STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Now for differentiability

$f(x) = \,|\,\,x\,\,|\,\, = \,\,|0|\,\, = 0$ and $f(0 + h) = f(h) = \,\,|h|$

$\therefore \,\,\mathop {\lim }\limits_{h \to 0 - } \,\frac{{f(0 + h) - f(0)}}{h} = \mathop {\lim }\limits_{h \to 0 - } \,\frac{{|h|}}{h} = - 1$

and $\mathop {\lim }\limits_{h \to 0 + } \,\frac{{f(0 + h) - f(0)}}{h} = \mathop {\lim }\limits_{h \to 0 + } \,\frac{{|h|}}{h} = 1$.

Therefore it is continuous and non-differentiable.