Question

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

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