Let $f(x) = x \left[ \frac{x}{2} \right]$,for $-10 < x < 10$,where $[t]$ denotes the greatest integer function. Then the number of points of discontinuity of $f$ is equal to

Find the values of $k$ so that the function $f$ is continuous at the indicated point.
$f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & \text{if } x \neq \frac{\pi}{2} \\ 3, & \text{if } x = \frac{\pi}{2} \end{cases}$ at $x = \frac{\pi}{2}$

If $f(x) = \begin{cases} x, & x > 1 \\ x^2, & x < 1 \end{cases}$,then $\lim_{x \to 1} f(x) = $

If the function $f$ defined as $f(x) = \frac{1}{x} - \frac{k - 1}{e^{2x} - 1}$,$x \neq 0$,is continuous at $x = 0$,then the ordered pair $(k, f(0))$ is equal to?

In order that the function $f(x) = (x + 1)^{\cot x}$ is continuous at $x = 0$,$f(0)$ must be defined as

The function $f(x)=\frac{\tan \{\pi[x-\frac{\pi}{2}]\}}{2+[x]^{2}}$, where $[x]$ denotes the greatest integer $\leq x$, is