If $f(x) =\left\{\begin{matrix}\frac{\tan(\frac{\pi}{4}-x)}{\cot 2x }&,x≠\frac{\pi}{4}\\2K+1&, x=\frac{\pi}{4}\end{matrix}\right.$ is continuous at $x=\frac{\pi}{4}$, then the value of K is equal to

$-\frac{1}{4}$

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

Need continuity at $x=\frac{\pi}{4}$, so

$\displaystyle \lim_{x\to\frac{\pi}{4}}\frac{\tan\!\left(\frac{\pi}{4}-x\right)}{\cot 2x}=2K+1$

Let $t=x-\frac{\pi}{4}$, then as $x\to\frac{\pi}{4}$, $t\to 0$ and

$\tan\!\left(\frac{\pi}{4}-x\right)=\tan(-t)=-\tan t$

$\cot 2x=\cot\!\left(\frac{\pi}{2}+2t\right)=-\tan 2t$

Hence the limit is

$\displaystyle \lim_{t\to 0}\frac{-\tan t}{-\,\tan 2t}=\lim_{t\to 0}\frac{\tan t}{\tan 2t}=\frac{1}{2}$

So $2K+1=\frac{1}{2}\Rightarrow 2K=-\frac{1}{2}\Rightarrow K=-\frac{1}{4}$

The value of $K$ is $-\frac{1}{4}$.