Question

Let $f\left(x\right)={\int }_0^x{e}^{t}f\left(t\right)dt+e^x$ be a differentiable function for all $x\in R$. Then $f\left(x\right)$ equals.

• A. $2e^{(e^x-1)}-1$
• B. $e^{(e^x-1)}$
• C. $2e^{e^x}-1$
• D. $e^{e^x}-1$

Answer 1

The correct option is A $2e^{(e^x-1)}-1$

Given, $f\left(x\right)={\int }_0^x{e}^{t}f\left(t\right)dt+e^x...\left(1\right)$
Differentiating both sides w.r.t $x$
$f^{\prime }\left(x\right)=e^x.f\left(x\right)+e^x$
(Using Newton Leibnitz Therom)
$\Rightarrow \frac{f^{\prime }\left(x\right)}{f\left(x\right)+1}=e^x$
Integrating w.r.t $x$
$\int \frac{f^{\prime }\left(x\right)}{f\left(x\right)+1}dx=\int e^{x}dx$
$\Rightarrow \ln \left(f\right(x)+1)=e^x+c$
Put $x=0$
In $\ln 2=1+c$
$\left(Qf\right(0)=1,$ from equation $\left(1\right))$
$\therefore \ln \left(f\right(x)+1)=e^x+\ln 2-1$
$\Rightarrow f\left(x\right)+1=2\cdot e^{e^{x-1}-1}$
$\Rightarrow f\left(x\right)=2e^{e^x-1}-1$