7/18/2020 Chapter 4.
Transcript of 7/18/2020 Chapter 4.
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7/18/2020 Chapter 4. PARTIAL FRACTIONS
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MATHEMATICS Science Group 10th
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Contents Exercise 4.1 ............................................................................. 1
Exercise 4.2 ............................................................................. 3
Exercise 4.3 ............................................................................. 6
Exercise 4.4 ............................................................................. 9
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Fraction: A fraction is an indicated quotient of two numbers or algebraic expressions. Or The quotient of two numbers or algebraic expression is called a fraction. The quotient is indicated by a bar (____________). Rational Fraction:
An expression of the form π(π₯)
π·(π₯), π€ππ‘β π·(π₯) β 0
and π(π₯)πππ π·(π₯)πππ ππππ¦πππππππ in π₯ expressed with real coefficients, is called a rational fraction. Every fraction expression can be expressed as a quotient of two polynomials. For example:
2π₯
(π₯ β 1)(π₯ + 2)
Proper Fraction:
A rational fraction π(π₯)
π·(π₯), with π·(π₯) β π is called a
proper fration if degree of the polynomial π(π₯) in the numenator is less than the degree of the polynomial π·(π₯) in the denominator. For example:
2
π₯ + 1
Improper Fraction:
A rational fraction π(π₯)
π·(π₯), π€ππ‘β π·(π₯) β 0 is called
improper fraction if degree of the polynomial π(π₯) is greater or equal to the degree of the polynomial π·(π₯). For example:
5π₯
π₯ + 2,
6π₯4
π₯3 + 1
Identity: An identity is an equation, which is satisfied by all the valves of the variables involved. For example
2(π₯ + 1) = 2π₯ + 2
And 2π₯2
2= 2π₯ πππ πππππ‘ππ‘πππ , as these equations are
satisfied by all valves of π₯
Exercise 4.1 Resolve into partial fractions.
Question N0.1 ππ β π
(π + π)(π β π)
Solution:
Let 7π₯β9
(π₯+1)(π₯β3)=
π΄
(π₯+1)+
π΅
(π₯β3)β (π)
Multiplying equation (π)ππ¦ (π₯ + 1)(π₯ β 3)
7π₯ β 9 = π΄(π₯ β 3) + π΅(π₯ + 1) β (ππ)
πΉππ πππ‘ππππππ π‘βπ π£πππ’π ππ π΄ πππ π΅
π₯ β 3 = 0 β π₯ = 3 πππ π₯ + 1 = 0 β π₯ = β1 ππ’π‘π‘πππ π₯ = 3 πππ π₯ = β1 ππ’π‘ ππ ππ(ππ)π€π πππ‘
πΉππ π₯ = 3 7(3) β 9 = π΅(3 + 1)
21 β 9 = 4π΅ 12 = 4π΅
β
πΉππ π₯ = β1 7(β1) β 9 = π΄(β1 β 3)
β7 β 9 = β4π΄ β16 = β4π΄
β
ππ’π‘π‘πππ π‘βπ π£πππ’π ππ π΄ πππ π΅ ππ πππ’ππ‘πππ (π)ππ
πππ‘ π‘βπ ππππ’ππππ ππππ‘πππ πππππ‘ππππ 4
(π₯ + 1)+
3
(π₯ β 3)
Thus 7π₯ β 9
(π₯ + 1)(π₯ β 3)=
4
(π₯ + 1)+
3
(π₯ β 3)
Question No.2 π₯ β 11
(π₯ β 4)(π₯ + 3)
Solution:
Let π₯β11
(π₯β4)(π₯+3)=
π΄
(π₯β4)+
π΅
(π₯+3)β (π)
Multiplying equation (π)ππ¦ (π₯ β 4)(π₯ + 3) ππ
Both sides, we get
π₯ β 11 = π΄(π₯ + 3) + π΅(π₯ β 4) β (ππ)
βπ₯ + 3 = 0 β π₯ = β3 πππ π₯ β 4 = 0 β π₯ = 4 ππ’π‘π‘πππ π₯ = β3 πππ π₯ = 4 ππ’π‘ ππ ππ(ππ)π€π πππ‘
πΉππ π₯ = β3 β3 β 11 = π΅(β3 β 4)
β14 = β7π΅ β
πΉππ π₯ = 4 4 β 11 = π΄(4 + 3)
β7 = 7π΄ β
ππ’π‘π‘πππ π‘βπ π£πππ’π ππ π΄ πππ π΅ ππ πππ’ππ‘πππ (π)ππ
πππ‘ π‘βπ ππππ’ππππ ππππ‘πππ πππππ‘ππππ β1
(π₯ β 4)+
2
(π₯ + 3)
Hence the required partial fraction are
π΅ = 3 π΄ = 4
π΅ = 2 π΄ = β1
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π₯ β 11
(π₯ β 4)(π₯ + 3)=
β1
(π₯ β 4)+
2
(π₯ + 3)
Question N0.3 ππ β π
ππ β π
Solution: 3π₯ β 1
π₯2 β 1=
3π₯ β 1
(π₯ β 1)(π₯ + 1)
πΏππ‘ 3π₯ β 1
(π₯ β 1)(π₯ + 1)=
π΄
π₯ β 1+
π΅
π₯ + 1β (π)
ππ’ππ‘ππππ¦πππ πππ‘β π ππππ ππ¦(π₯ β 1)(π₯ + 1)
π€π πππ‘ 3π₯ β 1 = π΄(π₯ + 1) + π΅(π₯ β 1) β (ππ)
πΏππ‘ π₯ + 1 = 0 π. ππ₯ = β1 πππ π₯ β 1 = 0 π. π π₯ = 1 ππ’π‘π‘πππ π₯ = β1 πππ π₯ = 1 ππ’π‘ ππ ππ(ππ)π€π πππ‘
πΉππ π₯ = 1 3(1) β 1 = π΄(1 + 1)
3 β 1 = 2π΄ 2 = 2π΄
