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Saturday, 31 October

JEE Main 2021 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 4 August 2021, Shift 2, Maths Q69

Question 69 of 90 in this shift, Maths question 9 of 30, Section A.

If ∫sin⁡−1(2x+24x2+8x+13)dx=Atan⁡−1(2x+23)+Blog⁡e(4x2+8x+13)+C\int\sin^{-1}\left(\frac{2x+2}{\sqrt{4x^2+8x+13}}\right)dx=A\tan^{-1}\left(\frac{2x+2}{3}\right)+B\log_e\left(4x^2+8x+13\right)+C (C is a constant of integration), then the ordered pair (A, 4B) is equal to
  1. (1)(32(x+1),−3)\left(\frac{3}{2}(x+1),-3\right)
  2. (2)(32(x+1),3)\left(\frac{3}{2}(x+1),3\right)
  3. (3)(x+1,−3)(x+1,-3)Official answer
  4. (4)(x+1,−6)(x+1,-6)

Official answer

Option 3

NTA final key.

Chapter
Integrals

Same idea in other shifts

Asked 2× in all
  1. 3 Sep 2020, Shift 2 · Q60If ∫sin⁡−1(x1+x)dx=A(x)tan⁡−1(x)+B(x)+C\int \sin^{-1}\left(\sqrt{\frac{x}{1+x}}\right) dx = A(x)\tan^{-1}(\sqrt{x}) + B(x) + C, where C is a constant of integration, then…HardSingle 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.