1st year math chapter 5 partials friction

Class 11 Chapter5
Rational fraction: iv Solve the resulting equations for the
𝑃(𝑥) coefficients.
The quotient of two polynomials 𝑄(𝑥)
𝑤ℎ𝑒𝑟𝑒 𝑄(𝑥) ≠ 0
Case I
with no common factor is called rational fraction. 𝑷(𝒙)
For example Resolution of 𝑸(𝒙) in to partial fractions when Q(x)
𝑥2 + 1 𝑥4 has only repeated linear factors.
,
𝑥2 − 1 𝑥2 + 1 The polynomial 𝑄(𝑥) may be written as
Proper rational fraction: 𝑄(𝑥) = (𝑥 − 𝑎1 )(𝑥 − 𝑎2 ) … (𝑥 − 𝑎𝑛 ) where
𝑃(𝑥)
A rational fraction 𝑄(𝑥) is called a proper rational 𝑎1 ≠ 𝑎2 ≠ ⋯ ≠ 𝑎𝑛
𝑃(𝑥) 𝐴1 𝐴2 𝐴𝑛
fraction is the degree of polynomial 𝑃(𝑥) is less than ∵ = + + ⋯+
𝑄(𝑥) 𝑥 − 𝑎1 𝑥 − 𝑎2 𝑥 − 𝑎𝑛
the degree of polynomial 𝑄(𝑥)
Are numbers to be found.
For example
Note:
2𝑥 − 5 3
, Which can be factorize, first of all can be factorized it
𝑥2 + 4 𝑥 + 1
Improper rational fraction:  We uses partial fraction when the fraction
𝑃(𝑥) 𝑷(𝒙)
A rational fraction 𝑄(𝑥) is called an improper ration 𝑸(𝒙)
is proper rational fraction.

fraction if the degree of the polynomial 𝑃(𝑥) is greater  If we are given improper fraction (division is
than or equal to the degree of polynomial 𝑄(𝑥) possible) then first of all divide the fraction
3𝑥 2 + 1 𝑥 4 and make it proper fraction. After this sues




pk
, partial fraction.
𝑥 − 1 𝑥2 − 1
Partial fraction:
To express a single rational fraction as a sum of two or s. Exercise 5.1
more single rational fractions is called partial fraction.
Resolve the following into partial fractions.
Partial fraction resolution:
Question No.1
te
Expressing a rational fraction as a sum of partial 1
fraction is called partial fraction resolution. 2
𝑥 −1
Conditional equation:
no

Solution:
It is an equation which is true for a particular valves of 1 1
=
𝑥 2 −1 (𝑥−1)(𝑥+1)
the variable
1 𝐴 𝐵
For example: = + …………… (Z)
(𝑥−1)(𝑥+1) (𝑥−1) (𝑥+1)
sy



3 Multiply both sides by (x-1)(x+1)
2𝑥 = 3 𝑖𝑠 𝑡𝑟𝑢𝑒 𝑜𝑛𝑙𝑦 𝑥 =
2 1=A(x+1)+B(x-1) ……………(1)
For simplicity, a conditional equation is called an
Put x-1=0 ⟹ x=1 in equation (1)
ea




equation.
1=A(1+1)+B(1-1)
Identity: 1
1=A(2) ⟹ A=2
It as an equation which holds good for all valves of
variable. Now put x+1=0 ⟹ x= -1 in equation (1)
For example 1=A(-1+1)+B(-1-1)
1
(𝑎 + 𝑏)𝑥 = 𝑎𝑥 + 𝑏𝑥 1=B(-2) ⟹B=− 2
The symbol " = " be used both for equation and Now put A and B in equation (Z)
identity. 1 𝐴 𝐵
Hence (𝑥−1)(𝑥+1) = (𝑥−1) + (𝑥+1)
𝑷(𝒙)
Resolution of a rational fraction in to partial 1 1
𝑸(𝒙) = −
2(𝑥−1) 2(𝑥+1)
fractions.
Following are the main points of resolving a rational
𝑃(𝑥)
Question No.2
fraction in to partial fraction. 𝑥2 + 1
𝑄(𝑥)
i The degree of 𝑃(𝑥) must be less than that of (𝑥 + 1)(𝑥 − 1)
𝑄(𝑥). If not, divide and and work with the 𝑥 2 +1 𝑥 2 +1
Solution:- (𝑥+1)(𝑥−1) = 𝑥 2 −1
remainder theorem.
𝑥 2 +1 2
ii Clear the given equation of fractions. 𝑥 2 −1
=1+𝑥 2 −1
iii Equate the coefficients of like term s (power of 2 𝐴 𝐵
Now consider (𝑥+1)(𝑥−1)= (𝑥−1) + (𝑥+1) …………… (Z)
x)
Multiply both sides by (x-1)(x+1)

