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

If the function $f$ defined on $\left(\frac{\pi}{6}, \frac{\pi}{3}\right)$ by $f(x)=\begin{cases} \frac{\sqrt{2} \cos x-1}{\cot x-1}, & x \neq \frac{\pi}{4} \\ k, & x=\frac{\pi}{4} \end{cases}$ is continuous,then $k$ is equal to

At $x=0$,the function $f(x) = \begin{cases} \frac{x}{|x|+2x^2}, & x \neq 0 \\ k, & x=0 \end{cases}$ is:

The value of $k$ so that the function $f(x) = \begin{cases} k(2x - x^2), & x < 0 \\ \cos x, & x \ge 0 \end{cases}$ is continuous at $x = 0$,is

In order that the function $f(x) = (x + 1)^{1/x}$ is continuous at $x = 0$,$f(0)$ must be defined as

If $f(x) = \begin{cases} \frac{\log (1 + 2ax) - \log (1 - bx)}{x}, & x \neq 0 \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$,then the value of $k$ is