If the function $f(x)=\begin{cases} \frac{1-\cos x}{x^{2}}, & \text{for } x \neq 0 \\ k, & \text{for } x=0 \end{cases}$ is continuous at $x=0$,then the value of $k$ is

Match the items given in List $A$ with those of the items of List $B$:

$A$. $|x| + |x - 2|$	$I$. Right hand limit does not exist at $x = 2$.
$B$. $\text{cosech } x$	$II$. Continuous only for non-zero real values of $x$.
$C$. $x - [x]$	$III$. Limit is zero for all real $x$.
$D$. $\sqrt{2 - x}$	$IV$. Continuous for all real value of $x$.
	$V$. Discontinuous at all integral values of $x$.

The correct match is:

If $f(x) = \begin{cases} mx^2 + n, & x < 0 \\ nx + m, & 0 \leq x \leq 1 \\ nx^3 + m, & x > 1 \end{cases}$
For what integers $m$ and $n$ do $\lim_{x \to 0} f(x)$ and $\lim_{x \to 1} f(x)$ exist?

$f(x) = \begin{cases} \frac{72^x - 9^x - 8^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}}, & x \neq 0 \\ k \log 2 \log 3, & x = 0 \end{cases}$ Find the value of $k$ for which the function $f$ is continuous.

Let the function $f(x) = \begin{cases} \frac{\log_{e}(1+5x) - \log_{e}(1+\alpha x)}{x} & \text{if } x \neq 0 \\ 10 & \text{if } x = 0 \end{cases}$ be continuous at $x = 0$. The value of $\alpha$ is equal to:

If $f(x) = \begin{cases} \frac{1}{x} \sin(x^2), & x \ne 0 \\ 0, & x = 0 \end{cases}$,then