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 $a^2+b^2$.

If a function $f(x)$ defined on $[a, b]$ is discontinuous at $x=\alpha \in(a, b)$,then

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

Consider $f(x) = [x] + \sqrt{\{x\}}$,where $[.]$ denotes the greatest integer function and $\{.\}$ denotes the fractional part function. Identify the correct statement.

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

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