trial addmate spm 2010 johor paper 2 answer

8/8/2019 Trial Addmate Spm 2010 Johor Paper 2 Answer

1/19

SULIT 3472/2

3472/2 Hak Cipta JPNJ 2010 [Lihat sebelahSULIT

JABATAN PELAJARAN NEGERI JOHORPEPERIKSAAN PERCUBAAN SPM 2010

ADDITIONAL MATHEMATICSKertas 2

Kertas soalan ini mengandungi 17 halaman bercetak

3472/2AdditionalMathematicsKertas 2September 2010

2 Jam

MARKING SCHEME


2/19

SULIT 3472/2


2

BAHAGIAN A

No Solution Submarks

Totalmarks

1x = 1 + 2y Or y =

2

1xOr y =

1

4

xOr x = 1

4

y

Eliminate x or y*(1 + 2y)y + y = 4Or

x + 1 =

)2

1(*

4

x)

Or

or equivalent

2y2 + 2y 4 = 0x2 9 = 0

(x 3)(x + 3) = 0(2y 2)(y + 2) = 0

x = 3, - 3y = - 2 , y = 1or

y = - 2 , y = 1or

x = 3, - 3

Note :

OW-1 if the working of solving quadratic equation is not shown.

5

5

Solve the quadraticequation by using thefactorization @quadratic formula @completing the square

P1

K1

K1

N1

N1


3/19

SULIT 3472/2


3


Totalmarks

2a)

b)

c)

2( x 2)2 7

m = 2

k = - 7.

OR equivalent method

-2(x 2 )2 + 7 Or equivalent

3

2

1 6

3a) i)

ii)

d = 50

Use Tn = a + (n 1 )d

T6 = 500 + 5(50)

750OR other valid method.

Use Tn = a + (n 1 )d = 1000

150

5001000

11.Or November 2009.

1

2

2

1

x

y

2, - 7

Shape :

Minimum point : (2, -7)

N1

P1

P1

K1

N1

N1

N1

N1

K1

K1

P1


4/19

SULIT 3472/2


4


Totalmarks

b) Use Sn =2

n[2a + (n 1 )d]

2

12 [2(500) + (11)(50)]

9300

*9300 x 0.08 = RM 744

Note :If listing method is used all terms must be correctly listed, accept forcorrect answer.

3

8

4a)

b)

*L = 19.5 or *F = 21 + p or *fm = 10

Use median formula

21.5 = *19.5 + 5)10*

)21(*)2

55(

(

pp

With *fm and F corresponding to *L

8 = 13 p

p = 5

Draw histogram with scale given.Sekurang-kurangnya 6 bars.

Find the mode from his histogram.

17.75Accept in the range (17.50 18.00)

4

3 7

N1

N1

K1

N1

K1

K1

K1

K1

N1


5/19

SULIT 3472/2


5

2

4

6

8

0

2

4

6

8

4.5 9.5 14.5 19.5 24.5 29.5 34.5


6/19

SULIT 3472/2


6


Totalmarks

5a)

b) i)

ii)

Use identityCos2x Sin2x = Cos2xOr 2sinxcosx = sin2x

LHS = RHSNo mistake allowed

2

Graph Sin2 period in 0 x 2

Amplitude 2

Max = 2 and min = - 2

Drawing of the straight line from the equation involvingx and y, either gradient OR y intercept of straight

Line must be correct.y = 1 -

2

x

Straight line drawn correctly andNumber of solutions = 4

All must be correct

2

3

3

8

K1

N1

0

2

P1

P1

K1

N1

N1

-2

P1

1


7/19

SULIT 3472/2


7


Totalmarks

6a) i)

ii)

b)

MBC = - 2

Use y y1 = m(x x1)Or equivalent method,and substitute x = 7 and y = 2

y - 2 = - 2(x 7)y = - 2x + 16

Solve simultaneous equation

y = x + 6y = - 2x + 16

x + 6 = - 2x + 16

x = 4, y = 8

B (4, 8)

Use C (7, 2) = ]4

)8(1)(3,

4

)4(1)(3[

yx

24

)8(1)(37

4

)4(1)(3

yOR

x

D (8, 0)

OW 1 for correct answer without working.

3

2

38

K1

K1

N1

N1

K1

K1


8/19

SULIT 3472/2


8

BAHAGIAN B


Totalmarks

7a)

b)

c)

y = 4 x2

dx

dy= - 2x

= -2

mPQ =

01

3

k=

k =

2

5

Integrate (4 - x2 )

Use limit 2*

1in to *[

3

)(4

3x

x ]

A1 =3

5

A2 = find the area ofTrapezium

= (25 + 3)(1) =

411

ORArea of shaded region = A1 + A2.

=12

53 4.42

Integrate x2

[4y - 2

)(2y

] Use limit

4

3 in

*[ 2

)(

4

2y

y ]

3

4

3 10

K1

K1

K1

K1

N1

K1

K1

N1

K1

N1

8


9/19

SULIT 3472/2


9



10/19

SULIT 3472/2


10

1 2 3 4 5 8760

2

4

6

8

10

12

14

16

18

20

xy

x

10


11/19

SULIT 3472/2


11



12/19

SULIT 3472/2


12



13/19

SULIT 3472/2


13



14/19

SULIT 3472/2


14



15/19

SULIT 3472/2


15



16/19

SULIT 3472/2


16



17/19

SULIT 3472/2


17



18/19

SULIT 3472/2


18

20 40 60 80 100 120 140 160

20

40

60

80

100

120

140

160

180

200

180 200

x = 120

(60, 140)

R

18


19/19

SULIT 3472/2


19

trial addmate spm 2010 johor paper 2 answer

Documents