Question

If $\frac{1}{x(x^2+a^2)}=\frac{A}{x}+\frac{Bx+C}{x^2+a^2}$, then ${\tan }^{-1}(\frac{A}{B})=$

• A. $\frac{3\pi }{4}$
• B. $\frac{\pi }{4}$
• C. $-\frac{\pi }{4}$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is C $-\frac{\pi }{4}$

$LHS=\frac{1}{x(x^2+a^2)};RHS=\frac{A}{x}+\frac{Bx+C}{x^2+a^2}$
$=\frac{A(x^2+a^2)+(Bx+C)x}{(x^2+a^2)x}$
$LHS=RHS\Rightarrow 1=A(x^2+a^2)+(Bx+C)x$
comparing the co-efficients of powers of ${}^{\prime }{x}^{\prime }$
co-efficients of $x^2$
$o=A+B$
co-efficients of $x$
$C=o$
Co-efficient of 1
$A{a}^2=1$
$A=1/a^2$
$A=-B$
$A/B=-1$
$Ta{n}^{-1}\left(\frac{A}{B}\right)=$
$=Ta{n}^{-1}(-1)$
$=-\pi /4$
$Ta{n}^{-1}\left(\frac{A}{B}\right)=-\pi /4$