If $f(x)=\left\{\begin{array}{cc}\frac{1-\sqrt{2} \sin x}{\pi-4 x}, & x \neq \frac{\pi}{4} \\ a, & x=\frac{\pi}{4}\end{array}\right.$ is continuous at $x=\frac{\pi}{4}$, then a =

1/4

It is given that f(x) is continuous at $x=\frac{\pi}{4}$

∴  $\lim\limits_{x \rightarrow \pi / 4} f(x)=f\left(\frac{\pi}{4}\right)$

$\Rightarrow \lim\limits_{x \rightarrow \pi / 4} \frac{1-\sqrt{2} \sin x}{\pi-4 x}=a$

$\Rightarrow \sqrt{2} \lim\limits_{x \rightarrow \pi / 4} \frac{\frac{1}{\sqrt{2}}-\sin x}{\pi-4 x}=a$

$\Rightarrow \frac{\sqrt{2}}{4} \lim\limits_{x \rightarrow \pi / 4} \frac{\sin \frac{\pi}{4}-\sin x}{\frac{\pi}{4}-x}=a$

$\Rightarrow \frac{1}{4 \sqrt{2}} \lim\limits_{x \rightarrow \pi / 4} \frac{2 \sin \left(\frac{\pi}{8}-\frac{x}{2}\right) \cos \left(\frac{\pi}{8}+\frac{x}{2}\right)}{\left(\frac{\pi}{8}-\frac{x}{2}\right)}=a$

$\Rightarrow \frac{1}{2 \sqrt{2}} \cos \frac{\pi}{4}=a \Rightarrow a=\frac{1}{4}$

Alternative 

$\lim\limits_{x \rightarrow \pi / 4} f(x)=f\left(\frac{\pi}{4}\right)$

$\Rightarrow \lim\limits_{x \rightarrow \pi / 4} \frac{1-\sqrt{2} \sin x}{\pi-4 x}=a$

$\Rightarrow \lim\limits_{x \rightarrow \pi / 4} \frac{-\sqrt{2} \cos x}{-4}=a$          [Using De L' Hospital's Rule]

$\Rightarrow \frac{1}{2 \sqrt{2}} \cos \frac{\pi}{4}=a \Rightarrow a=\frac{1}{4}$