If $f(x) = \operatorname{sgn} \left( 3\cos x - \frac{a}{3} \right)$ is continuous for all $x$,then the least positive integral value of $a$ is - (where $\operatorname{sgn}(x)$ denotes the signum function of $x$)

Let $f(x) = \begin{cases} \frac{1 - \cos 4x}{x^2}, & x < 0 \\ a, & x = 0 \\ \frac{\sqrt{x} + \sqrt{2}}{\sqrt{16 + \sqrt{x}}}, & x > 0 \end{cases}$
If $f(x)$ is continuous at $x = 0$,then the value of $a$ is

The function $f$ is defined by $f(x) = \begin{cases} 2x - 1, & \text{if } x > 2 \\ k, & \text{if } x = 2 \\ x^2 - 1, & \text{if } x < 2 \end{cases}$. If $f$ is continuous at $x = 2$,then the value of $k$ is equal to:

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

$A$ point in the domain of a function where the discontinuity cannot be removed by redefining the function at that point is called:

Consider the function $f(x) = \begin{cases} \frac{P(x)}{\sin(x-2)}, & x \neq 2 \\ 7, & x = 2 \end{cases}$ where $P(x)$ is a polynomial such that $P''(x)$ is always a constant and $P(3) = 9$. If $f(x)$ is continuous at $x = 2$,then $P(5)$ is equal to: