If the function f defined on left(frac{pi}{6} | vedclass

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(A) Since $f(x)$ is continuous on $\left(\frac{\pi}{6}, \frac{\pi}{3}\right)$,it must be continuous at $x=\frac{\pi}{4}$.Therefore,$f\left(\frac{\pi}{

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$\frac{1}{2}$

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(A) Since $f(x)$ is continuous on $\left(\frac{\pi}{6}, \frac{\pi}{3}\right)$,it must be continuous at $x=\frac{\pi}{4}$.Therefore,$f\left(\frac{\pi}{4}\right) = \lim_{x \rightarrow \frac{\pi}{4}} f(x) = \lim_{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2} \cos x-1}{\cot x-1}$.This is a $\frac{0}{0}$ indeterminate form.Applying $L$'Hospital's rule,we differentiate the numerator and denominator with respect to $x$:$k = \lim_{x \rightarrow \frac{\pi}{4}} \frac{\frac{d}{dx}(\sqrt{2} \cos x - 1)}{\frac{d}{dx}(\cot x - 1)} = \lim_{x \rightarrow \frac{\pi}{4}} \frac{-\sqrt{2} \sin x}{-\operatorname{cosec}^2 x} = \lim_{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2} \sin x}{\operatorname{cosec}^2 x}$.Substituting $x = \frac{\pi}{4}$:$k = \frac{\sqrt{2} \sin(\frac{\pi}{4})}{\operatorname{cosec}^2(\frac{\pi}{4})} = \frac{\sqrt{2} \cdot \frac{1}{\sqrt{2}}}{(\sqrt{2})^2} = \frac{1}{2}$.

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

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