Examine the differentiability of $f$, where $f$ is defined by $f(x) = \begin{cases} x[x], & \text{if } 0 \leq x < 2 \\ (x-1)x, & \text{if } 2 \leq x < 3 \end{cases} \text{ at } x = 2$.

Not differentiable at $x = 2$ because $\text{LHD} = 1$ and $\text{RHD} = 3$.

The correct answer is Option (3) → Not differentiable at $x = 2$ because $\text{LHD} = 1$ and $\text{RHD} = 3$ ##

We have,

$f(x) = \begin{cases} x[x], & \text{if } 0 \leq x < 2 \\ (x-1)x, & \text{if } 2 \leq x < 3 \end{cases} \text{ at } x = 2$

At $x = 2$,

$Lf'(2) = \lim\limits_{h \to 0} \frac{f(2-h) - f(2)}{-h}$

$= \lim\limits_{h \to 0} \frac{(2-h)[2-h] - (2-1)2}{-h} \quad [∵[a-h] = a-1, \text{ where } a \text{ is any positive number}]$

$= \lim\limits_{h \to 0} \frac{(2-h)(1) - 2}{-h}$

$= \lim\limits_{h \to 0} \frac{2-h-2}{-h} = \lim\limits_{h \to 0} \frac{-h}{-h} = 1$

$Rf'(2) = \lim\limits_{h \to 0} \frac{f(2+h) - f(2)}{h}$

$= \lim\limits_{h \to 0} \frac{(2+h-1)(2+h) - (2-1) \cdot 2}{h}$

$= \lim\limits_{h \to 0} \frac{(1+h)(2+h) - 2}{h} = \lim\limits_{h \to 0} \frac{2 + h + 2h + h^2 - 2}{h}$

$= \lim\limits_{h \to 0} \frac{h^2 + 3h}{h} = \lim\limits_{h \to 0} \frac{h(h+3)}{h} = 3$

$∴Lf'(2) \neq Rf'(2)$

So, $f(x)$ is not differentiable at $x = 2$.