Question

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

• A. $1$
• B. $0$
• C. $-1$
• D. $4$

Answer 1

The correct option is B $0$

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