Question

The function
$f\left(x\right)=\frac{\cos x-\sin x}{\cos 2x}$ is not defined at $x=\frac{\pi }{4}$. The value of $f\left(\frac{\pi }{4}\right)$ so that $f\left(x\right)$ is continuous everywhere, is

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

Answer 1

The correct option is D $\frac{1}{\sqrt{2}}$

The function will be continuous at $x=\frac{\pi }{4}$ if $f\left(\frac{\pi }{4}\right)=\underset{x\to \frac{\pi }{4}}{lim}f\left(x\right)$.
$f\left(\frac{\pi }{4}\right)=\underset{x\to \frac{\pi }{4}}{lim}f\left(x\right)=\underset{x\to \frac{\pi }{4}}{lim}\frac{\cos x-\sin x}{{\cos }^{2}x-{\sin }^{2}x}=\underset{x\to \frac{\pi }{4}}{lim}\frac{\cos x-\sin x}{(\cos x-\sin x)(\cos x+\sin x)}$

For $x\ne \frac{\pi }{4},\cos x-\sin x\ne 0$, so the limit becomes

$=\underset{x\to \frac{\pi }{4}}{lim}\frac{1}{\cos x+\sin x}=\frac{1}{\cos \frac{\pi }{4}+\sin \frac{\pi }{4}}=\frac{1}{\sqrt{2}}$