Question

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

• A. $\frac{1}{2-e}$
• B. $\frac{-1}{2-e}$
• C. $\frac{-e}{2-e}$
• D. $\frac{e}{2-e}$

Answer 1

The correct option is D $\frac{e}{2-e}$

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

Now $f\left(x\right)=e^x+e^x.a=e^x(1+a)f\left(t\right)=e^t(1+a)$

since $a={\int }_0^{1}f\left(t\right)dt$

$a={\int }_0^1{e}^t(1+a)dt$

$a=(1+a){\int }_0^1{e}^{t}dt$

$a=(1+a)(e-1)$

$a(1-e+1)=e-1$

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

now,

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

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