If $f(x) = \begin{cases} \frac{x^3 + x^2 - 16x + 20}{(x-2)^2}, x \neq 2 \\ k, x = 2 \end{cases}$ is continuous at $x = 2$, then $k = \rule{1cm}{0.15mm}$

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

Let $f(x) = \begin{cases} 1 + 6x - 3x^2, & x \leq 1 \\ x + \log_2(b^2 + 7), & x > 1 \end{cases}$. Then the set of all possible values of $b$ such that $f(1)$ is the maximum value of $f(x)$ is

If $f(x) = \begin{cases} 3 + x; & x \geqslant 0 \\ 2 - 3x; & x < 0 \end{cases}$,then $\lim_{x \to 0} f(f(x))$ is equal to -

For the function $f(x) = \frac{1}{x + 2^{\frac{1}{x - 2}}}$,$x \neq 2$,which of the following holds?

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