1. We are given $f(x)=\frac{1-\tan x}{4x-\pi}, \qquad x\ne \frac{\pi}{4}, \quad x\in \left[0,\frac{\pi}{2}\right].$

   For $f(x)$ to be continuous on the entire interval $\left[0,\frac{\pi}{2}\right]$, it must also be continuous at $x=\frac{\pi}{4}$.

2. Since the formula is not defined at $x=\frac{\pi}{4}$, we must define $f\left(\frac{\pi}{4}\right)=\lim_{x\to \pi/4} \frac{1-\tan x}{4x-\pi}.$

3. Evaluate the limit: $\lim_{x\to \pi/4} \frac{1-\tan x}{4x-\pi}.$

   Substitute $x=\frac{\pi}{4}$ in numerator and denominator: $1-\tan\frac{\pi}{4}=1-1=0,$ $4\cdot \frac{\pi}{4}-\pi=\pi-\pi=0.$

   So this is a $\frac{0}{0}$ form, and we apply L'Hospital's Rule.

4. Differentiate numerator and denominator: $\frac{d}{dx}(1-\tan x)=-\sec^2 x,$ $\frac{d}{dx}(4x-\pi)=4.$

   Therefore,

   =\lim_{x\to \pi/4} \frac{-\sec^2 x}{4}.$$

5. Now evaluate at $x=\frac{\pi}{4}$: $\sec^2\frac{\pi}{4}=2.$

   Hence, $\lim_{x\to \pi/4} \frac{-\sec^2 x}{4}=-\frac{2}{4}=-\frac{1}{2}.$

6. Therefore, for continuity, $f\left(\frac{\pi}{4}\right)=-\frac{1}{2}.$

7. Checking options:

   • A: $-1$ ❌
   • B: $\frac{1}{2}$ ❌
   • C: $-\frac{1}{2}$ ✅
   • D: $1$ ❌

So the correct answer is Option C.