β
πΉππ π₯ = β1 3(β1) β 1 = π΅(β1 β 1)
β3 β 1 = β2π΅ β4 = β2π΅
β
Hence the required partial fraction are 3π₯ β 1
(π₯ β 1)(π₯ + 1)=
1
π₯ β 1+
2
π₯ + 1
Question No.4 π β π
ππ + ππ β π
Solution: π₯ β 5
π₯2 + 2π₯ β 3=
π₯ β 5
π₯2 + 3π₯ β π₯ β 3
=π₯ β 5
π₯(π₯ + 3) β 1(π₯ + 3)=
π₯ β 5
(π₯ β 1)(π₯ + 3)
π₯ β 5
(π₯ β 1)(π₯ + 3)=
π΄
π₯ β 1+
π΅
π₯ + 3β (ππ)
ππ’ππ‘ππππ¦πππ πππ‘β π ππππ ππ¦ (π₯ β 1)(π₯ + 3), π€π πππ
π₯ β 5 = π΄(π₯ + 3) + π΅(π₯ β 1) β (ππ)
πΏππ‘ π₯ + 3 = 0 β π₯ = β3 πππ π₯ β 1 = 0 β π₯ = 1 ππ’π‘π‘πππ π₯ = β3πππ π₯ = 1 ππ πππ’ππ‘πππ(ππ)π€π πππ‘
πΉππ π₯ = β3 β3 β 5 = +π΅(β3 β 1)
β8 = β4π΅
π΅ =β8
β4
β
πΉππ π₯ = 1 1 β 5 = π΄(1 + 3)
β4 = 4π΄
π΄ =β4
4
β
Hence the required partial fractions are π₯ β 5
(π₯ β 1)(π₯ + 3)=
β1
π₯ β 1+
2
π₯ + 3
Question No.5 3π₯ + 3
(π₯ β 1)(π₯ + 2)
Solution:
πΏππ‘ 3π₯ + 3
(π₯ β 1)(π₯ + 2)=
π΄
π₯ β 1+
π΅
π₯ + 2 β (π)
ππ’ππ‘ππππ¦πππ πππ‘β π ππππ ππ¦ (π₯ β 1)(π₯ + 2)π€π πππ‘
3π₯ + 3 = π΄(π₯ + 2) + π΅(π₯ β 1) β (ππ)
πΏππ‘ π₯ β 1 = 0 π. π π₯ = 1 πππ π₯ + 2 = 0 π. π π₯ = β2 ππ’π‘π‘πππ π₯ = 1 πππ π₯ = β2 ππ πππ’ππ‘πππ (ππ)
πΉππ π₯ = 1 3(1) + 3 = π΄(1 + 2)
3 + 3 = 3π΄ 6 = 3π΄ 6
3= π΄
β
πΉππ π₯ = β2 3(β2) + 3 = π΅(β2
β 1) β6 + 3 = π΅(β2) β6 + 3 = β3π΅
π΅ =β3
β3
β
Hence the required partial fractions are 3π₯ + 3
(π₯ β 1)(π₯ + 2)=
2
π₯ β 1+
1
π₯ + 2
Question N0.6 7π₯ β 25
(π₯ β 4)(π₯ β 3)
Solution:
πΏππ‘ 7π₯ β 25
(π₯ β 4)(π₯ β 3)=
π΄
π₯ β 4+
π΅
π₯ β 3
Multiplying both sides by π₯ β 4)(π₯ β 3), π€π πππ‘
7π₯ β 25 = π΄(π₯ β 3) + π΅(π₯ β 4) β (ππ)
πΏππ‘ π₯ β 3 = 0 π. π π₯ = 3 πππ π₯ β 4 = 0 π. π π₯ = 4 ππ’π‘π‘πππ π₯ = 3 πππ π₯ = 4 ππ πππ’ππ‘πππ (ππ)π€π πππ‘
πΉππ π₯ = 3 7(3) β 25 = π΅(3 β 4)
21 β 25 = βπ΅ β4 = βπ΅
β
πΉππ π₯ = 4 7(4) β 25 = π΄(4 β 3)
28 β 25 = 1π΄ 3 = π΄
β
Hence the required partial fractions are 7π₯ β 25
(π₯ β 4)(π₯ β 3)=
3
π₯ β 4+
4
π₯ β 3
Question No.7
ππ + ππ + π
(π β π)(π + π
Solution: π₯2+2π₯+1
(π₯β2)(π₯+3 ππ ππ ππππππ‘πππ‘ πππππ‘πππ.
First we resolve it into proper fraction.
By log division we get 1
π₯2 + π₯ β 6βπ₯2 + 2π₯ + 1
Β±π₯2 Β± π₯ β 1
π₯ + 7
We haveπ₯2+2π₯+1
π₯62+π₯β6= 1 +
π₯+7
π₯2+π₯β6
πΏππ‘ π₯ + 7
(π₯ β 2)(π₯ + 3)=
π΄
π₯ β 2+
π΅
π₯ + 3β (π)
Multiplying both sides by (π₯ β 2)(π₯ + 3) π€π πππ‘
π₯ + 7 = π΄(π₯ + 3) + π΅(π₯ β 2) β (ππ)
πΏππ‘ π₯ + 3 = 0 π. π π₯ = β3
πππ π₯ β 2 = 0 π. π π₯ = 2
π΄ = 1 π΅ = 2
π΅ = 2 π΄ = β1
π΄ = 2 π΄ = 1
π΅ = 4 π΄ = 3
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πΉππ π₯ = β3 β3 + 7 = π΅(β3 β 2)
4 = β5π΅ β
πΉππ π₯ = 2 2 + 7 = π΄(2 + 3)
9 = 5π΄ β
Hence the required partial fractions are
π₯2 + 2π₯ + 1
π₯62 + π₯ β 6= 1 +
9
5(π₯ β 2)β
4
5(π₯ + 3)
Question No.8
πππ + πππ β π
πππ β ππ β π
Solution:
6π₯3+5π₯2β7
3π₯2β2π₯β1 is an improper fraction.
First we resolve in to proper fraction.
2π₯ + 3
3π₯2 β 2π₯ β 1β6π₯3 + 5π₯2 β 7
Β±6π₯3 Β± 4π₯2 β 2π₯
9π₯2 + 2π₯ β 7Β±9π₯2 β 6π₯ β 3 8π₯ β 4
6π₯3 + 5π₯2 β 7
3π₯2 β 2π₯ β 1= (2π₯ + 3) +
8π₯ β 4
(3π₯ + 1)(π₯ β 1)
Now, Let 8π₯β4
(3π₯+1)(π₯β1)=
π΄
3π₯+1+
π΅
π₯β1
Multiplying both sides by (3π₯ + 1)(π₯ β 1, π€π πππ‘
8π₯ β 4 = π΄(π₯ β 1) + π΅(3π₯ + 1) β (ππ)
πΏππ‘ π₯ β 1 = 0 π. π π₯ = 1
πππ 3π₯ + 1 = 0 π. π π₯ = β1
3
ππ’π‘π‘πππ π₯ = 1 πππ π₯ = β1
3 ππ πππ’ππ‘πππ (ππ)
π€π πππ‘
πΉππ π₯ = 1 8(1) β 4 = π΅[3(1) + 1]
8 β 4 = 4π΅ 4 = 4π΅
4π΅ = 4
π΅ =4
4
β
πΉππ π₯ = β1
3
8 (β1
3) β 4 = π΄ (β
1
3
β 1)
β8
3β 4 = π΄ (
β1 β 3
3)
β8 β 12
3=
π΄(β4)
3
β20
3=
π΄(β4)
3
β
Hence the required of a fraction when π·(π₯)ππππ ππ π‘π
Of repeated linear factors are
6π₯3 + 5π₯2 β 7
3π₯2 β 2π₯ β 1= 2π₯ + 3 +
5
3π₯ + 1+
1
π₯ β 1
Resolution of a fraction when π·(π₯) consists of repeated linear factors. Rule II: If a linear factor (ππ₯ + π) occurs π times as a factor of π·(π₯), then there are π partial fractions of the form.
