model answer for combined mathematics i -2013 al paper.pdf

7/27/2019 Model Answer for Combined Mathematics I -2013 AL Paper.pdf

1/18

- 1 -Done By : Chandima Peiris (B.Sc - Special)

General Certificate of Education (Adv. Level) Examination, August 2013

Combined Mathematics I - Part B

Model Answers

11.(a)Let ( ) 61123 ++= xbxaxxf , where a , b .

If ( )1x is a factor of ( )xf , and the remainder when ( )xf is divided by ( )4x is 6 , find the values ofa and b. Also, find the other two linear factors of ( )xf .

(b)Let and be the roots of the equation 02 =++ cbxx , and and be the roots of the equation02 =++ nmxx , where b, c, m , n .

(i) Find ( )2 in terms ofb and c, and hence write down ( )2 in terms ofm and n.Deduce that if +=+ , then nmcb 44 22 = .

(ii) Show that ( )( )( )( ) ( ) ( )( )cmbnmbnc += 2 .Deduce that the equations 02 =++ cbxx and 02 =++ nmxx have a common root if and only if

( ) ( )( )cmbnbmnc = 2 .

The equations 0102 =++ kxx and 0102 =++ kxx have a common root, where kis a real

constant. Find the values ofk.

Answer

(a) ( ) 61123 ++= xbxaxxf ; a , b Since ( )1x is a factor of ( )xf , ( ) 01 =f (QAccording to the factor theorem )

0611 =++ ba ( )15 =+ ba

Since the remainder when ( )xf is divided by ( )4x is 6 , ( ) 64 =f (QAccording to the remainder

theorem )

66441664 =++ ba

321664 =+ ba

( )224 =+ ba

( ) ( )21 ; 33 = a 1= a From ( )1 ; 615 =+=b

1=a

6=b

( ) 6116 23 ++= xxxxf

( ) ( )206222482 =++= xf is also a factor of ( )xf

( ) ( )3063354273 =++= xf is also a factor of ( )xf

( ) 6116 23 += xxxxf

( ) ( )( )( )321 = xxxxf


2/18


(b)(i)

02 =++ cbxx 02 =++ nmxx

b=+ and c= m=+ and n=

( ) 2222 +=

( ) 42

+= ( ) cb 422 =

Similarly, ( ) = 2 nm 42

If +=+ then =

( ) ( )22 =

nmcb 44 22 =

(ii) ( )( )( )( ) ( ){ } ( ){ } ++++= 22

nmnm ++++= 22 ( m=+Q and )n= ( )( )nmcbnmcb ++++= (Qand are the roots of the equation )02 =++ cbxx ( ) ( ){ } ( ) ( ){ }bmcnbmcn ++=

( ) ( )( )( ) ( )22 bmcnbmcn +++=

( ) ( )( ) ( )22 bmccnbmbnc +=

( ) ( )( ) ( )22

mbccnmbbnc ++= ( ) ( )[ ]cmbcbcbnmbnc ++= 2

( ) ( )( )cmbnmbnc += 2

( )( )( )( ) ( ) ( )( )cmbnmbnc += 2

If the equations 02 =++ cbxx and 02 =++ nmxx have a common root, then = (we say)

Then ( ) 0=

( ) ( )( ) 02 =+ cmbnmbnc

( ) ( )( )cmbnbmnc = 2

0102 =++ kxx 0102 =++ kxx

Here 10=b and kc = Here km = and 10=n

Since 0102 =++ kxx and 0102 =++ kxx have a common root,

( ) ( )( )cmbnbmnc = 2 ( ) ( )( )22 1001010 kkk = ( ) ( )( ) 01001010 22 = kkk ( ) 01001010 2 =+ kkk

10= k or 01102 =+ kk


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10= k or ( )( ) 01011 =+ kk 10=k or 11=k

12.(a)A student council of 15 students consists of 3 Science students, 5 Arts students and 7 Commerce

students. It is required to select 6 students from this student council to work on a project. Find the

number of different ways in which this can be done, if(i) all 15 students are eligible to be selected,

(ii) two particular students are not permitted to work together,(iii) two students from each of the subject streams need to be selected.Also, find the number of different ways in which a group selected under (iii) above can be seated around

a circular table, if the two students from the Science stream in the group are not permitted to sit next to

each other.

(b)Let ( )( ) ( )22 2313

163

+

+=

rr

rUr for

+r and let =

=n

r

rnUS

1

for +n . Find the values of the constantsA

andB such that ( ) ( )22 2313 ++= rB

r

A

Ur for

+

r .

