Right answer is (b) They are equal

To explain I would SAY: The given function is f(x) = ∣x − 1∣, x ∈ R.

It is known that a function f is differentiable at point x = c in its domain if both

\(\lim\limits_{h \rightarrow 0^-}\) hf(c + h) – f(c)

And

\(\lim\limits_{h \rightarrow 0^+}\) hf(c + h) – f(c) are FINITE and equal.

To check the differentiability of the function at x = 1,

LHS,

Consider the left hand limit of f at x=1

\(\lim\limits_{h \rightarrow 0^-}\frac{|1+h-1|-|1-1|}{h}\)

= \(\lim\limits_{h \rightarrow 0^-}\frac{|h|}{h}\)

= \(\lim\limits_{h \rightarrow 0^-}\frac{-h}{h}\)

= −1

RHS,

Consider the right hand limit of f at x − 1

\(\lim\limits_{h \rightarrow 0^+}\frac{|1+h-1|-|1-1|}{h}\)

= \(\lim\limits_{h \rightarrow 0^+}\frac{|h|}{h}\)

= 1

Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x = 1.

As, LHS = -1 and RHS = 1, it is clear that, f’(1) < 0 on the left of x = 1 and f’(x) > 0 on the right of the point x = 1.

Hence, f’(x) changes sign, from negative on the left to positive on the right of the point x = 1.

Therefore, f(x) has a local minima at x = 1.