π΄1
(ππ₯ + π)+
π΄2
(ππ₯ + π)2+ β― +
π΄π
(ππ₯ + π)π
Where π΄1, π΄2, β¦ , π΄π are constant and π β₯ 2 is a positive integer.
β΅π(π₯)
π·(π₯)=
π΄1
(ππ₯ + π)+
π΄2
(ππ₯ + π)2+ β― +
π΄π
(ππ₯ + π)π
Exercise 4.2 Resolve into partial fractions:
Question No.1
ππ β ππ + π
(π β π)π(π β π)
Solution:
πΏππ‘ π₯2 β 3π₯ + 1
(π₯ β 1)2(π₯ β 2)=
π΄
π₯ β 1+
π΅
(π₯ β 1)2+
πΆ
π₯ β 2
β (π)
Multiplying both sides by (π₯ β 1)2(π₯ β 2) π€π πππ‘
π₯2 β 3π₯ + 1 = π΄(π₯ β 1)(π₯ β 2) + π΅(π₯ β 2) + πΆ(π₯ β 1)2
β (ππ)
π₯2 β 3π₯ + 1 = π΄(π₯2 β 3π₯ + 2) + π΅(π₯ β 2) + πΆ(π₯2
β 2π₯ + 1)
ππ’π‘π‘πππ π₯ β 1 = 0 π. ππ₯ = 1 ππ (ππ)π€π πππ‘
(1)2 β 3(1) + 1 = (1 β 2) 1 β 3 + 1 = π΅(β1)
β1 = βπ΅
β π΅ = 1
ππ’π‘π‘πππ π₯ β 2 = 0 π. π π₯ = 2 ππ (ππ)π€π πππ‘
(2)2 β 3 + 1 = πΆ(2 β 1)2 4 β 6 + 1 = πΆ
β1 = πΆ β β1
β πΆ = β1
Equating the coefficient of π₯2 in (ii) we get
1 = π΄ + πΆ
1 = π΄ β 1
π΄ = 1 + 1
β π΄ = 2
Hence the required partial fractions are
π₯2 β 3π₯ + 1
(π₯ β 1)2(π₯ β 2)=
2
π₯ β 1+
1
(π₯ β 1)2+
1
π₯ β 2
Question No.2
ππ + ππ + ππ
(π + π)π(π + π)
Solution:
πΏππ‘ π₯2+7π₯+11
(π₯+2)2(π₯+3)=
π΄
π₯+2+
π΅
(π₯+2)2 +πΆ
π₯+3β (π)
Multiplying both sides by (π₯ + 2)2(π₯ + 3)
βπ₯2 + 7π₯ + 11 = π΄(π₯ + 2)(π₯ + 3) + π΅(π₯ + 3) +
πΆ(π₯ + 2)2
π΅ = β4
5 π΄ =
9
5
π΅ = 1
π΄ = 5
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βπ₯2 + 7π₯ + 11 = π΄(π₯2 + 5π₯ + 6) + π΅(π₯ + 3) +
πΆ(π₯2 + 4π₯ + 4) β (ππ) ππ’π‘π‘πππ π₯ + 2 = 0 π. ππ₯ = β2 ππ(ππ)π€π πππ‘
(β2)2 + 7(β2) + 11 = π΅(β2 + 3)
4 β 14 + 11 = π΅
β π΅ = 1
Putting π₯ + 3 = 0 π. π π₯ = β3 ππ(ππ)π€π πππ‘
(β3)2 + 7(β3) + 11 = πΆ(β3 + 2)2
9 β 21 + 11 = πΆ(β1)2 20 β 21 = πΆ(1)
β1 = πΆ
β πΆ = β1
Equating coefficient of π₯2 in (ii) we get
π΄ + πΆ = 1
π΄ β 1 = 1
π΄ = 1 + 1
β π΄ = 2
Hence the required partial fractions are
π₯2 + 7π₯ + 11
(π₯ + 2)2(π₯ + 3)=
2
π₯ + 2+
1
(π₯ + 2)2β
1
π₯ + 3
Question No.3 π
(π β π)(π + π)π
Solution:
πΏππ‘9
(π₯ β 1)(π₯ + 2)2=
π΄
π₯ β 1+
π΅
π₯ + 2+
πΆ
(π₯ + 2)2β (π)
Multiplying both sides by (π₯ β 1)(π₯ + 2)2 π€π πππ‘
9 = π΄(π₯ + 2)2 + π΅(π₯ β 1)(π₯ + 2) + πΆ(π₯ β 1) β (ππ) ππ’π‘π‘πππ π₯ β 1 = 0 π. π π₯ = 1 ππ(ππ)π€π πππ‘
9 = π΄(1 + 2)2
9 = π΄(3)2 9 = 9π΄
β π΄ = 1
ππ’π‘π‘πππ π₯ + 2 = 0 π. π π₯ = β2 ππ (ππ)π€π πππ‘
9 = πΆ(β2 β 1)
9 = β3πΆ
β πΆ = β3
Equating the coefficient of π₯2 in (ii) we get
π΄ + π΅ = 0
π΅ = βπ΄
β π΅ = β1
Hence the partial fractions are 9
(π₯ β 1)(π₯ + 2)2=
1
π₯ β 1β
1
π₯ + 2+
3
(π₯ + 2)2
Question No.4
ππ + π
ππ(π β π)
Solution:
π₯4 + 1
π₯2(π₯ β 1)=
π₯4 + 1
π₯3 β π₯2 ππ ππ ππππππππ πππππ‘πππ.
πΉπππ π‘ π€π πππ πππ£π ππ‘ πππ‘π ππππππ πππππ‘πππ.
π₯3 β π₯2 βπ₯4 + 1
Β±π₯4 β π₯3
π₯3 + 1Β±π₯3 β π₯2
π₯3 + 1
π₯4 + 1
π₯2(π₯ β 1)= (π₯ + 1) +
π₯2 + 1
π₯2(π₯ β 1)β (π)
πΏππ‘ π₯2 + 1
π₯2(π₯ β 1)= (π₯ + 1) +
π₯2 + 1
π₯2(π₯ β 1) β (ππ)
Multiplying both sides by π₯2(π₯ β 1)we get
π₯2 + 1 = π΄(π₯)(π₯ β 1) + π΅(π₯ β 1) + ππ₯2 β (πππ) ππ’π‘π‘π‘πππ π₯ = 0 ππ (πππ)π€π πππ‘
0 + 1 = π΅(0 β 1)
1 = βπ΅
β π΅ = 1
Putting π₯ β 1 = 0 π. π π₯ = 1 ππ (πππ)π€π πππ‘
(1)2 + 1 = πΆ(1)2 1 + 1 = πΆ(1)
2 = πΆ
β πΆ = 2
Equating the coefficient of π₯2 ππ (πππ) we get
π΄ + πΆ = 1
π΄ + 2 = 1
π΄ = 1 β 2
β π΄ = β1
Putting the value of π΄, π΅, πΆ ππ πππ’ππ‘πππ (ππ)
Thus required partial fraction are
π₯4 + 1
π₯2(π₯ β 1)= (π₯ + 1) β
1
π₯β
1
π₯2+
2
π₯ β 1
Question No.5 ππ + π
(ππ + π)(π + π)π
Solution:
πΏππ‘ 7π₯ + 4
(3π₯ + 2)(π₯ + 1)2=
π΄
3π₯ + 2+
π΅
π₯ + 1+
πΆ
(π₯ + 1)2 β (π)
Multiplying both sides by (3π₯ + 2)(π₯ + 1)2 π€π πππ‘
7π₯ + 4 = π΄(π₯ + 1)2 + π΅(3π₯ + 2)(π₯ + 1) + πΆ(3π₯ + 2) β (ππ)
ππ’π‘π‘πππ 3π₯ + 2 = 0 π. π π₯ = β2
3 ππ (ππ)π€π πππ‘
7 (β2
3) + 4 = π΄ (β
2
3+ 1)
2
β14
3+ 4 = π΄ (
β2 + 3
3)
2
β14 + 12
3= π΄ (
1
3)
2
β2
3=
1
9π΄
β18 = 3π΄
π΄ = β18
3
β π΄ = β6
Putting π₯ + 1 = 0 π. π π₯ = β1 ππ (ππ) π€π πππ‘
7(β1) + 4 = πΆ(3(β1) + (+2)
β7 + 4 = βπΆ
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β3 = βπΆ
β πΆ = 3
πΈππ’ππ‘πππ π‘βπ π£πππ’π ππ π΄, π΅ πππ πΆ ππ πππ’ππ‘πππ (π)
π€π πππ‘ ππππ’ππππ ππππ‘πππ πππππ‘ππππ . 7π₯ + 4
(3π₯ + 2)(π₯ + 1)2=
β6
3π₯ + 2+
2
π₯ + 1+
3
(π₯ + 1)2
Question No.6 π
(π β π)π(π + π)
Solution:
πΏππ‘1
(π₯ β 1)2(π₯ + 1)=
π΄
π₯ β 1+
π΅
(π₯ β 1)2+
πΆ
π₯ + 1β (ππ)
Multiplying both sides by (π₯ β 1)(π₯ β 1)2 π€π πππ‘
1 = π΄(π₯ β 1)(π₯ + 1) + π΅(π₯ + 1) + πΆ(π₯ β 1)2 β (ππ)
ππ’π‘π‘πππ π₯ β 1 = 0 π. π π₯ = 1 ππ (ππ)π€π πππ‘
1 = π΅(1 + 1)
1 = 2π΅
β π΅ =
1
2
ππ’π‘π‘πππ π₯ + 1 = 0 π. π π₯ = β1 ππ (ππ) π€π πππ‘
1 = πΆ(β1 β 1)2
1 = πΆ(β1 β 1)2
1 = πΆ(β2)2 1 = 4πΆ
β πΆ =
1
4
Equating the coefficient of π₯2 ππ (ππ)π€π πππ‘
π΄ + πΆ = 0
π΄ = βπΆ
π΄ = β (1
4)
β π΄ = β
1
4
Putting the value of
π΄, π΅, πππ πΆ ππ πππ’ππ‘πππ (π)π€π πππ‘ ππππ’ππππ ππππ‘πππ
πππππ‘ππππ 1
(π₯ β 1)2(π₯ + 1)
=β1
4(π₯ β 1)+
1
2(π₯ β 1)2+
1
4(π₯ + 1)
Question No.7
πππ + πππ + ππ
(π + π)π
Solution:
3π₯2 + 15π₯ + 16
(π₯ + 2)2=
3π₯2 + 15π₯ + 16
π₯2 + 4π₯ + 4
The given fraction is improper fraction.
By long division, 3
π₯2 + 4π₯ + 4 β3π₯2 + 15π₯ + 16
Β±3π₯2 Β± 12π₯ Β± 12
3π₯ + 4
3π₯2 + 15π₯ + 16
(π₯ + 2)2= 3 +
3π₯ + 4
π₯2 + 4π₯ + 4β (π)
πΏππ‘3π₯ + 4
(π₯ + 2)2=
π΄
π₯ + 2+
π΅
(π₯ + 2)2β (ππ)
Multiplying both sides by (π₯ + 2)2 π€π πππ‘
3π₯ + 4 = π΄(π₯ + 2) + π΅ β (πππ)
ππ’π‘π‘πππ π₯ + 2 = 0 π. π π₯ = β2 ππ (πππ)π€π πππ‘
3(β2) + 4 = π΅
β6 + 4 = π΅
β π΅ = β2
Equating the coefficient of "π₯" we get
3 = π΄
β π΄ = 3
Putting the value of π΄ πππ π΅ ππ πππ’ππ‘πππ (ππ) and
π’π πππ πq.(π) π€π πππ‘ ππππ’ππππ ππππ‘πππ πππππ‘ππππ . 3π₯2+15π₯+16
(π₯+2)2 = 3 +3
π₯+2β
2
(π₯+2)2
Question No.8 π
(ππ β π)(π + π)
Solution: 1
(π₯2 β 1)(π₯ + 1)=
1
(π₯ β 1)(π₯ + 1)(π₯ + 1)
=1
(π₯ β 1)(π₯ + 1)2
πΏππ‘ 1
(π₯ β 1)(π₯ + 1)2=
π΄
π₯ β 1+
π΅
π₯ + 1+
πΆ
(π₯ + 1)2
Multiplying both sides by (π₯ β 1)(π₯ + 1)2 π€π πππ‘
1 = π΄(π₯ + 1)2 + π΅(π₯ + 1)(π₯ β 1) + πΆ(π₯ β 1)β (ππ)
ππ’π‘π‘πππ π₯ β 1 = 0 π. π π₯ = 1 ππ (ππ) π€π πππ‘
1 = π΄(1 + 1)2
1 = π΄(2)2
1 = 4π΄
β π΄ =1
4
Putting π₯ + 1 = π π. π π₯ = β1 ππ (ππ) π€π πππ‘
1 = πΆ(β1 β 1)
1 = β2πΆ
β πΆ =β1
2
Equating the coefficient of π₯2 ππ πππ’ππ‘πππ (ππ)
We get π΄ + π΅ = 0
π΅ = βπ΄
π΅ = β (1
4)
β π΅ = β1
4
ππ’π‘π‘πππ π‘βπ π£πππ’π ππ π΄ πππ π΅ ππ πππ’ππ‘πππ (ππ)
We get required partial fractions. 1
(π₯ β 1)(π₯ + 1)2
=1
4(π₯ β 1)β
1
4(π₯ + 1)β
1
2(π₯ + 1)2
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Resolution of fraction when π«(π) consists of non-repeated irreducible quadratic factors: Rule III If a quadratic
(πππ + ππ +factor π) with π β π occur once as a factor of π«(π). The partial fraction is of
the form π¨π+π©
(πππ+ππ+π) where A and B are constants to
be found.
