The number of solutions of the equation

(a) 2

We know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$

$\therefore \tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}$

$\Rightarrow \tan ^{-1}\left(\frac{2 x+3 x}{1-2 x \times 3 x}\right)=\frac{\pi}{4}$

$\Rightarrow \frac{2 x+3 x}{1-2 x \times 3 x}=\tan \frac{\pi}{4}$

$\Rightarrow \frac{5 x}{1-6 x^{2}}=1$

$\Rightarrow 5 x=1-6 x^{2}$

$\Rightarrow 6 x^{2}+5 x-1=0$

Therefore, there are two solutions.