Aspire Faculty ID #17952 · Topic: JEE Main 2026 (22 January Morning Shift) · Just now
JEE Main 2026 (22 January Morning Shift)

The number of solutions of $\tan^{-1}(4x) + \tan^{-1}(6x) = \frac{\pi}{6}$, where $-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$, is equal to

Solution

$\tan^{-1}(4x) + \tan^{-1}(6x) = \frac{\pi}{6}$

$\Rightarrow \tan^{-1}\left(\frac{4x + 6x}{1 - 24x^2}\right) = \frac{\pi}{6}$

$\Rightarrow \frac{10x}{1 - 24x^2} = \frac{1}{\sqrt{3}}$

$\Rightarrow 24x^2 + 10\sqrt{3}x - 1 = 0$

$x = \frac{-10\sqrt{3} \pm \sqrt{300 + 96}}{48}$

$x = \frac{-10\sqrt{3} \pm \sqrt{396}}{48}$

Only one solution lies in interval $\left(-\frac{1}{2\sqrt{6}}, \frac{1}{2\sqrt{6}}\right)$

$\Rightarrow$ number of solutions $= 1$

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