Exercise 4.3 Resolve into partial fraction.
Question No.1 ππ β ππ
(π + π)(ππ + π)
Solution: ππ β ππ
(π + π)(ππ + π)
πΏππ‘ ππ β ππ
(π + π)(ππ + π)=
π¨
π + π+
π©π + πͺ
π₯2 + 1
Multiplying both sides (π₯ β 3)(π₯2 + 1), π€π πππ‘
3π₯ β 11 = π΄(π₯2 + 1) + (π΅π₯ + πΆ)(π₯ + 3) β (π)
3π₯ β 11 = π΄(π₯2 + 1) + π΅π₯(π₯ + 3) + πΆ(π₯ + 3)
β (ππ) ππ’π‘π‘πππ π₯ + 3 = 0 π. π π₯ = β3, π€π πππ‘
3(β3) β 11 = π΄[(β3)2 + 1]
β9 β 11 = π΄(9 + 1)
β20 = 10π΄
π΄ =β20
10
β π΄ = β2
Now equating the coefficient of π₯2 and π₯ we get from
equation (πππ)
π΄ + π΅ = 0 β2 + π΅ = 0
π΅ = 2
β π΅ = 1
3π΅ + πΆ = 3 3(2) + πΆ = 3
6 + πΆ = 3 πΆ = 3 β 6
β πΆ = β3
Putting the value
ππ π΄, π΅ πππ πΆ ππ πππ’ππ‘πππ (π)π€π πππ‘
Required partial fractions. 3π₯ β 11
(π₯ + 3)(π₯2 + 1)=
β2
π₯ + 3+
2π₯ β 3
π₯2 + 1
Question No.2
ππ + π
(ππ + π)(π + π)
Solution:
πΏππ‘ ππ + π
(ππ + π)(π + π)=
π¨π + π©
ππ + π+
πͺ
π + πβ (ππ)
Multiplying both sides by π₯2 + 1)(π₯ + 3)
3π₯ + 7 = (π΄π₯ + π΅)(π₯ + 3) + πΆ(π₯2 + 1)
3π₯ + 7 = π΄π₯(π₯ + 3) + π΅(π₯ + 3) + πΆ(π₯2 + 1) β (ππ) ππ’π‘π‘πππ π₯ + 3 = 0 π. π π₯ = β3 ππ (ππ)π€π πππ‘
3(β3) + 7 = πΆ[(β3)2 + 1] β9 + 7 = πΆ(9 + 1)
β2 = 10πΆ
πΆ = β2
10
Now equation the coefficient of π₯2 πππ π₯ in equation
(πππ) we get.
π΄ + πΆ = 0
π΄ + (β1
5) = 0
π΄ β1
5= 0
β π΄ =
1
5
3π΄ + π΅ = 3
3 (1
5) + π΅ = 3
π΅ = 3 β3
5
π΅ = 3 β3
5
π΅ =15 β 3
5
π΅ =12
5
β πΆ = β3
Putting the value of
π΄, π΅ πππ πΆ ππ πππ’ππ‘πππ (π) π€π πππ‘
ππππ’ππππ ππππ‘πππ πππππ‘πππ. 3π₯ + 7
(π₯2 + 1)(π₯ + 3)=
π₯ + 12
5(π₯2 + 1)β
1
5(π₯ + 3)
Question No.3 π
(π + π)(ππ + π)
Solution:
πΏππ‘1
(π₯ + 1)(π₯2 + 1)=
π΄
π₯ + 1+
π΅π₯ + πΆ
π₯2 + 1β (ππ)
Multiplying both sides by (π₯ + 1)(π₯2 + 1)π€π πππ‘
1 = π΄(π₯2 + 1) + (π΅π₯ + πΆ)(π₯ + 1)
1 = π΄(π₯2 + 1) + π΅π₯(π₯ + 1) + πΆ(π₯ + 1) β (ππ)
ππ’π‘π‘πππ π₯ + 1 = 0 π. π π₯ = β1 ππ (ππ)π€π πππ‘
1 = π΄[(β1)2 + 1] 1 = π΄(1 + 1)
1 = 2π΄
Equation the coefficients of
π₯2 πππ π₯ ππ πππ’ππ‘πππ (ππ) We get
β πΆ =
β1
5
β π΄ =
1
2
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π΄ + π΅ = 0 1
2+ π΅ = 0
π΅ = β1
2
β π΅ = β
1
2
π΅ + πΆ = 0
β1
2+ πΆ = 0
πΆ =1
2
β πΆ =
1
2
Putting the value of π΄, π΅ πππ πΆ ππ πππ’ππ‘ππππ (π)π€π
Get required
partial
fractions. 1
(π₯ + 1)(π₯2 + 1)=
1
2(π₯ + 1)+
π₯ β 1
2(π₯2 + 1)
Question No.4 ππ β π
(π + π)(ππ + π)
Solution:
πΏππ‘ 9π₯ β 7
(π₯ + 3)(π₯2 + 1)=
π΄
π₯ + 3+
π΅π₯ + πΆ
π₯2 + 1β (π)
Multiplying both sides by (π₯ + 3)(π₯2 + 1)π€π πππ‘
9π₯ β 7 = π΄(π₯2 + 1) + (π΅π₯ + πΆ)(π₯2 + 1)
9π₯ β 7 = π΄(π₯2 + 1) + π΅π₯(π₯ + 3) + πΆ(π₯ + 3) β (ππ)
Putting π₯ + 3 = 0 π. π π₯ = β3 ππ (ππ) π€π πππ‘
9(β3) β 7 = π΄[(β3)2 + 1]
β27 β 7 = π΄(9 + 1) β34 = 10π΄
π΄ =β34
10
Equating coefficient of
π₯2 πππ π₯ ππ πππ’ππ‘πππ (ππ)π€π get
π΄ + π΅ = 0 β17
5+ π΅ = 0
π΅ =17
5
β π΅ =
β17
5
3π΅ + πΆ = 9
3 (17
2) + πΆ = 9
51
5+ πΆ = 9
πΆ = 9 β51
5
πΆ =45 β 51
5
β πΆ =
β6
5
Putting the value
ππ π΄, π΅ πππ πΆ ππ πππ’ππ‘πππ (π)π€π πππ‘
ππππ’ππππ ππππ‘πππ πππππ‘πππ. 9π₯ β 7
(π₯ + 3)(π₯2 + 1)=
β17
5(π₯ + 3)+
17π₯ β 6
5(π₯2 + 1)
