Question

Let $f\left(x\right)=\frac{\tan (\frac{\pi }{4}-x)}{\cot 2x},x\ne \frac{\pi }{4}.$ If $f\left(x\right)$ is continuous at $x=\frac{\pi }{4},$ then the value of $f\left(\frac{\pi }{4}\right)$ is

• A. $1$
• B. $\frac{1}{2}$
• C. $2$
• D. None of these

Answer 1

The correct option is B $\frac{1}{2}$

As $f\left(x\right)$ is continuous at $x=\frac{\pi }{4}$
$\therefore f\left(\frac{\pi }{4}\right)=\underset{x\to \frac{\pi }{4}}{lim}f\left(x\right)$
$\underset{x\to \frac{\pi }{4}}{lim}f\left(x\right)=\underset{x\to \frac{\pi }{4}}{lim}\frac{\tan (\frac{\pi }{4}-x)}{\cot 2x}$
Using L Hospital's rule
$=\underset{x\to \frac{\pi }{4}}{lim}\frac{-{\sec }^2(\frac{\pi }{4}-x)}{-2{\text{cosec}}^{2}2x}=\frac{1}{2}$
$\therefore f\left(\frac{\pi }{4}\right)=\frac{1}{2}$