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

Answer

Correct option: A.
$2 e ^{\left( e ^{ x }-1\right)}-1$
a
$f( x )=\int_{0}^{ x } e ^{ t } f( t ) dt + e ^{ x } \Rightarrow f(0)=1$

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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