Question No.5 ππ + π
(π + π)(ππ + π)
Solution:
πΏππ‘ 3π₯ + 7
(π₯ + 3)(π₯2 + 4)=
π΄
π₯ + 3+
π΅π₯ + πΆ
π₯2 + 4β (π)
Multiplying both sides by (π₯ + 3)(π₯2 + 4)π€π πππ‘
3π₯ + 7 = π΄(π₯2 + 4) + (π΅π₯ + πΆ)(π₯ + 3)
3π₯ + 7 = π΄(π₯2 + 4) + π΅π₯(π₯ + 3) + πΆ(π₯ + 3) β (ππ) ππ’π‘π‘πππ π₯ + 3 = 0 π. π π₯ = β3 ππ (ππ)π€π πππ‘
3(β3) + 7 = π΄((β3)2 + 4) β9 + 7 = π΄(9 + 4)
β2 = 13π΄
Equating the coefficient ππ π₯2 πππ π₯ ππ πππ’ππ‘πππ (ππ)
we get
π΄ + π΅ = 0 β2
13+ π΅ = 0
π΅ =2
13
β π΅ =
2
13
3π΅ + πΆ = 3
3 (2
13) + πΆ = 3
6
13+ πΆ = 9
πΆ = 3 β6
13
πΆ =39 β 6
13
β πΆ =
33
13
Putting the value of π΄, π΅ πππ πΆ ππ πππ’ππ‘πππ (π)π€π πππ‘
Required partial fractions. 3π₯ + 7
(π₯ + 3)(π₯2 + 4)=
β2
13(π₯ + 3)+
2π₯ + 33
13(π₯2 + 4)
Question No.6
ππ
(π + π)(ππ + π)
Solution:
πΏππ‘π₯2
(π₯ + 2)(π₯2 + 4)=
π΄
π₯ + 2+
π΅π₯ + πΆ
ππ + πβ (π)
Multiplying both sides by (π₯ + 2)(π₯2 + 4) π€π πππ‘
π₯2 = π΄(π₯2 + 4) + (π΅π₯ + πΆ)(π₯ + 2)
π₯2 = π΄(π₯2 + 4) + π΅π₯(π₯ + 2) + πΆ(π₯ + 2) β (ππ) Putting π₯ + 2 = 0 π. π π₯ = β2 ππ (ππ)π€π πππ‘
(β2)2 = π΄[(β2)2 + 4]
4 = π΄(4 + π΄)
4 = 8 + π΄
Equating the coefficients of
π₯3 πππ π₯ ππ πππ’ππ‘πππ (ππ) We get
β π΄ =
β17
5
β π΄ =
β2
13
β π΄ =
1
2
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π΄ + π΅ = 1 1
2+ π΅ = 1
π΅ = 1 β1
2
β π΅ =
1
2
2π΅ + πΆ = 0
2 (1
2) + πΆ = 0
1 + πΆ = 0 πΆ = 0 β 1 = β1
β πΆ = β1
Putting the value of
π΄, π΅ πππ πΆ ππ πππ’ππ‘πππ (π)π€π πππ‘
Required partial fractions.
π₯2
(π₯ + 2)(π₯2 + 4)=
1
2(π₯ + 2)+
π₯ β 2
2(π₯2 + 4)
Question No.7 π
ππ + π [π―πππ:
π
ππ + π
=π
(π + π)(ππ β π + π)]
Solution:
πΏππ‘ 1
(π₯ + 1)(π₯2 β π₯ + 1)=
π΄
π₯ + 1+
π΅π₯ + πΆ
π₯2 β π₯ + 1β (π)
Multiplying both sides by (π₯ β 1)(π₯2 β π₯ +
1), π€π πππ‘
1 = π΄(π₯2 β π₯ + 1) + (π΅π₯ + πΆ)(π₯ + 1)
1 = π΄(π₯2 β π₯ + 1) + π΅π₯(π₯ + 1) + π(π₯ + 1) β (ππ)
Putting π₯ + 1 = 0 π. π π₯ = β1 ππ (ππ)π€π πππ‘
1 = π΄[(β1)2 β (β1) + 1]
1 = π΄[(β1)2 β 1(β1) + 1]
1 = π΄(1 + 1 + 1)
1 = 3π΄
Comparing the coefficients of
π₯2 πππ π₯ ππ πππ’ππ‘πππ (ππ)π€π πππ‘
π΄ + π΅ = 0 1
3+ π΅ = 0
π΅ =β1
3
β π΅ =
β1
3
βπ΄ + π΅ + πΆ = 0
(β1
3) β
1
3+ πΆ = 0
β2
3+ πΆ = 0
β πΆ =
2
3
Putting the value of
π΄, πππ π΅ πππ πΆ ππ πππ’ππ‘πππ (π)π€π πππ‘
Required partial fractions. 1
(π₯ + 1)(π₯2 β π₯ + 1)=
1
3(π₯ + 1)β
π₯ β 2
3(π₯2 β π₯ + 1)
Question No.8
ππ + π
ππ + π
Solution:
π₯2 + 1
[π₯3 + 1=
π₯2 + 1
(π₯ + 1)(π₯2 β π₯ + 1)
πΏππ‘ π₯2 + 1
(π₯ + 1)(π₯2 β π₯ + 1)=
π΄
π₯ + 1+
π΅π₯ + πΆ
π₯2 β π₯ + 1β (π)
Multiplying both sides by (π₯ + 1)(π₯2 β π₯ +
1), π€π πππ‘
π₯2 + 1 = π΄(π₯2 β π₯ + 1) + (π΅π₯ + πΆ)(π₯ + 1)
π₯2 + 1 = π΄(π₯2 β π₯ + 1) + π΅π₯(π₯ + 1) + πΆ(π₯ + 1)
β (ππ)
Putting π₯ + 1 = 0 π. π π₯ = β1 ππ (ππ)π€π πππ‘
(β1)2 + 1 = π΄[(β1)2 β (β1) + 1] 1 + 1 = π΄(1 + 1 + 1)
2 = 3π΄
Equating the coefficients of
π₯2 πππ π₯ ππ πππ’ππ‘πππ (ππ)π€π πππ‘
π΄ + π΅ = 1 2
3+ π΅ = 1
π΅ = 1 β2
3
β π΅ =
1
3
βπ΄ + π΅ + πΆ = 0
(β2
3) +
1
3+ πΆ = 0
β1
3+ πΆ = 0
β πΆ =
1
3
Putting the value of
π΄, π΅ πππ πΆ ππ πππ’ππ‘πππ (π) π€π πππ‘
ππππ’ππππ ππππ‘πππ πππππ‘ππππ .
