$f(x) = \begin{cases} 3x - 8 & \text{if } x \leq 5 \\ 2k & \text{if } x > 5 \end{cases}$ is continuous,find $k$.

If the function $f(x) = \begin{cases} \frac{\log_{e}(1-x+x^{2}) + \log_{e}(1+x+x^{2})}{\sec x - \cos x}, & x \in (-\frac{\pi}{2}, \frac{\pi}{2}) - \{0\} \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$,then $k$ is equal to.

The set of points of discontinuity of the function $f(x) = \frac{\tan x \cdot \tan^{-1}\left(\frac{1}{x-1}\right)}{x(x-3)(x-5)}$ is

$f(x) = \begin{cases} \frac{(2x^2 - ax + 1) - (ax^2 + 3bx + 2)}{x + 1} & ; x \neq -1 \\ k & ; x = -1 \end{cases}$ is a real-valued function. If $a, b, k \in R$ and $f$ is continuous on $R$,then $k =$

The function defined by $f(x) = \begin{cases} (x^2 + e^{\frac{1}{2-x}})^{-1} & x \neq 2 \\ k & x = 2 \end{cases}$ is continuous from the right at the point $x = 2$. Then $k$ is equal to:

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 ..........