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

For $f(x)$ to be continuous at $x=\pi /4,$ $f(\pi /4)$ must be equal to the limit as $x\to \pi /4$.
$f(\pi /4)={\lim }_{x\to \frac{\pi }{4}}\frac{1-\tan x}{4x-\pi }$.
This is a $0/0$ form. Applying L'Hopital's rule:
$f(\pi /4)={\lim }_{x\to \frac{\pi }{4}}\frac{-{\sec }^{2}x}{4}$.
$f(\pi /4)=\frac{-{\sec }^2(\pi /4)}{4}=\frac{-(\sqrt{2}{)}^2}{4}=\frac{-2}{4}=-\frac{1}{2}$.