The number of positive solutions satisfying the equation $ tan^{-1}\left(\frac{1}{2x+1}\right) + tan^{-1}\left(\frac{1}{4x+1}\right) = tan^{-1}\left(\frac{2}{x^2}\right)$, is
Answer & explanation
Correct answer: option 1
We have,
$ tan^{-1}\left(\frac{1}{2x+1}\right) + tan^{-1}\left(\frac{1}{4x+1}\right) = tan^{-1}\left(\frac{2}{x^2}\right)$
$⇒ tan^{-1}\begin{Bmatrix}\frac{\frac{1}{2x+1}+\frac{1}{4x+1}}{1-\frac{1}{(2x+1)(4x+1)}}\end{Bmatrix}=tan^{-1}(\frac{2}{x^2})$
$⇒ tan^{-1}\left(\frac{3x+1}{4x^2+3x}\right) = tan^{-1}(\frac{2}{x^2})$
$⇒ \frac{3x+1}{4x^2+3x}=\frac{2}{x^2}$
$⇒3x^2 - 7x - 6 = 0 ⇒ x = -\frac{2}{3},3$
3 is the only positive solution.