If $f(x) = \begin{cases} \frac{\sin 2x}{5x}, & x \ne 0 \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$,then the value of $k$ will be

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:

Define $f: R \rightarrow R$ by $f(x) = [x] + \sqrt{x - [x]}$ for $x \in R$,where $[x]$ denotes the greatest integer function. Then the set of points at which $f$ is continuous is

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 $/5$)

For the function $f(x) = \frac{1}{x + 2^{\frac{1}{x - 2}}}$,$x \neq 2$,which of the following holds?

Consider $f(x) = [x]|x^3 - 2x^2 - x + 2|$ in $[-\frac{3}{2}, \frac{9}{2}]$. The number of points where $f(x)$ is discontinuous is (where $[.]$ denotes the greatest integer function).