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

Option: 1

\sqrt{e}


Option: 2

1


Option: 3

2


Option: 4

none of these


Answers (1)

best_answer

fis continuous at \mathrm{x=\frac{\pi}{4}} if \mathrm{\lim _{x \rightarrow \frac{\pi}{4}} f(x)=f\left(\frac{\pi}{4}\right)}

\mathrm{L=\lim _{x \rightarrow \frac{\pi}{4}}(\sin 2 x)^{\tan ^2 2 x}}

Now

\mathrm{\Rightarrow \log \mathrm{L} =\lim _{x \rightarrow \frac{\pi}{4}} \tan ^2 2 x \log \sin 2 x=\lim _{x \rightarrow \frac{\pi}{4}} \frac{\log \sin 2 x}{\cot ^2 2 x} \quad\left(\frac{\infty}{\infty}\right)}
\mathrm{=\lim _{x \rightarrow \frac{\pi}{4}} \frac{2 \cot 2 x}{2 \cot 2 x \operatorname{cosec} 2 x \cdot 2}=-\frac{1}{2}}

\mathrm{\therefore \quad \mathrm{L}=\mathrm{e}^{-1 / 2} \Rightarrow f\left(\frac{\pi}{4}\right)=e^{-1 / 2}=\frac{1}{\sqrt{e}}}

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