Question

The function $f\left(x\right)={\left(sin2x\right)}^{ta{n}^{2}2x}$ 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 $\underset{x\to \pi /4}{\lim }f(x)=f(\pi /4)$.
Now, $=\underset{x\to \pi /4}{\lim }(sin2x{)}^{ta{n}^{2}2x}$
$\Rightarrow logL=\underset{x\to \pi /4}{\lim }ta{n}^{2}2xlogsin2x$
$=\underset{x\to \pi /4}{\lim }$$\frac{logsin2x}{co{t}^{2}2x}\left(\frac{\infty }{\infty }\right)$
$=\underset{x\to \pi /4}{\lim }$$\frac{2cot2x}{2cot2xcose{c}^{2}2x.2}=-\frac{1}{2}$
or $L=e^{-1/2}\therefore f\left(\pi /4\right)=e^{-1/2}=1/\sqrt{e}$