The function $f(x)=(\sin 2x)^{\tan^22x}$ is not defined at $x=\frac{π}{4}$. Then value of $f(\frac{π}{4})$ so that f is continuous at $x=\frac{π}{4}$ is
Answer & explanation
Correct answer: option 4
f is continuous at $x=\frac{π}{4}$ if $\underset{x→\frac{π}{4}}{\lim}f(x)=f(\frac{π}{4})$
Now $L=\underset{x→\frac{π}{4}}{\lim}(\sin 2x)^{\tan^22x}$
$⇒\log L=\underset{x→\frac{π}{4}}{\lim}tan^22x\log\sin 2x=\underset{x→\frac{π}{4}}{\lim}\frac{\log\sin 2x}{\cot^22x}$ $(\frac{∞}{∞})$
$=\underset{x→\frac{π}{4}}{\lim}\frac{2\cot 2x}{-2\cot 2x\,cosec^22x.2}=-\frac{1}{2}$
$∴L=e^{-1/2}⇒f(\frac{π}{4})=e^{-1/2}=\frac{1}{\sqrt{e}}$