Question

If $f\left(x\right)=\frac{\tan (\pi /4-x)}{\cot 2x}$ then it cuts out at $x=\frac{\pi }{4}$

• A. True
• B. False

Answer 1

The correct option is A True

We have,

$f\left(x\right)=\frac{\tan (\pi /4-x)}{\cot 2x}$

Since, it cuts at $x=\frac{\pi }{4}$

Then,

${lim}_{x\to \frac{\pi }{4}}f\left(x\right)$

Therefore,

${lim}_{x\to \frac{\pi }{4}}\frac{\tan (\pi /4-x)}{\cot 2x}$

Apply L-hospital rule,

${lim}_{x\to \frac{\pi }{4}}\frac{-{\sec }^2(\pi /4-x)}{-2{\csc }^{2}2x}$

${lim}_{x\to \frac{\pi }{4}}\frac{{\sec }^2(\pi /4-x)}{2{\csc }^{2}2x}$

$=\frac{{\sec }^2(\pi /4-\pi /4)}{2{\csc }^2\frac{\pi }{2}}$

$=\frac{{\sec }^{2}0}{2{\csc }^2\frac{\pi }{2}}$

$=\frac{1^2}{2\times 1^2}$

$=\frac{1}{2}$

Hence, this is the answer.