At $x=0$, the function $f(x) = \begin{cases} \frac{x}{|x|+2x^2}, & x \neq 0 \\ k, & x=0 \end{cases}$ is:

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 $a, b \in R, (a \ne 0)$. 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

If the function $f(x) = \frac{e^{x}(e^{\tan x-x}-1)+\log_{e}(\sec x+\tan x)-x}{\tan x-x}$ is continuous at $x=0$, then the value of $f(0)$ is equal to

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{\cos 3x - \cos x}{x^2}, & \text{for } x \neq 0 \\ \lambda, & \text{for } x = 0 \end{cases}$ and if $f$ is continuous at $x = 0$,then $\lambda$ is equal to

Let $f(x) = \begin{cases} x^2 + k, & \text{when } x \ge 0 \\ -x^2 - k, & \text{when } x < 0 \end{cases}$. If the function $f(x)$ is continuous at $x = 0$,then $k =$