Hence, show that( )223

1

4

1

+=

nSn for

+n .

Is the infinite series

=1rr

U convergent? Justify your answer.

Find the smallest value of +n such that 6104

1


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selected if two particular students are not permitted to work = !6!7

!13

together!7123456

!78910111213

=

6626=

= 1716

(iii) Number of different ways in which 6 students can be 272523 CCC = selected if two students from each of the subject streams

!5!2

!7

!3!2

!5

!1!2

!3=

need to be selected!2!2!2

!234567 =

= 630 Let us consider two students from the Science stream in the group are not permited to sit next to

each other as a single object.

Now we have 5 objects.

Number of differrent ways in which 5 objects can be arranged in ( )!15 = a circular table !4=

But two students from Science stream can be arranged in !2 different ways.

The number of required different ways ( ) !2!4!16 =

!2!4!5 =

( )25!4 = = 72

(b)( )

( ) ( )22 2313163

+

+=

rr

rUr

( )( ) ( ) ( ) ( )2222 23132313

163

++

=

+

+

r

B

r

A

rr

r ( ) ( ) ( )22 1323163 ++=+ rBrAr

When 3

1

=r When 3

2

=r

( )22113

163 +=

+ A ( )21216

3

23 =

+ B

99 =A 99 =B

1=A 1=B

( )( ) ( )

=+

+=

222313

163

rr

rUr ( ) ( )22 23

1

13

1

+

rr


5/18


Let ( )( )213

1

=

rrf .

( )( )223

11

+=+

rrf

( ) ( )1+= rfrfUr

( ) ( )211 ffU =

( ) ( )322 ffU =

( ) ( )433 ffU =

M M M

( ) ( )122 = nfnfUn

( ) ( )nfnfUn = 11

( ) ( )1+= nfnfUn

( ) ( )11 += nffSn

( ) ( )22 23

1

13

1

+

=

nSn

( )2231

4

1

+=

nSn for

+n .

( )

+=

223

1

4

1limlim

nS

nn

n

4

1=S (This is a finite value)

The infinite series

=1rrU is convergent. ( SQ is a finite value)

( )2231

4

1

+=

nSn

6104

1


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( )6

210

23

1 + n

10023 >+ n

67.332> n

The smallest value of +n such that 6104

1 if and only if 1 and

4Arg

=z on an Argand diagram.

For A/L Combined Maths (Group/Individual) Classes 2014/ 2015

Contact :0772252158P.C.P.Peiris

B.Sc (Maths Special)

University of Sri Jayewardenepura


7/18


Answer

(a)

=

11

11Q

IQQT =

=

0

0

11

11

11

11

=

0

02002

2=

( )12 = IQQT

QP2

1

11

11

2

1

2

1

2

12

1

2

1

=

=

=

11

2

1 = QP ( )22 11 = PQ

( ) ( )21 ; 11 22 = IPQQQT

TQP =122 ( IQQ =1Q and )11 = PIP

=

11

11

22

11P

( ) 13 P ; 11 = PDPAPP 1= PDPA

=

11

11

22

1

80

02

11

11

2

1A

=

11

11

82

82

4

1A

=

106

610

4

1A

=

2

5

2

32

3

2

5

A

(b)Modulus of the complex number iyxz += , where yx, is defined as 22 yxz += and its conjugateis defined as iyxz = .

( )( )22222

yxyixiyxiyxzz+==+=

( )3= PDAP


8/18


2

zzz =

( ) iyiyxiyxzz 2=+=

zizz Im2=

( )( )iziziz 333 2 = ( )( )iziziz 313131 2 ++=+

( )( )iziz 33 = ( )( )iziz 3131 ++= ( )( )iziz 33 += ( )( )ziiz 3131 +=

933 ++= ziizzz zzizzi 9331 ++=

( ) 932 ++= zziz ( ) 139 2 ++= zziz

( ) 9Im232

++= ziiz ( ) 1Im2392

++= ziiz

9Im6322

+= zziz ( )1 1Im6931 22 +=+ zziz ( )2

( ) ( )12 ; 18331 222 =+ ziziz

( )( )118 += zz

If 1

033122

ziziz

4


9/18


14.(a)Let ( )

13

2

=

x

xxf for 1x .

Show that ( )( )

( )233

1

2

+=

x

xxxf

lfor 1x , and deduce that the graph of ( )xfy = has turning points at

( )0,0 and

3

4,2

31

31

.

Sketch the graph of ( )xfy = indicating the turning points and asymptotes.

