Answer:
Option A
Explanation:
We have
f(x)=∣x−π∣.(e∣x∣−1)sin∣x∣
{(x−π)(e−x−1)sinx,x<0−(x−π)(ex−1)sinx,0≤x<π(x−π)(ex−1)sinx,x≥π
We check the differentiability at x=0 and π
We have,
{(x−π)(e−x−1)cosx+(e−x−1)sinx+(x−π)sinxe−x(−1),x<0−[(x−π)(e−x−1)cosx+(e−x−1)sinx+(x−π)sinxe−x],0<x<π(x−π)(e−x−1)cosx+(e−x−1)sinx+(x−π)sinxe−x,x>π
Clearly, limx→0−f′(x)=0=limx→0+f′(x)
and limx→π−f′(x)=0=limx→π+f′(x)
∴ f is differentiable at x=0 and x= \pi
Hence, f is differentiable for all x