Johor Addmaths SPM 2010 Trial P2 Marking Scheme

SULIT 3472/2

3472/2 Hak Cipta JPNJ 2010 [Lihat sebelah SULIT

JABATAN PELAJARAN NEGERI JOHOR PEPERIKSAAN PERCUBAAN SPM 2010

ADDITIONAL MATHEMATICS Kertas 2

Kertas soalan ini mengandungi 17 halaman bercetak

3472/2 Additional Mathematics Kertas 2 September 2010 2 ½ Jam

MARKING SCHEME

SULIT 3472/2


2

BAHAGIAN A

No Solution Sub marks

Total marks

1

x = 1 + 2y Or y = 2

1x Or y =

14x

Or x = 14

y

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

x + 1 = )

21(*

4x

)

Or or equivalent 2y2 + 2y – 4 = 0 x2 – 9 = 0 (x – 3)(x + 3) = 0 (2y – 2)(y + 2) = 0 x = 3, - 3 y = - 2 , y = 1 or y = - 2 , y = 1 or x = 3, - 3 Note : OW-1 if the working of solving quadratic equation is not shown.

5

5

Solve the quadratic equation by using the factorization @ quadratic formula @ completing the square

P1

K1

K1

N1

N1

SULIT 3472/2


3


Total marks

2 a)

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

3

a) i)

ii)

d = 50 Use Tn = a + (n – 1 )d T6 = 500 + 5(50) 750 OR 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 : Ụ

Minimum point : (2, -7)

N1

P1

P1

K1

N1

N1

N1

N1

K1

K1

P1

SULIT 3472/2


4


Total marks

b) Use Sn =

2n

[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 for correct answer.

3

8

4 a)

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. 17.75 Accept in the range (17.50 – 18.00)

4

3

7

N1

N1

K1

N1

K1

P1

K1

K1

K1

N1

SULIT 3472/2


5

2

4

6

8

10

12

14

16

18

4.5 9.5 14.5 19.5 24.5 29.5 34.5

SULIT 3472/2


6

No Solution Sub

marks Total

marks

5 a)

b) i)

ii)

Use identity Cos2x – Sin2x = Cos2x Or 2sinxcosx = sin2x LHS = RHS No mistake allowed

2 Graph Sin

2 period in 0 ≤ x ≤ 2 Amplitude 2 Max = 2 and min = - 2 Drawing of the straight line from the equation involving x and y, either gradient OR y intercept of straight Line must be correct.

y = 1 - 2x

Straight line drawn correctly and Number of solutions = 4 All must be correct

2

3

3

8

K1

N1

0

2

P1

P1

K1

N1

N1

-2

P1

1

SULIT 3472/2


7

No Solution Sub

marks Total

marks

6 a) 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 = ½x + 6 y = - 2x + 16 ½x + 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)(374

)4(1)(3

yORx

D (8, 0) OW – 1 for correct answer without working.

3

2

3

8

P1

N1

K1

K1

N1

N1

K1

K1

SULIT 3472/2


8

BAHAGIAN B

No Solution Sub

marks Total

marks

7 a)

b)

c)

y = 4 – x2

dxdy

= - 2x

= -2 mPQ = ½

013 k

= ½

k = 25

Integrate (4 - x2 )

Use limit 2*

1in to *[

3)(4

3xx ]

A1 = 35

A2 = find the area of Trapezium

= ½ (25

+ 3)(1) = 4

11

OR Area of shaded region = A1 + A2.

= 1253

⁄⁄ 4.42

Integrate x2

[4y - 2)( 2y

] Use limit 4

3in *[

2)(4

2yy ]

½

3

4

3

10

K1

K1

K1

K1

N1

K1

K1

N1

K1

N1

SULIT 3472/2


9

16.500


9

SULIT 3472/2


10

1 2 3 4 5 8 7 6 0

2

4

6

8

10

12

14

16

18

20

xy

x

SULIT 3472/2


11

4n – 5m = -1


11

SULIT 3472/2


12


12

SULIT 3472/2


13

- 1.281 // - 1.282

P(Z>10160h

) = -1.281 // -1.282


0.3085

13

SULIT 3472/2


14


14

SULIT 3472/2


15


100

y = 45.36

15

SULIT 3472/2


16


<BAD = 35.42°

16

SULIT 3472/2


17


17

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

Johor Addmaths SPM 2010 Trial P2 Marking Scheme

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