Polynomials F ()( )x2 −αx +1 x
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−
4π
2π
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