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

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(D) For $f(x)$ to be continuous at $x = \frac{\pi}{4}$, we must have $\lim_{x 	o \frac{\pi}{4}} f(x) = f\left(\frac{\pi}{4}ight)$.Given $f\left(\fr

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$1/4$

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

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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