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 ..........

If the function $f(x) = \begin{cases} \frac{72^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}} & , x \neq 0 \\ a \ln 2 \ln 3 & , x=0 \end{cases}$ is continuous at $x=0$,then the value of $a^2$ is equal to

Find the values of $k$ so that the function $f$ is continuous at the indicated point. $f(x) = \begin{cases} kx + 1, & \text{if } x \le 5 \\ 3x - 5, & \text{if } x > 5 \end{cases}$ at $x = 5$. (in $/5$)

Suppose $f(x) = \begin{cases} a + bx, & x < 1 \\ 4, & x = 1 \\ b - ax, & x > 1 \end{cases}$ and if $\lim_{x \to 1} f(x) = f(1)$,what are the possible values of $a$ and $b$?

If a function $f$ is defined by $f(x) = \begin{cases} \frac{1-\sqrt{2} \sin x}{\pi-4 x}, & x \neq \frac{\pi}{4} \\ k, & x = \frac{\pi}{4} \end{cases}$ and is continuous at $x = \frac{\pi}{4}$,then $k = $