Let $f(\mathrm{x})=\int_{0}^{x} \mathrm{e}^{\mathrm{t}} f(\mathrm{t}) \mathrm{dt}+\mathrm{e}^{\mathrm{x}}$ be a differentiable function for all $\mathrm{x} \in \mathrm{R}$. Then $f(\mathrm{x})$ equals :
Correct Option: 1
$f(x)=\int_{0}^{x} e^{t} f(t) d t+e^{x} \Rightarrow f(0)=1$
differentiating with respect to $x$
$f^{\prime}(x)=e^{x} f(x)+e^{x}$
$f^{\prime}(\mathrm{x})=\mathrm{e}^{\mathrm{x}}(f(\mathrm{x})+1)$
$\int_{0}^{x} \frac{f^{\prime}(x)}{f(x)+1} d x=\int_{0}^{x} e^{x} d x$
$\left.\ell \mathrm{n}(f(\mathrm{x})+1)\right|_{0} ^{\mathrm{x}}=\left.\mathrm{e}^{\mathrm{x}}\right|_{0} ^{\mathrm{x}}$
$\ell \mathrm{n}(f(\mathrm{x})+1)-\ell \mathrm{n}(f(0)+1)=\mathrm{e}^{\mathrm{x}}-1$
$\ell \mathrm{n}\left(\frac{f(\mathrm{x})+1}{2}\right)=\mathrm{e}^{\mathrm{x}}-1$
$\{$ as $f(0)=1\}$
$f(x)=2 e^{\left(e^{x}-1\right)}-1$