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. $e^{e^{x-1}}-1$
• C. $2e^{e^x}-1$
• D. $e^{e^x}-1$

Answer 1

The correct option is A

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

Explanation for the correct answer

Given that: $f\left(x\right)={\int }_0^x{e}^{t}f\left(t\right)dt+e^x$

Differentiating the expression with respect to $x$.

$$\begin{gathered} \Rightarrow f\text{'}\left(x\right)=e^{x}f\left(x\right)+e^x \\ \Rightarrow f\text{'}\left(x\right)=e^x\left(f\right(x)+1) \\ \Rightarrow {\int }_0^x\frac{f\text{'}\left(x\right)}{\left(f\right(x)+1)}dx={\int }_0^x{e}^{x}dx \\ \Rightarrow {\left[\log \left(f\right(x)+1)\right]}_0^x={\left[e^x\right]}_0^x\left[\because \int \frac{f\text{'}\left(x\right)}{\left(f\right(x)+c)}=log\left(\right(f\left(x\right)+c\left)\right)\right] \\ \Rightarrow \log \left(f\right(x)+1)-\log \left(f\right(0)+1)=e^x-e^0 \\ \Rightarrow \log \left(f\right(x)+1)-\log \left(2\right)=e^x-1\mathbf{}\mathbf{[}\mathit{a}\mathit{s}\mathbf{,}\mathbf{}\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{)}\mathbf{=}\mathbf{1}\mathbf{]} \\ \Rightarrow \log \left(\frac{f\left(x\right)+1}{2}\right)=e^x-1\left[\because loga-logb=log\left(\frac{a}{b}\right)\right] \end{gathered}$$

Taking anti-log on both sides:

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

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

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

Hence, the correct answer is option (A).