Mathematics

# If $\displaystyle \frac{(x-1)^2}{x^3+x}=\frac{A}{x}+\frac{Bx+C}{x^2+1}$, then $A=..., B=..., C=...$

$A=1,B=0,C=-2$

##### SOLUTION
$\displaystyle \frac { (x-1)^{ 2 } }{ x^{ 3 }+x } =\frac { A }{ x } +\frac { Bx+C }{ x^{ 2 }+1 }$

$\Rightarrow { \left( x-1 \right) }^{ 2 }=A\left( { x }^{ 2 }+1 \right) +\left( Bx+C \right) x$

$\Rightarrow \left( { x }^{ 2 }-2x+1 \right) =A\left( { x }^{ 2 }+1 \right) +B{ x }^{ 2 }+Cx$

$\Rightarrow \left( { x }^{ 2 }-2x+1 \right) =(A+B){ x }^{ 2 }+Cx+A$

On comparing coefficients

$A+B=1,C=-2,A=1\Rightarrow A=1,B=0,C=-2$

Hence, option 'A' is correct.

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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