The function defined by $f(x) = \begin{cases} (x^2 + e^{\frac{1}{2-x}})^{-1} & x \neq 2 \\ k & x = 2 \end{cases}$ is continuous from the right at the point $x = 2$. Then $k$ is equal to:

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

If $f(x) = \begin{cases} 1 + x, & \text{when } x \le 2 \\ 5 - x, & \text{when } x > 2 \end{cases}$,then which of the following is true?

The number of points of discontinuity of the function $f(x) = \frac{1}{1 - e^{\frac{-x-1}{x-2}}}$ is

Let $f:(0,1) \rightarrow R$ be a function defined as $f(x) = \sqrt{n}$ if $x \in \left[\frac{1}{n+1}, \frac{1}{n}\right)$ where $n \in N$. Let $g:(0,1) \rightarrow R$ be a function such that $\int_{x^2}^x \sqrt{\frac{1-t}{t}} dt < g(x) < 2\sqrt{x}$ for all $x \in (0,1)$. Then $\lim_{x \rightarrow 0} f(x)g(x)$

If $f(x) = \begin{cases} mx+1, & x \leq \frac{\pi}{2} \\ \sin x+n, & x > \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,where $m, n \in \mathbb{Z}$,then: