2014-2-JOHOR-SMKYongPeng_MATHS QA by TAN LEE MENG

Section A [45 marks]

Answer all questions in this section.

1. The function f is defined by

xxx

xf ,2

)( .

(a) Find )(lim0

xfx

and )(lim0

xfx

. [4]

(b) Determine whether f is continuous at 0. [1]

(c) Sketch the graph of ).(xfy [2]

2. Show that the tangent at the point ),( 3tt on the curve 3xy has the equation

)23(2 txty . Prove that the tangents to the curve 3xy at the points ),( 3tt and

),( 3tt are parallel. [6]

3. Use the substitution tant to show that

dt

td

22 1

1

1cos2

1

.

Hence, show that 32ln2

1

1cos2

16

0 2

d . [10]

4. Find the integrating factor for the differential equation xxydx

dy 3coscot . Solve

the differential equation, given that 0y when 4

1x . [9]

5. Use the series expansions given below to show that sincos iei .

.....!

.....!2

12

r

xxxe

rx

.....)!2(

)1(.....!4!2

1cos242

r

xxxx

rr

.....)!12(

)1(.....!5!3

sin1253

r

xxxxx

rr [5]

6. The diagram below shows the graph 22 2 xy in the first quadrant. The region R

is bounded by the curve, the axes and the line 1x . Use the trapezium rule with 5

ordinates to estimate the area of R, giving your answer correct to 3 significant figures.

State with a reason, whether the estimated area of R is greater or less than the exact

value. y [8]

R

x

Section B [15marks]

Answer any one question in this section.

7. Find the turning point(s) of the graph xexy 22)1( . Sketch the graph, showing clearly the

turning point(s), the point(s) of inflexion (you need not find the y-coordinates here). Indicate

the behaviour of the graph as x . Label the point(s) of inflexion as A and B.

If 0)1( 22 kex x has 3 distinct real roots, what are the values of k?

[15]

8. (a) Using Maclaurin theorem, show that if x is so small that x6 and higher power of x may be

neglected, 5432

)1ln(5432 xxxx

xx .

Deduce that for such values of x, )5

1

3

1(2

1

1ln 53 xxx

x

x

. [8]

(b) Show that the equation xex 1cot2 can be written as

xex

1

2tan 1 .

Use 7.01 x and the iterative formula

n

xne

x1

2tan 1

1 to find this root correct

to 2 decimal places. [7]

********************GOOD LUCK********************

MARKING SCHEME:

1.(a)

0,

0,0)(

xx

xxf

0)(lim0

xfx

0)(lim0

xfx

(b) yes, because

)(lim0

xfx

0)(lim0

xfx

(c)

2. 3xy

23xdx

dy

at the point ),( 3tt , 23tdx

dy

the equation of the tangent is )(3 23 txtty

)23(2 txty

at the point ),( 3tt , 2)(3 tdx

dy

= 3t2

since gradient at the points ),( 3tt and ),( 3tt is the same, so the tangents are parallel.

3. tant

2secd

dt =

2cos

1 =

2

21

1

1

t

= 21 t

21 t

dtd

22

2

2 11

1

12

1

1cos2

1

t

dt

t

d

=

2

2

2 1

1

)1(2

1

t

dt

t

t =

22

2

11

1

t

dt

t

t

= dt

t 21

1 (shown)

let 222 111

1

t

B

t

A

t

y

x

)1()1(1 tBtA

2

1,1 Bt

2

1,1 At

ttt 1

1

1

1

2

1

1

12

when 3

3,

6 t

0,0 t

3

3

0 26

0 2 1

1

1cos2

1dt

td

=

3

3

0 1

1

1

1

2

1dt

tt

= )1ln()1ln( 3

3

02

1tt

=

33

33

33

33ln

2

1

= )32ln(2

1

4. xdx

eIcot =

dxx

x

e sin

cos

= xe sinln = sinx

xxxydx

d 3cossin)sin(

xdxxxy 3cossinsin

= xdxx 3cossin

cx

xy 4

cossin

4

when 4

1, xoy

16

1c

16

1

4

cossin

4

x

xy

5. 432

24

1

6

1

2

11 xxxxe x +……

.......)(24

1)(

6

1)(

2

1)(1 432 iiiiei = .....

24

1

6

1

2

11 432 ii

= .....)6

1(......)

24

1

2

11( 342 i

= sincos i

6. 25.04

01

h

x 0 0.25 0.50 0.75 1

22 xy 1.4142 1.4361 1.5000 1.6008 1.7321

area of R = ]7321.1)6008.15.14361.1(24142.1)[25.0(2

1 = 1.53

estimated area > exact value since area of trapezium > area under the curve y

7. xexy 22)1(

)1)(1(2.2.)1( 222 xeexdx

dy xx = )1(2 2 xxe x

when ,0dx

dy

)1(2 2 xxe x = 0

0x or 1x

when 0,1

1,0

yx

yx

the turning points are (0, 1) and (1, 0)

)(2 22 xxedx

dy x

xx exxxedx

yd 222

2

2

4).()12.(2

= )12(2 22 xe x

when 0,02

2

dx

ydx , (0, 1) is a maximum point

when ,1x 02

2

dx

yd, (1, 0) is a minimum point

when 02

2

dx

yd, 0)12(2 22 xe x

since 02 xe , 012 2 x

2

1x

when yx ,

0, yx

0)1( 22 kex x

kex x 22)1(

ky

since 0)1( 22 kex x has 3 distinct real roots,

the set of values of k = },10:{ kkk

8.(a) let 0)0(),1ln()( fxxf

1)0(',1

1)('

f

xxf

1)0('',)1(

1)(''

2

f

xxf

2)0(''',)1(

2)('''

3

f

xxf

6)0(,)1(

6)( 4

4

4

fx

xf

24)0(,)1(

24)( 5

5

5

fx

xf

.....6

)0('''

2

)0('')0(')0()( 32 x

fx

fxffxf

5432

5

1

4

1

3

1

2

1)1ln( xxxxxx

)1ln()1ln(1

1ln xx

x

x

= )5

1

4

1

3

1

2

1( 5432 xxxxx ])(

5

1)(

4

1)(

3

1)(

2

1[ 5432 xxxxx

= 53

5

2

3

22 xxx

= )5

1

3

1(2 53 xxx

(b) xex 1cot2

2

1cot

xex

xe

x

1

2tan

xex

1

2tan 1

7.01 x

7.0

1

21

2tan

ex =0.5859

5859.0

1

31

2tan

ex =0.6208

6208.0

1

41

2tan

ex = 0.6102

6102.0

1

51

2tan

ex = 0.6134

61.0x (2 d.p)