trial stpm jitsin maths paper 1_2011

8/3/2019 TRIAL STPM JitSin Maths Paper 1_2011

1/13

CONFIDENTIAL*

1. Solve the equation

xx 734 21

= [4 marks]

2. The function f is defined forx 1 by

f(x) =x

x1

+

(a) Show that f(x) increases asx increases. [3 marks]

(b) State the range of f. [1 marks]

3. The function f and g are defined as follows :

f :x ,1,22 + xxxg :x + xx ,4

(a) Find the value ofx such that gf(x) = 7. [2 marks]

(b) Find an expression for )(f1x

. [3 marks]

4. Find

(a) + .316

2 dxx

x[2 marks]

(b) .4dxxe

x

[3 marks]

5. A spherical balloon is being inflated in such a way that its volume is increasing at

a constant rate of 150 -13scm . At time tseconds, the radius of the balloon is r

cm.

(a) Find

dt

drwhen r= 50. [4 marks]

(b) Find the rate of increase of the surface area of the balloon when its radius is

50 cm. [2 marks]

[The formulae for the volume and surface area of a sphere are3

3

4rV = ,

.4 2rA = ]

950/1, 954/1*This question paper is CONFIDENTIAL until the examination is over.

CONFIDENTIAL*

2


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6. Evaluate dxx

x

ln

4

2 by using

(a) the substitutiony = lnx, [4 marks]

(b) the trapezium rule with five ordinates. [3 marks]

Giving each answer correct to three decimal places.

7. (a) When the polynomial f(x) = 54 234 ++++ bxaxxx is divided by 12 x , aremainder of 2x + 3 is obtained. Find the values ofa and b. [4 marks]

(b) Given that the roots of the equation 022 =++ qpxx are and.

Find in terms of and, the roots of the equation

0222222

=++ qxqxpxq . [5marks]

8. The complex number 31 i+ is denoted by u. Show that u is a root of the

equation ,0422 =+ zz and state the other root of this equation.

Express u in the form r(cos +i sin ), where r> 0 and .


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CONFIDENTIAL*

10. A straight line of gradient mpasses through the point (2, 1) and cuts thex-axis

andy-axis atA andB respectively. A pointPlies onAB such thatAP:PB = 1 : 2.

(a) Show that as m varies, the locus ofPis a curve with equation043 = yxxy

[6 marks]

(b) Find the perpendicular distance from the point (2, 1) to the tangent of the

curve 043 = yxxy at the origin. [5 marks]

11. Given matrix M =

12

311

102

k

(a) Find the value ofkif M is a singular matrix. [2 marks]

(b) Ifk= 1, find the values ofp, q and rsuch that 0IMMM 23 =+++ rqp where I is a 3 3 identity matrix. Deduce the inverse of matrix M, .M -1

Find the determinant of M and hence write down the adjoin of matrix M.

[11 marks]

12. Find the turning point of the curvex

xy

ln= and determine whether this is a

maximum or minimum point. [6 marks]

Sketch the graph ofx

xy

ln= . [4 marks]

Hence, show that xe ex for all positive values ofx. [5 marks]


CONFIDENTIAL*

4


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CONFIDENTIAL*

No Working/AnswerPartial

marks

Total

marks

1 xx 734 2

1

=

Let n = ,21

x 2734 nn =

0347 2 =+ nn 0)1)(37( =+ nn

7

3=n or n = 1 (N.A. 0>n )

7

32

1

=x

49

9=x

B1

M1

A1

A1

4

2 1,

1)( += x

xxxf

(a) 21

1)('x

xf = > 0 for x > 1

f(x) increases asx increases.

(b) Whenx = 1, f(x) = 2Since f(x) increases asx increases, f(x) 2

M1, M1

A1

B1

3

1

3 f :x ,1,22 + xxx

g :x + xx ,4

(a) gf(x) = 7g( )22 xx + = 7,x 1

( )22 xx + + 4 = 7 0322 =+ xx

0)3)(1( =+ xx x = 1 (sincex 1)

(b) Let f(x) =yxxy 22 +=

Completing the square, 1)1(2

yx +=+

yx +=+ 11 (sincex 1)

