The function $f(x) = \begin{cases} \frac{2}{5-x}, & x < 3 \\ 5-x, & x \geq 3 \end{cases}$ is

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

If $f(x)=\begin{cases} \frac{x-3}{|x-3|}+a & , x<3 \\ a+b & , x=3 \\ \frac{|x-3|}{x-3}+b & , x>3 \end{cases}$ is continuous at $x=3$,then the value of $a-b$ is

Define $f: R \to R$ by $f(x) = \begin{cases} \frac{1-\cos 4x}{x^2}, & x < 0 \\ a, & x = 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x > 0 \end{cases}$ Then the value of $a$ so that $f$ is continuous at $x = 0$ is:

Find all points of discontinuity of $f$,where $f$ is defined by $f(x) = \begin{cases} 2x + 3, & \text{if } x \le 2 \\ 2x - 3, & \text{if } x > 2 \end{cases}$

If the function $f(x) = \frac{\tan(\tan x) - \sin(\sin x)}{\tan x - \sin x}$ is continuous at $x = 0$,then $f(0)$ is equal to . . . . . . .