(b)A garden whose boundary consists of eight straight line segmentsmeeting at right angles is shown in the diagram. The dimensions in

metres of the garden are indicated there. The area of the garden is

given to be 2800m . Expressy in terms ofx and show that the perimeter

P of the garden, measured in metres, is given by xx

P 10800

+=

and this formula for the perimeter is valid only for 100


10/18


0=x or 31

2=x

( ) 00 =f and3

4

12

22

31

32

31

=

=

f

The graphs of ( )xfy = has turning points at ( )0,0 and

3

4,2

31

31

.

Sign of ( )xfl

The graphs of ( )xfy = has the relative minimum point at

3

4,2

31

31

and the relative maximum

point at( )0,0 .

( ) =

xfx 1lim 1=x is the vertical asymptote of the graph ( )xfy =

( )

3

11

1

x

xxf

=

( ) =

0lim xfx

0=y is the horizontal asymptote of the graph ( )xfy = .

13

2

=

x

xy

3

4,2

31

31

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

x

y


11/18


x4

x2 x2

y

x

(b)Area of the garden xxxy += 822

284800 xxy +=

x

xy

22200 =

Since0>y

,02200 2 > x

1002


12/18


15.(a)Using integration by parts, find dxxx sin 12 .(b)Using partial fractions, find

( )( ).

11

4322

2

+++

dxxx

xx

(c)Let a , b such that 122 >+ ba andlet +++

+

=

2

0

22 sincos

cos

dxxbxaba

xa

I and dxxbxaba

xqb

J ++++

=

2

0

22 sincos

sin

.

Show that2

=+JI .

By considering aJbI , find the values ofIandJ.

Answer

(a) ( ) = dxxdxd

xxdx11 sinsin

=

dxx

xxx 2

1

1sin

( )dx

x

xxx

+=

2

1

1

2

2

1sin

Cxxxxdx ++= 211

1sinsin , where Cis a constant.

Let = xdxxI 12 sin

dxdx

xxxdxI

+=

212 1sin

dxxxxxxxxx ++= 21212 1sin21sin

+=

dxxxdxxxxxxx2122213 12sin21sin

JIxxxxI 221sin2213 += ( )1 , where = dxxxJ

21

= dxxxJ21

Let sin=x cossin1122

==x ddx cos=

= dJ cossin2

( )= coscos2 d

3

cos3=

( )3

11 22 xxJ

=


13/18


From the equation (1),

( ) 222213 113

21sin3 xxxxxxI ++=

3

12

3

1sin

22213 xxxxx

+

+=

=

dxxx sin 12 Kxxxxx

+

+

+

9

12

9

1

3

sin22213

, where Kis a constant

(b)( )( ) ( ) ( ) ( ) ( )3222

2

111111

43

++

++

++

+

++

x

D

x

C

x

B

x

A

xx

xx

( ) ( )( ) ( )( ) ( )11111143 232 ++++++++ xDxxCxxBxAxx

When 1=x When 1=x When 0=x When 2=x

4312 += D 4318 ++=A 4= DCBA 4643927 ++=+++ DCBA

1=D 1=A ( )12 =+ CB 1413927 =++ CB

( )243 =+ CB

( ) ( );12 22 =B

1=B 1=C

( )( ) ( ) ( ) ( ) ( )32222

1

1

1

1

1

1

1

1

11

43

+

+

+

+

++

xxxxxx

xx

( )( ) ( ) ( ) ( ) ( )

dxx

dxx

dxx

dxx

dxxx

xx +

+

+

+

++3222

2

1

1

1

1

1

1

1

1

11

43

( )( ) ( )222

2

121

111ln1ln

1143

+++++=+

++

xxxxdxxx xx

( )( )=

+

++ dx

xx

xx22

2

11

43 ( )( ) ( )

Cxxx

x+

++

++

+

2

12

1

1

1

1

1ln , where Cis a constant

(c) ++++

=2

0

22 sincos

cos

dxxbxaba

xaI

dxxbxaba

xqbJ +++

+=

2

0

22 sincos

sin

[ ] 202

0

2

0

22

22

1sincos

sincos xdxdx

xbxaba

xbxababJaI ==

+++

+++=+

( )12

=+

bJaI

[ ] 20

222

0

22

2

0

22sincosln

sincos

sincos

sincos

sincos

xbxabadxxbxaba

xaxbdx

xbxaba

xaabxbabaJbI +++=

+++

=

+++

+=

ababba ++++= 2222 lnln


14/18


( )2ln22

22

++

++=

aba

bbaaJbI

( ) ( ) ba + 21 ; ( ) ( ) ab 21 ;

( )aba

bbab

aIba

++

+++=+

22

2222

ln2

( )

aba

bbaa

bJba

++

++=+

22

2222

ln2

=I

++

++

++ aba

bba

b

a

ba 22

22

22 ln2

1

=J

++

++

+ aba

bba

a

b

ba 22

22

22 ln2

1

16. Find the coordinates of the centre and the radius of the circle Swhose equation is given by012222 =++ yxyx , and sketch the circle Sin thexy-plane.

