If the function $f(x) = \begin{cases} (\cos x)^{1/x}, & x \ne 0 \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$,then the value of $k$ is

If the function $f: R \rightarrow R$ defined by $f(x) = \begin{cases} \frac{a(1-\cos 2x)}{x^2}, & x < 0 \\ b, & x = 0 \\ \frac{\sqrt{x}}{\sqrt{4+\sqrt{x}}-2}, & x > 0 \end{cases}$ is continuous at $x = 0$, then $a+b=$

If $f(x) = \left(\frac{1+x}{1-x}\right)^{\frac{1}{x}}$ is continuous at $x = 0$,then $f(0) = $

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:

The number of discontinuities of the greatest integer function $f(x) = [x]$ for $x \in \left(-\frac{7}{2}, 100\right)$ is:

Define $f: R \to R$ by $f(x) = \begin{cases} \frac{1-\cos 4x}{x^2}, & x < 0 \\ a, & x = 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x > 0 \end{cases}$ Then the value of $a$ so that $f$ is continuous at $x = 0$ is: