Polynomials F ()( )x2 −αx +1 x

23
- 1 - Done By : Chandima Peiris (B.Sc - Special) General Certificate of Education (Adv. Level) Examination, August 2015 Combined Mathematics I - Part B Model Answers 11. (a) Polynomials () x F , () x G and () x H of degree 4 in x are given as follows: () ( )( ) 1 1 2 2 + + = x x x x x F β α , where α and β are real constants. () 6 35 62 35 6 2 3 4 + + = x x x x x G , () 1 2 4 + + = x x x H (i) If both () 0 = x F and () 0 = x G have the same roots, show that the quadratic equation with α and β as its roots is 0 50 35 6 2 = + x x . Hence find all the roots of the equation () 0 = x G . (ii) If () () x H x F , find possible values of α and β , and show that the roots of the equation () 0 = x H are not real. (b) (i) Let () 1 2 3 4 + + + x x x x f δ γ , where γ and δ are real constants. Given that 0 2 1 = f and ( ) 21 2 = f , find the two real linear factors of () x f . (ii) Find the two linear expressions () x P and () x Q satisfying the equation ( ) () ( ) () x x Q x x P x x 3 1 1 2 2 = + + + , for all real x. Model Answer (a) (i) () ( )( ) 1 1 2 2 + + = x x x x x F β α , where α and β are real constants. () 6 35 62 35 6 2 3 4 + + = x x x x x G , () 1 2 4 + + = x x x H If both () 0 = x F and () 0 = x G have the same roots. Let it a. () () 0 = = a G a F ( )( ) 0 1 1 2 2 = + + β α a a a a and () 2 0 6 35 62 35 6 2 3 4 = + + a a a a ( ) ( ) ( ) () 1 0 1 2 2 3 4 = + + + + + β α αβ β α a a a a () () ; 2 6 1 × ( ) ( ) ( ) 0 6 6 35 6 62 2 6 35 6 2 2 3 3 = + + + + + + + a a a a a a β α αβ β α ( ) ( ) ( ) a a a a a a 35 62 35 6 2 6 6 2 3 2 3 + + + + + β α αβ β α Comparing the coefficients of; ( ) 35 6 3 = + β α a ( ) 62 2 6 2 = + αβ a 6 35 = + β α 3 31 2 = + αβ 3 25 = αβ

Transcript of Polynomials F ()( )x2 −αx +1 x

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

General Certificate of Education (Adv. Level) Examination, August 2015

Combined Mathematics I - Part B

Model Answers

11. (a) Polynomials ( )xF , ( )xG and ( )xH of degree 4 in x are given as follows:

( ) ( )( )11 22 +−+−= xxxxxF βα , where α and β are real constants.

( ) 63562356 234 +−+−= xxxxxG ,

( ) 124 ++= xxxH (i) If both ( ) 0=xF and ( ) 0=xG have the same roots, show that the quadratic equation with α and β

as its roots is 050356 2 =+− xx . Hence find all the roots of the equation ( ) 0=xG .

(ii) If ( ) ( )xHxF ≡ , find possible values of α and β , and show that the roots of the equation ( ) 0=xH are not real.

(b)

(i) Let ( ) 12 34 +++≡ xxxxf δγ , where γ and δ are real constants. Given that 021 =⎟⎠⎞

⎜⎝⎛−f and

( ) 212 =−f , find the two real linear factors of ( )xf . (ii) Find the two linear expressions ( )xP and ( )xQ satisfying the equation

( ) ( ) ( ) ( ) xxQxxPxx 311 22 =−+++ , for all real x. Model Answer

(a) (i)

( ) ( )( )11 22 +−+−= xxxxxF βα , where α and β are real constants.

( ) 63562356 234 +−+−= xxxxxG ,

( ) 124 ++= xxxH If both ( ) 0=xF and ( ) 0=xG have the same roots. Let it a.

( ) ( ) 0== aGaF

( )( ) 011 22 =+−+−⇒ βα aaaa and ( )2063562356 234 −−−−=+−+− aaaa

( ) ( ) ( ) ( )1012234 −−−−=++−+++−⇒ βααββα aaaa

( ) ( );261 −× ( ) ( ) ( ) 0663566226356 2233 =−+++−−++++− aaaaaa βααββα

( ) ( ) ( ) aaaaaa 3562356266 2323 +−≡+++−+ βααββα Comparing the coefficients of;

( ) 3563 =+⇒ βαa ( ) 62262 −=+−⇒ αβa

6

35=+∴ βα 3312 =+αβ

325=∴αβ

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

The quadratic equation whose roots are α and β is given by ( ) 02 =++− αββα xx

0325

6352 =+−⇒ xx

⇒ 050356 2 =+− xx

050356 2 =+− xx 05015206 2 =+−− xxx

( ) ( ) 010351032 =−−− xxx

( )( ) 052103 =−− xx3

10=⇒ x or 25=x

∴ If 3

10=α then 25=β

Substituting 3

10=α and 25=β in ( )xF

( ) ⎟⎠⎞

⎜⎝⎛ +−⎟⎠⎞

⎜⎝⎛ +−= 1

251

310 22 xxxxxF

⎟⎟⎠

⎞⎜⎜⎝

⎛ +−⎟⎟⎠

⎞⎜⎜⎝

⎛ +−=2

2523

1103 22 xxxx

( ) ( )( )( )( )21231361 −−−−= xxxxxF

∴The roots of the equation ( ) 0=xF are31=x , 3=x ,

21=x and 2=x .

As the equations ( ) 0=xF and ( ) 0=xG have the same roots, the roots of the equation ( ) 0=xG are31=x ,

3=x , 21=x and 2=x .

(ii)

If ( ) ( )xHxF ≡ , ( )( ) 111 2422 ++=+−+− xxxxxx βα

( ) ( ) ( ) 112 24234 ++≡++−+++−⇒ xxxxxx βααββα Comparing the coefficients of;

⇒3x 0=+ βα 122 =+⇒αβx

1−=∴αβ Possible values of α and β are ( )1 , 1 −== βα and ( )1 , 1 =−= βα Let yx =2

( ) 12 ++= yyyH For the roots of ( ) 0=yH 012 =++⇒ yy Discriminant of ( ) 0=yH is 0341 <−=−=Δ ∴The roots of 012 =++ yy are not real. i.e y is not real. Thus the roots of the equation ( ) 0=xH are not real.

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

(b) (i) ( ) 12 34 +++≡ xxxxf δγ

0121

81

16120

21 =+⎟

⎠⎞

⎜⎝⎛−×+⎟

⎠⎞

⎜⎝⎛−×+×⇒=⎟

⎠⎞

⎜⎝⎛− δγf

0841 =+−−⇒ δγ ( )194 −−−−=+ δγ

( ) 21128162212 =+−−×⇒=− δγf ( )21228 −−−−=+ δγ ( ) ( ); 122 −× 92416 −=− γγ 1=γ

From ( )1 , 2=δ .

( ) 122 34 +++= xxxxf

Since 021 =⎟⎠⎞

⎜⎝⎛−f , according to the factor theorem ( )12 +x is a factor of ( )xf .

( ) ( )( )112 3 ++=∴ xxxf

( ) ( )( )( )1112 2 +−++= xxxxxf

∴ ( )12 +x and ( )1+x are the two linear factors of ( )xf

(ii) Let ( ) baxxP += and ( ) dcxxQ += , where ℜ∈ , , , dcba .

( )( ) ( )( ) xdcxxbaxxx 311 22 ≡+−++++

( ) ( ) ( ) ( ) xdbxcbaxdbaxca 323 ≡−+−++++++ Comparing the coefficients of;

⇒3x ( )10 −−−−=+ ca

⇒2x ( )20 −−−−=++ dba ⇒x ( )33 −−−−=−+ cba

Constants⇒ ( )40 −−−−=− db From ( )4 , db = ( )⇒2 ( )502 −−−−=+ ba ( ) ( )31 + ; ( )632 −−−−=+ ba ( ) ( )526 −× ; ⇒= 63a 2=a

( )⇒5 1−=b ∴ 1−=d

( )⇒1 2−=c ∴ ( ) 12 −= xxP and ( ) 12 −−= xxQ

13 +x

122 34 +++ xxx342 xx +

12 +x12 +x

0

12 +x

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

12. (a) A panel of four members is to be formed in order to serve as judges in a talent show competition. This

panel is to be selected from a group consisting of three sportswoman, two sportsmen, six female singers, five male singers, two actresses and four actors. The head judge has to be a sportsman or a sportswoman. Other three members of the panel have to be selected from the group excluding sportsmen and sportswomen. Find the number of different ways in which the panel can be formed, in each of the following cases:

(i) if at least one female singer and one male singer must be included in the panel, (ii) if two males and two females, including the head judge, must be in the panel,

(iii) if the head judge must be a sportswoman.

(b) Find the values of the constants A, B and C such that ( ) ( ) CrrBrA +=+−+ 22 15 for +∈ Zr .

Hence, show that the rth term ( ) ( )( )22 531

8+++

=rrr

U r of an infinite series can be expressed as

( ) ( )2+− rfrf , where ( )rf is a function to be determined.

Find the sum of the series∑=

n

rrU

1

, and deduce that the series ∑∞

=1rrU converges to the sum 22 15

181 + .

Model Answer (a) Let SM- Sportsmen (2), SW-Sportswomen (3) , MS-Male singers (5), FS- Female singers(6),

ASS-Actresses (2) and AR-Actor (4)

(i) If at least one female singer and one male singer must be included in the panel; when the head judge is a sportsmen when the head judge is a sportsmen

Number of different panels

( ) ( ) ( ) ( )13

12

16

25

13

12

26

15

13

12

14

16

15

13

12

12

16

15 CCCCCCCCCCCCCCCCCC +×++×++××++××=

( ) ( ) ( ) ( )32610321553246532265 +××++××++×××++×××= ( )6075120605 +++=

= 1575

(ii) When head judge is a sportsman,

Head judge can be selected in 12C ways. i.e. 2 ways.

Another male person should be selected from male singers and actors. This can be done in 19C

ways. Remaining female persons should be selected from female singers and actresses. This can be done in 2

8C ways.

Combinations SM(2) SW(3) MS(5) FS(6) ASS(2) AR(4)

0 1 1 1 1 0 0 1 1 1 0 1 0 1 1 2 0 0 0 1 2 1 0 0

Combinations SM(2) SW(3) MS(5) FS(6) ASS(2) AR(4)

1 0 1 1 1 0 1 0 1 1 0 1 1 0 1 2 0 0 1 0 2 1 0 0

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

When head judge is a sportswoman, Head judge can be selected in 1

3C ways. i.e. 3 ways. Another female person should be selected from female singers and actresses. This can be done in

18C ways. Remaining male persons should be selected from male singers and actors. This can be done in

29C ways.

∴ Number of different panels, if two males and two females, including the head judge, must be in the panel is 2

91

82

81

9 32 CCCC ××+×× 36832892 ××+××=

864504 += =1368

(iii) When head judge is a sportswoman, she can be selected in 13C ways. i.e. 3 ways.

The remaining 3 members of the panel should be selected from 17 members including male singers, female singers, actresses and actors. This can be done in 3

17C ways.

∴ Number of different panels, if the head judge must be a sportswoman is 317

13 CC ×

6803×= = 2040

(b) ( ) ( ) CrrBrA +=+−+ 22 15

( ) ( ) CrrrBrrA +≡++−++ 122510 22

( ) ( ) ( ) CrBArBArBA +≡−+−+− 252102 Comparing the coefficient of ;

⇒2r BABA =⇒=− 0 ⇒r 181210 =⇒=− ABA ( )BA =Q

81=A ⇒

81=B

Constants⇒ 32425 ==⇒=− ACCBA

∴81=A ,

81=B and 3=C

∴ ( ) ( ) 31815

81 22 +≡+−+ rrr ( )*−−−−

Dividing ( )* by ( ) ( ) ( )222 531 +++ rrr

( ) ( ) ( )( )

( ) ( ) ( )( )

( ) ( ) ( )222

2

222

2

222 5311

81

5315

81

5313

++++−

++++≡

++++

rrrr

rrrr

rrrr

( ) ( )( ) ( ) ( ) ( ) ( )222222 531

311

5318

++−

++≡

+++ rrrrrrr

( ) ( ) ( ) ( )2222 531

311

++−

++≡

rrrrU r

This is of the form ( ) ( )2+−= rfrfU r , where ( )( ) ( )22 31

1++

=rr

rf .

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

( ) ( )2+−= rfrfUr

When 1=r ; ( ) ( )311 ffU −= When 2=r ; ( ) ( )422 ffU −= When 3=r ; ( ) ( )533 ffU −=

When 4=r ; ( ) ( )644 ffU −= ................................................ ................................................ When 3−= nr ; ( ) ( )133 −−−=− nfnfU n

When 2−= nr ; ( ) ( )nfnfU n −−=− 22

When 1−= nr ; ( ) ( )111 +−−=− nfnfU n

When nr = ; ( ) ( )2+−= nfnfU n

( ) ( ) ( ) ( )21211

+−+−+=∑=

nfnfffUn

rr

=∑=

n

rrU

1

( ) ( ) ( ) ( )22222222 53

142

15.3

14.2

1++

−++

−+nnnn

Let ∑=

=n

rrn US

1

.

( ) ( ) ( ) ( )22222222 531

421

5.31

4.21

++−

++−+=

nnnnSn

( ) ( ) ( ) ( ) ⎟⎟⎠

⎞⎜⎜⎝

++−

++−+=

∞→∞→ 22222222 531

421

5.31

4.21limlim

nnnnS

nnn

22 151

81 +=∞S

∴The infinite series ∑∞

=1rrU converges to the sum 22 15

181 + .

+

For A/L Combined Maths (Group/Individual) Classes 2016/ 2017 Contact :0772252158 P.Chandima Peiris B.Sc (1st Class- Maths Special) University of Sri Jayewardenepura

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

13. (a) Three matrices A, B and C are given by

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=

210320

A , ⎟⎟⎠

⎞⎜⎜⎝

⎛=

00

dcba

B and ⎟⎟⎟

⎜⎜⎜

⎛=

213243

C

(i) Show that ⎟⎟⎠

⎞⎜⎜⎝

⎛==

1001

2IAC . Also find the productCA .

(ii) Find the values of a, b, c and d such that 2IBC = . (iii) If ( ) 2ICBA =+ μλ , obtain an equation connecting λ and μ . Express the matrix

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−−=

452683

D in terms of A and B, and hence find the product DC.

(b) A complex number z is given as θθ sincos iz += , where θ ( )πθπ ≤<− is a real parameter.

Find the locus C of the point representing z, on an Argand diagram.

Obtain expressions for θcos and θsin , in terms of z and z1 .

Let 1

22 +

=z

zw and 11

2

2

+−=

zzt , where z is on C such that iz ±≠ .

(i) Show that ( ) 0Im =w and ( ) 0Re =t . Hence, or otherwise, show further that 122 =+ tw . (ii) Find the complex numbers z satisfying the equation 2=w .

(iii) Find the complex numbers z satisfying the equation it = .

Model Answer (a)

(i) ⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=

210320

A , ⎟⎟⎠

⎞⎜⎜⎝

⎛=

00

dcba

B and ⎟⎟⎟

⎜⎜⎜

⎛=

213243

C

⎟⎟⎠

⎞⎜⎜⎝

⎛+−+−

−−=

⎟⎟⎟

⎜⎜⎜

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=

43226634

213243

210320

AC

21001

IAC =⎟⎟⎠

⎞⎜⎜⎝

⎛=

=⎟⎟⎠

⎞⎜⎜⎝

⎛−

⎟⎟⎟

⎜⎜⎜

⎛=

210320

213243

CA⎟⎟⎟

⎜⎜⎜

⎛ −

100010120

(ii) ⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎞⎜⎜⎝

⎛++++

=⎟⎟⎟

⎜⎜⎜

⎟⎟⎠

⎞⎜⎜⎝

⎛=

1001

34233423

213243

00

dcdcbaba

dcba

BC

( )1123 −−−−=+ ba ( )3023 −−−−=+ dc ( )2034 −−−−=+ ba ( )4134 −−−−=+ dc

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

( ) ( ) 3122 ×−× ; 3−=− a ( ) ( ) 3324 ×−× , 2=− c ∴ 3=a ∴ 2−=c ( )⇒1 4−=b ( )⇒3 3=d

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=∴

032043

B

(iii) ⎟⎟⎠

⎞⎜⎜⎝

⎛+−−

−−=⎟⎟

⎞⎜⎜⎝

⎛−

−+⎟⎟

⎞⎜⎜⎝

⎛−

−=+

λμλμλμλμ

μμμμ

λλλλ

μλ2323423

032043

20320

BA

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛++−−++−−−−+−−+

=⎟⎟⎟

⎜⎜⎜

⎟⎟⎠

⎞⎜⎜⎝

⎛+−−

−−=+

λμλμλμλμλμλμλμλμ

λμλμλμλμ

μλ493826266126123849

213243

2323423

CBA

⎟⎟⎠

⎞⎜⎜⎝

⎛+

+=

μλμλ

00

Since ( ) 2ICBA =+ μλ , ⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎞⎜⎜⎝

⎛+

+1001

00

μλμλ

∴ 1=+ μλ

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=

210320

A , ⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=

032043

B

DBA =⎟⎟⎠

⎞⎜⎜⎝

⎛−

−−=−

452683

2

∴ BAD −= 2 ( ) 222222 IIIBCACCBADC =−=−=−=

∴ ⎟⎟⎠

⎞⎜⎜⎝

⎛==

1001

2IDC

(b) ( )1sincos −−−−+= θθ iz

1sincos 22 =+= θθz

12 =∴ z ⇒The locus of C is the circle whose centre is the origin and the radius 1 unit.

( )( )θθ

θθθθ sincos

sincossincos

11ii

iz −−

+=

θθ

θθ22 sincos

sincos+−= i

( )2sincos1 −−−−−= θθ iz

( ) ( )21 + ; z

z 1cos2 +=θ ( ) ( )21 − ;z

zi 1sin2 −=θ

∴ ⎟⎠⎞

⎜⎝⎛ +=

zz 1

21cosθ ⎟

⎠⎞

⎜⎝⎛ −=

zz

i1

21sinθ

Re

Im

CofLocus

( )0,0 ( )0,1( )0,1−

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

⎟⎠⎞

⎜⎝⎛ −−=

zzi 1

2

∴ ⎟⎠⎞

⎜⎝⎛ −= iz

zi

21sinθ

(i)

122 +

=z

zw and 11

2

2

+−=

zzt

θθθθ sincos2sincos 222 iz +−= θθ 2sin2cos2 iz +=

122 +

=z

zw 11

2

2

+−=

zzt

( )θθ

θθ2sin12cos

sincos2i

i++

+= ( )θθθθθ

sincoscos22sin12cos

ii+

+−=

( )θθθ

θθcossin211cos2

sincos22 i

i++−+= ( )θθθ

θθθsincoscos2

cossin21sin21 2

ii

++−−=

( )( )θθθ

θθsincoscos2

sincos2i

i+

+= ( )( )θθθ

θθθsincoscos2cossinsin2

ii

+−−=

θcos1=w ( )( )

( )( )θθθθθθθθθθ

sincossincoscossincoscossinsin

iiii

−+−−−=

θsec=w ( )( )θθ

θθθθθθθ22

222

sincoscossinsincoscossintan

+++−−= ii

∴ ( ) 0Im =w θtanit =

∴ ( ) 0Re =t

θθ 22222 tansec itw +=+

θθ 22 tansec −= ( )12 −=iQ 122 =+ tw

(ii) 2sec2 =⇒= θw

21cos =⇒ θ

3

2 ππθ ±= n , +∈Zn

⎭⎬⎫

⎩⎨⎧ −=

3,

3ππθ

∴Complex numbers are i23

21 + and i

23

21 −

(iii) ⇒= it 1tan =θ

4ππθ += n , +∈Zn

⎭⎬⎫

⎩⎨⎧ −=

43,

4ππθ ∴Complex numbers are i

21

21 + and i

21

21 −−

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

14.

(a) Let ⎟⎠⎞

⎜⎝⎛=

xxy 1sin for 0≠x . Show that

(i) ⎟⎠⎞

⎜⎝⎛−=

xy

dxdyx 1cos and

(ii) 02

24 =+ y

dxydx

(b) Let ( )( )2

2

112

−+=

xxxf for 1≠x .

Find the first derivative and the turning point of ( )xf . Sketch the graph of ( )xfy = , indicating the turning point and the asymptotes.

(c) In the given figure, ABCD is a trapezium with parallel sides BC and AD. Lengths of the sides, measured in centimetres are given by aCDAB == , bBC = and xbAD 2+= , where ax <<0 . BE and CF are perpendiculars drawn from the vertices B and C, respectively, on to the side AD.

Show that the area ( )xS of the trapezium ABCD is given by ( ) ( ) 22 xaxbxS −+= in square

centimetres. If 6=a and 4=b , show further that ( )xS is maximum for a certain value of x, and find this value of x and the maximum area of the trapezium.

Model Answer

(a) ⎟⎠⎞

⎜⎝⎛=

xxy 1sin for 0≠x .

Differentiating y with respect to x

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛−×⎟

⎠⎞

⎜⎝⎛=

xxxx

dxdy 1sin11cos 2

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛−=

xxxdxdy 1sin1cos1

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛−=

xx

xdxdyx 1sin1cos

(i) ⎟⎠⎞

⎜⎝⎛−=

xy

dxdyx 1cos ( )1−−−−

Differentiating (1) with respect to x

⎟⎠⎞

⎜⎝⎛−×⎟

⎠⎞

⎜⎝⎛+=+ 22

2 11sinxxdx

dydxdy

dxydx

A

B Cb

a

x E F D

a

x

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

⎟⎠⎞

⎜⎝⎛−=

xxdxydx 1sin1

22

2

⎟⎠⎞

⎜⎝⎛−=

xdxydx 1sin2

23

⎟⎠⎞

⎜⎝⎛−=

xx

dxydx 1sin2

24

(ii) 02

24 =+ y

dxydx

(b) ( ) ( )2

2

112

−+=

xxxf for 1≠x .

( ) ( ) ( ) ( )( )4

22

11121241

−×−×+−×−=

xxxxxxf l

( ) ( ) ( )[ ]

( )4

2

1121212

−+−−−=

xxxxx

[ ]

( )3

22

112222

−−−−=

xxxx

( ) ( )( )31

122−

+−=x

xxf l

Critical point of ( )xfy = is given by ( ) 0=xf l .

21−=⇒ x

32

94

23

121

121

21

2 =×=

⎟⎠⎞

⎜⎝⎛ −−

+=⎟

⎠⎞

⎜⎝⎛−f

Sign of ( )xf l

21−<x 1

21 <<− x

Test value -1 0 ( )12 +x ( )− ( )+

( )31−x ( )− ( )−

( ) ( )( )31

122−

+−=x

xxf l ( )

( ) ( )−=−−−

( )

( ) ( )+=−+−

∴ ⎟⎠⎞

⎜⎝⎛−

32,

21 is a relative minimum point of ( )xfy = .

( )− ( )+

21−=x

1=x

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

( ) 1110 ==f

( )2

2

2

2

121

12lim

1212limlim

xx

xxx

xxfxxx

+−

+=

+−+=

∞→∞→∞→

( ) 2lim =∞→

xfx

⇒ 2=y is the horizontal asymptote of the graph of ( )xfy = .

( ) ∞=→

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

(c)

22 xah −=

( ) ( )hxbbxS 221 ++=

( ) ( ) 22 xabxxS −+=

If 6=a and 4=b then ( ) ( ) 264 xxxS −+= Differentiating ( )xS with respect to x

( ) ( ) ( ) 2

262

6214 xx

xxxS l −+−×

−×+=

( ) ( )2

2

6

46

xxxxxS l

+−−=

x

21−

32

1

2=y

1=x

y

A

B Cb

a

x E F D

a

x

h

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

( )2

2

6

642

xxxxS l

+−−=

For the critical points of ( )xS , ( ) 0=xS l .

0642 2 =−+⇒ xx 0322 =−+⇒ xx

( )( ) 013 =−+⇒ xx 3−=⇒ x or 1=x

As x is a length, 0>x . ∴ cmx 1=

( ) ( )2

2

6

322

xxxxS l

−+−=

When 0=x , ( ) 0660 >+=lS and when 2=x , ( ) 0

2102 <−=lS .

∴When cmx 1= , ( )xS is maximum. The maximum area of the trapezium ( )1S=

( ) 1641 −+=

= 255 cm

1=x

( )+ ( )−

For A/L Combined Maths (Group/Individual) Classes 2016/ 2017 Contact :0772252158 P.Chandima Peiris B.Sc (1st Class- Maths Special) University of Sri Jayewardenepura

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

15.

(a) Show that ( ) ( )∫∫ −=ππ

π00

dxxfdxxf .

Show also that4

sin2

0

2 ππ

=∫ xdx .

Hence show that4

sin2

0

2 ππ

=∫ xdxx .

(b) Using a suitable substitution and the method of integration by parts, find ∫ dxex x23 .

(c) Find the values of the constants A, B and C such that 111

123 ++

++−

=− xx

CBxx

Ax

.

Hence integrate 1

13 −x

with respect to x.

