If $[x]$ denotes the greatest integer not exceeding the number $x$,then $f(x)$ defined by $f(x) = \begin{cases} [x], & \text{if } x < 2 \\ [x]-1, & \text{if } x \geq 2 \end{cases}$ is continuous in the interval.

Find all points of discontinuity of $f$,where $f$ is defined by $f(x) = \begin{cases} \frac{x}{|x|}, & \text{if } x < 0 \\ -1, & \text{if } x \ge 0 \end{cases}$. Is $f$ a continuous function?

Let $f: R \rightarrow R$ be defined as $f(x) = \begin{cases} [e^x], & x < 0 \\ a e^x + [x - 1], & 0 \leq x < 1 \\ b + [\sin(\pi x)], & 1 \leq x < 2 \\ [e^{-x}] - c, & x \geq 2 \end{cases}$ where $a, b, c \in R$ and $[t]$ denotes the greatest integer less than or equal to $t$. Then,which of the following statements is true?

Let $f(x) = \begin{cases} x^2 - a & x < 3 \\ b\sqrt{x - 2} + a & 3 \leqslant x < 6 \\ 2x + b & x \geqslant 6 \end{cases}$. If $f(x)$ is continuous $\forall x \in R$,then find the value of $\frac{f(1) - f(3)}{4}$.

Let $[t]$ represent the greatest integer not more than $t$. Then the number of discontinuous points of $f(x) = [x^{1/x}]$ in $(0, \infty)$ is

If $f(x) = \begin{cases} \frac{x - 1}{2}, & 0 \leqslant x < 1 \\ 1/2, & 1 \leqslant x < 2 \end{cases}$ and $g(x) = (2x + 1)(x - k) + 3$ for $0 \leqslant x < \infty$,then $g(f(x))$ will be continuous at $x = 1$ if $k$ is equal to: