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 ————

Discuss the continuity of the function $f$,where $f$ is defined by $f(x) = \begin{cases} -2, & \text{if } x \le -1 \\ 2x, & \text{if } -1 < x \le 1 \\ 2, & \text{if } x > 1 \end{cases}$. Is it continuous at $x=3$?

If $f(x) = |x|/x$ for $x \neq 0$ and $1$ for $x = 0$,then the function is

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

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{2 \sin x-\sin 2 x}{2 x \cos x}, & \text{if } x \neq 0 \\ a, & \text{if } x=0 \end{cases}$, then the value of $a$ so that $f$ is continuous at $x=0$ is

Let $f:[0, \pi] \rightarrow R$ be defined as $f(x)=\begin{cases} \sin x, & \text{if } x \text{ is irrational and } x \in[0, \pi] \\ \tan^2 x, & \text{if } x \text{ is rational and } x \in[0, \pi] \end{cases}$. The number of points in $[0, \pi]$ at which the function $f$ is continuous is