Discuss the continuity of the function $f$ given by $f(x) = |x|$ at $x = 0$.

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:

If $f(x) = \begin{cases} x, & \text{if } x \text{ is irrational} \\ 0, & \text{if } x \text{ is rational} \end{cases}$,then $f$ is

Let $a, b \in R, b \neq 0$. Define a function $f(x) = \begin{cases} a \sin \frac{\pi}{2}(x-1), & \text{for } x \leq 0 \\ \frac{\tan 2x - \sin 2x}{bx^3}, & \text{for } x > 0 \end{cases}$. If $f$ is continuous at $x = 0$,then $10 - ab$ is equal to ...... .

Suppose that $f$ is continuous on $[a, b]$ and that $f(x)$ is an integer for each $x$ in $[a, b]$. Then in $[a, b]$

Assertion $(A)$: $f(x)=|x-a|+|x-b|$ is continuous on $R$. Reason $(R)$: $\frac{|x-\alpha|}{x-\alpha}$ is continuous at $x \in R-\{\alpha\}$. The correct option among the following is: