The number of solutions of tan -1 4 x+tan -1 6 x=π/6 , where -frac 1 2√(6) < x < frac 1 2√(6) is equal to

Mathematics · Inverse Trigonometric Functions

JEE Main 2026 — 22 January, Morning Shift — Question 17

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

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

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

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

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

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

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

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

Answer key and solution verified before publishing.

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