If $f(x) = \begin{cases} kx + 1, & x \leq \frac{\pi}{2} \\ \sin x, & x > \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,then $k = $ . . . . . . .

The function $f(x) = \begin{cases} sgn([x]) & x \notin I \\ [sgn(x)] & x \in I \end{cases}$ is (where $sgn()$ denotes the signum function and $[.]$ denotes the greatest integer function):

If the function $f(x) = \begin{cases} x + a \sqrt{2} \sin x & \text{if } 0 \leq x \leq \frac{\pi}{4} \\ 2x \cot x + b & \text{if } \frac{\pi}{4} < x \leq \frac{\pi}{2} \\ a \cos 2x - b \sin x & \text{if } \frac{\pi}{2} < x \leq \pi \end{cases}$ is continuous in $[0, \pi]$,then $a - b = $

$f(x) = \begin{cases} \frac{x-4}{|x-4|} + a, & x < 4 \\ a+b, & x=4 \\ \frac{x-4}{|x-4|} + b, & x > 4 \end{cases}$
If $f(x)$ given above is continuous at $x=4$,then find the values of '$a$' and '$b$'.

If $f(x) = \frac{4}{x^4} \left[ 1 - \cos \frac{x}{2} - \cos \frac{x}{4} + \cos \frac{x}{2} \cdot \cos \frac{x}{4} \right]$ is continuous at $x = 0$,then $f(0)$ is

Let $f(x) = \frac{1 - x(1 + |1 - x|)}{|1 - x|} \cos \left(\frac{1}{1 - x}\right)$ for $x \neq 1$. Then