Question

Let $f$ be a continuous function satisfying $lnx={\int }_1^{\tan x}f\left(t\right)dt$. Then the value of $f\left(1\right)$, is

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

Answer 1

Answer: (C)

$\frac{2}{\pi }$

$lnx=\underset{1}{\overset{\tan x}{\int }}f\left(t\right)dt$

On differentiating both side w.r.t. $x$

$\frac{1}{x}=f\left(tanx\right)\times se{c}^{2}x-f\left(1\right)\times 0$

putting $x=\frac{\pi }{4}$ on both sides

$\frac{4}{\pi }=f\left(1\right)\times 2$

$f\left(1\right)=\frac{2}{\pi }$