Question

Let $f\left(x\right)$ be a differentiable function defined on $[0,2]$ such that $f^{\prime }\left(x\right)=f^{\prime }(2-x)$ for all $x\in (0,2),$ $f\left(0\right)=1$ and $f\left(2\right)=e^2.$ Then the value of $\underset{0}{\overset{2}{\int }}f\left(x\right)dx$ is

• A. $1+e^2$
• B. $1-e^2$
• C. $2(1-e^2)$
• D. $2(1+e^2)$

Answer 1

The correct option is A $1+e^2$

$f^{\prime }\left(x\right)=f^{\prime }(2-x)$
On integrating both sides, we get
$f\left(x\right)=-f(2-x)+c$
Put $x=0$
$f\left(0\right)+f\left(2\right)=c$
$\Rightarrow c=1+e^2$
$\therefore f\left(x\right)+f(2-x)=1+e^2$
$I=\underset{0}{\overset{2}{\int }}f\left(x\right)dx$
$=\underset{0}{\overset{1}{\int }}\left\{f\right(x)+f(2-x\left)\right\}dx=1+e^2$