JEE Advanced 2023 Paper 2 Q03 Mathematics Trigonometry Inverse Trigonometric Functions Hard

JEE Advanced 2023 Paper 2 · Q03 · Inverse Trigonometric Functions

For any $y \in \mathbb{R}$, let $\cot^{-1}(y) \in (0, \pi)$ and $\tan^{-1}(y) \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$. Then the sum of all the solutions of the equation $\tan^{-1}\left(\dfrac{6y}{9 - y^2}\right) + \cot^{-1}\left(\dfrac{9 - y^2}{6y}\right) = \dfrac{2\pi}{3}$ for $0 < |y| < 3$ is equal to

  1. A. $2\sqrt{3} - 3$
  2. B. $3 - 2\sqrt{3}$
  3. C. $4\sqrt{3} - 6$
  4. D. $6 - 4\sqrt{3}$
Reveal answer + step-by-step solution

Correct answer:C

Solution

When $0 < y < 3$: $9 - y^2 > 0$ and $6y/(9-y^2)$ is positive, so $\tan^{-1} + \cot^{-1}$ of positive reciprocals = $\pi/2$, but here we get $2\pi/3$, which means the relation $\cot^{-1}(z) = \tan^{-1}(1/z)$ must be adjusted by $\pi$ in one branch. Reduces to $\tan^{-1}(6y/(9-y^2)) = \pi/3$ in one case and a related equation in the other. Solving: $6y/(9-y^2) = \sqrt{3}$ gives $\sqrt{3}y^2 + 6y - 9\sqrt{3} = 0$, with positive root. The other case (negative $y$) gives a similar quadratic with one root in $(-3, 0)$. Sum of valid roots = $4\sqrt{3} - 6$.

Want to drill deeper? Open this question inside SolveKar AI and tap "Why?" on any step — the AI Mentor reads your full solution context and explains until it clicks. Download SolveKar AI →