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Mathematics · Inverse Trigonometric Functions

JEE Advanced 2023 — Paper 2 — Question 3

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∣<30<|y|<3, is equal to

  1. Option A:

    23−32 \sqrt{3}-3

  2. Option B:

    3−233-2 \sqrt{3}

  3. Option C:

    43−64 \sqrt{3}-6

    Correct
  4. Option D:

    6−436-4 \sqrt{3}

Answer: C

Step-by-step solution

If 0<y<30<y<3 then given equation can be written as

tan⁡−1(6y9−y2)+tan⁡−1(6y9−y2)=2π3⇒tan⁡−1(6y9−y2)=π3\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 & \tan ^{-1}\left(\frac{6 y}{9-y^{2}}\right)=\frac{\pi}{3} \end{aligned} ⇒tan⁡(tan⁡−16y9−y2)=tan⁡π3⇒6y9−y2=3⇒3y2+6y−93=0⇒y=−6+36+10823=623=3\begin{aligned} & \Rightarrow \quad \tan \left(\tan ^{-1} \frac{6 y}{9-y^{2}}\right)=\tan \frac{\pi}{3} \\& \Rightarrow \quad \frac{6 y}{9-y^{2}}=\sqrt{3} \Rightarrow \sqrt{3} y^{2}+6 y-9 \sqrt{3}=0 \\& \Rightarrow \quad y=\frac{-6+\sqrt{36+108}}{2 \sqrt{3}}=\frac{6}{2 \sqrt{3}}=\sqrt{3} \end{aligned}

If −3<y<0-3<y<0 then given equation can be written as

tan⁡−1(6y9−y2)+tan⁡−1(6y9−y2)+π=2π3⇒tan⁡−1(6y9−y2)=−π6⇒tan⁡(tan⁡−1(6y9−y2))=−13⇒63y=−9+y2⇒y2−63y−9=0⇒y=63−1442=63−122=33−6\begin{aligned} & \tan ^{-1}\left(\frac{6 y}{9-y^{2}}\right)+\tan ^{-1}\left(\frac{6 y}{9-y^{2}}\right)+\pi=\frac{2 \pi}{3} \\ \Rightarrow & \tan ^{-1}\left(\frac{6 y}{9-y^{2}}\right)=-\frac{\pi}{6} \\ \Rightarrow & \tan \left(\tan ^{-1}\left(\frac{6 y}{9-y^{2}}\right)\right)=-\frac{1}{\sqrt{3}} \\ \Rightarrow & 6 \sqrt{3} y=-9+y^{2} \Rightarrow y^{2}-6 \sqrt{3} y-9=0 \\ \Rightarrow & y=\frac{6 \sqrt{3}-\sqrt{144}}{2}=\frac{6 \sqrt{3}-12}{2}=3 \sqrt{3}-6 \end{aligned}

Therefore sum of all the solutions

=33−6+3=43−6=3 \sqrt{3}-6+\sqrt{3}=4 \sqrt{3}-6

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2023
Paper
Paper 2
Subject
Mathematics
Chapter
Inverse Trigonometric Functions
Topic
Properties related to Inverse Trigonometric Functions