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

The function $f(x) = [x]$,where $[x]$ is the greatest integer function,is continuous at:

If the function $f(x) = \begin{cases} (1+|\cos x|)^{\frac{\lambda}{|\cos x|}} & , 0 < x < \frac{\pi}{2} \\ \mu & , x = \frac{\pi}{2} \\ e^{\frac{\cot 6x}{\cot 4x}} & , \frac{\pi}{2} < x < \pi \end{cases}$ is continuous at $x = \frac{\pi}{2}$,then $9\lambda + 6 \log_{e} \mu + \mu^6 - e^{6\lambda}$ is equal to

If the function $f(x) = \frac{\log(1 + ax) - \log(1 - bx)}{x}$,$x \neq 0$ is continuous at $x = 0$,then $f(0) = $ . . . . . .

If $f(x) = \begin{cases} [x] + [-x], & x \neq 2 \\ \lambda, & x = 2 \end{cases}$ is continuous at $x = 2$,then $\lambda = $ (where $[.]$ denotes the greatest integer function).

Find all points of discontinuity of $f$,where $f$ is defined by $f(x) = \begin{cases} x + 1, & \text{if } x \ge 1 \\ x^2 + 1, & \text{if } x < 1 \end{cases}$. Is $f$ a continuous function?