If the function $f(x) = \begin{cases} k_{1}(x-\pi)^{2}-1, & x \leq \pi \\ k_{2} \cos x, & x>\pi \end{cases}$ is twice differentiable,then the ordered pair $(k_{1}, k_{2})$ is equal to

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:

Statement $1$: $A$ function $f: R \to R$ is continuous at $x_0$ if and only if $\lim_{x \to x_0} f(x)$ exists and $\lim_{x \to x_0} f(x) = f(x_0)$.
Statement $2$: $A$ function $f: R \to R$ is discontinuous at $x_0$ if and only if $\lim_{x \to x_0} f(x)$ exists and $\lim_{x \to x_0} f(x) \neq f(x_0)$.

The function $f(x) = \begin{cases} x, & \text{if } 0 \le x \le 1 \\ 1, & \text{if } 1 < x \le 2 \end{cases}$ is

If the real valued function $f(x) = \begin{cases} \frac{\cos 3x - \cos x}{x \sin x} & \text{if } x < 0 \\ p & \text{if } x = 0 \\ \frac{\log(1 + q \sin x)}{x} & \text{if } x > 0 \end{cases}$ is continuous at $x = 0$, then $p + q =$

The function $f(x) = [x]^2 - [x^2]$,(where $[y]$ is the greatest integer less than or equal to $y$),is discontinuous at