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 at $x=\frac{\pi }{4}$ is

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

Answer 1

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

It is given that the function is not defined at x = $\frac{\pi }{4}$
So, $f\left(\frac{\pi }{4}\right)$ =$\underset{x\to \frac{\pi }{4}}{Lim}\frac{cosx-sinx}{cos2x}$
Since, this is of the $\frac{0}{0}$ form, we apply L-Hospital's Rule,
$\underset{x\to \frac{\pi }{4}}{Lim}\frac{-sinx-cosx}{-2sin2x}$
Now, applying the limit,
$f\left(\frac{\pi }{4}\right)$ = $\frac{1}{2^{0.5}}$