Question

If $ta{n}^{-1}(x-1)+ta{n}^{-1}(x+1)+ta{n}^{-1}x=ta{n}^{-1}3x$, then in the interval $[-1,1]$, then x has :

• A. one value
• B. 2 values
• C. 3 values
• D. No value

Answer 1

The correct option is C 3 values

Given equation $\Rightarrow ta{n}^{-1}(x-1)+ta{n}^{-1}(x+1)=ta{n}^{-1}3x-ta{n}^{-1}x$
$\Rightarrow ta{n}^{-1}\frac{(x-1)+(x+1)}{1-(x-1)(x+1)}=ta{n}^{-1}\frac{3x-x}{1+3x.x}$
$\Rightarrow \frac{2x}{2-x^2}=\frac{2x}{1+3x^2}$
$\Rightarrow 2x(4x^2-1)=0\Rightarrow x=0,\pm \frac{1}{2}$
All three solutions satisfy the given equation.