Question

Let $f\left(x\right)=\underset{n\to \infty }{lim}\frac{1}{{\left(\frac{3}{\pi }{\tan }^{-1}2x\right)}^{2n}+5}.$ Then the complete set of values of $x$ for which $f\left(x\right)=0$ is

• A. $|2x|>\sqrt{3}$
• B. $|2x|<\sqrt{3}$
• C. $|2x|\ge \sqrt{3}$
• D. $|2x|\le \sqrt{3}$

Answer 1

The correct option is A $|2x|>\sqrt{3}$

$f\left(x\right)=\underset{n\to \infty }{lim}\frac{1}{{\left(\frac{3}{\pi }{\tan }^{-1}2x\right)}^{2n}+5}=0$
$\Rightarrow {\left[{\left(\frac{3}{\pi }{\tan }^{-1}2x\right)}^2\right]}^n\to \infty$ as $n\to \infty$
$\Rightarrow {\left(\frac{3}{\pi }{\tan }^{-1}2x\right)}^2>1\Rightarrow \left|\frac{3}{\pi }{\tan }^{-1}2x\right|>1\Rightarrow |{\tan }^{-1}2x|>\frac{\pi }{3}$
$\Rightarrow {\tan }^{-1}2x<-\frac{\pi }{3}$ or ${\tan }^{-1}2x>\frac{\pi }{3}$
$\Rightarrow 2x<-\sqrt{3}$ or $2x>\sqrt{3}$
$\Rightarrow |2x|>\sqrt{3}$