answer zone c p2 add maths 2009

8/14/2019 Answer Zone C P2 Add Maths 2009

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PEPERIKSAAN PERCUBAAN BERSAMA

JAWATANKUASA KURIKULUM ZON C KUCHING

Model answer

No. Solution Marks

1 x + 2 y = 1

x = 1 2 y ( substitute intox2

+ 2y2

= 5 x y)Therefore,

( 1 2 y ) 2 + 2 y 2 = 5 ( 1 2 y ) y

1 4 y + 4 y 2 + 2 y 2 = 5 y + 2 y 2

1 4 y + 6 y 2 - 5 + y - 2 y 2 = 0

4 y 2 - 3 y 4 = 0

)4(2

)4)(4(4)3()3( 2 =y

y = 1.443 y = -0.693

x = 1 2 ( 1.443 ) x = 1 2 ( -0.693 ) = - 1.886 = 2.386

52(a) 6x -y = k

y = 6x k

The gradient of the tangent, m = 62(3 2) 3y x=

29 12 1 y x x= +

The gradient of the tangent,

tangent 18 12dy

m xdx

= =

18 12 6x =

x = 1

2(3(1) 2) 3y =

y = -2

Hence, coordinatesP= (1, -2) 5

(b) At coordinatesP= (1, -2)

6(1) (-2) = kk= 8

1

1

3 (a) Length of first arc = r = 2

Length of second arc = 5

Length of third arc = 8

This forms an AP with a = 2 and d = 3

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2

[2 ( 1) ]2

[4 ( 1)(3 )] 772

( 3 ) 154

3 154 0

(3 22)( 7) 0

227

3

( )

n

nS a n d

nn

n n

n n

n n

n or n

ignored

= +

+ =

+ =

+ =+ =

= =

(b) Radius:

2, 5, 8, a = 2, d = 3

T7 = 2 + 6(3) = 20Radius of the seventh semicircle = 20 cm

Area of the seventh semicircle =21 ( )(20)

2

= 200 cm2

7

4(a) LHS =

x

xx2sin1

sin1sin1

++

=x2cos

2

= x2sec22

(b) (i)xy 2sin=

The shape of sinx

Two period for the range .3600 x

Modulus of the graphThe amplitude is 1

(ii) Find the equation =y2

x

Draw the straight line =y2

xin the graph

Number of solutions = 86

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5(a)

18.0

20.0

16.0

14.0

12.0

10.0

8.0

6.0

0

2.0

4.0

89.569.549.529.59.5

57. 5

Histogram

Correct scale P1

Shape of Histogram P1

Correct value N1


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5 (a) Histogram

modal mark = 57.5

(b)

5(19.5) 11(39.5) 18(59.5) 8(79.5)

42

2239

53.3142

mean mark+ + +

=

= = 6

6 a) i) BD = BE + ED

= yx 103 +

ii) AD = AB + BD

1(10 ) ( 3 10 )

5

12 3

y x y

y x

= + +

=

b) CD = (m-1) AD

( 1)(12 3 )12( 1) 3( 1)m y x

m y m x= =

CD = nEB + BD

yxn

yxxn

10)33(

)103()3(

+=

++=

Comparing the coefficient of y , 12m 12 = 10, m =11

6Comparing the coefficient ofx , 3n 3 = 3 3m.

3n = 6 3(11

6

)

n =1

6

7

7 (a)

(2x-1) 1 3 5 7 9

logy 0.46 0.8

6

1.2

4

1.6

0

1.97

(b) Graph

(c )2 1xy pk =

logy = (log k) (2x-1)+ logp

From the graph,

(i) log k= gradient of the graph

=1.97 0.3

9 0

= 0.1856

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k=1.533

(ii) logp = 0.3

P = 1.995

10

8 (a) Intersection of curve and straight line;2( 3) 9x x = +

x = 0 or 7

Hence, n = 7

(b) Area of trapezium OABC =1 175

(9 16)(7)2 2

+ =

Area under curve AB =7

2

0( 3) x dx

=7

2

0( 6 9)x x dx +

=73

2

0

3 93

xx x +

=

32(7) 3(7) 9(7) 0

3

+

= 30.33

Area of shaded region, P

= Area of trapezium OABC area under curve AB

=175

2-30.33

= 2157 unit6

(c ) Generated volume =3

2

0y dx

=23

2

0 ( 3)x dx

=

35

0

( 3)

5

x

=

5 5(3 3) (0 3)

5 5

=

3348 unit5

10

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9

10

10 (a) MAB

MCE

Equation of DA

3

(b)

Equation of BC is

Equation of perpendicular bisector of AB is

4

(c ) Area of ABCD

= [ ]1

(8 42 3) (6 7 4)2

+ + +

=25 unit2 3

11 (a) (i) Let X be the student from Science Stream

P(X= number of student selected) = = 0.6

a)

b)

K

1

1

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1

1

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c)

K

1

1

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1

1

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1

1K

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For this binomial distribution, n = 6 , ,

[6C0 +6C1

5]

]

= 0.95904

(ii) For this binomial distribution, n = 280 , ,

Standard deviation,

= 8.198

(b) (i) Let X be the diameter of a golf ball.

Given and

0.3085

Percentage = 30.85%.

(ii)

= 0.3721

n = 107510

12(a) t

dt

dva 1015 ==

215,0 == msat

(b)3

5

2

15 32 tts =

,0=s 03

5

2

15 32=

tt

0)29(52

= tt

st2

9=

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1

1K1

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1

1N1

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(c)

3,0

0)3(5

0515

0

2

==

=

=

=

tt

tt

tt

dt

ds

At ,3=t )3(1015=

a= -15 (< 0) maximum

3

)3(5

2

)3(15 32=s

= 22.5 m

10

13 (a) (i)

(a) (ii)

Price index ofPfor 2005 based on the year 2001

(b) (i) Given the composite index of the drink for 2005 based on the year

2003 was 123

x = 122.5

(ii)

Price of a box of drink in 2003 if its corresponding price in the year 2005

= RM 10.1610

14 (a) Using the cosine rule,

BD = 7 + (18.2) - 2 (7)(18.2) cos 52

2

K

1

1K

1

1K

1

1

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1

1N1

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1

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2

1

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2001 2003 2005

-- 100 125

100 120 ?


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BD = 14.95 cm

(b) Using the sine rule,

sin < ABD = sin 52

7 14.946

< ABD = 21 40'

2

(c )sin < B'DC = sin 110

6.3 14.946

< BDC = 2320'


10/12

maximum point occur at point (100, 400) and point (400, 100)

By testing the point (100,400):20(100) + 40(400) = k

k= RM18000

By testing point(400, 100):

20(400) + 40(100) = kk = RM12000

So, maximum sales is RM18000.

(ii) Whenx = 200

100 y 300

The fee collection at (200, 300) is maximum=20(200) + 40(300)

=4000+12000

=RM16000

log y

7b

0 2 4 6 8 10 (2x

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

0.3

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11/12500

y = 100

x +y = 500

y - x = 300

(100,400)

(400,100)

R

0 100 200 300 400

100

200

300

400

500

y15b

0100200 4003002001000

Label axes correctly K1

Draw scale correctly N1

Plot all the points correctly N1

Draw the best line fit N1


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answer zone c p2 add maths 2009

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