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Saturday, 31 October

JEE Main 2022 · MathsNumerical answerNumerical valueHardMulti-step

JEE Main 27 July 2022, Shift 1, Maths Q21

Question 21 of 90 in this shift, Maths question 21 of 30, Section B.

For k∈Rk \in \mathbb{R}, let the solutions of the equation cos⁡(sin⁡−1(xcot⁡(tan⁡−1(cos⁡(sin⁡−1x)))))=k\cos\left(\sin^{-1}\left(x\cot\left(\tan^{-1}\left(\cos\left(\sin^{-1}x\right)\right)\right)\right)\right) = k, 0<∣x∣<120 < |x| < \dfrac{1}{\sqrt{2}} be α\alpha and β\beta, where the inverse trigonometric functions take only principal values. If the solutions of the equation x2−bx−5=0x^2 - bx - 5 = 0 are 1α2+1β2\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2} and αβ\dfrac{\alpha}{\beta}, then bk2\dfrac{b}{k^2} is equal to ________.

Official answer

12

NTA final key.

Chapter
Inverse Trigonometric Functions
Topic
Inverse Trig Identities
Idea tested
Evaluating Trig of Inverse Trig

Same idea in other shifts

Asked 3× in all
  1. 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
  2. 26 Jun 2022, Shift 2 · Q19If the inverse trigonometric functions take principal values, then…EasySingle correct

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 108 of the 174 JEE Main shifts held from 2017 to 2026.