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

Question

Let $f(x) = {{1 - \tan x} \over {4x - \pi }}$, $x \ne {\pi \over 4}$, $x \in \left[ {0,{\pi \over 2}} \right]$.

If $f(x)$ is continuous in $\left[ {0,{\pi \over 2}} \right]$, then $f\left( {{\pi \over 4}} \right)$ is