yx ++= 11

M1

A1

M1

2

3


CONFIDENTIAL*

5


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CONFIDENTIAL*

)(f1x

= x++ 11 ,x 3

A1

A1

4(a) ++=+ cxdxx

x)31ln(

31

6 22

(b) dxxe

x4

Let u =x andxe

dx

dv 4=

1=dx

du

xev 4

4

1=

dxxex4

= dxexe xx 44 41

4

1

= cexexx + 44

16

1

4

1

M1, A1

M1, A1

A1

2

3

5 (a) Let Vbe the volume of the spherical balloon and rbe the

radius. Then

3

3

4rV =

)3(3

4 2rdr

dV = = 24 r

dt

dr

dr

dV

dt

dV=

150 = 24 rdtdr

24

150

rdt

dr

= or 22

75

r

When r= 50, 2)50(4

150

=

dt

dr

015.0= =

200

3or 0.00477

B1

M1

M1

A1

4


CONFIDENTIAL*

6


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CONFIDENTIAL*

5 (b) LetAbe the surface area of the spherical balloon. Then

24 rA =

rdr

dA8=

dt

dr

dr

dA

dt

dA

=

r8=

015.0

When r= 50, =dt

dA)50(8

015.0 6=

the surface area is increasing at a rate of 6 12 scm

M1

A1

2

6(a)dx

x

x

ln

4

2

Lety = lnx

dy = dxx

1

whenx = 2, y = ln 2

whenx = 4, y = ln 4

dxx

x

ln

4

2 dyy4ln

2ln=

4ln

2ln

2

2

=

y

])2(ln)4[(ln2

122 =

721.0

B1

M1

M1

A1

4

6(b)Let ,

ln

x

xu = 5 ordinates = 4 strips each ofwidth = 0.5

nx 2 2.5 3 3.5 4

nu 0.3466 0.3665 0.3662 0.3579 0.3466

By using trapezium rule,

dxx

x

ln

4

2

=

+++

+3579.03662.03665.0

2

3466.03466.05.0

719.0

B1

M1

A1

3

7(a) f(x) = 54 234 ++++ bxaxxx 4


CONFIDENTIAL*

7


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CONFIDENTIAL*

f(1) = 541 ++++ ba2 (1) + 3 = 10 + a + b

5=+ba .f(1) = 5)1(41 +++ ba

2 (1) + 3 = 2 + ab

ab +=1

Substitute into,51 =++ aa62 =a3=a

From, 2)3(1 =+=b

3=a , 2=b

M1

M1

M1

A1

7(b) 022 =++ qpxx

Roots are andp=+ 2q=

)( +=p =2q

02 22222 =++ qxqxpxq

Substitute )( +=p and =2q

02)(22 =+++ xxx

0]2)[(22 =++ xx

0)( 222 =++ xx0))(( = xx

,

=x

=x

The roots are

and

.

B1

M1

A1

M1

A1

5

8 0422 =+ zz

2

1642 =z

31 iz =

31 i+ is a root (u) [Shown]

M1

10


CONFIDENTIAL*

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CONFIDENTIAL*

31 i is the other root.

++=

3sin

3cos31

iu =

+

3sin

3cos2

i

iu 32312 += = i322+

arg3

2)3(tan)(

12 == u

41242 =+=u

8623 ==u

arg =)( 3u ,83 =u

A1

B1

B1

M1

A1

A1

M1

A1

A19(a)

Sum of first n terms, =

+=n

r

r

n rS1

1 )23(

==

+=n

r

n

r

rr

11

123

++

= )1(2

12

13

)13(nn

n

)1(2

13nn

n

++

=

M1A1,M1

A1

4

9(b) Let common difference be d, d = 10.

Given 00010=nS .

[ ] 00010)1(22

=+ dnan

n

na00020

10)1(2 =+

)1(1000020

2 = nn

a

)1(500010

= nn

a

dnaTn )1( +=

)1(10)1(500010

+= nnn

m

B1

M1

7


CONFIDENTIAL*

9


9/13

CONFIDENTIAL*

)1(500010

+= nn

[Shown]

Given .500


10/13

CONFIDENTIAL*

11(a)

M =

12

311

102

k

M is a singular matrix

0

12

311

102

=

k

02

111

1

310

12

312 =

++

kk

0)2(1)61(2 =++ kk= 16

M1

A1

2

11(b)

=2M

121

311

102

121

311

102

=

603

176

123

=3M

121

311

102

603

176

123

=

91212

16518

843

0IMMM 23 =+++ rqp

91212

16518

843

+

603

176

123

p +

121

311

102

q +

100

010

001

r =0

024 =+ p 0212 =+ q 069 =+++ rqp

2=p 6=q 15=r

0IMMM 23 =+++ rqp1M

, =+++ -12 MIMM rqp I1M = (

1

rI)MM

2 qp

B1

B1

M1

M1

A2

M1


CONFIDENTIAL*

11


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CONFIDENTIAL*

= (15

1I)6M2M

2 +

15

1= I)6M2(M 2 +

151=

+

100

010

001

6

121

311

102

2

603

176

123

15

1=

241

714

127

=

15

2

15

4

15

115

7

15

1

15

415

1

15

2

15

7

(ii)21

1110

12

312M

+

=

)12()61(2 += 15=

MAdj(M)M =

1MMAdj(M) =

=

241

714

127

M1

A1

B1

B1


CONFIDENTIAL*

12


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CONFIDENTIAL*

12

x

xy

ln=

Using2v

dx

dvu

dx

duv

dx

dy

= with u = lnx and v =x

2

ln1

)(

x

xx

x

dx

dy

=

2

ln1

x

x=

0=dx

dy 0

ln12

=

x

x

1 lnx = 0

lnx = 1 x =e

whenx =eee

ey

1ln==

The turning point is .1,

ee

4

2

2

2

2)ln1(1

x

xxx

x

dxyd

=

4

ln22

x

xxxx +=

4

ln23

x

xxx += or 3

ln23

x

x+

x =e =2

2

dx

yd0

13

trial stpm jitsin maths paper 1_2011

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