=0 This possible only when p>0........ (i) f′(0)=
lim
h→0
f(h)−f(0)
h
=
lim
h→0
hp‌sin‌
1
h
−0
h
=
lim
h→0
hp−1‌sin‌
1
h
⇒f′(0) will exist only when p>0 ∴f(x) will not be differentiable if p≤1 .........(ii) From Eqs. (i) and (ii), for f(x) to be not differentiable but continuous at x=0, possible values of p are given by 0<p≤1. Hence option (c) is correct.