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

If a real valued function $f(x) = \begin{cases} e^{\frac{\sin a(x-[x])}{x-[x]}}, & \text{if } x < 1 \\ b+1, & \text{if } x = 1 \\ \frac{|x^2+x-2|}{x-1}, & \text{if } x > 1 \end{cases}$ is continuous at $x = 1$,then $b \sin a =$ ([x] denotes the greatest integer function)

The function $f: R - \{0\} \to R$,given by $f(x) = \frac{1}{x} - \frac{2}{e^{2x} - 1}$ can be made continuous at $x = 0$ by defining $f(0)$ as:

The function $f(x) = \begin{cases} x + 2, & 1 \le x \le 2 \\ 4, & x = 2 \\ 3x - 2, & x > 2 \end{cases}$ is continuous at

The function $f(x) = [x]^2 - [x^2]$,(where $[y]$ is the greatest integer less than or equal to $y$),is discontinuous at

If the function $f(x) = \begin{cases} (\cos x)^{1/x}, & x \ne 0 \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$,then the value of $k$ is