Solve the Following Questions

Question:

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 :

  1. $2 e^{\left(e^{x}-1\right)}-1$

  2. $e^{e^{x}}-1$

  3. $2 e^{e^{x}}-1$

  4. $e^{\left(e^{x}-1\right)}$


Correct Option: 1

Solution:

$f(x)=\int_{0}^{x} e^{t} f(t) d t+e^{x} \Rightarrow f(0)=1$

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 $f(0)=1\}$

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

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