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If $f(x)= \begin{cases}\frac{1-\sqrt{2} \sin x}{\pi-4 x} & \text { if } x \neq \frac{\pi}{4} \\ a & \text { if } x=\frac{\pi}{4}\end{cases}$ is continuous at $\frac{\pi}{4}$, then $a$ is equal to
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$1 / 4$
$\because \quad f(x)= \begin{cases}\frac{1-\sqrt{2} \sin x}{\pi-4 x}, & \text { if } x \neq \frac{\pi}{4} \\ a, & \text { if } x=\frac{\pi}{4}\end{cases}$
$\lim _{x \rightarrow \frac{\pi}{4}} f(x)=\lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\sqrt{2} \sin x}{\pi-4 x}$
$=\lim _{x \rightarrow \frac{\pi}{4}} \frac{-\sqrt{2} \cos x}{-4}$
(By L' Hospital's rule)
$=\frac{\sqrt{2} \cdot \frac{1}{\sqrt{2}}}{4}=\frac{1}{4}$
Since $f(x)$ is continuous at $x=\frac{\pi}{4}$.
$\therefore \quad \lim _{x \rightarrow \frac{\pi}{4}} f(x)=f\left(\frac{\pi}{4}\right)$
$\Rightarrow \quad \frac{1}{4}=a$
$\lim _{x \rightarrow \frac{\pi}{4}} f(x)=\lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\sqrt{2} \sin x}{\pi-4 x}$
$=\lim _{x \rightarrow \frac{\pi}{4}} \frac{-\sqrt{2} \cos x}{-4}$
(By L' Hospital's rule)
$=\frac{\sqrt{2} \cdot \frac{1}{\sqrt{2}}}{4}=\frac{1}{4}$
Since $f(x)$ is continuous at $x=\frac{\pi}{4}$.
$\therefore \quad \lim _{x \rightarrow \frac{\pi}{4}} f(x)=f\left(\frac{\pi}{4}\right)$
$\Rightarrow \quad \frac{1}{4}=a$
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