Let f(x) =


Let $f(x)=\int_{0}^{x} e^{t} f(t) d t+e^{x}$ be a differentiable function for all $x \in R$. Then $\mathrm{f}(\mathrm{x})$ equals.

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

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

  3. (3) $2 e^{e^{3}}-1$

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

Correct Option: 1


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

Differentiating both sides w.r.t $x f^{\prime}(x)=e^{x} \cdot f(x)+e^{x}$

(Using Newton Leibnitz Theorem) $\Rightarrow \frac{f^{\prime}(z)}{f(x)+1}=e^{x}$

Integrating w.r.t $x \int \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})+1} \mathrm{dx}=\int \mathrm{e}^{\mathrm{X}} \mathrm{dx}$

$\Rightarrow \ell n(f(x)+1)=e^{x}+c$

Put $x=0$

$2=1+c \quad(\because \mathrm{f}(0)=1$, from equation $(1))$

$\therefore \ln (f(x)+1)=e^{x}+\ln 2-1$

$\Rightarrow f(x)+1=2 \cdot e^{e^{x}-1}$

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

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now