CONFIDENTIAL* 2014-2-PAHANG-SAB Kuantan_MATHS QA

Section A [45 marks ]

Answer all questions in this section

1. The function f is defined by

3, 3

3

( ) 2, 3

3, 3

xx

x

f x x

x x

− < −

= = − >

(a) Without sketching graph, determine whether f is a continuous function. [4 marks]

(b) Sketch the graph of f. [2 marks]

2. Find the gradient of the normal to the curve 3x2y + 2xy2 = 2 at the point (1, –2). [6 marks]

3.(a) Find 2

d3 1

xx

x +∫ [2 marks]

(b) Using substitution 3sinx θ= , show that2 1 29 1

9 sin 92 3 2

xx dx x x c− − = + − +

∫ . [5 marks]

4. Determine the integrating factor of the first order first degree differential equation

12

d

d−=+ z

xx

z. [1 mark]

Using the substitutiony

z1

= , show that differential equation 22d

dxyy

x

yx =− may be reduced

to 12

d

d−=+ z

xx

z. [3 marks]

Hence, find the particular solution y in terms of x for the differential equation given that

y = 3 when x = 1. [6 marks]

5.(a) The series expansion of sin x is given by ∑∞

=

+

+−=

0

12

)!12()1(sin

n

nn

n

xx .

By writing out the first three terms of the series, find the first three terms of the Maclaurin

series of f(x) = sin x2. [3 marks]

(b) Using the Maclaurin series of f(x) = sin x2 in (a), estimate the value of the integral ∫

1

0

2sin dxx

by giving your answer correct to four places of decimals. [4 marks]

6. Sketch the graphs of 3xy −= and 1+= xy on the same axes. [2 marks]

Hence, explain how the sketched graphs can be used to show that the equation 013 =++ xx has only one real root. [2 marks]

The real root of 013 =++ xx lies in this interval the interval ),( ba .

State the interval ),( ba if a and b are the successive integers. [1 mark]

By using –0.5 as the first approximated root, find the real root of 013 =++ xx correct to

three decimal places by the Newton-Raphson’s iterative approach.. [4 marks]

CONFIDENTIAL*

2

Section B [15 marks ]

Answer any one question in this section

7.(a) A particle moves in a straight line in a positive direction from a fixed point O.

The velocity v m s−1, at time t seconds, where 0≥t , satisfies the differential equation

( )

.50

1

d

d2vv

t

v +−=

(i) Explain why dx

dvv

dt

dv= , where x metres is the displacement from O.

Hence, show that, when 0>v , the motion of this particle can also be described by the

differential equation ( )50

1

d

d 2v

x

v +−= .

(ii) Given that 1ms10 −=v when x = 0 m, solve the differential equation expressing x in terms

of v. [10 marks]

(b) Diagram 1 shows the curves x

y4

= and intersect at the point (2 , 2).

Calculate the volume of the solid of revolution

when the region bounded by ,

, and is rotated through

about the y-axis. [5 marks]

8.(a) Using Maclaurin’s theorem, find the first three non-zero terms of the series expansion for

x2sin [3 marks]

Hence, find x

x

x

2sin

0

lim

→. [2 marks]

(b) Show that [ ]2

1

1

1tan

xx

dx

d

+=− .

Hence, find the first four terms of the power series representation for x1tan − in ascending

power of x given that ⋯++++=−

3211

1xxx

x. [5 marks]

(c) A sphere, of radius 26cm, has circular cylinder inscribed within it such that edges of the two

circular ends of the cylinder are always on the surface of the sphere.

Find the radius of the cylinder when the curved surface area of the cylinder is maximum.

[5 marks]

1 x

y

0

(2 , 2)

Diagram 1

CONFIDENTIAL*

3

MARKING SCHEME

Section A [45 marks ]

1. The function f is defined by

(a) Without sketching graph, determine whether f is a continuous function. [4 marks]

(b) Sketch the graph of f. [2 marks]

No. SCHEME MARKS

1.(a)

3 3

3 3

3 3

3

lim ( ) lim( 3) 0

3lim ( ) lim 1

3

lim ( ) lim ( )

, lim ( ) .

x x

x x

x x

x

f x x

xf x

x

f x f x

Hence f x does not exist

+ +

− −

+ −

→ →

→ →

→ →

= − =

−= = −

−

≠

( ) 3. f x is not continuous at x =

M1

Find both one sided limit

M1

Make correct conclusion

based on two limits

A1

Correct conclusion from

two correct limits

A1

A1

1.(b)

D1 : correct graph of

y = –1

or

y = x – 3

seen

D2 : both correct graphs

seen

2. Find the gradient of the normal to the curve 3x2y + 2xy

2 = 2 at the point (1, –2). [6 marks]

No. SCHEME MARKS

2. Attempting to differentiate implicitly

3x2y + 2xy

2 = 2 0

d

d42

d

d36 22 =+++⇒

y

yxyy

x

yxxy

Substituting x = 1 and y = –2

–12 + 3 0d

d88

d

d=−+

x

y

x

y

5

4

d

d4

d

d5 −=⇒=−⇒

x

y

x

y

Gradient of normal is

4

5

M1A1(one implicit

differentiation correct)

A1 (all correct)

M1A1

A1

f(x)

x

2

3 0

y=f(x)

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3.(a) Find 2

d3 1

xx

x +∫ [2 marks]

(b) Using substitution 3sinx θ= , show that2 1 29 1

9 sin 92 3 2

xx dx x x c− − = + − +

∫ .

[5 marks]

No. SCHEME MARKS

3(a)

2

1 6d

6 3 1

xx

x +∫

M1

( )21ln 3 1

6x c+ + A1

3.(b) 3sin , sin

3

xx θ θ= =

d

3cos d 3cos dd

xx= ⇒ =θ θ θ

θ

B1

∫∫ −=− )cos3()sin3(99 22 θθθ ddxx

= ∫ θθ d2cos9

M1

= ∫

+θ

θd

2

12cos9

= ∫ + θθ d)12(cos2

9

M1

= c+

+θθ

2

2sin

2

9

A1

= c++ )cos(sin

2

9θθθ

= c+−+ )sin1sin(2

9 2 θθθ

= cxx

x +

−

+−2

1

31

3sin

2

9

= cxx

x +

−

+− 21 99

sin2

9

2 1 29 19 sin 9

2 3 2

xx dx x x C− − = + − +

∫ A1

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4. Determine the integrating factor of the first order first degree differential equation

12

d

d−=+ z

xx

z. [1 mark]

Using the substitutiony

z1

= , show that differential equation 22d

dxyy

x

yx =− may be reduced

to 12

d

d−=+ z

xx

z. [3 marks]

Hence, find the particular solution y in terms of x for the differential equation given that

y = 3 when x = 1. [6 marks]

No. SCHEME MARKS

4. Integrating factor =

∫ dxxe

2

= 2x

B1

211

21

d

d

=

−

z

xzzx

x M1

Substitute y in terms of z

2

22

21

z

x

zdx

dz

zx =

−

− A1

22

2

d

dz

z

x

zxz

x=−−

−=

−−

x

z

z

x

x

z

zx

2

2

22

d

dz

1

2

d

d−=+ z

xx

z

A1

( ) ( ) ( )222 12

d

dxxz

xx

zx −=+

[ ] 22

d

dxzx

x−=

M1

Multiply integrating

factor and simplify LHS

∫−= dxxzx 22

M1

Reverse process

c

xzx +

−=

3

32

A1

z and x relation

c

xx

y+−=

3

1 32

A1

y and x relation

c+−=

3

)1()1(

3

13

2 3

2=⇒ c

3

2

3

32

+−=x

y

x

M1

Find c and attempt to

write y in terms of x

3

2

32

3

x

xy

−=

A1

CONFIDENTIAL*

6

5.(a) The series expansion of sin x is given by ∑∞

=

+

+−=

0

12

)!12()1(sin

n

nn

n

xx .

By writing out the first three terms of the series, find the first three terms of the Maclaurin

series of f(x) = sin x2. [3 marks]

(b) Using the Maclaurin series of f(x) = sin x2 in (a), estimate the value of the integral ∫

1

0

2sin dxx

by giving your answer correct to four places of decimals. [4 marks]

No. SCHEME MARKS

5.(a) sin x =∑

∞

=

+

+−

0

12

)!12()1(

n

nn

n

x

=)!1)0(2(

)1(1)0(2

0

+−

+x+

)!1)1(2()1(

1)1(21

+−

+x+

)!1)2(2()1(

1)2(22

+−

+x+....

sin x = 53

120

1

6

1xxx +− +....

