Question

Let $f\left(x\right)=\left(\frac{\tan (\frac{\pi }{4}-x)}{\cot 2x}\right)(x\ne \frac{\pi }{4})$ The value which should be assigned to f at $\frac{\pi }{4}$ so that it is continuous everywhere, is

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

Answer 1

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

By applying L hospital's rule we get

$\underset{x\to \frac{\pi }{4}}{lim}f\left(x\right)=\frac{{\sec }^2(\frac{\pi }{4}-x)}{2{\csc }^{2}2x}=\frac{1}{2}$