Let $f(x)=\frac{1-\tan x}{4 x-\pi}, x \neq \frac{\pi}{4}, x \in\left[0, \frac{\pi}{2}\right)$ is continuous in $\left[0, \frac{\pi}{2}\right)$ then $f\left(\frac{\pi}{4}\right)$, is

$-\frac{1}{2}$

It is given that f(x) is continuous in $[0, \pi / 2)$. So, it is continuous at $x=\pi / 4$.

∴ $f\left(\frac{\pi}{4}\right)=\lim\limits_{x \rightarrow \pi / 4} f(x)$

$\Rightarrow f\left(\frac{\pi}{4}\right)=\lim\limits_{x \rightarrow \pi / 4} \frac{1-\tan x}{4 x-\pi}$

$\Rightarrow f\left(\frac{\pi}{4}\right) =\lim\limits_{x \rightarrow \pi / 4} \frac{\cos x-\sin x}{4 \cos x(x-\pi / 4)}$

$\Rightarrow f\left(\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{4} \lim\limits_{x \rightarrow \pi / 4} \frac{\left(\frac{1}{\sqrt{2}} \sin x-\frac{1}{\sqrt{2}} \cos x\right)}{\left(x-\frac{\pi}{4}\right) \cos x}$

$\Rightarrow f\left(\frac{\pi}{4}\right)=-\frac{1}{2 \sqrt{2}} \lim\limits_{x \rightarrow \pi / 4} \frac{\sin \left(x-\frac{\pi}{4}\right)}{\left(x-\frac{\pi}{4}\right)} \times \frac{1}{\cos x}$

$\Rightarrow f\left(\frac{\pi}{4}\right)=-\frac{1}{2 \sqrt{2}} \times 1 \times \sqrt{2}=-\frac{1}{2}$