Question

The function $f \left(x\right)=\left(sin \,2x\right)^{tan^2\,2x}$ is not defined at $x=\frac{\pi}{4}$. The value of $f \left(\frac{\pi}{4}\right)$ so that f is continuous at $x=\frac{\pi}{4}$ is

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

Answer 1

Answer: (A)

f is continuous at $x=\pi /4$, if $\displaystyle \lim_{x \to \pi/4}f (x)=f (\pi/4)$.
Now, $=\displaystyle \lim_{x \to \pi/4}(sin\,2x)^{tan^2\,2x}$
$\Rightarrow log L=\displaystyle \lim_{x \to \pi/4}tan^2\,2x\,log\,sin\,2x$
$=\displaystyle \lim_{x \to \pi/4}$$\frac{log\,sin\,2x}{cot^{2}\,2x}\left(\frac{\infty}{\infty}\right)$
$=\displaystyle \lim_{x \to \pi/4}$$\frac{2\,cot\,2x}{2\,cot\,2x\,cos\,ec^{2}\,2x.2}=-\frac{1}{2}$
or $L=e^{-1/2}\, \therefore f \left(\pi/4\right)=e^{-1/2}=1/\sqrt{e}$