Question

If $f\left(x\right)=1+\frac{1}{x}\underset{1}{\overset{x}{\int }}f\left(t\right)dt$ for $x>0,$ then the value of $f\left(e^{-1}\right)$ is

• A. $1$
• B. $0$
• C. $-1$
• D. $2$

Answer 1

The correct option is B $0$

Given, $f\left(x\right)=1+\frac{1}{x}\underset{1}{\overset{x}{\int }}f\left(t\right)dt\cdots \left(1\right)$
$\Rightarrow xf\left(x\right)=x+\underset{1}{\overset{x}{\int }}f\left(t\right)dt$
On differentiating both the sides,
$f\left(x\right)+x{f}^{\prime }\left(x\right)=1+f\left(x\right)\Rightarrow f^{\prime }\left(x\right)=\frac{1}{x}\Rightarrow f\left(x\right)=\ln \left|x\right|+C$
From eqn. $\left(1\right),$ $f\left(1\right)=1$
$\therefore f\left(x\right)=\ln \left|x\right|+1\Rightarrow f\left(e^{-1}\right)=-1+1=0$