Question

Let $S_n={\cot }^{-1}(3x+\frac{2}{x})+{\cot }^{-1}(6x+\frac{2}{x})+{\cot }^{-1}(10x+\frac{2}{x})+\cdots +\text{upto}n\text{terms},$ where $x>0.$ If $\underset{n\to \infty }{lim}{S}_n=1,$ then $x$ equals to

• A. $\frac{\pi }{4}$
• B. $1$
• C. $\tan 1$
• D. $\cot 1$

Answer 1

The correct option is D $\cot 1$

Here, $T_n={\cot }^{-1}(\frac{(n+1)(n+2)}{2}x+\frac{2}{x})$
$\Rightarrow T_n={\tan }^{-1}\left(\frac{2x}{(n+2)(n+1)x^2+4}\right)$
$\Rightarrow T_n={\tan }^{-1}\left(\frac{\left(\frac{n+2}{2}\right)x-\left(\frac{n+1}{2}\right)x}{1+\left(\frac{n+1}{2}\right)\left(\frac{n+2}{2}\right)x^2}\right)$
$\therefore T_n={\tan }^{-1}\left(\left(\frac{n+2}{2}\right)x\right)-{\tan }^{-1}\left(\left(\frac{n+1}{2}\right)x\right)$

$S_n=\sum T_n$
So, $S_n={\tan }^{-1}\left(\left(\frac{n+2}{2}\right)x\right)-{\tan }^{-1}x$
$\Rightarrow \underset{n\to \infty }{lim}{S}_n=\frac{\pi }{2}-{\tan }^{-1}x={\cot }^{-1}x=1$
$\Rightarrow x=\cot 1$