics Paper 2 SULIT9

9

P1

N1

K1

K1

N1

K1 Substitute correctlimit or

N1

P1

K1

K1Substitutecorrect limit

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2007 SPM TRIAL EXAMINATION Marking Scheme

Qn. Solution and Marking SchemeSub-

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Full

Mark

9(a)

(b)

(c)

(d)

2=PQm

Equation ofAB :

y 1 = 2 (x 3 )

y =2x + 7 or equivalent

3

2==PRAC mm

2

3=m

y 7 =2

3 (x 4 )

2y + 3x = 26 or equivalentPQRABC = 4

Using formula

1

3

7

4

3

6

1

3

2

1=PQR

32

1

2

4

3 10

JPNKedah 3472/2 [Lihat sebelah]

Additional Mathematics Paper 2 SULIT10

10

P1

K1

N1

Find eqn. of straight line

P1

Find eqn. of locusUsing m

1m

2= 1

K1

N1

K1

N1

Find area of

triangle

K1

K1

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2007 SPM TRIAL EXAMINATION Marking Scheme

Qn. Solution and Marking SchemeSub-

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Full

Mark

9(a)

(b)

(c)

(d)

Alternative solution :

2=

PQm

Equation ofAB :

y 1 = 2 (x 3 )

y =2x + 7 or equivalent

Equation ofAC:

3

2==PRAC

mm

y 7 =3

2(x 4 )

3y = 2x + 13

Solving simultaneous equations :

y =2x + 7 and 3y = 2x + 13

A(1, 5)

Using mid-point,

C(7, 9)

Using distance formula :2222 )9()7()5()1( +=+ yxyx

2y + 3x = 26 or equivalent

B(5, -3)

Using formula

51

97

35

51

21

=ABC

32

1

2

4

3

10

JPNKedah 3472/2 [Lihat sebelah]

Additional Mathematics Paper 2 SULIT11

11

P1

K1

N1

Find eqn. of straight line

Find eqn. of AC

or BC and solvesimultaneous eqn.

for A or C

N1

Find area oftriangle

K1

P1

N1

K1

K1

N1 A or C

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2007 SPM TRIAL EXAMINATION Marking Scheme

Qn. Solution and Marking SchemeSub-

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Mark

10(a)

(b)

(c )

= 10

8cos

1APQ or equivalent

radAPB 287.1=

radAQB 855.1=

AQ = 6cm

Perimeter = 8 ( 1.287 ) + 6 ( 1.855 )

= 21.43 cm

)287.1sin287.1()8(2

1 2

+ )855.1sin855.1()6(21 2

= 26.57 cm2

= 3.9149 cm

2

4

4

10

Qn. Solution and Marking SchemeSub-

mark

Full

Mark

JPNKedah 3472/2 [Lihat sebelah]

Additional Mathematics Paper 2 SULIT12

12

N1

K1

P1

P1

Use formula

s = r and

add up twoarcs

N1

K1

Using formula

Lsegment

=r2( -sin )K1

K1

K1

Add up twosegments

N1

Using formula

Lsector

=r2

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2007 SPM TRIAL EXAMINATION Marking Scheme

11

(a)(i)

(ii)

(b)(i)

(ii)

P( X = 2 )

= 5C2(3

2)2(3

1)3

=81

40or0.4938

P( X 1 )

= 1 P( X = 0 )

= 1 5C0(3

2)0(3

1)5

=243

242or 0.9959

P ( X x ) = 0.9

= P ( Z 24

42x) = 0.9

Selesaikan

24

42x= 1.28

x = 11.28 or 11

P ( X 80)

= P ( Z 24

4280)

= 0.0571

Number of candidates = 0.0571 1000

= 23

2

3

3

210

JPNKedah 3472/2 [Lihat sebelah]

Additional Mathematics Paper 2 SULIT13

13

K1

N1

N1

use

Z =

X

N1

Use P(X = r) = n Cr prqnr ,

p + q = 1

K1

N1

K1

K1

K1

K1Use

P(X = r) = n Cr prqnr ,

p + q = 1