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Saturday, 31 October

JEE Main 2021 · MathsMultiple choiceSingle correctMediumCalculation

JEE Main 26 August 2021, Shift 1, Maths Q70

Question 70 of 90 in this shift, Maths question 10 of 30, Section A.

The value of lim⁡n→∞1n∑r=02n−1n2n2+4r2\lim_{n\to\infty} \frac{1}{n}\sum_{r=0}^{2n-1} \frac{n^2}{n^2+4r^2} is :
  1. (1)14tan⁡−1(4)\frac{1}{4}\tan^{-1}(4)
  2. (2)12tan⁡−1(4)\frac{1}{2}\tan^{-1}(4)Official answer
  3. (3)tan⁡−1(4)\tan^{-1}(4)
  4. (4)12tan⁡−1(2)\frac{1}{2}\tan^{-1}(2)

Official answer

Option 2

NTA final key.

Chapter
Integrals
Topic
Integral as Limit of Sum
Idea tested
Fractional or Multiple Upper Bounds

Same topic in other shifts

All Integral as Limit of Sum questions
  1. 9 Apr 2024, Shift 1 · Q28Let…HardNumerical value
  2. 30 Jan 2024, Shift 1 · Q10The value of lim⁡n→∞∑k=1nn3(n2+k2)(n2+3k2)\lim_{n\to\infty}\sum_{k=1}^{n}\frac{n^3}{(n^2+k^2)(n^2+3k^2)} is :MediumSingle correct
  3. 13 Apr 2023, Shift 1 · Q7Among (S1) : lim⁡n→∞1n2(2+4+6+……+2n)=1\lim_{n\to\infty}\frac{1}{n^2}(2+4+6+\ldots\ldots+2n)=1 (S2) :…MediumTwo statements
  4. 25 Jul 2022, Shift 1 · Q27If…MediumNumerical value
  5. 25 Jul 2022, Shift 2 · Q7…MediumSingle correct
  6. 26 Jul 2022, Shift 1 · Q10If a=lim⁡n→∞∑k=1n2nn2+k2\mathrm{a} = \lim_{n \to \infty} \sum_{k=1}^{n} \dfrac{2n}{n^2 + k^2} and f(x)=1−cos⁡x1+cos⁡xf(x) = \sqrt{\dfrac{1 - \cos x}{1 + \cos x}},…MediumSingle 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.