Question

The function $f\left(x\right)=\frac{\tan (\frac{\pi }{4}-x)}{\cot 2x},x\ne \frac{\pi }{4}$ then the value which should be assigned to $f$ at $x=\frac{\pi }{4}$ 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 $\frac{1}{2}$

Given $f\left(x\right)=\frac{\tan (\frac{\pi }{4}-x)}{\cot 2x}$ $x\ne \frac{\pi }{4}$
$f\left(x\right)$ to be continuous at $x=\frac{\pi }{4}$
$f(\frac{\pi }{4}-h)=f(\frac{\pi }{4}-h)=f\left(\frac{\pi }{4}\right)$
$RHL={lim}_{x\to \pi /4^{+}}f\left(x\right)$
$={lim}_{h\to 0}f(\frac{\pi }{4}-h)$
$={lim}_{h\to 0}\frac{\tan (\frac{\pi }{4}-(\frac{\pi }{4}-h\left)\right)}{\cot 2(\frac{\pi }{4}+h)}$
${lim}_{h\to 0}\frac{-\tan h}{\cot (\frac{\pi }{2}+2h)}$
$={lim}_{h\to 0}\frac{-\tan h}{-\tan 2h}$
${lim}_{h\to 0}=\frac{\frac{\tan h}{h}}{2\left(\frac{\tan 2h}{2h}\right)}=\frac{1}{2}$
$\Rightarrow k=\frac{1}{2}$