Question

The set of values of $x$ such that ${\text{Tan}}^{-1}\left(\frac{x}{x-2}\right)-{\text{Tan}}^{-1}\left(\frac{x}{2x-1}\right)={\text{Tan}}^{-1}\left(\frac{2}{3}\right)$ is

• A. $\varphi$
• B. $\left\{\frac{1}{2}\right\}$
• C. $\left\{\frac{1}{3},2\right\}$
• D. $\left\{\frac{1}{3},4\right\}$

Answer 1

Answer: (D)

We have,
${\tan }^{-1}\left(\frac{x}{x-2}\right)-{\tan }^{-1}\left(\frac{x}{2x-1}\right)={\tan }^{-1}\left(\frac{2}{3}\right)$
${\tan }^{-1}\frac{\left(\frac{x}{x-2}-\frac{x}{2x-1}\right)}{1+\left(\frac{x}{x-2}\right)\left(\frac{x}{2x-1}\right)}$
$={\tan }^{-1}\frac{2}{3}$
$\Rightarrow \frac{2x^2-x-x^2+2x}{2x^2-4x-x+2+x^2}=\frac{2}{3}$
$\Rightarrow \frac{x^2+x}{3x^2-5x+2}=\frac{2}{3}$
$\Rightarrow 3x^2+3x=6x^2-10x+4$
$\Rightarrow 3x^2-13x+4=0$
$\Rightarrow (3x-1)(x-4)=0$
$\Rightarrow x=\frac{1}{3},4,x\in \left\{\frac{1}{3},4\right\}$