Verified Solution • 2025–26 Session moderate

QJEEAdv23Ma3b38: JEE Advanced 2023 Mathematics - Class 12 JEEAdv23Ma3b38 Solution: For any , let and . Then the sum of all the solutions of the…

For any y∈Ry \in \mathbb{R} , let cot⁡−1(y)∈(0,π)\cot ^{-1}(y) \in(0, \pi) and tan⁡−1(y)∈(−π2,π2)\tan ^{-1}(y) \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) . Then the sum of all the solutions of the equation

tan⁡−1(6y9−y2)+cot⁡−1(9−y26y)=2π3\tan ^{-1}\left(\frac{6 y}{9-y^2}\right)+\cot ^{-1}\left(\frac{9-y^2}{6 y}\right)=\frac{2 \pi}{3} for 0<|y|<3 , is equal to :

Multiple Choice Options

  • A 23−32 \sqrt{3}-3
  • B 3−233-2 \sqrt{3}
  • C 43−64 \sqrt{3}-6
  • D 6−436-4 \sqrt{3}

Step-by-Step Verified Solution

Concept :

cot⁡−1x={π+tan⁡−11x;amp;xlt;0tan⁡−11xamp;;xgt;0\cot ^{-1} x=\left\{\begin{aligned}\pi+\tan ^{-1} \frac{1}{x}; & x<0 \\\tan ^{-1} \frac{1}{x} & ; x>0\end{aligned}\right.

Solution : Given, 0 < |y| < 3

⇒\Rightarrow y∈(−3,3)−y \in(-3,3)- {0}

9−y29 - {y^2} is always positive when y∈(−3,3)−y \in(-3,3)- {0}

And 6y6y is positive when y∈(0,3)y \in(0,3)

And 6y6y is negative when y∈(−3,0)y \in(-3,0)

∴\therefore In overall, \frac{6 y}{9-y^2} > 0 when y∈(0,3)y \in(0,3)

And \frac{6 y}{9-y^2} < 0 when y∈(−3,0)y \in(-3,0)

Case - 1 :

When -3 < y < 0

amp;tan⁡−1(6y9−y2)+π+tan⁡−1(6y9−y2)=2π3amp;⇒2tan⁡−1(6y9−y2)=−π3amp;⇒6y9−y2=−13amp;⇒y2−63y−9=0amp;⇒y2−63y−9=0\begin{aligned}& \tan ^{-1}\left(\frac{6 y}{9-y^2}\right)+\pi+\tan ^{-1}\left(\frac{6 y}{9-y^2}\right)=\frac{2 \pi}{3} \\\\& \Rightarrow 2 \tan ^{-1}\left(\frac{6 y}{9-y^2}\right)=\frac{-\pi}{3} \\\\& \Rightarrow \frac{6 y}{9-y^2}=\frac{-1}{\sqrt{3}} \\\\& \Rightarrow y^2-6 \sqrt{3} y-9=0 \\\\& \Rightarrow y^2-6 \sqrt{3} y-9=0\end{aligned}

amp;⇒y=63±108+362=63±122=33±6amp; as y∈(−3.0)∴y=33−6\begin{aligned}& \Rightarrow y=\frac{6 \sqrt{3} \pm \sqrt{108+36}}{2}=\frac{6 \sqrt{3} \pm 12}{2}=3 \sqrt{3} \pm 6 \\\\& \text { as } y \in(-3.0) \therefore y=3 \sqrt{3}-6\end{aligned}

Case-2 : When 0 < y < 3

amp;tan⁡−1(6y9−y2)+tan⁡−1(6y9−y2)=2π3amp;⇒2tan⁡−1(6y9−y2)=2π3amp;⇒6y9−y2=3\begin{aligned}& \tan ^{-1}\left(\frac{6 y}{9-y^2}\right)+\tan ^{-1}\left(\frac{6 y}{9-y^2}\right)=\frac{2 \pi}{3} \\\\& \Rightarrow 2 \tan ^{-1}\left(\frac{6 y}{9-y^2}\right)=\frac{2 \pi}{3} \\\\& \Rightarrow \frac{6 y}{9-y^2}=\sqrt{3}\end{aligned}

amp;⇒6y=93−3y2amp;⇒3y2+6y−93=0amp;⇒3y2+9y−3y−93=0amp;⇒3y(y+33)−3(y+33)=0amp;⇒(y+33)−(3y−3)=0\begin{aligned}& \Rightarrow 6 y=9 \sqrt{3}-\sqrt{3} y^2 \\\\& \Rightarrow \sqrt{3} y^2+6 y-9 \sqrt{3}=0 \\\\& \Rightarrow \sqrt{3} y^2+9 y-3 y-9 \sqrt{3}=0 \\\\& \Rightarrow \sqrt{3} y(y+3 \sqrt{3})-3(y+3 \sqrt{3})=0 \\\\& \Rightarrow(y+3 \sqrt{3})-(\sqrt{3} y-3)=0\end{aligned}

y≠−33∴y=3 as y∈(0,3)y \neq-3 \sqrt{3} \quad \therefore y=\sqrt{3} \text { as } y \in(0,3)

∴ Sum of solutions =3+(33−6)=43−6\therefore \text { Sum of solutions }=\sqrt{3}+(3 \sqrt{3}-6)=4 \sqrt{3}-6

Related Learning Resources & Exam Preparation

JEE Advanced PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring