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{5 x}{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 $
$\therefore x=\{-1,0,1\} \Rightarrow 3 \text { solution }$