Let $f(x) = \begin{cases} \frac{\sin \pi x}{5x}, & x \ne 0 \\ k, & x = 0 \end{cases}$. If $f(x)$ is continuous at $x = 0$,then $k =$

For every pair of continuous functions $f, g: [0, 1] \rightarrow \mathbb{R}$ such that $\max \{f(x): x \in [0, 1] \} = \max \{g(x): x \in [0, 1] \} = \lambda$,the correct statement$(s)$ is (are):

If the function $f(x) = \begin{cases} \frac{k\cos x}{\pi - 2x}, & x \neq \frac{\pi}{2} \\ 3, & x = \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,then $k = $

If $f(x) = \begin{cases} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}, & \text{for } -1 \leq x < 0 \\ 2x^2+3x-2, & \text{for } 0 \leq x \leq 1 \end{cases}$ is continuous at $x=0$,then $k$ 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:

If $f(x)$ is continuous at $x = 0$, where $f(x) = \frac{8^x - 2^x}{k^x - 1}$ for $x \neq 0$ and $f(0) = 2$, then the value of $k$ is ...