Correct Option(4) h is not differentiable at two values of x
Explanation:
h(x) = min {x, x2 } for all x implies
h(x) is continuous at 0 and 1 and hence for all x.
h(x) is not differentiable at x = 0 and at x = 1
h′ (0 − ) = 1 and h′(0 + ) = 0
h′ (1 − ) = 2 and h′(1 + ) = 1
Also h′(x) = 1 for all x > 1