Question

Let $x_i,i=1,2,3,\dots ,n$ be the solutions of the equation ${\tan }^{-1}x+{\cot }^{-1}(-\left|x\right|)=2{\tan }^{-1}6x.$ Then $6\sum _{i=1}^n{x}_i$ is equal to

Answer 1

Case $1:$ When $x>0$
${\tan }^{-1}x+{\cot }^{-1}(-x)=2{\tan }^{-1}6x$
$\Rightarrow {\tan }^{-1}x+\pi -{\cot }^{-1}x=2{\tan }^{-1}6x$
$[\because {\cot }^{-1}(-x)=\pi -{\cot }^{-1}x\forall x\in R]$
$\Rightarrow \frac{\pi }{2}+2{\tan }^{-1}x=2{\tan }^{-1}6x$
$\Rightarrow {\tan }^{-1}6x-{\tan }^{-1}x=\frac{\pi }{4}$
$\Rightarrow {\tan }^{-1}\left(\frac{6x-x}{1+6x\cdot x}\right)=\frac{\pi }{4}$
$\Rightarrow \frac{5x}{1+6x^2}=1$
$\Rightarrow 6x^2-5x+1=0$
$\Rightarrow x=\frac{1}{2},\frac{1}{3}$

Case $2:$ When $x<0$
${\tan }^{-1}x+{\cot }^{-1}x=2{\tan }^{-1}6x$
$\Rightarrow {\tan }^{-1}6x=\frac{\pi }{4}$
This is not possible as $x<0$

$6\sum _{i=1}^n{x}_i=6(\frac{1}{2}+\frac{1}{3})=5$