The value of $k$ so that the function $f(x) = \begin{cases} k(2x - x^2), & x < 0 \\ \cos x, & x \ge 0 \end{cases}$ is continuous at $x = 0$,is

Given $f(x) = \begin{cases} \frac{1-\cos 4x}{x^2}, & \text{if } x < 0 \\ a, & \text{if } x = 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & \text{if } x > 0 \end{cases}$
If $f(x)$ is continuous at $x=0$,then the value of $a$ is:

If $f(x) = \begin{cases} -x^2, & \text{when } x \le 0 \\ 5x - 4, & \text{when } 0 < x \le 1 \\ 4x^2 - 3x, & \text{when } 1 < x < 2 \\ 3x + 4, & \text{when } x \ge 2 \end{cases}$,then:

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{\cos 3x - \cos x}{x^2}, & \text{for } x \neq 0 \\ \lambda, & \text{for } x = 0 \end{cases}$ and if $f$ is continuous at $x = 0$,then $\lambda$ is equal to

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 ...... .

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: