Question

Suppose $f\left(x\right)$ and $g\left(x\right)$ are two continuous functions defined for $0\le x\le 1$. Given, $f\left(x\right)={\int }_0^1{e}^{x+1}.f\left(t\right)dt$ and $g\left(x\right)={\int }_0^1{e}^{x+t}.g\left(t\right)dt+x$ The value of $f\left(1\right)$ equals

• A. $0$
• B. $1$
• C. $e^{-1}$
• D. $e$

Answer 1

The correct option is A $0$

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

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

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

$f\left(x\right)=k{e}^{x+1}$

Now since ${\int }_0^{1}f\left(t\right)dt=k$, ${\int }_0^{1}k{e}^{t+1}dt=k$

$k(e^2-e-1)=0$

$k=0$

Hence $f\left(x\right)=0$ so for $f\left(1\right)=0$