1

3472/2 Form 4 Additional Mathematics Paper 2 2010

PEPERIKSAAN AKHIR TAHUN 2010

TINGKATAN 4

ADDITIONAL MATHEMATICS

Paper 2

MARKING SCHEME

This marking scheme consists of 11 printed pages.

2

Number

Solution and marking scheme

Submarks

Full marks

1

x = 6 – 2y 2(6 – 2y)2 – y2 + y(6 –2y) = 2

)5(2)70)(5(4)42()42( 2

y

y = 6.108 , y = 2.292 x = -6.216 , x = 1.416 atau setara.

1

1

1

1 1

5

2

( a ) g -1 = g -1(4) = 3 ( b ) = 6

a = 5

2

1 1

2

6

3

a i) 20 ii) 106.45 b i) 19 4(3) + 7 ii) 47.61 2.3 2 (9)

1 1

2 B1

3

B2

7

3

4

a) 9,3 hp

039

0

hpdxdy

@ p – h = 12

b) 2

2

2

2

)13()3)(()2)(13(

xxxx

dxyd

2

2

)13(23

xxx

2, 2

B1

B1

1

1

6

( a ) y + 2x - 4 = 0 y = -2x + 4 Gradient of a tangent = -2

1 1

2

5

( b ) y = x3 + 3x2 -11x + 9

dxdy = 3x2 + 6x - 11

-2 = 3x2 + 6x - 11 3x2 + 6x - 9 = 0 x2 + 2x - 3 = 0 ( x-1)(x+3) = 0 x=1 or x = -3 When x= 1 , y=2 When x=-3 , y=42 Substitut (1,2) into y = -2x + 4 2=2 Substitut (-3,42) into y = -2x + 4 42 ≠10

1

1

1

4

4

Thus the coordinate of point P is ( 1, 2 )

1

( c ) y - 42 = -2 ( x + 3 ) y = -2x + 36

1 1

2

( a ) log3 m√n = log3 m + ½ log3 n = x + ½ y

1 1

2

( b ) 2x 3x = 62x – 6

( 2 X 3 )x = 62x – 6

6x = 62x – 6 x = 6

1

1

2

6

(c ) 2 log9 x = log34 log9 x2 = log34

9log

log

3

23 x = log34

23

23

3loglog x = log34

log3 x2 = log342

x2 = 42 x = 4

1

1

1

1

4

Answer four questions from this section

7

( a ) i. PG = GA 22 )3()2( yx = 22 )03()22( x2+y2 -4x -6x – 12 =0 ii. B (5 , t ) , 52 +t2 -4(5) – 6t -12 = 0 t2 -6t – 7= 0 ( t + 1 )( t – 7 ) = 0 t = - 1 or 7

1 1 1 1 1 , 1

5

( b ) Gradient GA =

2230 =

43

The equation of tangent ; y – 0 = 3

4 ( x + 2 )

y = 3

4 x - 38

x = 0 , y = - 38 T ( 0 , -

38 )

Area of triangle 0AH = ½ x 2 x 38

= 38 unit2

1 1 1 1

10

8

( a ) Gradient TU =

0657

= 31

Gradient UV = - 3 3

67

pq

3p + q -25 = 0………………….(1) ( b ) Area triangle TUV = ½

575060

qp

= ½ )56()730( pqp 15 + p – 3q shown ( c ) Area of TUVW = 40 units2

Area triangle TUV = 20 units2

15 + p – 3q = 20 p = 5 + 3q …………….( 2) sub (2) into ( 1) 3 [ 5 +3q ] + q – 25 = 0 q = 1 , p = 8 Koordinat V =( 8 , 1 )

1 1 1 1 1 1 1 1

10

6

( d ) Gradient TW = UV = -3 Equation TW ; y = -3x + 5 y + 3x = 5

5y 1

53

x

1 1

9

a)

4rad

b) 2 22 2OA = 2.828 cm c) OQ = 3OB = 3(2) = 6 BQ = 6 – 2 = 4 AP = OP – OA = 6 – 2.828 = 3.172 cm

Length of arc PQ = 64

= 4.713 cm

Perimeter of the shaded region = AP + AB + BQ + Arc PQ = 3.172 + 2 + 4 + 4.713 = 13.89 cm d) Area of the shaded region = Area sector OPQ – Area ofOAB

= 21 1(6 ) (2)(2)2 4 2

= 12.14 2cm

1

1 1

1

1

1

1

1,1

1

10

10

a) 5x + 5x + 6x + y + y = 120 y = 60 – 8x ………….. (1 )

b) A = 16 ( ) (6 )(4 )2

x y x x

= 26 12xy x ……………………(2) Substitude (1) into (2)

1 1 1 1

10

7

26 (60 8 ) 12

36 (10 )A x x x

x x

c) 360 72dA xdx

For A is a maximum

0dAdx

360 – 72x = 0 x = 5 when x = 5 then y = 60 – 8(5) = 20 Maximum area = 36(5)(10 – 5) = 900 2cm

1 1 1 1 1 1

11

(a) (i) =48.25

p = 17

(ii) m =

= 47.23

(b) Graph Draw the histogram with the uniform scale – x-axis & y-axis

2 1

2 1

2

10

8

Draw – find mode Mode = 45.5

1 1

Answer two questions from this section

12 a) x = 140, y = RM 11.25 , z = RM8.00 b) 131.50

100

)15(150)25(130)10(125)20(140)30(120

1, 1, 1

3

B2

10

9

a) RM 110.46

80.157

10012050.131

8.15710070

07/05

05

0705

XI

XPPXP

4

B3

B2

13

a) (i) 110100220

xx

x = RM 200 (ii) y = 100

1505.187 x

y = 125 (iii) 130100

400xz

z = RM 520 (b)

WIW

I

280

)40130()60105()80125()100110( xxxx

116.07 c) 07.116100

75000xU

U = RM 87 053.57 (d) 07.116

100140 x

162.5

1 1 1

1 1 1 1 1,1 1

10

14

a)

2 2 2

2 2 0

2( ( )cos12 8 2(12)(8)cos65126.86

11.26

AC AB BC AB BC ABC

AC cm

1 1

10

10

b) i)

0

0 '

sin sin 7511.26 12.8

58 11

ADC

ADC

ii) 0 0 0 '180 75 58 11ACD

= 0 '46 49 c) Area of quadrilateral ABCD = Area of ABC + Area of ACD

0 0 '1 1(12)(8)sin 65 (11.26)(12.8)sin 46 492 2

= 96.05 2

cm

1 1 1 1 1,1,1

1

11

15

Sin ADC = 23

ADC = 60 0 ( a ) AC 2 = (6.6)2 + (5.4)2 -2(6.6)(5.4)cos 60 0 = 43.56 + 29.16 - 71.28(0.5) = 37.08 AC = 6.089 cm

(b) 089.6

sin = 3.950sin 0

= 30.100

(c) 9.99sin

AB = 50sin3.9

AB = 11.96 cm ½ X t X 11.96 = ½ X 6.089 X 9.3 X sin 99.9 t = 4.664 cm OR EC = AC sin 50 = 4.664 cm

1

1

1

1

1

1

1

1

1 1

1 1

10