If $f(x)=\left\{\begin{matrix}\frac{1-\tan x}{4x-π}&x≠\frac{π}{4}\\k&x=\frac{π}{4}\end{matrix}\right.$ is continuous at $x=\frac{π}{4}$ then the value of $k$ is

$\frac{-1}{2}$

The correct answer is Option (4) → $\frac{-1}{2}$

Given:

$f(x) = \begin{cases} \frac{1 - \tan x}{4x - \pi} & \text{if } x \ne \frac{\pi}{4} \\ k & \text{if } x = \frac{\pi}{4} \end{cases}$

To ensure that $f(x)$ is continuous at $x = \frac{\pi}{4}$, we must have:

$ \lim\limits_{x \to \frac{\pi}{4}} f(x) = f\left(\frac{\pi}{4}\right) = k $

Step 1: Compute the limit as $x \to \frac{\pi}{4}$

Let us simplify the expression:

$\lim\limits_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{4x - \pi}$

At $x = \frac{\pi}{4}$:

$\tan \left( \frac{\pi}{4} \right) = 1$, so numerator $1 - \tan x \to 0$ and $4x - \pi \to 0$

This is an indeterminate form $\frac{0}{0}$, so apply L'Hôpital’s Rule:

$ = \lim\limits_{x \to \frac{\pi}{4}} \frac{ \frac{d}{dx} (1 - \tan x) }{ \frac{d}{dx}(4x - \pi) } = \frac{ -\sec^2 x }{4} $

Now evaluate at $x = \frac{\pi}{4}$:

$ \sec \left( \frac{\pi}{4} \right) = \sqrt{2} \Rightarrow \sec^2 \left( \frac{\pi}{4} \right) = 2 $

$\Rightarrow \frac{ -\sec^2 x }{4} = \frac{ -2 }{4 } = -\frac{1}{2}$

Step 2: Use continuity condition

$k = \lim\limits_{x \to \frac{\pi}{4}} f(x) = -\frac{1}{2}$