π₯2 + 1
(π₯ + 1)(π₯2 β π₯ + 1)=
2
3(π₯ + 1)+
π₯ + 1
3(π₯2 β π₯ + 1)
Resolution of a fraction when π·(π₯) has repeated irreducible quadratic factor: Rule IV If a quadratic factor (ππ₯2 + ππ₯ + π) with π β 0 occures twice in the denominator the corresponding partial fractions are,
π΄π₯ + π΅
ππ₯2 + ππ₯ + π+
πΆπ₯ + π·
(ππ₯2 + ππ₯ + π)2
The constants π΄, π΅, πππ πΆ πππ πππ’ππ in the usual.
β π΄ =1
3
β π΄ =
2
3
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Exercise 4.4 Resolve into Partial Fractions.
Question No.1
ππ
(ππ + π)π
Solution:
πΏππ‘π₯3
(π₯2 + 4)2=
π΄π₯ + π΅
π₯2 + 4+
πΆπ₯ + π·
(π₯2 + 4)2β (π)
ππ’ππ‘ππππ¦πππ πππ‘βπ ππππ ππ¦ (π₯2 + 4)2, π€π πππ‘
π₯3 = (π΄π₯ + π΅)(π₯2 + 4) + (πΆπ₯ + π·)
π₯3 = π΄π₯(π₯2 + 4) + π΅(π₯2 + 4) + (πΆπ₯ + π·) β (πππ)
Equating the coefficients of π₯3, π₯2, π₯ πππ ππππ π‘πππ‘, we
get
Coefficients ππ π₯3: π΄ = 1
Coefficients of π₯2: π΅ = 0
Coefficients of π₯: 4π΄ + πΆ = 0
πΆ = β4
Constants:4π΅ + π· = 0
4(0) + π· = 0
π· = 0
Putting the value of π΄, π΅, πππ πΆ ππ πππ’ππ‘πππ (π)
We get required partial fractions.
π₯3
(π₯2 + 4)2=
π₯
π₯2 + 4β
4π₯
(π₯2 + 4)2
Question No.2
ππ + πππ + π + π
(π + π)(ππ + π)π
Solution:
πΏππ‘ π₯4 + 3π₯2 + π₯ + 1
(π₯ + 1)(π₯2 + 1)2=
π΄
π₯ + 1+
π΅π₯ + πΆ
π₯2 + 1+
π·π₯ + πΈ
(π₯2 + 1)2
β (π)
Multiplying both sides by (π + π)(ππ + π)π
ππ πππ
π₯4 + 3π₯2 + π₯ + 1
= π΄(π₯2 + 1)2 + (π΅π₯ + πΆ)(π₯ + 1)(π₯2 + 1) +(π·π₯ + πΈ)(π₯ + 1) β (ππ)
π₯4 + 3π₯2 + π₯ + 1 = π΄(π₯4 + 2π₯2 + 1)
+π΅π₯(π₯3 + π₯2 + π₯ + 1) + πΆ(π₯3 + π₯2 + π₯ + 1) +π·π₯(π₯ + 1) + πΈ(π₯ + 1)
π₯4 + 3π₯2 + π₯ + 1 = π΄(π₯4 + 2π₯2 + 1)
+π΅(π₯4 + π₯3 + π₯2 + π₯) + πΆ(π₯3 + π₯2 + π₯ + 1)
π·(π₯2 + π₯) + πΈ(π₯ + 1) β (πππ) Putting π₯ + 1 = 0 π. π π₯ = β1 ππ ππ(ππ)π€π πππ‘
(β1)4 + 3(β1)2 + (β1) + 1 = π΄[(β1)2 + 1]2
1 + 3(1) β 1 + 1 = π΄(1 + 1)2 4 = 4π΄
π΄ = 1
Now equating the coefficients of π₯4, π₯3, π₯2, π₯ πππ
ππππ π‘πππ‘π , π€π πππ‘ ππππ πππ’ππ‘πππ (πππ)
πππππππππππ‘π ππ π₯4: π΄ + π΅ = 1
1 + π΅ = 1
π΅ = 1 β 1
β π΅ = 0
Coefficients of π₯3: π΅ + πΆ = 0
0 + πΆ = 0
β πΆ = 0
Coefficients of π₯2: 2π΄ + π΅ + πΆ + π· = 3
2(1) + 0 + 0 + π· = 3
π· = 3 β 2
π· = 1
πΆππππππππππ‘π ππ π₯: π΅ + πΆ + π· + πΈ = 1
0 + 0 + 1 + πΈ = 1
πΈ = 1 β 1
β πΈ = 0
Putting the value ππ π΄, π΅, πΆ πππ π· ππ πππ’ππ‘πππ (π)
We get required partial fractions.
πΏππ‘ π₯4 + 3π₯2 + π₯ + 1
(π₯ + 1)(π₯2 + 1)2=
1
π₯ + 1+
π₯
(π₯2 + 1)2
Question No.3
ππ
(π + π)(ππ + π)π
Solution:
πΏππ‘ ππ
(π + π)(ππ + π)π=
π΄
π₯ + 1+
π΅π₯ + πΆ
π₯2 + 1+
π·π₯ + πΈ
(π₯2 + 1)2
β (π)
Multiplying both sides by (π + π)(ππ + π)π
ππ πππ
π₯2 = π΄(π₯2 + 1)2 + (π΅π₯ + πΆ)(π₯ + 1)(π₯2 + 1) +(π·π₯ + πΈ)(π₯ + 1) β (ππ)
π₯2 = π΄(π₯4 + 2π₯2 + 1) + π΅π₯(π₯3 + π₯2 + π₯ + 1) +
πΆ(π₯3 + π₯2 + π₯ + 1) +π·π₯(π₯ + 1) + πΈ(π₯ + 1)
π₯2 = π΄(π₯4 + 2π₯2 + 1) +π΅(π₯4 + π₯3 + π₯2 + π₯)
+πΆ(π₯3 + π₯2 + π₯ + 1)+π·(π₯2 + π₯) + πΈ(π₯ + 1) β (πππ)
Putting π₯ + 1 = 0 π. π π₯ = β1 ππ ππ(ππ)π€π πππ‘
(β1)2 = π΄[(β1)2 + 1]2
1 = π΄(1 + 1)2 1 = 4π΄
β
π΄ =1
4
Now equating the coefficients of π₯4, π₯3, π₯2, π₯ πππ
ππππ π‘πππ‘π , π€π πππ‘ ππππ πππ’ππ‘πππ (πππ)
πππππππππππ‘π ππ π₯4: π΄ + π΅ = 0 1
4+ π΅ = 1
β π΅ = β
1
4
Coefficients of π₯3: π΅ + πΆ = 0
β1
4+ πΆ = 0
β πΆ =
1
4
Coefficients of π₯2: 2π΄ + π΅ + πΆ + π· = 1
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2 (1
4) β
1
4+
1
4+ π· = 1
1
2+ π· = 1
π· = 1 β1
2
β π· =2 β 1
2
β π· =
1
2
πΆππππππππππ‘π ππ π₯: π΅ + πΆ + π· + πΈ = 0
β1
4+
1
2+
1
2+ πΈ = 0
1
2+ πΈ = 0
β πΈ = β
1
2
Putting the value ππ π΄, π΅, πΆ πππ π· ππ πππ’ππ‘πππ (π)
We get required partial fractions.
