MCQ
Let $f(x)=\int_{0}^{x} e^{t} f(t) d t+e^{x}$ be a differentiable function for all $x \in R$. Then $f(x)$ equals ..... .
- ✓$2 e ^{\left( e ^{ x }-1\right)}-1$
- B$e ^{ e ^{ x }}-1$
- C$2 e ^{ e ^{ x }}-1$
- D$e ^{\left( e ^{ x }-1\right)}$
differentiating with respect to $x$
$f^{\prime}(x)=e^{x} f(x)+e^{x}$
$f^{\prime}(x)=e^{x}(f(x)+1)$
$\int_{0}^{x} \frac{f^{\prime}(x)}{f(x)+1} d x=\int_{0}^{x} e^{x} d x$
$\left.\ell n(f(x)+1)\right|_{0} ^{x}=\left.e^{x}\right|_{0} ^{x}$
$\ell \ln (f(x)+1)-\ell n(f(0)+1)=e^{x}-1$
$\ell\left(\frac{f(x)+1}{2}\right)=e^{x}-1 \quad\{$ as $f(0)=1\}$
$f(x)=2 e^{\left(e^{x}-1\right)}-1$
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