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(A) Given that the function $f(x)$ is continuous at $x = \frac{\pi}{4}$,we must have $f(\frac{\pi}{4}) = \lim_{x 	o \frac{\pi}{4}} f(x)$.Since $f(\fr

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$\frac{1}{4}$

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(A) Given that the function $f(x)$ is continuous at $x = \frac{\pi}{4}$,we must have $f(\frac{\pi}{4}) = \lim_{x 	o \frac{\pi}{4}} f(x)$.Since $f(\frac{\pi}{4}) = k$,we evaluate the limit:$k = \lim_{x 	o \frac{\pi}{4}} \frac{1-\sqrt{2} \sin x}{\pi-4 x}$.This is a $\frac{0}{0}$ indeterminate form. Applying $L$'$H$ôpital's Rule by differentiating the numerator and denominator with respect to $x$:$k = \lim_{x 	o \frac{\pi}{4}} \frac{\frac{d}{dx}(1-\sqrt{2} \sin x)}{\frac{d}{dx}(\pi-4 x)}$$k = \lim_{x 	o \frac{\pi}{4}} \frac{-\sqrt{2} \cos x}{-4}$Substituting $x = \frac{\pi}{4}$:$k = \frac{-\sqrt{2} \cos(\frac{\pi}{4})}{-4} = \frac{-\sqrt{2} 	imes \frac{1}{\sqrt{2}}}{-4} = \frac{-1}{-4} = \frac{1}{4}$.

If a function $f$ is defined by $f(x) = \begin{cases} \frac{1-\sqrt{2} \sin x}{\pi-4 x}, & x \neq \frac{\pi}{4} \\ k, & x = \frac{\pi}{4} \end{cases}$ and is continuous at $x = \frac{\pi}{4}$,then $k = $

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