If $f: R \rightarrow R$ is defined by $f(x) = x - [x]$,where $[x]$ is the greatest integer not exceeding $x$,then the set of points of discontinuity of $f$ is

Let $f(x) = \begin{cases} x^3+8; x < 0 \\ x^2-4; x \ge 0 \end{cases}$ and $g(x) = \begin{cases} (x-8)^{1/3}; x < 0 \\ (x+4)^{1/2}; x \ge 0 \end{cases}$. Then the number of points,where the function $g \circ f$ is discontinuous,is ————

If $f(x) = \begin{cases} 1 + \cos x, & x \le 0 \\ a - x, & 0 < x < 2 \\ (x - b)^2, & x \ge 2 \end{cases}$ is continuous at $x=0$ and $x=2$,then find the value of $a^2+b^2$.

Consider the piecewise defined function $f(x) = \begin{cases} \sqrt{-x} & \text{if } x < 0 \\ 0 & \text{if } 0 \leqslant x \leqslant 4 \\ x - 4 & \text{if } x > 4 \end{cases}$. Choose the answer which best describes the continuity of this function.

Let $f(x) = \begin{cases} x \sin \left( \frac{1}{x} \right) \sin \left( \frac{1}{x \sin \left( \frac{1}{x} \right)} \right), & x \neq 0 \\ 0, & x = 0 \end{cases}$. Then $f(x)$ is:

The number of points of discontinuity of the function $f(x) = [\frac{x^2}{2}] - [\sqrt{x}]$ for $x \in [0, 4]$,where $[\cdot]$ denotes the greatest integer function,is . . . . . . .