SULIT3472/2Additional MathematicsPaper 2Mei 2008

SEKOLAH BERASRAMA PENUH KEMENTERIAN PELAJARAN MALAYSIA

PEPERIKSAAN PERTENGAHAN TAHUN TINGKATAN 5

2008

ADDITIONAL MATHEMATICS

Paper 2

MARK SCHEME

This mark scheme consists of 10 printed pages

SKEMA PERMARKAHAN MATEMATIK TAMBAHAN KERTAS 2 PEPERIKSAAN PERTENGAHAN TAHUN TINGKATAN 5, 2008

Number Solution and mark scheme Sub Marks Full Marks

1y = 2 – 2x or x =

(2 – 2x)2–10 = – 2x or

Solve the equation(2x + 1)(x – 3 ) or

x = – , 3 and y = 3 , – 4

P1

K1

K1

N1 N1 52

(a)

(b)

kx2 – 4 = –1 k(1)2 – 4 = –1 k = 3

K1K1N1 3

6

– 4x + c

3 = (1)3 – 4(1) + cc = 6y = x3– 4x + 6

K1

K1

N1 3

3(a)

(b)

(c)

324 = a + ( 40 – 1 ) 8 a = 12

K1N1 2

6

P = S40 = [2(12) + ( 40 – 1 )8]

= 6720 cm

172 = 12 + ( n – 1 )(8)n = 21

K1

N1 2

K1N1 2

4(a)

cos sin 2 = ) = cos ( 1 – 2 sin2 ) = cos cos 2

K1N1 2

(b)

2

0

y

x

2y = 2 cos2x

Number Solution and mark scheme Sub Marks Full Marks

Shape as in the above diagramAmplitude is 2 unitNumber of period is 1

Equation of straight line y = – 1

Straight line is drawn Number of solution 2

K1K1K1

K1K1N1 6

8

5 (a)

(b)

(c)

K1

N1 2

8

P1

K1

N1 3

Coordinate P =

Area of parallelogram PRMQ = 2 x Area of Δ PMR

K1

K1

N1 3

6(a) K1

N1

K1

3

–2

–1y = – 1

Number Solution and mark scheme Sub Marks Full Marks

(b)

Correct standard deviation = 12.71 kg

N1 4

K1

N1

P1 3 7

7 Refer to the graph 10

8

(b) (i)

(ii)

= (7 – 2k )u + ( 8 – 4k )v

(c) 7hu + 2hv = (7 – 2k )u + (8 – 4k )v or 7h =7 – 2k or 2h = 8 – 4k

Solve 7h + 2k = 7 and 2h + 4k = 8

,

(d) =

K1N1 2

P1

K1N1 3

K1

K1

N1 N1 4

N1 1

10

9(a)

(b)

sin POS =

POR = 13076POR = 2282 rad

K1

N1 2

10 or 2282*(11)10 + 2.282*(11)5652*

K1K1N1 3

4

Number Solution and mark scheme Sub Marks Full Marks

(c)

(d)

10

(11)(11)sin 13076*

4583

K1

N1 2

(11)(11)2282* or (10)(10)

(10)(10) [ (11)(11)2282* (11)(11)sin 13076*]

64.86

K1

K1

N1 3

10(a)

(b)

Area of rectangle, A1 = xy

A2 =

A3 =

Area of rectangles form a GP and the common ratio is

K1

K1

N1 3

10

(i) a = xy = 160 ( 80 ) = 12800

(ii) =

= 266.65

(iii)

= 17066.67 or or

K1K1

N1 3

K1

N1 2

K1

N1 2

5

Number Solution and mark scheme Sub Marks Full Marks

11

(a)

(b)

(c)

or 12

+ 12 = 20

K1

K1

N1 3

10

y = x2 – 4x + 7

2x – 4 = 0

Minimum point ( 2 , 3 )

Expansion of ( x2 – 4x + 7 )2 = x4 – 8x3 + 30x2 – 56x + 49

or or 40.4

K1K1

N1 3

K1

K1

K1

N1 4

6

Number Solution and mark scheme Sub Marks Full Marks

12 (a) = 5.054 cm

(b) or

or

or

(c) Area of

= 17.61 cm2

(d) Area of = 2 Area of

Height constant , base is doubled AC = CD = 9cm or AD = 18 cm

K1N1 2

K1

K1

N1 3

K1

N1 2

K1

K1

N1 310

13 K1

K1

N1 3

K1

N1 2

K1

K1

7

Number Solution and mark scheme Sub Marks Full Marks

N1 3

K1

N1 2

10

Number Solution and marking scheme Sub Marks Full Marks

14K1

N1 2

K1

K1

N1 3

K2

N1 3

K1

N1 2

10

8

Number Solution and marking scheme Sub Marks Full Marks

15

K1

N1 2

K1

N1 2

P1

K1

N1 3

P1

K1

N1 3

10

9

Number Solution and marking scheme Sub Marks Full Marks

10

x

4.0

5.0

y

3.0

2.0

1.0

0 0.2 0.4 0.6 0.8 1.0 1.2

6.0

7.0

8.0

9.0

10.0

1.4

x0.100.250.500.751.001.253013925446958.470999N

N1

*

2 2 P1

2gradient or 6.067 K1

2 9.99 5.44 K1

A 1.25 0.5A 0.3297 0.05 N1

2B2.35 K1

AB 0.3874 0

By x

A A

A

.05 N1

( 0.5, 5.44 )

(1.25, 9.99)

N1 fit best of Line

K1 correctly plotted are points All

K1 axesboth Correct

QUESTION 7

11