2014-2-PERAK-SMKAndersonIpoh_MATHS QA by TAN GUAN HIN

Section A [45 marks]

Answer all questions in this section.

1. A function f is defined as follow:

2,2

31

2,2

2,1092

42

)(

2

2

xx

x

x

xxx

x

xf

(a) Show that )(lim2

xfx

exists. [5 marks]

(b) Determine whether f is continuous at x = – 2. [2 marks]

2. The equation of a curve is 4

252

x

xy . Find the coordinates of the stationary points and use the sign of

dx

dyto determine its

nature. State all the equations of asymptotes and hence sketch the graph.

[8 marks]

3. A region R is bounded by a curve xy = 12 and a line 3x + 4y = 30.

(a) Find the coordinates of the points of intersection of the curve xy = 12 and the line 3x + 4y = 30.

[ 3 marks]

(b) Calculate the area of the region R. [ 3 marks]

(c) Calculate the volume of the solid of revolution formed when this region R is rotated through 360o

about the x-axis. [3 marks]

4. Find the particular solution of the differential equation

0)2)(1(3 22 yyxdx

dyx

for which y =0 when x = 1.Hence, express y in terms of x. [7 marks]

5. Using the Maclaurin theorem ,show that the first two non-zero terms of the expansion series of ln(cos x ) is

...............12

1

2

1 42 xx [4 marks]

Hence, evaluate 32

22

0 3

)ln(cos2lim

xx

xx

x

[2 marks]

6. Sketch on the same coordinates axes, the graphs x

ey 2

1

and y = 3 sin 2x for 0 x and show that the equation

02sin32

1

xex

has two positive real roots. Show that the largest positive root lies between ¼ and ½ . Use the

Newton-Raphson method with an initial approximation x0 = 1.2 to obtain the largest positive root correct to three decimal

places. [8 marks ]

Section B [15 marks]

Answer any one question in this section.

7. Given that y = e– x

sin 2x,

(a) Show that dx

dyy

dx

yd25

2

2

. [4 marks]

(b) Using the Maclaurin’s theorem , find the series for e– x

sin 2x in ascending powers of x up to and

including the term in x4. [5 marks]

(c) Hence deduce the Maclaurin’s series for e– x

cos 2x in ascending powers of x up to and

including the term in x3. [3 marks]

(d) Using the series to find an approximate value of dxex

xx

2

1

2

2sin. [3 marks]

8. The variables x and y, where x > 0 and y >0, satisfy the differential equation xy

yxy

dx

dy

2

2 2 . Using the substitution u = y

2x,

show that the given differential equation can be transformed into the linear differential equation

xudx

du22 [4 marks]

Solve this linear differential equation, and find the particular solution of the given differential equation which satisfies the condition

ey

1 when

2

1x , giving your answer in the form y

2 = f(x). [9 marks]

Find the limiting value of y as x tends to infinity. [2 marks]

ooooooooooooooooooooooooo000000000000oooooooooooooooooooooooooooooo

STPM Trial Examination 2014 - Mathematics T Paper 2 Marking Scheme

1. A function f is defined as follow:

2,2

31

2,2

2,1092

42

)(

2

2

xx

x

x

xxx

x

xf

(c) Show that )(lim2

xfx

exists. [5 marks]

(d) Determine whether f is continuous at x = – 2. [2 marks]

Marks

Solution :

)2)(52(

)2(2lim

1092

42lim

22

2

xx

x

xx

x

xx

52

2lim

2

xx

5)2(2

2

2

)31)(2(

)31)(31(lim

2

31lim

2

22

2

2

2

xx

xx

x

x

xx

M1

A1

M1

)31)(2(

4lim

2

2

2

xx

x

x

)31)(2(

)2)(2(lim

22

xx

xx

x

)3)2(1(

)]2(2[

2

2 A1

Since )(lim2

xfx

= 2)(lim2

xfx

Therefore )(lim2

xfx

exits and 2)(lim2

xfx

(b) Since 2)2()(lim2

fxfx

OR )(lim2

xfx

= )2()(lim2

fxfx

Conclusion : f is continuous at x = – 2.

A1

M1

A1

7

2. The equation of a curve is 4

252

x

xy . Find the coordinates of the stationary points and use the sign of

dx

dyto

determine its nature. State all the equations of asymptotes and hence sketch the graph. [8 marks]

Marks

Solution : 4

252

x

xy

22

2

)4(

)25(2)4(2

x

xxx

dx

dy

222222

2

)4(

)4)(1(2

)4(

)45(2

)4(

8102

x

xx

x

xx

x

xx

For stationary points : 0dx

dy ; (x – 1 )(x – 4) =0

x = 1 , x = 4

When x = 1 , 141

)1(252

y ; When x = 4 ,

4

1

44

)4(252

y

Therefore coordinates of stationary points are (1 ,– 1) and ( 4, – ¼).

x 1- 1 1

+

0.5 1.5

x 4- 4 4

+

3 5

dx

dy +ve 0 -ve

dx

dy -ve 0 +ve

(1,-1) is a local maximum

(4, - ¼ ) is a local minimum

Asymptotes : When y ; x2 – 4 =0

x =2 and x = – 2 are vertical asymptotes.

2

2

2 41

25

4

25

x

xx

x

xy

When x , y 0.

y = 0 is vertical asymptote.

M1

M1

A1

M1

A1

B1

(all correct)

Intersection on y-axis : x = 0 , y = -5/4

Intersection on x-axis : y= 0, x = 2.5

Graph of 4

252

x

xy

y

-2

0 2 2.5 x

(4, - ¼) (1, -1)

-5/4

D1

D1

8

3. A region R is bounded by a curve xy = 12 and a line 3x + 4y = 30.

(d) Find the coordinates of the points of intersection of the curve xy = 12 and the line 3x + 4y = 30. [ 3 marks]

(e) Calculate the area of the region R. [ 3 marks]

(f) Calculate the volume of the solid of revolution formed when this region R is rotated through

360o about the x-axis. [3 marks]

Marks

Solution :

(a)x

y12

-----------------------(1)

3x + 4y = 30 -------------------(2)

Substitute (1) into (2) :

3012

43

xx

3x2 – 30x + 48 =0

x2 – 10x + 16 =0

(x – 2 )(x – 8) =0

x= 2, x = 8

62

122, When yx

5.18

128, When yx

Coordinates of the points of intersection are :(2,6) and (8,1.5)

y

xy= 12

3x+ 4y = 30

R

2 8 x

M1

M1

A1

(b)Area of region R = dxx

dxx

8

2

8

2

12

4

330

82

8

2

2

ln122

330

4

1x

xx

2ln8ln122

)2(3)2(30

2

)8(3)8(30

4

1 22

2

8ln1266096240

4

1

= 22.5 – 12 ln4.

or 5.86 unit2.

Alternative :

Area of region R

= dxx

8

2

12 trapeziumof Area

= 82ln 12-61.562

1x

= 22.5 – 12ln4

or 5.86 unit2.

M1

M1

A1

(c)

volume of the solid of revolution formed = dxx

dxx

8

2

2

8

2

2 1144)330(

16

1

8

2

8

2

3 1144

9

)330(

16

1

x

x

M1

M1

2

1

8

1144)630()2430(

144

1 33

545.94

3unit 5.40

A1

(mark is not

given if

no)

9

4. Find the particular solution of the differential equation

0)2)(1(3 22 yyxdx

dyx

for which y =0 when x = 1.Hence, express y in terms of x. [7 marks]

Marks

Solution :

dxx

xdy

yy

2

2

1

)2(

3

dxxx

dyyy

1

)1)(2(

3

cx

xdyy

B

y

A

2ln

)1()2(3

2

Use partial fractions : 3

1 BA

c

xxdy

yy

2ln

)1(

3

1

)2(

3

1

32

cx

xdyyy

2ln

)1(

1

)2(

1 2

B1

M1

cx

xyy 2

ln)1ln()2ln(2

-----------------------------------(1)

