If $f(x) = \frac{x^2 - 10x + 25}{x^2 - 7x + 10}$ for $x \neq 5$ and $f$ is continuous at $x = 5$,then $f(5) = $

The values of $a$ and $b$,so that the function $f(x) = \begin{cases} x+a \sqrt{2} \sin x, & 0 \leq x \leq \frac{\pi}{4} \\ 2 x \cot x+b, & \frac{\pi}{4} < x \leq \frac{\pi}{2} \\ a \cos 2 x-b \sin x, & \frac{\pi}{2} < x \leq \pi \end{cases}$ is continuous for $0 \leq x \leq \pi$,are respectively given by

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

Let $f(x) = \begin{cases} \frac{2^x + 2^{3-x} - 6}{\sqrt{2^{-x}} - 2^{1-x}} & \text{if } x > 2 \\ \frac{x^2 - 4}{x - \sqrt{3x - 2}} & \text{if } x < 2 \end{cases}$. Determine the nature of the function at $x = 2$.

If the function $f$ defined as $f(x) = \frac{1}{x} - \frac{k - 1}{e^{2x} - 1}$,$x \neq 0$,is continuous at $x = 0$,then the ordered pair $(k, f(0))$ is equal to?

If $f(x)$ is continuous at point $x=0$ where $f(x) = \begin{cases} \frac{3 \sin x + 5 \tan x}{a^x - 1} & , x < 0 \\ \frac{2}{\log 2} & , x = 0 \\ \frac{8x + 2x \cos x}{b^x - 1} & , x > 0 \end{cases}$ then the values of $a$ and $b$,respectively,are