If $\lim\limits_{x \to 3}\left(\dfrac{x^2 - ax - 3b}{x - 3}\right) = 5$, then $a + b = $

If $f(x) = \begin{cases} ax + 7 & \text{if } x < 1 \\ 3x - 1 & \text{if } x = 1 \\ \dfrac{b}{x + 3} & \text{if } x > 1 \end{cases}$ is continuous at $x = 1$, then

$\lim _{x \rightarrow 1} \frac{x^4-\sqrt{x}}{\sqrt{x}-1}$ is

Match the following:

In the following, $[\mathrm{x}]$ denotes the greatest integer less than or equal to x .

Column - I		Column - II
(a)	$\mathrm{x}|\mathrm{x}|$$\mathrm{x}|\mathrm{x}|$x|x|	(i)	continuous in (-1, 1)
(b)	$\sqrt{|\mathrm{x}|}$$\sqrt{|\mathrm{x}|}$sqrt(|x|)	(ii)	differentiable in (-1, 1)
(c)	$\mathrm{x}+[\mathrm{x}]$$\mathrm{x}+[\mathrm{x}]$x+[x]	(iii)	strictly increasing in (-1, 1)
(d)	$|\mathrm{x}-1|+|\mathrm{x}+1|$$|\mathrm{x}-1|+|\mathrm{x}+1|$|x-1|+|x+1|	(iv)	not differentiable at, at least one point in (-1, 1)