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)