If the function f defined on \(\left(\frac{\pi}{6}, \frac{\pi}{3}\right)\) by
\(f(x)=\left\{\begin{array}{cc}
\frac{\sqrt{2} \cos x-1}{\cot x-1}, & x \neq \frac{\pi}{4} \
\mathrm{k}, & x=\frac{\pi}{4}
\end{array}\right.\)
is continuous, then k is equal to

\(\mathrm{f}(x)\) is continuous on \(\left(\frac{\pi}{6}, \frac{\pi}{3}\right)\).
\(\Rightarrow \mathrm{f}(x)\) is continuous at \(x=\frac{\pi}{4}\).
\(\therefore \quad \mathrm{f}\left(\frac{\pi}{4}\right)=\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2} \cos x-1}{\cot x-1}\)
Applying L'Hospital's rule on R.H.S., we get
\(\begin{aligned} f\left(\frac{\pi}{4}\right) & =\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2}(-\sin x)}{-\operatorname{cosec}^2 x} \\ & =\frac{\sqrt{2} \sin \frac{\pi}{4}}{\operatorname{cosec}^2 \frac{\pi}{4}} \\ & =\frac{\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)}{(\sqrt{2})^2} \\ & =\frac{1}{2}\end{aligned}\)