Question

Let $f: R \rightarrow R$ be a continuous function which satisfies $f(x)=\int\limits_{0}^{x} f(t) d t$. Then, the value of $f\left(\log _{e} 5\right)$ is

• A. 0
• B. 2
• C. -5
• D. 3

Answer 1

Answer: (A)

0

Answer 2

Given, $f(x)=\int\limits_{0}^{x} f(t) d t$ ...(i)
Using Leibnitz theorem, we get
$f'(x)=f(x) \Rightarrow f(x)=k e^{x}$
On putting $x=0$ in E (i), we get
$f(0)=\int\limits_{0}^{0} f(t) d t$
$\Rightarrow k \theta^{0} =0$
$\left[\because \int\limits_{a}^{a} f(x) d x=0\right]$
$\Rightarrow k=0 \left[\because e^{0}=1\right]$
$\therefore f(x)=0$
$\Rightarrow f\left(\log _{e} 5\right)=0$