(N/A) function $f$ is differentiable at $x = c$ if the left-hand derivative $(LHD)$ and right-hand derivative $(RHD)$ exist and are equal.
For $x = 1$:
$LHD$ $= \lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^-} \frac{[1+h] - [1]}{h} = \lim_{h \to 0^-} \frac{0 - 1}{h} = \lim_{h \to 0^-} \frac{-1}{h} = \infty$.
Since the limit is not finite,the function is not differentiable at $x = 1$.
For $x = 2$:
$LHD$ $= \lim_{h \to 0^-} \frac{f(2+h) - f(2)}{h} = \lim_{h \to 0^-} \frac{[2+h] - [2]}{h} = \lim_{h \to 0^-} \frac{1 - 2}{h} = \lim_{h \to 0^-} \frac{-1}{h} = \infty$.
Since the limit is not finite,the function is not differentiable at $x = 2$.