(d) Using the substitution2

tan xt = , show that61

sin3cos45

2

0

=++∫

π

xxdx .

Model Answer

(a) Want : ( ) ( )∫∫ −=ππ

π00

dxxfdxxf

Let xu −= π When 0=x , π=u dudxdxdu −=⇒−= When π=x , 0=u

( ) ( )( )duufdxxf −=− ∫∫0

0 π

π

π

( )duuf∫−=0

π

( )duuf∫=π

0

( ) ( )dxxfdxxf ∫∫ =−ππ

π00

QDefinite integral independent of the variable

( )∫∫ −=2

0

2

0

2 2cos121sin

ππ

dxxxdx

⎟⎟⎟

⎜⎜⎜

⎛−= ∫ ∫

2

0

2

0

2cos121

π π

xdxdx

[ ] [ ]20

20 2sin

41

21 ππ

xx −=

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

[ ]0sinsin41

4−−= ππ

0

∴4

sin2

0

2 ππ

=∫ xdx

Let ( ) xxxf 2sin= .

( ) ( ) ( )xxxf −−=− πππ 2sin

( ) ( ) xxxf 2sin−=− ππ From the previous proved result,

( ) ( )∫∫ −=ππ

π00

dxxfdxxf

( ) xdxxdxxx 2

00

2 sinsin ∫∫ −=ππ

π

dxxxxdxdxxx ∫∫∫ −=πππ

π0

2

0

2

0

2 sinsinsin

∴ ∫∫ =ππ

π0

2

0

2 sinsin2 xdxdxxx

⇒ ∫∫ =ππ π0

2

0

2 sin2

sin xdxdxxx

∫∫ ×=2

0

2

0

2 sin22

sin

ππ π xdxdxxx QSymmetric property of the function xy 2sin=

4

22

sin0

2 πππ

××=∫ dxxx

4

sin2

0

2 ππ

=∫ dxxx

(b) Let ∫= dxexI x23 .

Let 2xeu = , ux ln2 =

2

222 dudxxexe

dxdu xx =⇒=

∫ ⋅= dxxexI x22

∫= uduln21

( )duududu∫= ln

21

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

⎥⎦⎤

⎢⎣⎡ ×−= ∫ du

uuuu 1ln

21

[ ]∫−= duuu 1ln21

[ ] Cuuu +−= ln21

[ ] Ceex xx +−=222

21

( ) CxeIx

+−= 12

22

, where C is a constant.

(c) ( )( ) 11111

11

223 ++++

−=

++−=

− xxCBx

xA

xxxx

( ) ( )( )111 2 −++++≡ xCBxxxA When 1=x When 0=x When 2=x

13 =A 1=− CA 127 =++ CBA

31=A 1

31 =− C 1

322

37 =−+ B

32−=C

31−=B

∴ ( )( )

( )132

131

11

23 +++−

−=

− xxx

xx

( )( )

( )dxxx

xdxx

dxx ∫∫∫ ++

+−−

=− 1

231

11

31

11

23

dxxx

xx ∫ +++−−=

142

611ln

31

2

⎥⎦⎤

⎢⎣⎡

+++

+++−−= ∫∫ dx

xxdx

xxxx

13

112

611ln

31

22

( ) ∫⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎠⎞

⎜⎝⎛ +

−++−−= dx

x

xxx 22

2

23

21

1211ln

611ln

31

( )⎟⎟⎟⎟

⎜⎜⎜⎜

⎛+

×−++−−= −

2321

tan3

2211ln

611ln

31 12

xxxx

( ) Cxxxxdxx

+⎟⎟⎠

⎞⎜⎜⎝

⎛ +−++−−=−

−∫ 312tan

311ln

611ln

31

11 12

3 , where C is an arbitrary constant.

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

(d) When 2

tan xt = , 212sin

ttx

+= and 2

2

11cos

ttx

+−=

⎟⎠⎞

⎜⎝⎛ +==

2tan1

21

2sec

21 22 xx

dxdt When 0=x , 0=t

( )2121 t

dxdt += When

2π=x , 1=t

∴ ( )212

tdtdx

+=

( )( )∫∫

++

+−+

+=++

1

022

2

22

0

16

1145

12

sin3cos45tt

tt

tdt

xxdx

π

∫ +−++=

1

022 64455

2ttt

dt

∫ ++=

1

02 96

2tt

dt

( )∫ +=

1

023

2t

dt

1

0312 ⎥⎦

⎤⎢⎣⎡

+−=

t

⎥⎦⎤

⎢⎣⎡ −−=

31

412

⎟⎠⎞

⎜⎝⎛−×−=

1212

61

sin3cos45

2

0

=++∫

π

xxdx

For A/L Combined Maths (Group/Individual) Classes 2016/ 2017 Contact :0772252158 P.Chandima Peiris B.Sc (1st Class- Maths Special) University of Sri Jayewardenepura

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

16. Let the equations of two circles be 02222 =++++ cfygxyx and 02222 =++++ lll cyfxgyx .

If these circles intersect orthogonally, show that lll ccffgg +=+ 22 .

Show that the circle C, with the equation 0166822 =+−−+ yxyx touches the x-axis. Two circles, 1C of radius r and 2C of radius ( )rR > , with common centre at the origin, touch the circle C at the points A and B, respectively. Find the values of r and R, and the coordinates of A and B. Let S be a circle which intersects both the circles C and 1C orthogonally and which touches the y-axis. Find the two possible equations for S. The common tangent drawn to the two circles C and 2C at the point B, meets the x-axis at P and the y-axis at Q. Show that the common tangent is 4034 =+ yx , and that the equation of the circle with the line segment

PQ as a diameter is ( ) 040303 22 =−−+ yxyx . Model Answer

Let 02222 =++++≡ cfygxyxS , Centre ( )fgO −−≡ , , Radius cfgr −+= 22

02222 =++++≡ llll cyfxgyxS , Centre ( )lll fgO −−≡ , , Radius llll cfgr −+= 22

If the two circles 0=S and 0=lS intersect orthogonally, °= 90ˆ lOEO , where E is the point of intersection of the circles. ⇒ ( )222 ll OOrr =+

( ) ( )222222 lllll ffggcfgcfg −+−=−++−+⇒ lllllll fffgggfgcfgcfg 22 22222222 −++−+=−++−+⇒

⇒ lll ccffgg +=+ 22

Let 0166822 =+−−+≡ yxyxC .

Centre ( )3 , 41 ≡O and Radius 3169161 =−+=r

Let d be the perpendicular distance from ( )3 , 4 the x-axis ( )0=y .

13 rd ==

⇒The circle 0166822 =+−−+ yxyx touches the x- axis.

Let 02221 =−+≡ ryxC and 0222

2 =−+≡ RyxC , where rR > .

Common centre ( )0 , 0≡O

0=S

0=lSlO

O

r lr

E

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

When the circles 01 =C and 02=C touch

the circle 0=C at the points A and B,

11 OOrr =+ and 11 rOOR += .

( ) ( )221 0304 −+−=OO

51 =OO

∴ 53 =+r and 35 +=R

2=r and 8=R

∴ 04221 =−+≡ yxC and

064222 =−+≡ yxC

Let ( )11, yxA ≡ and ( )22 , yxB ≡

As 3:2: 1 =AOOA ,

58

320342

1 =+

×+×=x and 56

320332

1 =+

×+×=y

∴ ⎟⎠⎞

⎜⎝⎛≡

56,

58A

As 3:5: 11 =BOOO ,

532

350354 2

2 =⇒+

×+= xx

and 524

350353 2

2 =⇒+

×+= yy

∴ ⎟⎠⎞

⎜⎝⎛≡

524,

532B

04221 =−+≡ yxC and 06422

2 =−+≡ yxC

Let 02222 =++++≡ cfygxyxS .

If 0=S touches the y -axis, the radius of this circle should be equal to the modulus of the x-coordinate of the centre ( )fg −− , .

gr −= ( )122 −−−−=⇒ gr

If 0=S intersects both 0=C and 01 =C orthogonally,

( ) ( ) 163242 +=−×+−× cfg and 40202 −=×+× cfg

( )21668 −−−−+=−− cfg and 4=c

O

r

( )3 , 41 ≡O

B

A

3

01 =C

02 =C

0=C

3

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

( )⇒2 2068 =−− fg

( )31034 −−−−=−− fg

Since 222 rcfg =−+ , 222 4 gfg =−+

⇒ 42 =f

∴ 2=f or 2−=f

When 2=f , from ( )3 When 2−=f , from ( )3

41064 −=⇒=−− gg 11064 −=⇒=+− gg

∴Possible equations for the circle 0=S are

044822 =++−+≡ yxyxS and 044222 =+−−+≡ yxyxS

As the common tangent drawn at B is perpendicular to the line AB, gradient of the common tangent

34

431

1 −=−=−=

ABofGradient

∴The equation of the common tangent at B is ⎟⎠⎞

⎜⎝⎛ −−=⎟

⎠⎞

⎜⎝⎛ −

532

34

524 xy

( )32553

45

245 −×

−=−⇒ xy

128207215 +−=−⇒ xy

2001520 =+⇒ yx

⇒ 4034 =+ yx

When the line 4034 =+ yx meets the x-axis, 0=y 10=⇒ x

∴ ( )0 , 10≡P

When the line 4034 =+ yx meets the y-axis, 0=x340=⇒ y

∴ ⎟⎠⎞

⎜⎝⎛≡

340,0Q

∴ The equation of the circle with the line segment PQ as a diameter is ( )( ) ( ) 03400010 =⎟

⎠⎞

⎜⎝⎛ −−+−− yyxx

03

4010 22 =−+−⇒yyxx

⇒ ( ) 040303 22 =−−+ yxyx

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

17. (a) Show that ( ) ( ) 1coscoscos2coscoscos 222 =+−+++ βαβαβαβα (b) Let ( ) ( ) 1sincos22sin2cos ++++= xxxxxf . Express ( )xf in the form ( ) ( )α++ xxk sincos1 , where k

and α are constants to be determined.

Let ( )xg be such that ( ) ( ){ }12cos1

−=+

xgx

xf , where22ππ ≤≤− x .

Sketch the graph of ( )xgy = and hence show that the equation ( ) 0=xf has only one solution in the range given above.

(c) In the usual notation, using the Sine Rule for a triangle ABC, show that

( ) ( ) ⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛ −+=−

2sec

2tan

2cot

2cosec 2 CBCBcbAAcba .

Model Answer (a) ( ) ( ) βαβαβαβα coscoscos2coscoscos.. 222 +−+++=SHL

( ) ( ) ( )12cos12cos21coscoscos2cos2 +++++−+= βαβαβαβα

( ) ( ) ( ) 12cos2cos21coscoscos2cos2 ++++−+= βαβαβαβα

( ) ( ) ( ) ( )( ) 1coscos221coscoscos2cos2 +−+++−+= βαβαβαβαβα

( ) ( ) ( ) ( ) 1coscoscos2coscoscos2 ++−−+++= βαβαβαβαβα

( ) ( ) ( )[ ] 1coscos2coscoscos +−−+++= βαβαβαβα

( )[ ] 1coscos2coscos2cos +−+= βαβαβα

1= SHRSHL .... =

∴ ( ) ( ) 1coscoscos2coscoscos 222 =+−+++ βαβαβαβα

(b) ( ) ( ) 1sincos22sin2cos ++++= xxxxxf

( ) 1sincos2cossin21cos2 2 ++++−= xxxxx

( )xxxxx sincos2cossin2cos2 2 +++=

( ) ( )xxxxx sincos2sincoscos2 +++=

( )( )xxx sincoscos12 ++=

( ) ⎟⎠

⎞⎜⎝

⎛ ++= xxx cos2

12

1sincos122

( ) ( ) ⎟⎠⎞

⎜⎝⎛ ++=

4sincos122 πxxxf

This is of the form ( ) ( ) ( )α++= xxkxf sincos1 ,where 22=k and 4πα = .

( ) ( )

( ) ⎟⎠⎞

⎜⎝⎛ +=

+

⎟⎠⎞

⎜⎝⎛ ++

=+ 4

sin22cos1

4sincos122

cos1π

π

xx

xx

xxf

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

( ) ( ){ }12114

sin22cos1

−=⎥⎦

⎤⎢⎣

⎡ −+⎟⎠⎞

⎜⎝⎛ +=

+xgx

xxf π

∴ ( ) 14

sin2 +⎟⎠⎞

⎜⎝⎛ += πxxg

For all ℜ∈x , 24

sin22 ≤⎟⎠⎞

⎜⎝⎛ +≤− πx

314

sin21 ≤+⎟⎠⎞

⎜⎝⎛ +≤−⇒

πx

( ) 31 ≤≤−⇒ xg , for all ℜ∈x When ( ) 1−=xg When ( ) 3=xg

14

sin −=⎟⎠⎞

⎜⎝⎛ + πx 1

4sin =⎟

⎠⎞

⎜⎝⎛ + πx

( ) ⎟⎠⎞

⎜⎝⎛−−+=+

21

4πππ nnx ( )

421 πππ −−+= nnx ; Zn ∈

( )42

1 πππ −⎟⎠⎞

⎜⎝⎛−−+= nnx ; Zn ∈

⎭⎬⎫

⎩⎨⎧=

4πx

⇒No real solutions for x in the range 22ππ ≤≤− x

When ( ) 1=xg

04

sin =⎟⎠⎞

⎜⎝⎛ + πx

4ππ −= nx ; Zn ∈

⎭⎬⎫

⎩⎨⎧−=

4πx

( ) ( ) ( )( )1cos12 −+= xgxxf When ( ) 0=xf ,

1cos −=x and ( ) 1=xg

When 0cos <x , 2π>x

∴ No real solutions for 1cos −=x in the given range. ( ) 1=xg ⇒The solutions are given by the point of intersection of the graphs ( )xgy = and 1=y .

∴The equation ( ) 0=xf has only one solution in the range given above. It is 4π−=x .

x

y

3

2

1

0

1−

1=y

4π−

2π−

( )xgy =

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

(c) From the Sine Rule, kC

cB

bA

a ===sinsinsin

, where ℜ∈k .

⇒ Aka sin= , Bkb sin= and Ckc sin= ( )

( )( )

( )22

2

2 sinsinsinsinsinCBk

CBAkcb

cba+

−=+

2

2cos

2sin2

2sin

2cossin2

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛ +

⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛ +

=CBCB

CBCBA

⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛ −

=

2cos

22sin4

2sin

22cos

2cos

2sin4

22 CBA

CBAAA

π

π

⎟⎠⎞

⎜⎝⎛ −=+

⇒=++222ACBCBA ππQ

⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛ −

=

2cos

2cos

2sin

2sin

2cos

2sin

22 CBA

CBAAA

⎟⎠⎞

⎜⎝⎛ −

×⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛ −

×

×

=

2cos

1

2cos

2sin

2sin

1

2sin

2cos

1CBCB

CB

AA

A

( )

( )2

cosec2

cot

2sec

2tan

2 AA

CBCB

cbcba

⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛ −

=+

∴ ( ) ( ) ⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛ −+=−

2sec

2tan

2cot

2cosec 2 CBCBcbAAcba

For A/L Combined Maths (Group/Individual) Classes 2016/ 2017 Contact :0772252158 P.Chandima Peiris B.Sc (1st Class- Maths Special) University of Sri Jayewardenepura