JEE Advanced 2023 Paper 2, Mathematics Section 1 Q3: Inverse Trigonometric Functions

JEE Advanced2023Paper 2Mathematics Section 1

For any $y \in R$ , let $cot^{- 1} (y) \in (0, π)$ and $tan^{- 1} (y) \in (- \frac{π}{2}, \frac{π}{2})$ . Then the sum of all the solutions of the equation

$tan^{- 1} (\frac{6 y}{9 - y ^{2}}) + cot^{- 1} (\frac{9 - y ^{2}}{6 y}) = \frac{2 π}{3}$

for $0 < ∣ y ∣ < 3$ , is equal to

A
$23 - 3$
B
$3 - 23$
C
$43 - 6$
D
$6 - 43$

Solution

Write $u = \frac{6 y}{9 - y ^{2}}$ . Then the second term equals $cot^{- 1} (1/ u)$ , which is $tan^{- 1} (u)$ if $u > 0$ and $π + tan^{- 1} (u)$ if $u < 0$ .

Case 1: $y \in (0, 3)$ . Here $u > 0$ , so the equation becomes $2 tan^{- 1} u = \frac{2 π}{3}$ , giving $u = tan \frac{π}{3} = 3$ . Thus $3 y^{2} + 6 y - 93 = 0$ , whose roots are $y = 3$ (accepted) and $y = - 33$ (rejected).

Case 2: $y \in (- 3, 0)$ . Here $u < 0$ , so the equation becomes $2 tan^{- 1} u + π = \frac{2 π}{3}$ , giving $tan^{- 1} u = - \frac{π}{6}$ and $u = - \frac{1}{3}$ . Thus $y^{2} - 63 y - 9 = 0$ , whose roots are $33 \pm 6$ . Only $y = 33 - 6 \approx - 0.8$ lies in $(- 3, 0)$ .

Sum of solutions $= 3 + 33 - 6 = 43 - 6.$

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JEE Advanced 2023 Paper 2, Mathematics Section 1 Q3: Inverse Trigonometric Functions

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