Question

Let $S_n={\cot }^{-1}\left(3x+\frac{2}{x}\right)+{\cot }^{-1}\left(6x+\frac{2}{x}\right)+{\cot }^{-1}\left(10x+\frac{2}{x}\right)+\dots \dots \dots \dots +n$ terms, where $x>0$. If $\underset{n\to \infty }{\mathrm{Lim}}{S}_n=1$, then $x$ equals

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

Answer 1

Answer: (D)

As, $T_n={\cot }^{-1}\left[\frac{(n+1)(n+2)}{2}x+\frac{2}{x}\right]\Rightarrow T_n={\tan }^{-1}\left(\frac{2x}{(n+2)(n+1)x^2+4}\right)$
$\therefore T_n={\tan }^{-1}\left(\left(\frac{n+2}{2}\right)x\right)-{\tan }^{-1}\left(\left(\frac{n+1}{2}\right)x\right)$
So, $S_n={\tan }^{-1}\left(\left(\frac{n+2}{2}\right)x\right)-{\tan }^{-1}x$
$\Rightarrow \underset{n\to \infty }{\mathrm{Lim}}{S}_n=\frac{\pi }{2}-{\tan }^{-1}x={\cot }^{-1}x=1$ (Given $)\Rightarrow x=\cot 1$.