Solve this following
Question:

Let $\mathrm{f}(\mathrm{x})$ be a differentiable function defined on $[0,2]$ such that $f^{\prime}(x)=f^{\prime}(2-x)$ for all $x \in(0,2)$,

$f(0)=1$ and $f(2)=e^{2}$. Then the value of $\int_{0}^{2} f(x) d x$

1. $1-\mathrm{e}^{2}$

2. $1+e^{2}$

3. $2\left(1-\mathrm{e}^{2}\right)$

4. $2\left(1+\mathrm{e}^{2}\right)$

Correct Option: , 2

Solution:

$f^{\prime}(x)=f^{\prime}(2-x)$

$f(x)=-f(2-x)+c$

put $x=0$

$\mathrm{f}^{\prime}(0)=-\mathrm{f}^{\prime}(2)+\mathrm{c}$

$\mathrm{c}=\mathrm{f}(0)+\mathrm{f}(2)=1+\mathrm{e}^{2}$

so, $\mathrm{f}(\mathrm{x})+\mathrm{f}(2-\mathrm{x})=1+\mathrm{e}^{2}$

$I=\int_{0}^{2} f(x) d x$

$I=\int_{0}^{2} f(2-x) d x$

$2 I=\int_{0}^{2}(f(x)+f(2-x)) d x$

$2 I=\left(1+e^{2}\right) \int_{0}^{2} d x$

$I=1+e^{2}$