If $f(x) = [x]$ for $x \in (-1, 2)$,then $f$ is discontinuous at (where $[x]$ represents the floor function).

If $f(x)$ is continuous at $x = 0$, where $f(x) = \frac{8^x - 2^x}{k^x - 1}$ for $x \neq 0$ and $f(0) = 2$, then the value of $k$ is ...

The function $f(x) = [x] \cos \left( \frac{2x - 1}{2} \pi \right)$,where $[.]$ denotes the greatest integer function,is discontinuous at

The points at which the function $f(x) = \frac{x + 1}{x^2 + x - 12}$ is discontinuous,are

If the function $f(x) = \begin{cases} x - \frac{|x|}{x}, & x < 0 \\ x + \frac{|x|}{x}, & x > 0 \\ 1, & x = 0 \end{cases}$,then which of the following is true?

Consider the function $f(x) = [x] + |1 - x|$ for $-1 \le x \le 3$,where $[x]$ is the greatest integer function.
Statement $1$: $f$ is not continuous at $x = 0, 1, 2$ and $3$.
Statement $2$: $f(x) = \begin{cases} -1 - x, & -1 \le x < 0 \\ 1 - x, & 0 \le x < 1 \\ 1 - x, & 1 \le x < 2 \\ 2 + x - 2, & 2 \le x < 3 \\ 3, & x = 3 \end{cases}$ (Note: The provided Statement $2$ in the prompt is incorrect).