Question

The minimum value of the function $f(x)=\int\limits_0^2 e^{|x-t|} d t$ is :

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

Answer 1

The Correct Option is B

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