A1

When x = 1, y = 0.

c2

11ln)01ln()02ln(

2

M1

2

12ln c

Substitute 2

12ln c into (1) :

2

12ln

2ln

1

2ln

2

xx

y

y

eex

xy

yln

2

12lnln

2ln

1

2ln

2

)1(

2

1 2

ln2ln1

2ln

x

exy

y

)1(

2

1 2

21

2

x

xey

y

)1(

2

1 2

2)1(2

x

xeyy

)1(

2

1

)1(2

1

2

2

21

]1[2

x

x

xe

xey

A1

M1

A1

7

5. Using the Maclaurin theorem ,show that the first two non-zero terms of the expansion series of ln(cos x ) is

...............12

1

2

1 42 xx [4 marks]

Hence, evaluate 32

22

0 3

)ln(cos2lim

xx

xx

x

[2 marks]

Solution :

Let y= ln (cos x)

xxxdx

dyy tan)sin(

cos

1'

)tan1(sec'' 22

2

2

xxdx

ydy

xxxxxxdx

ydy 322

3

3

tan2tan2)tan1(tan2sectan2'''

xxxdx

ydyiv 222

4

4

sectan6sec2

x =0 , y = ln(cos0)= ln1 =0

x =0, y’ = – tan0 = 0

x = 0, y’’= – (1+tan20) = – 1

x =0, y’’’= – 2tan0 – 2tan30 = 0

200cos

20sec0tan60sec2 ,0

2

222

ivyx

Maclaurin series for y = ln(cos x) = .......!4

)0(

!3

)0('''

!2

)0(''

!1

)0(')0(

432

xyxyxyxy

yiv

= .......!4

)2(

!3

)0(

!2

)1()0(0

432

xxx

x

...............12

1

2

1 42 xx (first two non-zero terms)

32

422

032

22

0 3

........12

1

2

122

lim3

)ln(cos2lim

xx

xxx

xx

xx

xx

)3(

....6

11

lim2

22

0 xx

xx

x

x

x

x

3

.....6

11

lim

2

0

=3

1

M1

(y’correct)

A1

(all correct)

M1

A1

M1

A1

6

6. Sketch on the same coordinates axes , the graphs x

ey 2

1

and y = 3 sin 2x for 0 x and show that the

equation 02sin32

1

xex

has two positive real roots. Show that the largest positive root lies between ¼

and ½ . Use the Newton-Raphson method with an initial approximation x0 = 1.2 to obtain the largest positive

root correct to three decimal places. [8 marks ]

Marks

Solution :

y x

ey 2

1

y =3sin 2x

3

2

1

¼ ½ x

-1

-2

-3

From the graph, the two curves intersect at two points. From the two intersection points ,it is obvious that the

equation 02sin32

1

xex

has two positive real roots.

Let xexfx

2sin3)( 2

1

0519.14

2sin34

42

1

ef

D1

D1

B1

01932.22

2sin32

22

1

ef

.2

and 4

between liesroot thesign,different have 2

and 4

Since

ff

xexfx

2sin3)( 2

1

xexfx

2cos65.0)(' 2

1

Given that x0 = 1.2 :

2383.1]4.2cos65.0[

]4.2sin3[2.1

)('

)(6.0

6.0

0

001

e

e

xf

xfxx

2372.1]4766.2cos65.0[

]4766.2sin3[2383.1

)('

)(61915.0

61915.0

1

112

e

e

xf

xfxx

2372.1]4744.2cos65.0[

]4744.2sin3[2372.1

)('

)(6186.0

6186.0

2

223

e

e

xf

xfxx

B1

M1

M1

M1

(x2 = x3)

Therefore the root is 1.237 (3 decimal places) A1

8

7. Given that y = e– x

sin 2x,

(a) Show that dx

dyy

dx

yd25

2

2

. [4 marks]

(b) Using the Maclaurin’s theorem , find the series for e– x

sin 2x in ascending powers of x up to and

including the term in x4. [5 marks]

