Find the values of $k$ so that the function $f$ is continuous at the indicated point. $f(x) = \begin{cases} kx + 1, & \text{if } x \le \pi \\ \cos x, & \text{if } x > \pi \end{cases}$ at $x = \pi$.

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

If $f(x) = \begin{cases} \frac{\sin [x]}{[x] + 1}, & \text{for } x > 0 \\ \frac{\cos \frac{\pi }{2}[x]}{[x]}, & \text{for } x < 0 \\ k, & \text{at } x = 0 \end{cases}$; where $[x]$ denotes the greatest integer less than or equal to $x$,then in order that $f$ be continuous at $x = 0$,the value of $k$ is

If $f(x) = \max(\sin x, \sin^{-1}(\cos x))$,then

The function $f(x) = \cot x$ is discontinuous at every point of the set

If the function $f(x) = \frac{\sqrt{1+x}-1}{x}$ is continuous at $x=0$,then $f(0) = $