Let P be the point on the circle S, furtherest from the origin O. Write down the coordinates of the point P

and show that the equation of the tangent line l to the circle Sat the point P is given by 22 +=+yx .

A circle lS which touches the line l , also touches the circle Sexternally at a point distinct from P.

Let

( )kh , be the coordinates of the centre of the circle lS . By considering the position ofO and the centre of

lS with respect to the line l, show that 22 +


15/18


++

2

12,

2

12P

Differentiating 0=S with respect tox

02222 =+dx

dy

dx

dyyx

( ) ( )xdxdyy = 11

1

1

=

y

x

dx

dy

Gradient of the tangent drawn to 0=S at the point P

=

dx

dyat

2

12 +== yx

11

1

12

12

2

121

=

=

+

+

=

The equation of the line l : 1=

ax

ay, where

2

12 +=a

ayx 2=+

+=+

2

122yx

22 +=+ yxl

Since the circle 0=l

S touches the line 0=l , also touches the circle 0=S externally at a point distinctfrom P, the origin and the centre ( )kh , should lie on the same side of the line 0=l .

( )( ) 0222200 >++ cc yxyx , where ( )khC , 0222200 >++ kh

022


16/18


( ) ( ) 12

2211

22 ++

=+kh

kh

( ) ( )2

22

2

22211

+=+

khkh

( ) ( )( )2

22222222222

222

22 khkhkhkhkh

++++++=++

( ) 2222 22444288444422 khkhkhkhkhkh ++++++=++

( ) 128242 22 +=+++ khkhkh

17.(a)Prove the identity

( ) ( ) ( ) ( ) ++++++2

1sin

2

1sin

2

1cos4coscoscoscos

(b)Let ( )2

cos42

cos2

sin322

sin2 22xxxx

xf ++=

Express ( )xf in the form ( ) bxa ++sin , where ( ),0>a b and

> qp .The sidesBC, CA andAB of a triangleABCare of lengths qp + ,p and qp respectively.

Show that 0sinsin2sin =+ CBA and deduce that 2cos22cosCACA +

=

.

Answer

(a) ( )

+

++

+=+++

2cos

2

2cos2

2cos

2cos2coscoscoscos

++

+=

2

2cos

2cos

2cos2

+

+

+=

2sin

2sin2

2cos2

( ) ( ) ( ) ( ) ++++++21sin

21sin

21cos4coscoscoscos


Contact :0772252158

P.C.P.Peiris




17/18


(b) ( )2

cos42

cos2

sin322

sin2 22xxxx

xf ++=

2cos2

2cos

2sin32

2cos

2sin2 222

xxxxx++

+=

112

cos2sin32 2 +

++=

xx

xx cossin33 ++=

++= xx cos

2

1sin

2

323

+

+=

6sincos

6cossin23

xx

( ) 36

sin2 +

+=

xxf

This is of the form ( ) ( ) bxaxf ++= sin . Here 2=a , 3=b and 6 = .

( ) 36

sin2 +

+=

xxf

16

sin1

+

x for all x

3236

sin232 ++

++

x for all x

( ) 51 xf for all x

Points of intersection of the graphs 3=y and ( )xfy = are given by,

36

sin23 +

+=

x

06

sin =

+

x

nx =+6

6= nx

=6

11,

6

5,

6

x


18/18


(c)Applying sine rule to the triangleABC

kqp

C

p

B

qp

A=

==

+

sinsinsin, where +k

( )qpkA +=sin kpB =sin ( )qpkC =sin

( )qppqpkCBA ++=+ 2sinsin2sin 0sinsin2sin =+ CBA

0sinsin2sin =+ CBA

2cos

2sin22

2cos

2sin2

BBCACA=

+

2cos

2sin2

2cos

22sin

BBCAB=

2cos

2sin2

2cos

2cos

BBCAB=

( )

+=

22sin2

2cos

CACA

+=

2cos2

2cos

CACA


Contact :0772252158P.C.P.Peiris



6

65

611

34

3

A

B

C

p

qp +

qp

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-1

1

2

3

4

5

x

y

model answer for combined mathematics i -2013 al paper.pdf

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