(c) Hence deduce the Maclaurin’s series for e– x

cos 2x in ascending powers of x up to and

including the term in x3. [3 marks]

(d) Using the series to find an approximate value of dxex

xx

2

1

2

2sin . [3 marks]

Marks

Solution :

(a) y = e– x

sin 2x

xexedx

dy xx 2sin)()2cos2(

yxedx

dy x )2cos2( ----------------------- (1)

dx

dyxexe

dx

yd xx )2cos2)(()2sin4(2

2

dx

dyy

dx

dyy

dx

yd

4

2

2

dx

dyy

dx

yd25

2

2

shown

(b) 2

2

3

3

25dx

yd

dx

dy

dx

yd

3

3

2

2

4

4

25dx

yd

dx

yd

dx

yd

x = 0 , y =0

x= 0, y’ = e0(2cos0) – 0=2

x= 0, y’’ = – 5(0) – 2(2) = – 4

x= 0, y’’’= – 5(2) – 2( – 4) = – 10 + 8 = – 2

x= 0, yiv= – 5(– 4 ) – 2( – 2) = 20 + 4 = 24

Maclaurin series for y = e– x

sin 2x = .......!4

)0(

!3

)0('''

!2

)0(''

!1

)0(')0(

432

xyxyxyxy

yiv

= .......!4

)24(

!3

)2(

!2

)4(20

432

xxx

x

= .......3

22 43

2 xx

xx

[marks are given if using Maclaurin theorem]

(c) From (1) :

y

dx

dyxe x

2

12cos

=

.......

322.....442

2

1 43

232 xx

xxxxx

= .................6

11

2

31

32

xxx [marks are given if using deduction]

(d) dxxx

xxx

dxex

xx

43

2

2

1

2

2

1

2 322

12sindxxx

x

2

1

2

3

12

2

2

1

32

362ln2

xxxx

M1

M1

M1

A1

M1

M1

A1

M1

A1

M1

M1

A1

M1

3

1

6

1)1(21ln2

3

2

6

2)2(22ln2

3232

= 2ln2 – 1/6

= 1.22

M1

A1

15

8. The variables x and y, where x > 0 and y >0, satisfy the differential equation xy

yxy

dx

dy

2

2 2 . Using the

substitution u = y2x, show that the given differential equation can be transformed into the linear differential equation can

be transformed into the linear differential equation

xudx

du22 [4 marks]

Solve this linear differential equation, and find the particular solution of the given differential equation which satisfies

the condition e

y1

when2

1x , giving your answer in the form y

2 = f(x). [9 marks]

Find the limiting value of y as x tends to infinity. [2 marks]

Marks

Solution :

u = y2x

dx

dyxyy

dx

dyyxy

dx

du22 22

22 ydx

du

dx

dyxy ----------------------------------- (1)

Given that :

xy

yxy

dx

dy

2

2 2

Multiply by 2xy : 22 222 yxxydx

dyxy --------------------------- (2)

Substitute (1) into (2) : 222 22 yxxyydx

du

Substitute u = y2x : xu

dx

du22 shown

Integrating factor : xdx

ee 22

Multiply each term by e2x : xxx xeuedx

due 222 22

xx xeuedx

d 22 2

Integrating both sides : dxxeue xx 22 2

Integration by parts : u = 2x xedx

dv 2

2dx

du xev 2

2

1

dxeexue xxx 222

2

12

ce

xeuex

xx 2

222

Substitute u = y2x c

exexey

xxx

2

2222

xxe

c

xy

2

2

2

11

When 2

1x and

ey

1 ,

2

12

2

2

12

12

11

1

e

c

e

2

1c

Substitute 2

1c :

xxexy

2

2

2

1

2

11

When x , 02

1

x and 0

2

12

xxe , y

2 = 1

Limiting value of 11 y

M1

M1

M1

A1

B1

M1

M1

M1

A1

A1

M1

A1

A1

M1

A1

15

Prepared by : MR.TAN GUAN HIN/09.04.14/SMK ANDERSON IPOH