If $f(x)=\frac{[x]}{|x|}, x \neq 0$ where [.] denotes the greatest integer function, then f'(1) is
Answer & explanation
Correct answer: option 3
We have,
$f(x)=\frac{[x]}{|x|}= \begin{cases}0, & 0<x<1 \\ \frac{1}{x}, & 1 \leq x<2\end{cases}$
Clearly, $\lim\limits_{x \rightarrow 1^{-}} f(x)=0$ and $\lim\limits_{x \rightarrow 1^{+}} f(x)=1$
∴ $\lim\limits_{x \rightarrow 1^{-}} f(x) \neq \lim\limits_{x \rightarrow 1^{+}} f(x)$
Thus, f(x) is not continuous at x = 1.
Hence, f(x) is not differentiable at x = 1.
Consequently, f'(1) does not exist.