Mathematics

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

$\dfrac {\pi}6$

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

Consider, $\displaystyle \frac { (x+1)^{ 2 } }{ x^{ 3 }+x }$

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

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

$\displaystyle \frac { { x }^{ 2 }+2x+1 }{ x({ x }^{ 2 }+1) } =\frac { (A+B)x^{ 2 }+Cx+A }{ x({ x }^{ 2 }+1) }$

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

by comparing the coefficients we get

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

$\Rightarrow \sin^{-1}\left(\dfrac{A}{C}\right)= \sin^{-1}\left(\dfrac{1}{2}\right)=\displaystyle \dfrac{\pi}{6}$

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 84

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