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Saturday, 31 October

JEE Main 2024 · MathsMultiple choiceSingle correctMediumCalculation

JEE Main 30 January 2024, Shift 1, Maths Q10

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

The 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 :
  1. (1)13(23−3)π8\frac{13(2\sqrt{3}-3)\pi}{8}
  2. (2)(23+3)π24\frac{(2\sqrt{3}+3)\pi}{24}
  3. (3)π8(23+3)\frac{\pi}{8(2\sqrt{3}+3)}
  4. (4)13π8(43+3)\frac{13\pi}{8(4\sqrt{3}+3)}Official answer

Official answer

Option 4

NTA final key.

Chapter
Integrals
Topic
Integral as Limit of Sum
Idea tested
Summation to Riemann Form

Same idea in other shifts

Asked 9× in all
  1. 9 Apr 2024, Shift 1 · Q28Let…HardNumerical value
  2. 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
  3. 25 Jul 2022, Shift 1 · Q27If…MediumNumerical value
  4. 25 Jul 2022, Shift 2 · Q7…MediumSingle correct
  5. 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
  6. 24 Jun 2022, Shift 2 · Q7…HardSingle correct
  7. 30 Jun 2022, Shift 1 · Q11lim⁡n→∞∑r=1nr2r2−7rn+6n2\lim_{n\to\infty} \sum_{r=1}^{n} \frac{r}{2r^2 - 7rn + 6n^2} is equal to :MediumSingle correct
  8. 25 Feb 2021, Shift 2 · Q74lim⁡n→∞[1n+n(n+1)2+n(n+2)2+………+n(2n−1)2]\lim_{n\to\infty}\left[\frac{1}{n} + \frac{n}{(n+1)^2} + \frac{n}{(n+2)^2} + \ldots\ldots\ldots + \frac{n}{(2n-1)^2}\right] is equal to :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.