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 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 $|2x|>\sqrt{3}$

Given, $f\left(x\right)=\underset{n\to \infty }{lim}\frac{1}{{\left(\frac{3}{\pi }ta{n}^{-1}2x\right)}^{2n}+5}=0$

$or{\left[\left(\frac{3}{\pi }{\tan }^{-1}2x\right)\right]}^n\to \infty$

$or\left(\frac{3}{\pi }{\tan }^{-1}2x\right)>1$

$or|{\tan }^{-1}2x|>\frac{\pi }{3}$

$\Rightarrow {\tan }^{-1}2x<-\frac{\pi }{3}or{\tan }^{-1}2x>\frac{\pi }{3}$

$\Rightarrow 2x<-\sqrt{3}\text{or}2x>\sqrt{3},i.e.|2x|>\sqrt{3}$