Question

The minimum value of the function $f(x)=\underset{0}{\overset{2}{\int }}{e}^{|x-t|}dt$ is :

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

Answer 1

Answer: (B)

For $x\le 0$
$f(x)=\underset{0}{\overset{2}{\int }}{e}^{t-x}dt=e^{-x}\left(e^2-1\right)$
For $0<x<2$
$f(x)=\underset{0}{\overset{x}{\int }}{e}^{x-t}dt+{\int }_x^2{e}^{t-x}dt=e^x+e^{2-x}-2$
For $x\ge 2$
$f(x)=\underset{0}{\overset{2}{\int }}{e}^{x-t}dt=e^{x-2}\left(e^2-1\right)$
For $x\le 0,f(x)$ is $\downarrow$ and $x\ge 2,f(x)$ is $\uparrow$
$\therefore$ Minimum value of $f(x)$ lies in $x\in (0,2)$
Applying $A.M\ge G.M$,
minimum value of $f(x)$ is $2(e-1)$