Let $f(x) = \frac{1-\tan x}{4x-\pi}$,where $x \neq \frac{\pi}{4}$ and $x \in [0, \frac{\pi}{2}]$. If $f(x)$ is continuous in $[0, \frac{\pi}{2}]$,then $f(\frac{\pi}{4})$ is:

If the function $f(x) = \begin{cases} \frac{\sqrt{1+px}-\sqrt{1-px}}{x}, & \text{if } -1 \leq x < 0 \\ \frac{2x+1}{x-2}, & \text{if } 0 \leq x \leq 1 \end{cases}$ is continuous in the interval $[-1, 1]$,then $p = $

If the function $f(x) = \begin{cases} \frac{1}{x} \log_{e}\left(\frac{1+\frac{x}{a}}{1-\frac{x}{b}}\right), & x < 0 \\ k, & x = 0 \\ \frac{\cos^{2} x - \sin^{2} x - 1}{\sqrt{x^{2}+1}-1}, & x > 0 \end{cases}$ is continuous at $x = 0$,then $\frac{1}{a} + \frac{1}{b} + \frac{4}{k}$ is equal to:

Let $[t]$ denote the greatest integer less than or equal to $t$. Let $f:[0, \infty) \rightarrow \mathbb{R}$ be a function defined by $f(x) = [\frac{x}{2} + 3] - [\sqrt{x}]$. Let $S$ be the set of all points in the interval $[0, 8]$ at which $f$ is not continuous. Then $\sum_{a \in S} a$ is equal to:

If the function defined by $f(x) = \frac{\log (1+x)^{1+x}}{x^2} - \frac{1}{x}, x \neq 0$ is continuous at $x=0$,then $6 f(0)$ is equal to

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 $a^2+b^2$.