Koleksi Diff 2009

7/31/2019 Koleksi Diff 2009

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SULIT September, 2009

3472/2 Additional Mathematics Paper 2 [Lihat sebelah

SULIT

5

3.(a) Sketch the graph of 2 sin 2y x= - for 0 2x p .

(b) Hence, using the same axes, sketch a suitable straight line to find

the number of solutions for the equation sin22

xx

p

= for

0 2x p . State the number of solutions.

(a) Lakar graf bagi 2 sin 2y x= - untuk 0 2x p .

(b) Seterusnya, dengan menggunakan paksi yang sama, lakar satugaris lurus yang sesuai untuk mencari bilangan penyelesaian

bagi persamaan sin22

xx

p

= untuk 0 2x p .

Nyatakan bilangan penyelesaian itu.

[4 marks]

[3 marks]

[4 markah]

[3 markah]

4. Given that ( )2 1x x - is the gradient function of a curve which passes

through the point )1,1( -P . Find

(a) the gradient of the tangent to the curve at P,

(b) the equation of the curve,

(c) the coordinates of the turning point at x = 1 . Hence determine

whether the turning point is a maximum or a minimum point.

[1 mark]

[3 marks]

[3 marks]

Diberi ( )2 1x x - ialah fungsi kecerunan bagi suatu lengkung yang

melalui titik )1,1( -P . Cari

(a) kecerunan tangen kepada lengkung itu di P,

(b) persamaan lengkung itu,

(c) koordinat bagi titik pusingan pada x = 1 . Seterusnya tentukansama ada titik pusingan itu adalah titik maksimum atau titik

minimum.

[1 markah]

[3 markah]

[3 markah]

j2kk


5/23

N0. SOLUTION MARKS

4

(a)

(b)

(c)

( ) ( )2

1 1 1 2dy

dx= - - - = -

3 2

4 3

4 3

( )

4 3

1 11

4 3

5

12

5

4 3 12

y x x dx

x xy c

c

c

x xy

= -

= - +

= + +

=

= - +

( )2

22

2

2

2

1 0

0@1

3 2

1 1 0

1 1 5 1

4 3 12 3

dy

x xdx

x

d yx x

dx

d yx

dx

y

= - ==

= -

= = >

= - + =

11,3

Minimum point.

N1

K1 Integrate gradient

function

K1 Substi tute ( 1,1)

into equation y

N1

K1 Find2

2

d y

dx

N1 N1

7

j2kk


6/23

Percubaan Addmath Kelantan 2009 Paper 2

Section A

Bahagian A

[40 marks][40 markah]

Answerall questions.Jawab semua soalan.

1 Solve the simultaneous equations n1

2m + 1 = 0 and m

2 9 = 2n.

Give your answers correct to three decimal places.

[5 marks]

Selesaikan persamaan serentak n1

2m + 1 = 0 dan m

2 9 = 2n.

Beri jawapan anda betul kepada tiga tempat perpuluhan.

[5 markah]

2 A curve with gradient function2

8x

x has a turning point at (k, 6).

Suatu lengkung dengan fungsi kecerunan2

8x

x mempunyai titik pusingandi (k, 6).

(a) Find the value ofk. [2 marks]Cari nilai k. [2 markah]

(b) Determine whether the turning point is a maximum or minimum point. [2 marks]Tentukan sama ada titik pusingan ini adalah titik maksimum atau titik minimum. [2 markah]

(c) Find the equation of the curve. [3 marks]Cari persamaan lengkung itu. [3 markah]


7/23

ADDITIONAL MATHEMATICS

PEPERIKSAAN PERCUBAAN SPM 2009(KELANTAN)

MARKING SCHEME

PAPER 2

No Solution Marks Total

1n =

1

2m 1 @ 2n = m 2 @ m = 2n + 2

m2

9 = 2(1

2m 1) @ m

2 9 = m 2 (2n + 2)

2 9 = 2n

m2

m 7 = 0 4n2

+ 6n 5 = 0

2( 1) ( 1) 4(1)( 7)

2(1)m

= @

26 (6) 4(4)( 5)

2(4)n

=

m = 3.193, 2.193 n = 0.596, 2.096n = 0.597, 2.097 m = 3.192, 2.192

P1

K1

K1

N1(both)

N1(both)

5

2(a)

(b)

(c)

, k= 2

When x = 2,2

2 3

161 3(

2

d y

dx= + = > 0)

so, the turning point (2,6) is a minimum point

2

8y x dx

x=

2 8

2

xy c

x

= + +

Therefore,

K1

N1

K1

N1

K1

K1

N1

7


8/23

SULIT

3472/2

SULIT

Section A

[40 marks]Answerall questions in this section .

Jawab semua soalan

1 Solve the following simultaneous equations:

Selesaikan persamaan serentak berikut.

92

34

223

yx

yx

[5 marks]

[5 markah]

2 (a) The straight line 6y + x = kis a normal to the curve, y = (3x 2)2 2 at point P.

Find the coordinates of point P and the value ofk. [ 4 marks]

Garis lurus 6y + x = k ialah normal kepada lengkung y = (3x 2)2

2 pada titik P.Carikan koordinat titik P dan nilai k. [4 markah]

(b) Given2

2

dx

yd= 2x , where y =

3

8and

dx

dy= 2 when x = 1. Express y in terms of x

[ 3 marks]

Diberi2

2

dx

yd= 2x dimana y =

3

8dan

dx

dy= 2 apabila x = 1. Ungkapkan y dalam sebutan x.

[3 markah]

3 ( a) Prove that cosec 2x cot 2x = cot x .

Buktikan bahawa kosec 2x kot 2x = kot x .

[ 3 marks ][ 3 markah ]

( b) (i) Sketch the graph of y = cos 2x for 0 x 2.Lakarkan graf bagi y = kos 2x untuk 0 x 2.

(ii) Hence, using the same axes, draw a suitable straight line to find the number of solutions tothe equation 2 cos 2x = x for 0 x 2 . State the number of solutions.Seterusnya, menggunakan paksi yang sama, lukiskan garis lurus yang sesuai untuk mencari

bilangan penyelesaian bagi persamaan 2 kos 2x = x for 0 x 2 .Nyatakan bilangan penyelesaiannya.

[ 5marks ]

[ 5 markah ]

4


9/23

2

NO.

SOALAN

PENYELESAIAN PER

MARKAHAN

1

2 2

2 2 2 3

3 2

8 3 182 2 2 2 2 3 2 3

8 3( ) 18( ) @8( ) 3 18 ( )3 3 2 2

6 1 0 27 33 8 0

(3 1)(2 1) 0@ (9 8)(3 1) 0

1 1,

3 2

8 1,

9 3

y xx atau y

y x xyy y x x

y y x x

y y x x

y y x x

y y

x x

P 1

K 1

K 1( factor)

N 1( Value y)

N 1(Value x)

2a)

16

6n t

m m

6(3x-2) = 6

P (1,-1)

-5 = k

2 2

3

3

( ) 3

2 1 33

3

33

dyb x c y x

dx

xc atau x c

c

xy x

K1

K1

N1

N1

K1(Integrate)

K1(Value c)

N1

3

x

x

x 2sin

2cos

2sin

1

xx

xcossin2

1cos212

x

x

sin

cos

K1

K1

N1


10/23

www.cikgurohaiza.com


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12/23

CONFIDENTIAL 3472/28

(b)Given that k= 4, draw a histogram to represent the frequency distribution of themark by using a scale of 2 cm to 10 marks on the horizontal axis and 2 cm to 1

student on the vertical axis.

Diberi k = 4, lukis sebuah histogram untuk mewakili taburan kekerapan markah

dengan menggunakan skala 2 cm kepada 10 markah pada paksi ufuk dan 2 cm

kepada 1 pelajar pada paksi tegak.

Hence, find the modal mark. [3 marks]

Seterusnya, cari markah mod. [3 markah]

5. (a) Find the equation of the normal to the curve xxy

1

3 += at (1 , 4). [3 marks]

Cari persamaan normal kepada lengkungx

xy1

3 += pada (1 , 4). [3 markah]

(b) Diagram 2 shows a leaking hemispherical container with a radius ofrcm.

Rajah 2 menunjukkan sebuah bekas bocor yang berbentuk hemisfera dengan jejari

r cm.

Given that the radius of the water surface is decreasing at the rate of 0.1 cm s

1

, findin terms of, the rate of change of the volume of water in the container at the instant

the radius of the water surface is 20 cm. [3 marks]

Diberi jejari permukaan air menyusut dengan kadar0.1 cm s1, cari dalam sebutan

, kadar perubahan isipadu air dalam bekas itu pada ketika jejari permukaan air

adalah 20 cm. [3 markah]

Diagram 2

Rajah 2

r cm


13/23

CONFIDENTIAL 3472/222

Question Working/Solution Marks Total

4.(a) (i) Use median formula with at least two of the L, N, F, f

and c correctly substituted.

)10(10

102

22

5.305.34

+

+=

k

6=k

(ii) median mark = 42.5

K1

N1

N1

P1

4(b) Correct axes and uniform scales with all the lower and

upper boundaries correctly labeled and the Height of at

least three bars are proportional to the frequency

Correct way of finding the value of mode.

Modal mark = 35.0

K1

K1

N1 7

5(a)

xxy

13 +=

2

13

xdx

dy=

Find gradient of normal and use )( 11 xxmyy =

at (1 , 4), 2=dx

dy

Gradient of normal =2

1

Equation of normal :

)1(2

14 = xy

092 =+ yx or equivalent

P1

K1

N1

5(b)Get the expression for V and find

dr

dVto determine the

value of dr

dV

at r= 20 cm.

3

3

2rV =

22 rdr

dV=

Usedt

dr

dr

dV

dt

dV=

13

2

80

)1.0()20(2

=

=

scm

dt

dV

K1

K1

N1 6


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15/23

3

Mean, x =690

30

= 23

2 = 1749030

- 232

= 54

1

1

1

4a.) x

dx

dy6

pm gent 6tan

When x = p , y = 3p2 + 1The equation of the tangent is y (3p

2 + 1) = 6p ( x p )y = 6px 3p

2 + 1

b ) 34 xdx

dy

xdx

dyy

= ( 4(2) 3 )p= 5p

C ) y = 2x 3x-1

2

32

xx

dx

dy

1

11

1

11

1

1


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17/23


ANSWER TRIAL PAPER 2, 2009

SECTION A

1.3

434

xyoryx

== 1M

Substitute x or y into eqn. 2. 1M

0434

34=+

+

y

y

y

yor

04

)3

4(

)3

4(

=+

+

x

x

x

x

or

equivalent.

0842 =+ yy or

08202 =++ xx

)1(2

)8)(1(444 2 =y 1M

Or

)1(2

)8)(1(42020 2

=x

y = 1.464, 5.464 1M

x = 0.392, 20.392 1M

2. (a) 49 2 = xdx

dy1M

10)49()18( 2 xxx 1M= 3 1M

(b) 049 2 =x 1MDetermine max. point 1M

9

79,

3

2== yx 1M

3. (a) graph cos 1M,

Max 3 and min 1 1M

1 cycle 1MShift up 1 unit 1M

(b) 22

=

xy 1M,

Draw straight line graph 1M

No. of solution = 2 1M

4. (a) [ ] 17608)1()12(22

+ nn

1M

[Accept =]Solve 1M

(n 20)(n + 22) = 0

n = 20 1M(b) 12 + (n 1) 8 = 108 1M

n = 13 1M

(c) 12 + (20 1)(8) 1M

= 164 1M

5. (a) 1.047 1M

094.2=POQ /2.095 1M(b) QR = 1.047 x 10 1M10 + 10 + 10.47 1M

= 30.47 1M

(c) A1 = 094.2102

1 2 1M

A2 =22

510102

1 or

equivalent 1M

Area of shaded region

= 61.4 1M

[ 61.39 61.45]

6. (a) LQ1 = 45.5 and FQ1 = 18 or fQ1= 20

(any 2 values 1M)

Q1 = )5(20

184

126

5.45

+ 1M

= 48.875 1M

(b) Use correct boundaries or mid points1M

Histogram with all correct frequency

2M

Mode = 53.25 53.75 1M

Section B

7. a)

x

1 0.67 0.5 0.4 0.33 0.25 0.2 0.1

y

1 0.95 0.77 0.64 0.57 0.48 0.42 0.32

1M, 1M

1

3

2


18/23

SULIT 9

3472/2 @ 2009 Hak Cipta JPWPKL [Lihat sebelah

SULIT

5 The equation of the straight line kxy =+13 is normal to the curve 534 2 = xxy at the pointA.

Persamaan garis lurus kxy =+13 ialah normal kepada lengkung 534 2 = xxy

pada titikA.

