The function $f(x) = \begin{cases} x + 2, & 1 \le x \le 2 \\ 4, & x = 2 \\ 3x - 2, & x > 2 \end{cases}$ is continuous at

Which of the following function$(s)$ not defined at $x = 0$ has/have an irremovable discontinuity at $x = 0$?

Let $f(x) = \begin{cases} \frac{2^x + 2^{3-x} - 6}{\sqrt{2^{-x}} - 2^{1-x}} & \text{if } x > 2 \\ \frac{x^2 - 4}{x - \sqrt{3x - 2}} & \text{if } x < 2 \end{cases}$. Determine the nature of the function at $x = 2$.

Show that the function defined by $f(x)=|\cos x|$ is a continuous function.

The values of $p$ and $q$ such that the function $f(x) = \begin{cases} (1+|\sin x|)^{\frac{p}{|\sin x|}}, & \frac{-\pi}{6} < x < 0 \\ q, & x = 0 \\ e^{\frac{\sin 2x}{\sin 3x}}, & 0 < x < \frac{\pi}{6} \end{cases}$ is continuous at $x=0$ are:

Let $f(x) = \begin{cases} \operatorname{sgn}(x^2 - 3x + 2) & x \in \mathbb{Q} \\ 0 & x \notin \mathbb{Q} \end{cases}$,then the number of points where $f(x)$ is continuous is (where $\operatorname{sgn}(x)$ denotes the signum function of $x$).