JEE Main 2026 Jan 22 Shift 1, Mathematics Q18: Inverse Trigonometric Functions

The given interval $|x|<\frac{1}{2\sqrt{6}}$ ensures $24x^2<1$, so the addition formula gives ${\tan }^{-1}\left(\frac{4x+6x}{1-24x^2}\right)=\frac{\pi }{6}$.

So $\frac{10x}{1-24x^2}=\frac{1}{\sqrt{3}}$, giving $24x^2+10\sqrt{3}x-1=0$.

$x=\frac{-10\sqrt{3}\pm \sqrt{300+96}}{48}=\frac{-10\sqrt{3}\pm \sqrt{396}}{48}$.

Numerically: $10\sqrt{3}\approx 17.32$, $\sqrt{396}\approx 19.90$. Two roots: about $0.054$ and $-0.775$.

The interval bound is $\frac{1}{2\sqrt{6}}\approx 0.204$. Only the positive root lies inside the interval, so exactly 1 solution.