The value of $f(0)$ so that the function $f(x) = \frac{2^x - 2^{-x}}{x}$ for $x \neq 0$ is continuous at $x = 0$ is:

Let $f:(0, \pi) \rightarrow \mathbb{R}$ be a function given by
$f(x)=\begin{cases} (\frac{8}{7})^{\frac{\tan 8x}{\tan 7x}}, & 0 < x < \frac{\pi}{2} \\ a-8, & x=\frac{\pi}{2} \\ (1+|\cot x|)^{\frac{b}{a}|\tan x|}, & \frac{\pi}{2} < x < \pi \end{cases}$
Where $a, b \in \mathbb{Z}$. If $f$ is continuous at $x=\frac{\pi}{2}$,then $a^2+b^2$ is equal to ..........

If $f(x) = x\sqrt{1 - [x]^2}$,then (where $[.]$ denotes the greatest integer function):

Let $f(x) = \begin{cases} 2 - |x^2 + 5x + 6|, & x \neq -2 \\ a^2 + 1, & x = -2 \end{cases}$. Then the range of $a$ such that $f(x)$ has a maximum at $x = -2$ is

The value of $p$ for which the function $f(x) = \begin{cases} \frac{(4^x - 1)^3}{\sin(\frac{x}{p}) \log(1 + \frac{x^2}{3})}, & x \ne 0 \\ 12(\log 4)^3, & x = 0 \end{cases}$ is continuous at $x = 0$ is:

Let $f: R \rightarrow R$ be a function given by $f(x) = \begin{cases} \frac{1-\cos 2x}{x^2} & , x < 0 \\ \alpha & , x = 0 \\ \frac{\beta \sqrt{1-\cos x}}{x} & , x > 0 \end{cases}$. If $f$ is continuous at $x = 0$,then $\alpha^2 + \beta^2$ is equal to: