If $\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}$, then number of solution(s) of the given equation is :

Two

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

taking tan on both sides

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

$= \frac{2 x+3 x}{1-2 x \times 3 x}=1$

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

$5 x=1-6 x^2$

$6 x^2+5 x-1=0$

factorising equation 

$6 x^2+6 x-x-1=0$

so  $6x(x + 1) - 1(x + 1) = 0$

$(6x - 1)(x + 1) = 0$

$x = \frac{1}{6},  -1$

Hence option 2 has two solutions.