If the function $f(x) = \begin{cases} \frac{\sqrt{2 + \cos x} - 1}{(\pi - x)^2}, & x \neq \pi \\ k, & x = \pi \end{cases}$ is continuous at $x = \pi$,then $k$ equals:

Let $f:[-1,2] \rightarrow \mathbb{R}$ be given by $f(x)=2x^2+x+[x^2]-[x]$,where $[t]$ denotes the greatest integer less than or equal to $t$. The number of points where $f$ is not continuous is:

Let $[x]$ be the greatest integer less than or equal to $x$. At which of the following point$(s)$ is the function $f(x) = x \cos(\pi(x + [x]))$ discontinuous?
$[A]$ $x = -1$
$[B]$ $x = 0$
$[C]$ $x = 2$
$[D]$ $x = 1$

Find all points of discontinuity of $f$,where $f$ is defined by $f(x) = \begin{cases} x^{10} - 1, & \text{if } x \le 1 \\ x^2, & \text{if } x > 1 \end{cases}$.

If $f(x) = \frac{(e^{2x} - 1) \sin x^{\circ}}{x^2}, x \neq 0$ is continuous at $x = 0$,then $f(0) =$

Let $f(x) = \frac{1 - \tan x}{4x - \pi }, x \ne \frac{\pi }{4}, x \in [0, \frac{\pi }{2}]$. If $f(x)$ is continuous in $[0, \frac{\pi }{2}]$,then $f(\frac{\pi }{4})$ is