If $f(x) = \begin{cases} \frac{x-2}{|x-2|}+a & , x<2 \\ a+b & , x=2 \\ \frac{x-2}{|x-2|}+b & , x>2 \end{cases}$ is continuous at $x=2$, then $a+b=$

The set of points of discontinuity of the function $f(x) = x - [x]$, where $x \in R$, is:

Let $a, b \in R, (a \ne 0)$. If the function $f$ defined as
$f(x) = \begin{cases} \frac{2x^2}{a}, & 0 \le x < 1 \\ a, & 1 \le x < \sqrt{2} \\ \frac{2b^2 - 4b}{x^3}, & \sqrt{2} \le x < \infty \end{cases}$
is continuous in the interval $[0, \infty)$,then an ordered pair $(a, b)$ is

If $f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & x \neq \frac{\pi}{2} \\ \frac{1}{2}, & x = \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,then the value of $k$ is . . . . . . .

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

If $f(x) = [x]$ for $x \in (-1, 2)$,then $f$ is discontinuous at (where $[x]$ represents the floor function).