Find

Cari

(i) the coordinates of the pointA, [3 marks]koordinat titik A, [3 markah]

(ii) the value ofk, [2 marks]nilai k, [2 markah]

(iii) the equation of the tangent at the pointA. [2 marks]

persamaan tangen pada titikA. [2 markah]

6 (a) Prove that xx

x 2tan12cos

12cos=

+

. [2 marks]

Buktikan xx 2tan

12cos

12cos=

+

[2 markah]

(b)(i) Sketch the graph of xy 2cos

2

3= for 0 x 2. [3 marks]

Lakar graf bagi xy 2cos2

3= untuk0x 2.. [3 markah]

(ii)Hence, using the same axes, sketch a suitable straight line to find the numberof solutions for the equation 12cos

2

3=+

xx for 0 x 2.

State the number of solutions.

Seterusnya, dengan menggunakan paksi yang sama, lakar satu garis lurus

yang sesuai untuk mencari bilangan penyelesaian bagi persamaan

12cos2

3=+

xx untuk 0x 2.

Nyatakan bilangan penyelesaian.

[3 marks][3 markah]

SHARED BY PN. DING HONG ENG, SMK ALAM SHAH, KLSHARED BY PN. DING HONG ENG, SMK ALAM SHAH, KL


19/23

SULIT 3472/2

4. (a) List of perimeters ; 4x, 8x, 16x,2

2

3

1

2 ==T

T

T

T

This is Geometric Progression and r= 2 ..

(b) 9)2(410240 x= .......................................................................

x = 5................................................................................

(c) List of numbers of squares: 1, 4, 16,

Find14

)14(1

14

)14(1

14

)14(1 10

10

5

5

4

4

=

=

= SorSorS ......

=

14

14

14

14 410

410 SS ...........................................

= 349440...............................................................

1

1

1

1

1

1

1 7

5.(i) 38 = x

dx

dy..

2

.................................................................1338

=

=

x

x

A(2, 5)

(ii) 13(5) + 2 = k

k= 67 ..

(iii) 21)2(135 == corxy ..

y = 13x 21 .

1

1

1

1

1

1

1

7

6. (a) 1cos22cossin212cos 22 == xxorxx ..

x

xor

x

x2

2

2

2

cos

sin

cos2

sin2

x2tan=

1

1and ..


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22/23

SULIT 3472/2

3472/2 ZON A KUCHING 2009 SULIT

5

4 (a) If the volume of a cube decreases from 1253

cm to 124.4 cm3, find the small change

in the sides of the cube.

[3 marks]

(b) Given that2

3 4( )

3

xf x

x

+=

, find the value off(2). [3 marks]

5 (a) Prove that sin 2x = 2 sin2x cotx. [2 marks]

(b) Sketch the graph ofy = x2sin2 for 0 x . Hence, using the same axes,

sketch a suitable straight line to find the number of solutions of the equation x2sin2

=

x2for 0 x. State the number of solutions.

[6 marks]

6 (a)

A piece of wire is cut into 15 parts which are bent to form circles as shown inDiagram 2.

The radius of each circle increases by 3 cm consecutively.Calculate

(i) the radius of the last circle, [2 marks]

(ii) the area of the last circle. [1 mark]

(b) Diagram 3 shows a rectangular geometric pattern.

Diagram 2

The first rectangle isABCD and followed by MBNP andso on. The length and width

of the next rectangle is half of the length and width of the previous rectangle. GiventhatAB = 30 cm andBC= 20 cm. Find the perimeter of the seventh rectangle.

[3 marks]

5 cm2 cm 8 cm

A

D

B

C

M

P N

DIAGRAM 3

DIAGRAM 2

. . .


23/23

4

QUESTION

NO.SOLUTION MARKS

V= 0.6 ORdV

dx= 3x

2OR x = 5

20.6 3(5) x

0.008x

34

(a)

(b)

[ ]

2

2 2

2

22

(3 )(3) (3 4)( 2 )( )

(3 )

3 (2) (3) (3(2) 4)( 2(2))(2)

3 (2)

37

x x xf x

x

f

+ =

+

=

3

5

(a) 2 sin2x

cos

sin

x

x

= 2 sinx cosx

= sin 2x

2

(b)

xy

2=

Number of solutions = 4

6

K1

N1

K1

N1

K1

6

K1

N1

P1

8

xy 2sin2=

x

xy

2=

y

O

2

1

K1

Sine curve

Period

Amplitude

Modulus

P1

P1

P1

P1

Sketch straight

line correctly

N1

Koleksi Diff 2009

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