Question

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

• A. $\frac{1}{4}$
• B. $\frac{1}{2\sqrt{2}}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\sqrt{2}$

Answer 1

The correct option is A $\frac{1}{4}$

For continuity of $f\left(x\right)$ at $x=\frac{\pi }{4}$, we must have
$f\left(\frac{\pi }{4}\right)=\underset{x\to \frac{\pi }{4}}{lim}f\left(x\right)$
$\Rightarrow a=\underset{x\to \frac{\pi }{4}}{lim}\frac{1-\sqrt{2}\sin x}{\pi -4x}$
$=\underset{x\to \frac{\pi }{4}}{lim}\frac{-\sqrt{2}\cos x}{-4}=\underset{x\to \frac{\pi }{4}}{lim}\frac{\cos x}{2\sqrt{2}}$
$=\frac{1}{4}$