Question

A function $f\left(X\right)$ which satisfies the relation $f\left(X\right)=e^x+{\int }_0^1{e}^{x}f\left(t\right)dt$, then $f\left(X\right)$ is

• A. $\frac{e^x}{2-e}$
• B. $(e-2)e^x$
• C. $2e^x$
• D. $\frac{e^x}{2}$

Answer 1

The correct option is A $\frac{e^x}{2-e}$

Let, ${\int }_0^{1}f\left(t\right)dt=C$

Then,

$f\left(x\right)=e^x+e^x.C$.............(1)

Integrating both sides,

${\int }_0^{1}f\left(x\right)={\int }_0^1{e}^x(1+C)$

=> $C=(e-1)(1+C)$

On solving we get,

C=$\frac{e-1}{2-e}$

Put value of 'C' in (1).

$f\left(x\right)=e^x(1+\frac{e-1}{2-e})$

=>$f\left(x\right)=\frac{e^x}{2-e}$