Question

Let $f(x)$ 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 $\int_{0}^{2} f(x) d x$ is :

• A. $1- e ^{2}$
• B. $1+ e ^{2}$
• C. $2\left(1- e ^{2}\right)$
• D. $2\left(1+ e ^{2}\right)$

Answer 1

Answer: (B)

$f^{\prime}(x)=f^{\prime}(2-x)$
$f(x)=-f(2-x)+c$
put $x=0$
$f^{\prime}(0)=-f^{\prime}(2)+c$
$c=f(0)+f(2)=1+e^{2}$
so, $f(x)+f(2-x)=1+e^{2}$
$I=\int_{0}^{2} f(x) d x$
$I=\int_{0}^{2} f(2-x) d x$
$2 I=\int_{0}^{2}(f(x)+f(2-x)) d x$
$2 I=\left(1+e^{2}\right) \int_{0}^{2} d x$
$I=1+e^{2}$