Question

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

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is C $\frac{\pi }{3}$

$LHS\frac{(x+1{)}^2}{x(x^2+1)};RHS=\frac{A}{x}+\frac{Bx+C}{x^2+1}$
$=\frac{A(x^2+1)+(Bx+C)x}{x(x^2+1)}$
$LHS=RHS\Rightarrow x^2+2x+1=A(x^2+1)+(Bx+C)-x$
Comparing the co-efficients of the power of
$x$
co-efficient of $x^2$
$1=A+B$
co efficient of $x$
$c=2$
Co-efficient of 1
$1=A$
$co{s}^{-1}\left(\frac{A}{C}\right)=co{s}^1\left(1/2\right)={60}^0=\pi /3$
$co{s}^{-1}\left(A/C\right)=\pi /3$