Question

The number of solutions of the equation
$ta{n}^{-1}(x-1)+ta{n}^{-1}\left(x\right)+ta{n}^{-1}(x+1)=ta{n}^{-1}\left(3x\right)$ is :

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (C)

$3$

$ta{n}^{-1}x+ta{n}^{-1}(x+1)+ta{n}^{-1}(x-1)=ta{n}^{-1}\left(3x\right)$
$\Rightarrow ta{n}^{-1}x+ta{n}^{-1}\left(\frac{2x}{1-(x^2-1)}\right)=ta{n}^{-1}3x$
$\Rightarrow ta{n}^{-1}\left(\frac{2x}{2-x^2}\right)=ta{n}^{-1}3x-ta{n}^{-1}x$
$=ta{n}^{-1}\left(\frac{3x-x}{1+3x^2}\right)$
$\frac{2x}{2-x^2}=\frac{2x}{1+3x^2}$
if $x\ne 0$
$1+3x^2=2-x^2$
$4x^2=1$
$x=\pm \frac{1}{2}$
$\Rightarrow x=0,-\frac{1}{2},\frac{1}{2}$