Question

The number of solutions of the equation ${\tan }^{-1}\left(\frac{x}{3}\right)+{\tan }^{-1}\left(\frac{x}{2}\right)={\tan }^{-1}x$ is

• A. 3
• B. 2
• C. 1
• D. 0

Answer 1

Answer: (A)

${\tan }^{-1}\left(\frac{x}{3}\right)+{\tan }^{-1}\left(\frac{x}{2}\right)={\tan }^{-1}x$
or ${\tan }^{-1}\left(\frac{x/3+x/2}{1-x^2/6}\right)={\tan }^{-1}x$
where $x>0\&x^2/6<1\Rightarrow x^2<6\Rightarrow -\sqrt{6}<x<\sqrt{6}$
now, $\left(\frac{5x}{6-x^2}\right)=x\Rightarrow x\left[\frac{5}{6-x^2}-1\right]=0$
$\Rightarrow x=0\text{ or }{x}^2-1=0\Rightarrow x=\pm 1$
$\therefore x=\{-1,0,1\}\Rightarrow 3\text{ solution }$