2

2014-2-SARAWAK-SMKStJosephKuching_MATHS QA

Section A.

Answer all questions. [45 marks]

1.

The function f is defined by

0( )

ln 1 2 0

x xx

f x x

x x

.

Determine if f is continuous at x = 0. Sketch the graph of ( )y f x .

[6]

2. A piece of wire of length 16 cm is cut into two pieces, one of length x cm and the other of

length 16 x cm. The piece of length x cm is bent to form a circle with circumference x cm.

The other piece is bent to form a square with perimeter 16 x cm. Show that , as x varies, the

sum of the areas enclosed by these two pieces of wire is a minimum when the radius of the

circle is 8

4 cm.

[7]

3. Using the suitable substitution, evaluate

31

1

2 ln

e

dxx x

.

[4]

4. Show that

2ln tan

sin 2

dx

dx x . Hence, find the particular solution of the differential equation

)1(2)2(sin yydx

dvx , given that

1

3y when

4x

. Express y in terms of x.

[10]

5. If 2 1 siny x , show that

222

22 2 1 0

d y dyy y

dx dx

. Deduce an equation which has the

term in 3

3

d y

dx. Hence, obtain the expansion of 1 sin x in ascending powers of x up to the

term in x3.

[7]

3

6. Using the trapezium rule, with four ordinates estimate the value of

5

2ln x dx . Give your

answer to three decimal places. Explain with the aid of a sketch, why the trapezium rule under

estimates the value of the integral. By using integration by parts, evaluate the exact value of

5

2ln x dx . Hence, calculate the percentage error involved, correct to one significant figure.

[11]

Section B.

Answer any one question. [15 marks]

7. Sketch the curve with equation 2( 1)( 2)y x x .

a) Find the turning points and determine it nature. Hence sketch the curve.

b) Calculate the area of the finite region above the x-axis bounded by the curve and the x-axis.

c) Find the coordinates of the point of inflexion on the curve and the equation of the tangent at

this point.

[9]

[3]

[3]

8. A particle moves along the positive x-axis in the direction of increasing values of x. The

acceleration of the particle is 2 2dv ku v

dt v where k and u are positive constants and v is the

speed of the particle at the time t. Show that the time taken for the particle to accelerate from

speed 1

4u to speed

1

2u is

1 5ln

2 4k

.

Show that the acceleration of the particle may be expressed as dv

vdx

where x and v are the

displacement and velocity respectively. Prove that the distance travelled by the particle while it

accelerates from speed 1

4u to speed

1

2u is

92ln 1

4 5

u

k

.

[15]

4

MARKING SCHEME

1.

0 ,2)1ln(

0 ,)(

xx

xx

xx

xf

2-

2lim

lim

lim)(lim

0

0

00

x

x

xx

x

xx

x

xxxf

-2

2)1ln(lim)(lim00

xxfxx

-2

2)01ln()0(

f

2)0()(lim)(limsince 00

fxfxfxx

0at continuous is )( xxf

x

2. y

rx 2 xy 164

y

0 12 e

-2

5

2-8

4

168

1

2

1616

1

4

4

16

2

square, and circle of areas of sum

22

22

22

x

xx

dx

dA

xx

xx

yrA

8

42

2

dx

Ad

For minimum value , 0dx

dA, hence 02

8

4

x

4

16x

since ,so A is min at

4

16x

r2

4

16

4

8r with minimum area.

3.

xdx

duxu

1 ln2let 3, and 2,1when uexux

e du

du

dx

xx

1

3)ln2(

1

3

2 3)(

1xdu

ux

3

2

3

2

2

1

2

13

uduu

72

5 or 0.06944

6

(shown)2sin

2

cossin2

2

cossin

1

cossin

cos1

tan

sectanln .4

2

2

x

xx

xx

xx

x

x

xx

dx

d

x

xy

xxy

xxyy

xyxy

xy

y

xy

y

xy

y

c

c

yx

cxy

y

cxyy

dxx

dyyy

dxx

dyyy

yydx

dyx

tan2

tan

tan)tan2(

tantan2

tantan2

tan2

1

1

tan2

1ln

1ln

2

1lntanln

1ln

2

1ln

311

31

ln

4tanln

311

31

ln

3

1 ,

4when

tanln1

ln

tanln)1ln(ln

2sin

2

1

11

2sin

2

)1(

1

)1(22sin

7

5.

