Question

Let $f:R\to R$ be a continuous function which satisfies $f\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt$. Then the value of $f\left({\log }_{e}5\right)$ is

• A. $0$ (Zero)
• B. $2$
• C. $5$
• D. $3$

Answer 1

The correct option is A $0$ (Zero)

$f\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt\cdots \left(1\right)$
Differentiate the above equation, $\frac{d}{dx}f\left(x\right)=\frac{d}{dx}\underset{0}{\overset{x}{\int }}f\left(t\right)dt$
Using Newton-Leibnitz's rule, we get
$f^{\prime }\left(x\right)=f\left(x\right)\times 1-0=f\left(x\right)$
$\Rightarrow f\left(x\right)=A{e}^x$ where $A$ is constant.

From $\left(1\right)$, we get
$f\left(0\right)=0\Rightarrow A=0$
$\Rightarrow f\left(x\right)=0,\forall x\in R$
$\therefore f\left({\log }_{e}5\right)=0$