2014-2-PEN-JIT SIN MATHS T 954/2

Section A [45 marks]

Answer all questions in this section.

1. a) Evaluate 7

9lim

2

x

x

x. [3m]

b) Given f(x) =

, 2 x,10

2,2

83

x

x

x

i) find , f(x) lim2x

ii) determine whether f is continuous at x = 2. [5m]

2. a) Find dx

dy in terms of x if .

)(ln 2x

xy [2m]

b) By substituting ,sec ax find . 22

dxx

ax

[6m]

3. Find the area of the region bounded by .2 line theand 4)2( 2 yxyx [5m]

4. Water is poured into an inverted cone of depth 20 cm and radius 10 cm at

a rate of 10 cm3 s-1. Find the rate at which the radius of the water in the

cone is increasing when the depth is 4 cm. [7m]

5. Find the integrating factor for the differential equation

.cos x cot 3 xydx

dy

Solve the differential equation, giving that y = 0 when x = .4

1 [6m]

6. a) Given that y = tan2 x, express dx

dy in terms of tan x. Hence, show that

.6822

2

2

yydx

yd [4m]

b) By further differentiation of this result, show that if x is small enough

that powers of x higher than x4 may be neglected, then .3

2tan 422 xxx

Use your series to estimate . tan

1.0

0

2 dxx Give your answer correct to 3

significant figures [7m]

Section B [15 marks]

Answer any one question in this section.

7. a) In a Chemical reaction, hydrogen peroxide is continuously converted into water and

oxygen. At time t minutes after the reaction started, the quantity of hydrogen peroxide that

has not been converted is x cm3 and the rate at which x is decreasing is proportional to x,

i) Write a differential equation relating x and t. [1m]

ii) Solve the equation given that the initial amount of x is xo,

and it took 3 minutes for the hydrogen peroxide to be reduced

to half the original amount. [7m]

iii) Sketch a graph to show how x varies with t. [1m]

iv) Find the time required to reduce the hydrogen peroxide

to one tenth of the original amount. Give your answer

correct to two decimal places. [2m]

b) Use the trapezium rule with 5 ordinates to evaluate .1 1

1dxe x

Give your answer correct to three decimal places. [4m]

8. a) Find the equations of the asymptotes of the curve

)3)(1(

32

xx

xy [2m]

b) State the coordinates of points where the curve intersects

the axes. [1m]

c) Find the stationary points of the curve and determine the nature

of the stationary points. [7m]

d) Sketch the curve. [4m]

e) Determine the values of k for which the equation

2x 3 = k ( x 1) (x + 3) has only one real root. [1m]

2014-2-PEN-JIT SIN MATHS T 954/2 MARKING SCHEME

No Working/Answer Partial

marks Total marks

1 (a)

7

9lim

2

x

x

x

=

x

xx

x

x 7

9

lim2

2

M1

8

= 01

01

71

91

lim2

x

xx

M1

(without

this line,

NEW1)

= 1 A1

(b) i)

2

8lim

3

2

x

x

x 2

)42)(2(lim

2

2

x

xxx

x

M1

(without

lim

NWW1)

)42(x lim 22

xx

4)2(2)2( 2 M1

2

8lim

3

2

x

x

x= 12

A1

ii) 10)2( f

2at x continuousnot is f

f(2) f(x) lim2

x

M1

A1

2

(a)

3

4

2

2

ln

2 ln

)(ln

1)(ln2)(ln

)(ln

x)(

-x

x

xxxx

dx

dy

x

xy

M1

A1

8

(b)

dadx

ax

tan sec

sec

B1

) tan sec (sec

sec

222

22

da

a

aa

dxx

ax

M1

da

da

tan

tan tan

2

M1

da )1(sec2 A1

ca )][(tan M1

c

a

x

a

axa

)sec( 1

22

c

a

xaax )(sec 122

A1

3 2 line theand 4)2( 2 yxyx

Pts. of intersection

3 , 2

24)2( 2

y

yy

B1

5

3

2

2 )]84()2[( dyyyy M1(either)

M1 (diff)

= 3

2

2 )65( dyyy

= 32

23

|62

5

3y

yy

M1

=

)2(6

2

)2(5

3

)2()3(6

2

)3(5

3

)3( 2323

2

6

1units

A1

4 Let r = radius of the water in the cone

h = depth of the water in the cone

rhrh

21020

M1

7

2

3

22

2

3

2

)2(3

1

3

1

rdr

dV

rV

rrhrV

A1

M1

Given 1310 scmdt

dV

When h = 4, 4 = 2r, so r = 2 cm

B1

8)2)(6( 2 dr

dV

A1

1

4

5

810

cmsdt

dr

dt

dr

dt

dr

dr

dV

dt

dV

M1

Hence, when the radius is 4 cm, the rate at which the radius

of the water in the cone is increasing is .4

5 1cms

A1

alternative

10cm

20cm h cm

r cm

4 Let r = radius of the water in the cone

h = depth of the water in the cone

rhrh

21020

M1

7

3

22

3

2

)2(3

1

3

1

rV

rrhrV

A1

When h = 4, 4 = 2r, so r = 2 cm B1

dt

drr

dt

dV 22 M1A1

dt

dr2)2(210

M1

4

5

dt

dr cm s1

Hence, when the radius is 4 cm, the rate at which the radius

of the water in the cone is increasing is .4

5 1cms

A1

5 .cos x cot 3xy

dx

dy

Int. factor =

x

exdx

sin

cot

B1

6

.sinxcos x cot )(sinsinx 3xyxdx

dy

M1

(multiply

integrating

factor)

10cm

20cm h cm

r cm

cx

xdx

xdxxxy

dxxxydx

d

4

3

3

3

cos4

1

)(cos cos

cossinsin

cossin)sin(

M1

(integrate)

16

1

)2

1(

4

10

4,0

4

c

c

xy

M1 (subst.)

A1

16

1cos

4

1sin 4 xxy

A1

6 (a) y = tan2 x

)(sectan2 2 xxdx

dy

)tan1(tan2 2 xxdx

dy or xx 3tan2tan2

)sectan3(2sec2 2222

2

xxxdx

yd

)tan1((tan6)tan1(2 222 xxx

2

2

2

682 yydx

yd

M1

A1

M1

A1

11

(b)

)(12)(128

)(128

2

2

2

2

4

4

3

3

dx

dy

dx

dy

dx

ydy

dx

yd

dx

yd

dx

dyy

dx

dy

dx

yd

2

2

2

2

2

4

4

)(12)(128dx

dy

dx

ydy

dx

yd

dx

yd

M1

M1

By Maclaurin Series

y(0)= tan20 =0

0)0(' y

)0()2(y = 2+y(0)+6[y(0)]2 = 2

)0()3(y = 8(0)+12(0) = 0

)0()4(y = 8(2)+0+0 =16

........3

2

......)16(!4

)2(!2

tan)(

42

42

2

xx

xx

xxy

).3( 000335.0

53

)3

2(tan

)15

2

3(

1.0

0

4

1.0

0

2

1.0

0

2

fs

dxxxxdx

xx

A1 (all

correct)

M1

A1

M1

A1

7 (a) (i) x

dt

dx

kxdt

dx

(ii) kdtdxx1

cktx ln

ceA where ktAex

t=0 x=x0 A=x0

Hence x=x0 e-kt

t=3 x= 2

1 x0

k

oo exx3

2

1

B1

M1

M1

A1

A1

M1

15

ke 3

2

1

2

1ln

3

1k

2ln3

1

t

oexx)2ln

3

1(

A1

A1

(iii)

(iv)

(2d.p) minutes 9.97

2ln

10ln3

10

1ln)2ln

3

1(

10

1

10

1when

)2ln3

1(

t

t

e

xx

t

o

D1

M1

A1

(b) Using Trapezium Rule

4

)1(1 d =0.5

x -1 -0.5 0 0.5 1

xe1

1.1696 1.2675 1.4142 1.6275 1.9283

B1

B1

xo

O

x

t

(3d.p.) 2.929

)]6275.14142.1

2675.1(2)9283.11696.1[(5.02

11

1

1

dxex

M1

A1

8 (a)

)3)(1(

32

xx

xy

Vertical asymptotes, x = 1, x = 3

Horizontal asymptotes, y = 0

(b) (0,1) , (3/2,0)

(c)

22

2

22

)3()1(

)32(x

)3()1(

)22)(32()3)(1(2

xx

x

xx

xxxx

dx

dy

0)3()1(

)22)(32()3)(1(222

xx

xxxx or

0)3()1(

)32(x22

2

xx

x or 0

)3()1(

)32(x22

2

xx

x

x = 0, 3

stationary points = (0, 1), )4

1,3(

when x = 0, 02

2

dx

yd, 0,1)( is a mnimum point

when x = 3, 02

2

dx

yd)

4

1(3, is a mximum point

B1

B1

B1

M1A1

M1

A1

M1(use

2

2

dx

yd or

gradient of

tangent -

either)

A1A1

15

(d)

Label

correct D1

Divides

graph into

3 parts

(with

asymptotes

)

D1D1D1

(d) 2x 3 = k ( x 1) (x + 3)

kxx

x

)3)(1(

32

To have solution, the line y = k must cut the curve once only,

therefore k = 0, , 1

Accept k = 0

B1

y

x O 3 1

3

y = 0

x = 3

x = 1

(0, 1) (3, )