2 1 siny x

(shown) 0122

122

sin2

cos2

2

2

2

2

2

2

2

2

2

2

2

ydx

dy

dx

ydy

ydx

dy

dx

ydy

xdx

dy

dx

ydy

xdx

dyy

...48

1

8

1

2

11

...!3

8

1

!2

4

1

2

11sin1

8

1)0( 0

2

11

4

1

2

13)1(

,4

1)0( ,011

2

12)1(2

2

1)0(' ),0cos()1(2

1)0sin(1)0( ,0when

32

32

'''

3

3

"2

2

2

2

xxx

xxxx

fdx

yd

fdx

yd

fdx

dy

fx

8

636.31440ln2

1

)4ln3ln25ln2ln2

1ln

6094.15ln)( ,5

3863.14ln)( ,4

0986.13ln)( ,3

6391.02ln)( ,2

13

25 strips 3 , ordinatesfour For .6

5

2

xdx

xfx

xfx

xfx

xfx

d

Since the curve of f =lnx higher than the trapezium rule is under estimates of the value of the integral

%683.01003.661

3.636-3.661error of Percentage

3.661

2ln3-5ln5

1lnln

5

2

5

2

5

2

dxx

xxxxdx

2 3 4 5

9

7.a

point min. is 1.55,0.631-155.0 when 0.\

point max. is )11.2,215.0(215.0 when 0.

1.55,0.631- and )11.2,215.0( are points turning the

631.0 ,11.2

55.1, ,215.0

0143

2

2

2

2

2

xdx

yd

xdx

yd

yy

xx

xx

2

1

2

234

1-

2-

2

12

5

22

1

3

2

4

21 axis- xaboveregion finite of Area b.

unit

xxxx

dxxx

Max. pt

Min. pt

10

27

62

3

7

3

2

3

7

27

20y

inflexion ofpoint at the tangent theofEquation

3

7,

3

2when

inflexion ofpoint theis 27

20,

3

2

06,27

20,

3

2 when

,3

2046

0inflexion ofpoint For the c.

3

3

2

2

xy

x

dx

dyx

dx

ydyx

xx

dx

yd

2...............4

3ln

2

1

let ,2

1 when

1..............16

15ln

2

1

4

1ln

2

1

let ,4

1 when

ln2

1

2

1-

.8

2

2

2

1

2

1

2

2

1

22

22

22

22

cktu

ttuv

cktu

cktuu

ttux

cktvu

kdtdv

vu

v

kdtdv

vu

v

vuv

k

dt

dv

11

4

5ln

2

1 is

2

1 speed to

4

1 speed from particleby taken timeThe

(shown) 4

5ln

2

1)(

)(15

16

4

3ln

2

1-

)(16

15ln

4

3ln

2

1 )1()2(

12

12

12

22

kuu

ktt

ttk

ttkuu

vdx

dv

dt

dx

dx

dv

dt

dv

)3........(3

5ln

24

1

.

4

14

1

ln24

1

let and .4

1 when

ln2

)ln()ln(2

11

21

11

1

)(

)(

1

1

1

2

22

2

22

2

22

2

22

ckxu

u

ckx

uu

uuu

u

xxuv

ckxvu

vuuv

ckxvuvuu

v

kdxdvvuvuu

udv

kdxdvvu

udv

kdxdvvu

u

kdxdvvu

v

vuv

k

dx

dvv

1

12

15

9ln2

4 is distance thehence

4

1

5

9ln

2

1

)3........(3

5ln

24

1

)4......(3ln22

1

)3()4(

)4......(3ln22

1

.

2

12

1

ln22

1

let and .2

1when

12

1

2

2

2

2

k

u

k

uxx

ckxu

u

ckxu

u

ckxu

u

ckx

uu

uuu

u

xxuv