If $f(x)=\left\{\begin{matrix}\frac{tan(\frac{\pi}{4}-x)}{cot2x}, & x≠\frac{\pi}{4}\\K , &x=\frac{\pi}{4}\end{matrix}\right.$ is continuous at $x=\frac{\pi}{4}$, then the value of K is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\frac{1}{2}$
$f(x)=\left\{\begin{matrix}\frac{\tan(\frac{\pi}{4}-x)}{\cot 2x}, & x≠\frac{\pi}{4}\\K , &x=\frac{\pi}{4}\end{matrix}\right.$
$f(\frac{\pi}{4})=k$, $\lim\limits{x→\frac{\pi}{4}}\frac{\tan(\frac{\pi}{4}-x)}{\cot 2x}$
as $\frac{\tan(\frac{\pi}{4}-x)}{\cot 2x}=\frac{1-\tan x}{1+\tan x}\frac{2\tan x}{1-\tan^2x}=\frac{2\tan x}{(1+\tan x)^2}$
$=\lim\limits{x→\frac{\pi}{4}}\frac{2\tan x}{(1+\tan x)^2}=\frac{2}{2^2}=\frac{1}{2}$