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Friday, 9 October

JEE Main 2025 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 4 April 2025, Shift 2, Maths Q13

Question 13 of 75 in this shift, Maths question 13 of 25, Section A.

The sum of the infinite series cot⁡−1(74)+cot⁡−1(194)+cot⁡−1(394)+cot⁡−1(674)+…\cot^{-1}\left(\frac{7}{4}\right)+\cot^{-1}\left(\frac{19}{4}\right)+\cot^{-1}\left(\frac{39}{4}\right)+\cot^{-1}\left(\frac{67}{4}\right)+\ldots is :
  1. (1)π2−cot⁡−1(12)\frac{\pi}{2}-\cot^{-1}\left(\frac{1}{2}\right)
  2. (2)π2+tan⁡−1(12)\frac{\pi}{2}+\tan^{-1}\left(\frac{1}{2}\right)
  3. (3)π2−tan⁡−1(12)\frac{\pi}{2}-\tan^{-1}\left(\frac{1}{2}\right)Official answer
  4. (4)π2+cot⁡−1(12)\frac{\pi}{2}+\cot^{-1}\left(\frac{1}{2}\right)

Official answer

Option 3

NTA final key.

Chapter
Inverse Trigonometric Functions
Topic
Inverse Trig Identities
Idea tested
Telescoping Series with cot-1

Same topic in other shifts

All Inverse Trig Identities questions
  1. 4 Apr 2026, Shift 1 · Q16If y=tan⁡−1(3cos⁡x−4sin⁡x4cos⁡x+3sin⁡x)+2tan⁡−1(x1+1−x2)y=\tan^{-1}\left(\frac{3\cos x-4\sin x}{4\cos x+3\sin x}\right)+2\tan^{-1}\left(\frac{x}{1+\sqrt{1-x^2}}\right), then dydx\frac{dy}{dx}…MediumSingle correct
  2. 4 Apr 2026, Shift 2 · Q20The integral ∫01cot⁡−1(1+x+x2)dx\int_0^1\cot^{-1}\left(1+x+x^2\right)dx is equal to:HardSingle correct
  3. 5 Apr 2026, Shift 1 · Q24If π4+∑p=111tan⁡−1(2p−11+22p−1)=α\frac{\pi}{4} + \sum_{p=1}^{11} \tan^{-1}\left(\frac{2^{p-1}}{1 + 2^{2p-1}}\right) = \alpha, then tan⁡α\tan\alpha is equal to ________.MediumNumerical value
  4. 6 Apr 2026, Shift 1 · Q13Let 0<α<10<\alpha<1, β=13α\beta=\frac{1}{3\alpha} and tan⁡−1(1−α)+tan⁡−1(1−β)=π4\tan^{-1}(1-\alpha)+\tan^{-1}(1-\beta)=\frac{\pi}{4}. Then 6(α+β)6(\alpha+\beta) is equal to:MediumSingle correct
  5. 6 Apr 2026, Shift 2 · Q13If sin⁡(tan⁡−1(x2))=cot⁡(sin⁡−11−x2)\sin\left(\tan^{-1}\left(x\sqrt{2}\right)\right)=\cot\left(\sin^{-1}\sqrt{1-x^2}\right), x∈(0,1)x\in(0,1), then the value of xx is :MediumSingle correct
  6. 8 Apr 2026, Shift 2 · Q12Let α=3sin⁡−1(611)\alpha = 3\sin^{-1}\left(\frac{6}{11}\right) and β=3cos⁡−1(49)\beta = 3\cos^{-1}\left(\frac{4}{9}\right), where inverse trigonometric functions…HardTwo statements

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.