f(x)=|x−π|(e|x|−1)sin|x| Check differentiability of f(x) at x=π and x=0 at x=π:
R.H.D.=
lim
h→0
|π+h−π|(e|x+h|−1)sin|π+h|−0
h
L.H.D =
lim
h→0
|π−h−π|(e|π−h|−1)sin|π−h|−0
−h
=0
∵RHD=LHD Therefore, function is differentiable at x=π at x=0 :
R.H.D=
lim
h→0
|h−π|(e|h|−1)sin|h|−0
h
=0
L.H.D. =
lim
h→0
|−h−π|(e−h|−1)sin|−h|−0
−h
=0
∴ RHD = LHD Therefore, function is differentiable. at x = 0. Since, the function f(x) is differentiable at all the points including p and 0. i.e., f(x) is every where differentiable . Therefore, there is no element in the set S. ⇒S=ϕ( an empty set )