The function $f: R - \{0\} \to R$,given by $f(x) = \frac{1}{x} - \frac{2}{e^{2x} - 1}$ can be made continuous at $x = 0$ by defining $f(0)$ as:

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{x + 2}{x^2 + 3 x + 2}, & x \in R - \{-1, -2\} \\ -1, & x = -2 \\ 0, & x = -1 \end{cases}$ then $f$ is continuous on the set

Discuss the continuity of the function $f$ given by $f(x) = |x|$ at $x = 0$.

Consider the function $f(x) = \begin{cases} \frac{x}{[x]} & \text{if } 1 \leqslant x < 2 \\ 1 & \text{if } x = 2 \\ \sqrt{6-x} & \text{if } 2 < x \leqslant 3 \end{cases}$ where $[x]$ denotes the greatest integer function. At $x = 2$,the function:

If $f(x) = [x] - [\frac{x}{4}]$,$x \in R$,where $[x]$ denotes the greatest integer function,then

If $f(x) = \begin{cases} Kx^2, & x \leq 2 \\ 3, & x > 2 \end{cases}$ is continuous at $x = 2$,then the value of $K$ is: