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

Answer: (B)

$2(e-1)$

Answer 2

For x=0
f(x)=0?2?et-xdt=e-x(e2-1)
For 0<x<2
f(x)=0?x?ex-tdt+?x2?et-xdt=ex+e2-x-2
For x=2
f(x)=0?2?ex-tdt=ex-2(e2-1)
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)