Question

$If{\int }_{-1}^1\frac{g\left(x\right)}{1+t^2}dt=f\left(x\right),where,g\left(x\right)=\sin x$ , then $f^{\prime }\left(\frac{\pi }{3}\right)$ equals

• A. $\frac{\pi }{4}$
• B. does not exist
• C. $\frac{\pi \sqrt{3}}{4}$
• D. None of these

Answer 1

The correct option is A $\frac{\pi }{4}$

$f\left(x\right)={\int }_{-1}^1\frac{\sin x}{1+t^2}dt=\sin x{\int }_{-1}^1\frac{1}{1+t^2}dt$
$=\sin x{\left[{\tan }^{-1}t\right]}_{-1}^1=\sin x[\frac{\pi }{4}+\frac{\pi }{4}]=\sin x\left[\frac{\pi }{2}\right]$
$\therefore f^{\prime }\left(x\right)=\frac{\pi }{2}\cos x\Rightarrow f^{\prime }\left(\frac{\pi }{3}\right)=\frac{\pi }{2}\frac{1}{2}=\frac{\pi }{4}$