The function defined by $f(x) = \begin{cases} \frac{x-4}{|x-4|} + a, & x < 4 \\ a + b, & x = 4 \\ \frac{x-4}{|x-4|} + b, & x > 4 \end{cases}$ is continuous at $x = 4$,then:

If $f(x) = \begin{cases} 1 + x, & \text{when } x \le 2 \\ 5 - x, & \text{when } x > 2 \end{cases}$,then which of the following is true?

The set of points of discontinuity of the function $f(x) = \frac{\tan x \cdot \tan^{-1}\left(\frac{1}{x-1}\right)}{x(x-3)(x-5)}$ is

Let $f(x) = \frac{1-\tan x}{4x-\pi}$,where $x \neq \frac{\pi}{4}$ and $x \in [0, \frac{\pi}{2}]$. If $f(x)$ is continuous in $[0, \frac{\pi}{2}]$,then $f(\frac{\pi}{4})$ is:

If the function $f(x) = \begin{cases} x + a^2\sqrt{2} \sin x, & 0 \le x < \pi/4 \\ x \cot x + b, & \pi/4 \le x < \pi/2 \\ b \sin 2x - a \cos 2x, & \pi/2 \le x \le \pi \end{cases}$ is continuous in the interval $[0, \pi]$,then the values of $(a, b)$ are:

If $f(x) = \left[\tan \left(\frac{\pi}{4} + x\right)\right]^{\frac{1}{x}}$ for $x \neq 0$ and $f(x) = k$ for $x = 0$,is continuous at $x = 0$,then $k =$