ππ
(π + π)(ππ + π)π
=π΄
4(π₯ + 1)+
π₯ β 1
4(π₯2 + 1)+
π₯ β 1
2(π₯2 + 1)2
Question No.4
ππ
(π β π)(ππ + π)π
Solution:
ππ
(π β π)(ππ + π)π=
π΄
π₯ β 1+
π΅π₯ + πΆ
π₯2 + 1+
π·π₯ + πΈ
(π₯2 + 1)2
β (π)
Multiplying both sides by (π₯ + 1)(π₯2 + 1)2 π€π πππ‘
π₯2 = π΄(π₯2 + 1)2 + (π΅π₯ + πΆ)(π₯ β 1)(π₯2 + 1) +(π·π₯ + πΈ)(π₯ β 1) β (ππ)
π₯2 = π΄(π₯4 + 2π₯2 + 1) + π΅π₯(π₯ β 1)(π₯2 + 1) +
πΆ(π₯ β 1)(π₯2 + 1) +π·π₯(π₯ β 1) + πΈ(π₯ β 1)
π₯2 = π΄(π₯4 + 2π₯2 + 1) +π΅(π₯4 β π₯3 + π₯2 β π₯)
+πΆ(π₯3 β π₯2 + π₯ β 1)+π·(π₯2 β π₯) + πΈ(π₯ β 1) β (πππ)
Putting π₯ β 1 = 0 π. π π₯ = 1 ππ ππ(ππ)π€π πππ‘
(1)2 = π΄[(1)2 + 1]2
1 = π΄(1 + 1)2 1 = 4π΄
β
π΄ =1
4
Now equating the coefficients of π₯4, π₯3, π₯2, π₯ πππ
ππππ π‘πππ‘π , π€π πππ‘ ππππ πππ’ππ‘πππ (πππ)
πππππππππππ‘π ππ π₯4: π΄ + π΅ = 0 1
4+ π΅ = 1
β π΅ = β
1
4
Coefficients of π₯3: π΅ + πΆ = 0
β (β1
4) + πΆ = 0
β πΆ = β
1
4
Coefficients of π₯2: 2π΄ + π΅ β πΆ + π· = 1
2 (1
4) β
1
4β (β
1
4) + π· = 1
1
2β
1
4+
1
4+ π· = 1
π· = 1 β1
2
β π· =2 β 1
2
β π· =
1
2
πΆππππππππππ‘π ππ π₯: β π΅ + πΆ β π· + πΈ = 0
β (β1
4) β
1
4β
1
2+ πΈ = 0
1
4β
1
4β
1
2+ πΈ = 0
β1
2+ πΈ = 0
β πΈ =
1
2
Putting the value ππ π΄, π΅, πΆ πππ π· ππ πππ’ππ‘πππ (π)
We get required partial fractions.
ππ
(π β π)(ππ + π)π
=1
4(π₯ + 1)+
π₯ + 1
4(π₯2 + 1)+
π₯ + 1
2(π₯2 + 1)2
Question No.5
ππ
(ππ + π)π
Solution:
π₯4
(π₯2 + 2)2=
π₯4
π₯4 + 4π₯2 + 4 ππ ππ ππππππππ πππππ‘πππ
πΉπππ π‘ π€π πππ πππ£π ππ‘ πππ‘π ππππππ πππππ‘πππ
1
π₯4 + 4π₯2 + 4βπ₯4
Β±π₯4 Β± 4π₯2 Β± 4
β4π₯2 β 4
π₯4
(π₯2 + 2)2= 1 +
β4π₯2 β 4
(π₯2 + 2)2
πΏππ‘ β4π₯2 β 4
(π₯2 + 2)2=
π΄π₯ + π΅
π₯2 + 2+
πΆπ₯ + π·
(π₯2 + 2)2β (π)
Multiplying both sides by (π₯2 + 2)2 π€π πππ‘
β4π₯2 β 4 = (π΄π₯ + π΅)(π₯2 + 2) + (πΆπ₯ + π·)
β4π₯2 β 4 = π΄(π₯3 + 2π₯) + π΅(π₯2 + 2) + πΆπ₯ + π·β (ππ)
Equating the coefficients of π₯3, π₯2, π₯ πππ ππππ π‘πππ‘π
In equation (ii) we get
Coefficients of π₯2: π΅ = β4
πΆππππππππππ‘π ππ π₯: 2π΄ + πΆ = 0
2(0) + πΆ = 0
Constants : 2π΅ + π· = β4
2(β4) + π· = β4
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β8 + π· = β4
π· = 8 β 4
π· = 4
Putting the value of A, B, C and D in equation (i) we
get required partial fractions.
π₯4
(π₯2 + 2)2= 1 +
β4
π₯2 + 2+
4
(π₯2 + 2)2
π₯4
(π₯2 + 2)2= 1 β
4
π₯2 + 2+
4
(π₯2 + 2)2
Question No.6
ππ
(ππ + π)π
Solution:
π₯5
(π₯2 + 1)2=
π₯5
π₯4 + 2π₯2 + 1 ππ ππ ππππππππ πππππ‘πππ.
πΉπππ π‘ π€π πππ πππ£π ππ‘ πππ‘π ππππππ πππππ‘πππ.
π₯
π₯4 + 2π₯2 + 1βπ₯5
Β±π₯5 Β± 2π₯3 Β± π₯
β2π₯3 β π₯
π₯5
(π₯2 + 1)2= π₯ +
β2π₯3 β π₯
(π₯2 + 1)2
πΏππ‘ β2π₯3 β π₯
(π₯2 + 1)2=
π΄π₯ + π΅
π₯2 + 1+
πΆπ₯ + π·
(π₯2 + 1)2β (π)
Multiplying both sides by (π₯2 + 1)2 π€π πππ‘
β2π₯3 β π₯ = (π΄π₯ + π΅)(π₯2 + 1) + (πΆπ₯ + π·)
β2π₯ β π₯ = π΄(π₯3 + π₯) + π΅(π₯2 + 1) + πΆπ₯ + π·
Equating the coefficients of π₯3, π₯2, π₯ πππ ππππ π‘πππ‘π
We get
Coefficients of π₯3: π΄ = β2
Coefficients of π₯2: π΅ = 0
Coefficients of π₯: π΄ + πΆ = β1
β2 + πΆ = β1
πΆ = β1 + 2
β πΆ = 1
Constants: π΅ + π· = 0
0 + π· = 0
β π· = 0
Hence the required partial fractions are
π₯5
(π₯2 + 1)2= π₯ +
β2π₯
π₯2 + 1+
π₯
(π₯2 + 1)2
π₯5
(π₯2 + 1)2= π₯ β
2π₯
π₯2 + 1+
π₯
(π₯2 + 1)2
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