Question

If $ta{n}^{-1}(x-1)+ta{n}^{-1}x+ta{n}^{-1}(x+1)=ta{n}^{-1}3x$, then the values of $x$ are

• A. $\pm \frac{1}{2}$
• B. $0,\frac{1}{2}$
• C. $0,-\frac{1}{2}$
• D. $0,\pm \frac{1}{2}$

Answer 1

Answer: (D)

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