Question

Let $f:R\to R$ be a continuous function which satisfies $f\left(x\right)={\int }_0^{x}f\left(t\right)dt.$ Then, the value of $f\left({\log }_{e}5\right)$ is

• A. $0$
• B. $2$
• C. $5$
• D. $3$

Answer 1

Answer: (A)

$0$

Given,

$f\left(x\right)={\int }_0^{x}f\left(t\right)dt$ ...(i)

Using Leibnitz theorem, we get
$f^{\prime }\left(x\right)=f\left(x\right)\Rightarrow f\left(x\right)=k{e}^x$

On putting $x=0$ in Eq. (i), we get
$f\left(0\right)={\int }_0^{0}f\left(t\right)dt$
$\Rightarrow k{e}^0=0$

$[\therefore {\int }_a^{a}f\left(x\right)dx=0]$

$\Rightarrow k=0$ $[\therefore e^0=1]$

$\therefore f\left(x\right)=0\Rightarrow f\left({\log }_{e}5\right)=0$