If f(x) = begin{cases} frac{1-sqrt{2} sin x}{ | vedclass

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(D) Since $f(x)$ is continuous at $x = \frac{\pi}{4}$,we must have $\lim_{x \rightarrow \frac{\pi}{4}} f(x) = f\left(\frac{\pi}{4}\right)$.$\lim_{x \r

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$1/4$

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(D) Since $f(x)$ is continuous at $x = \frac{\pi}{4}$,we must have $\lim_{x \rightarrow \frac{\pi}{4}} f(x) = f\left(\frac{\pi}{4}\right)$.$\lim_{x \rightarrow \frac{\pi}{4}} f(x) = \lim_{x \rightarrow \frac{\pi}{4}} \frac{1-\sqrt{2} \sin x}{\pi-4x}$.This is a $\frac{0}{0}$ form,so we apply $L$'Hospital's rule:$\lim_{x \rightarrow \frac{\pi}{4}} \frac{\frac{d}{dx}(1-\sqrt{2} \sin x)}{\frac{d}{dx}(\pi-4x)} = \lim_{x \rightarrow \frac{\pi}{4}} \frac{-\sqrt{2} \cos x}{-4}$.Substituting $x = \frac{\pi}{4}$:$= \frac{-\sqrt{2} \cdot \frac{1}{\sqrt{2}}}{-4} = \frac{-1}{-4} = \frac{1}{4}$.Since $f\left(\frac{\pi}{4}\right) = a$,we have $a = \frac{1}{4}$.

If $f(x) = \begin{cases} \frac{1-\sqrt{2} \sin x}{\pi-4x} & \text{if } x \neq \frac{\pi}{4} \\ a & \text{if } x = \frac{\pi}{4} \end{cases}$ is continuous at $x = \frac{\pi}{4}$,then $a$ is equal to

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