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, Class 11 Chapter 5
2=A(x+1)+B(x-1) ……………(1) 3(−5)2 − 4(−5) − 5=B(-5-2)(-5+2)
Put x-1=0 ⟹ x=1 in equation (1) 90
90=B(21) ⟹B=21
2=A(1+1)+B(1-1) 30
B= 7
2=A(2) ⟹ A=1
Now put x+1=0 ⟹ x= -1 in equation (1) Now put x+2=0 ⟹ x= -2 in equation (1)
1=A(-1+1)+B(-1-1) 3(−2)2 − 4(−2) − 5=C(-2-2)(-2+5)
−15
2=B(-2) ⟹B=−1 15=C(-12) ⟹C= 12
Now put A and B in equation (Z) C=− 4
5
1 𝐴 𝐵
= + (𝑥+1)
(𝑥−1)(𝑥+1) (𝑥−1) Now put A,B and C in equation (Z)
1 1 3𝑥 2 −4𝑥−5 −1 30 5
= (𝑥−1) − (𝑥+1) Hence (𝑥−2)(𝑥+5)(𝑥+2) = 28(𝑥−2) + 7(𝑥+5) − 4(𝑥+2)
𝑥 2 +1 1 1
Hence (𝑥+1)(𝑥−1)= 2 + (𝑥−1) − (𝑥+1)
Question No.5
Question No.3
1
2𝑥 + 1
(𝑥 − 1)(2𝑥 − 1)(3𝑥 − 1)
(𝑥 − 1)(𝑥 + 2)(𝑥 + 3)
2𝑥+1
Solution:- (𝑥−1)(𝑥+2)(𝑥+3) 1
Solution:- (𝑥−1)(2𝑥−1)(3𝑥−1)
Now consider
2𝑥+1 𝐴 𝐵 𝑐 Now consider
= + (𝑥+2) + 𝑥+3 …………… (Z) 1 𝐴 𝐵 𝑐
(𝑥−1)(𝑥+2)(𝑥+3) (𝑥−1) =
(𝑥−1)(2𝑥−1)(3𝑥−1) (𝑥−1)
+ (2𝑥−1) + (3𝑥−1) …………… (Z)




pk
Multiply both sides by (x-1)(x+2)(x+3)
Multiply both sides by (𝑥 − 1)(2𝑥 − 1)(3𝑥 − 1)
2x+1= A(x+2)(x+3)+B(x-1)(x+3)+C(x-1)(x+2) ……………(1)
1= A(2𝑥 − 1)(3𝑥 − 1)+B(𝑥 − 1)(3𝑥 − 1)+C(𝑥 −
Put x-1=0 ⟹ x=1 in equation (1)
1)(2𝑥 − 1)……………(1)
2(1)+1=A(1+2)(1+3) s.
1 Put x-1=0 ⟹ x=1 in equation (1)
2(1)+1=A(12) ⟹ A=4 1=A(2(1)-1)(3(1)-1)
Now put x+2=0 ⟹ x= -2 in equation (1)
te
1
1=A(12) ⟹ A=2
2(-2)+1=B(-2-1)(-2+3) 1
-3=B(-3) ⟹B=1 Now put 2x-1=0 ⟹ x= 2 in equation (1)
no

Now put x+3=0 ⟹ x= -3 in equation (1) 1 1
1=B(2-1)(3(2)-1)
2(-3)+1=C(-3-1)(-3+2) 1
−5 1=B(-4) ⟹B=−4
-5=C(4) ⟹C= 1
4
Now put 3x-1=0 ⟹ x= 3 in equation (1)
sy



Now put A,B and C in equation (Z)
1 −5 2 1
2𝑥+1 1 1=C(− 3) (− 3)
Hence (𝑥−1)(𝑥+2)(𝑥+3)= (𝑥−1)
4
+ (𝑥+2) + 𝑥+3
4
2 9
1=(9)C ⟹C=2
ea




2𝑥+1 1 1 5
⟹ (𝑥−1)(𝑥+2)(𝑥+3)= 4(𝑥−1) + (𝑥+2) − 4 (𝑥+3)
Now put A,B and C in equation (Z)
Question No.4 1 1 4 9
Hence (𝑥−1)(2𝑥−1)(3𝑥−1)= 2(𝑥−1) − (2𝑥−1) + 2(3𝑥−1)
3𝑥 2 − 4𝑥 − 5
(𝑥 − 2)(𝑥 2 + 7𝑥 + 10) Question No.6
𝑥
3𝑥 2 −4𝑥−5 3𝑥 2 −4𝑥−5
(𝑥 − 𝑎)(𝑥 − 𝑏)(𝑥 − 𝑐)
Solution:- AS (𝑥−2)(𝑥2 +7𝑥+10)=(𝑥−2)(𝑥+5)(𝑥+2) ∴
Solution:-
𝑥
(𝑥−𝑎)(𝑥−𝑏)(𝑥−𝑐)
𝑥 2 + 7𝑥 + 10=𝑥 2 + 5𝑥 + 2𝑥 + 10 Now consider
𝑥 𝐴 𝐵 𝑐
= + (𝑥−𝑏) + (𝑥−𝑐) …………… (Z)
x(x+5)+2(x+5)=(x+5)(x+2) (𝑥−𝑎)(𝑥−𝑏)(𝑥−𝑐) (𝑥−𝑎)
Now consider Multiply both sides by (𝑥 − 𝑎)(𝑥 − 𝑏)(𝑥 − 𝑐)
3𝑥 2 −4𝑥−5 𝐴 𝐵 𝐶 x= A(𝑥 − 𝑏)(𝑥 − 𝑐) +B(𝑥 − 𝑎)(𝑥 − 𝑐)+C(𝑥 − 𝑏)(𝑥 −
(𝑥−2)(𝑥+5)(𝑥+2)
= (𝑥−2) + (𝑥+5) + 𝑥+2 …………… (Z)
𝑎)……………(1)
Multiply both sides by (x-2)(x+5)(x+2) Put x-a=0 ⟹ x=a in equation (1)
3𝑥 2 − 4𝑥 − 5= A(x+5)(x+2)+B(x-2)(x+2)+C(x-2)(x+5) a=A(a-b)(a-c)
……………(1) 𝑎
⟹ A=(𝑎−𝑏)(𝑎−𝑐)
Put x-2=0 ⟹ x=2 in equation (1)
3(2)2 − 4(2) − 5=A(2+5)(2+2) Now put x-b=0 ⟹ x= b in equation (1)
−1 b=B(b-a)(b-c)
−1=A(28) ⟹ A= 28 𝑏
⟹B=(𝑏−𝑎)(𝑏−𝑐)
Now put x+5=0 ⟹ x= -5 in equation (1)
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