If $f(x) = \begin{cases} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}, & -1 \leq x < 0 \\ 2x^2+3x-2, & 0 \leq x \leq 1 \end{cases}$ is continuous at $x = 0$, then $k$ is equal to

Let $a, b \in R, (a \ne 0)$. If the function $f$ defined as
$f(x) = \begin{cases} \frac{2x^2}{a}, & 0 \le x < 1 \\ a, & 1 \le x < \sqrt{2} \\ \frac{2b^2 - 4b}{x^3}, & \sqrt{2} \le x < \infty \end{cases}$
is continuous in the interval $[0, \infty)$,then an ordered pair $(a, b)$ is

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

If the function $f(x) = \frac{\sqrt{1+x}-1}{x}$ is continuous at $x=0$,then $f(0) = $

The number of points of discontinuity of the function $f(x) = [\frac{x^2}{2}] - [\sqrt{x}]$ for $x \in [0, 4]$, where $[\cdot]$ denotes the greatest integer function, is:

The values of $p$ and $q$ such that the function $f(x) = \begin{cases} (1+|\sin x|)^{\frac{p}{|\sin x|}}, & \frac{-\pi}{6} < x < 0 \\ q, & x = 0 \\ e^{\frac{\sin 2x}{\sin 3x}}, & 0 < x < \frac{\pi}{6} \end{cases}$ is continuous at $x=0$ are: