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

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Inverse Trigonometric Functions

The number of solutions of

tan^{- 1} 4 x + tan^{- 1} 6 x = \frac{π}{6}

, where

- \frac{1}{2 6} < x < \frac{1}{2 6}

is equal to

Answer: (C)

In the given interval

24 x^{2} < 1

, so

tan^{- 1} 4 x + tan^{- 1} 6 x = tan^{- 1} \frac{10 x}{1 - 24 x ^{2}}

. Setting this to

\frac{π}{6}

gives

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

, i.e.

24 x^{2} + 103 x - 1 = 0

, with roots

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

. The negative root

\approx - 0.78

lies outside

(- \frac{1}{2 6}, \frac{1}{2 6})

and would make the left side negative, while the positive root

\approx 0.054

lies inside and satisfies the equation. There is exactly one solution.

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Single correct option

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