M1

Write out first three terms

and simplify

sin x

2 =

52322 )(120

1)(

6

1)( xxx +− +....

M1

Substitute x2 in to the

series obtained

=

1062

120

1

6

1xxx +− +....

A1

5.(b)

∫1

0

2sin dxx ≈ ∫ +−1

0

1062 ]120

1

6

1[ dxxxx

M1

≈

1

0

1173

1320423

+−xxx

A1

≈ [ ]0

1320

)1(

42

)1(

3

)1( 1173

−

+−

≈ 9240

2867

M1

≈ 0.3103 A1

6. Sketch the graphs of 3xy −= and 1+= xy on the same axes. [2 marks]

Hence, explain how the sketched graphs can be used to show that the equation 013 =++ xx has only one real root. [2 marks]

The real root of 013 =++ xx lies in this interval the interval ),( ba .

State the interval ),( ba if a and b are the successive integers. [1 mark]

By using –0.5 as the first approximated root, find the real root of 013 =++ xx correct to

three decimal places by the Newton-Raphson’s iterative approach.. [4 marks]

CONFIDENTIAL*

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No. SCHEME MARKS

6

D1 : graph of y = x + 1

showing the intercepts

or

graph of y = –x3

passing through (0,0)

D2 : Both graphs correct

013 =++ xx 31 xx −=+

1+= xy and 3xy −=

The roots of 013 =++ xx are same as the x-coordinate of the intersecting points of 1+= xy and 3xy −= .

From the graph ,there is only one point of intersection

the equation has only one root.

M1

A1

(Apply NWP-1 if the roots

are not related to the x-

coordinate of the point of

intersection)

And the real root lies in the interval ( -1 , 0 ) B1

( ) 2f 3 1x x′ = +

( )( )1

f 0.50.5

f 0.5x

−= − −

′ −

M1

Find f’(x) and use the

iterative formula

1 0.7143x = − A1

( )( )2

f 0.71430.7143

f 0.7143

x−

= − −′ −

⋮

( )( )4

f 0.68230.6823

f 0.6823x

−= − −

′ −

M1

Keep using formula and

check when to stop

4 0.6823x = −

0.682x = − A1

Section B [15 marks ]

Answer any one question in this section

y

CONFIDENTIAL*

8

7.(a) A particle moves in a straight line in a positive direction from a fixed point O.

The velocity v m s−1, at time t seconds, where 0≥t , satisfies the differential equation

( )

.50

1

d

d2vv

t

v +−=

(i) Explain why dx

dvx

dt

dv= .

Hence, show that, when 0>v , the motion of this particle can also be described by the

differential equation ( )50

1

d

d2v

x

v +−= where x metres is the displacement from O.

(ii) Given that 1ms10 −=v when x = 0 m, solve the differential equation expressing x in terms

of v. [10 marks]

(b) Diagram 1 shows the curves x

y4

= and

intersect at the point (2 , 2).

Calculate the volume of the solid of revolution

when the region bounded by ,

, and is rotated through

about the y-axis. [5 marks]

No. SCHEME MARKS

7.(a)

(i) From

dt

dx

dx

dv

dt

dv×= ⇒

dx

dvv

dt

dv=

B1

50

)1( 2vv

dt

dv +−=

50

)1( 2vv

dx

dvv

+−=

( )50

1

d

d2v

x

v +−=

M1

A1

7.(a)

(ii) Either attempting to separate variables or inverting to obtain

v

x

d

d

M1

∫ ∫−=+

xv

vd

50

1

1

d2

(or equivalent)

A1

Attempting to integrate both sides

M1

cxv +−=−

50

1tan 1

Note: Award A1 for a correct LHS and A1 for a correct RHS that

must include C

A1(RHS)

A1(LHS)

No. SCHEME MARKS

When x = 0, v = 10 and so 10tan 1−=c

10tan50

1tan 11 −− +−= xv

M1

1 x

y

0

(2 , 2)

Diagram 1

CONFIDENTIAL*

9

)tan10(tan50 11 vx −− −= A1

7.(b)

Volume =

=

=

=

=

=

M1

Difference of two

volumes

M1

Correct integration

A1

M1

Find value

A1

8.(a) Using Maclaurin’s theorem, find the first three non-zero terms of the series expansion for

x2sin [3 marks]

Hence, find x

x

x

2sin

0

lim

→. [2 marks]

(b) Show that [ ]2

1

1

1tan

xx

dx

d

+=− .

Hence, find the first four terms of the power series representation for x1tan − in ascending

power of x given that ⋯++++=−

3211

1xxx

x. [5 marks]

(c) A sphere, of radius 26 cm, has circular cylinder inscribed within it such that edges of the two

circular ends of the cylinder are always on the surface of the sphere.

Find the radius of the cylinder when the curved surface area of the cylinder is maximum.

[5 marks]

No. SCHEME MARKS

8.(a) xxf 2sin)( = 0)0( =⇒ f

xxf 2cos2)(' = 2)0(' =⇒ f

xxf 2sin4)('' −= 0)0('' =⇒ f

xxf 2cos8)(''' −= 8)0(''' −=⇒ f

xxf 2sin16)(4 = 0)0(4 =⇒ f

xxf 2cos32)(5 = 32)0(5 =⇒ f

M1

Find first non-zero

derivative values

No. SCHEME MARKS

⋯++++++= 5

54

432

!5

)0(

!4

)0(

!3

)0('''

!2

)0(''

!1

)0(')0(2sin x

fx

fx

fx

fx

ffx

⋯+++−

+++= 5422

!5

32

!4

0

!3

8

!2

0

!1

20 xxxxx

⋯++−= 53

15

4

3

422sin xxxx

M1

Use Maclaurin

Theorem

A1

CONFIDENTIAL*

10

x

xxx

xx

x

x

⋯++−

→=

→

53

15

4

3

42

0

lim2sin

0

lim

++−→

= ⋯42

15

4

3

42

0

limxx

x

22sin

0

lim=

→ x

x

x

M1

Substitute and

simplify

A1

8.(b) [ ]dx

dyx

dx

d=−1tan , xy

1tan

−=

xy =tan

1sec2 =dx

dyy

1)tan1( 2 =+dx

dyy

1)1( 2 =+dx

dyx

21

1

xdx

dy

+=

[ ]2

1

1

1tan

xx

dx

d

+=∴ −

M1

A1

)(1

1

1

122xx −−

=+

⋯+−+−+−+=−−

32222

2)()()(1

)(1

1xxx

x

⋯+−+−= 6421 xxx

From [ ]2

1

1

1tan

xx

dx

d

+=−

dxx

x ∫ +=−2

1

1

1tan

dxxxx )1( 642∫ +−+−= ⋯

⋯+−+−=∴ −

753tan

7531 xxx

xx

B1

M1

A1

No. SCHEME MARKS

8.(c) r = radius of cylinder

h = height of cylinder

22

2

262

=+

r

h

22 42704 rh −=

A = curved surface area of the cylinder

hrA π2=

M1

Find relation of h

and r

and substitute to

obtain the relation

A and r

r

h

26

CONFIDENTIAL*

11

2222 4 hrA π=

)42704(42222rrA −= π

)8(4)42704(82 2222 rrrrdr

dAA −+−= ππ

)82704(82 22 rrdr

dAA −= π

For maximum A, 0=dr

dA

)82704(8)0(222rrA −= π

)82704(80 22 rr −= π 0827040 2 =−⇒> rr

3382 =r

213=r cm

)8)(82704()16(822 222

2

2

ππ rrrdr

dA

dr

dA

dr

AdA −+−=

+

)]82704(16[822 222

2

2

2

rrdr

dA

dr

AdA −+−=

+ π

)242704(822 22

2

2

2

rdr

dA

dr

AdA −=

+ π

When 213=r , 3382 =r and 0=dr

dA

( ) )]338(242704[8022 22

2

2

−=+ πdr

AdA

2

2

2

432642 π−=dr

AdA

Adr

Ad 2

2

2 21632π−= < 0

213=r maximises A

The curved surface area of the cylinder is maximum when 213=r cm

A1

M1

Find dr

dA and use

0=dr

dA to get r

M1

Find 2

2

dr

Ad and

check the sign of

2

2

dr

Ad for r such

that 0=dr

dA

A1

CONFIDENTIAL*

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