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Saturday, 31 October

JEE Main 2024 · MathsNumerical answerNumerical valueHardMulti-step

JEE Main 9 April 2024, Shift 1, Maths Q28

Question 28 of 90 in this shift, Maths question 28 of 30, Section B.

Let lim⁡n→∞(nn4+1−2n(n2+1)n4+1+nn4+16−8n(n2+4)n4+16+…+nn4+n4−2n⋅n2(n2+n2)n4+n4)\lim_{n\to\infty}\left(\frac{n}{\sqrt{n^4+1}}-\frac{2n}{(n^2+1)\sqrt{n^4+1}}+\frac{n}{\sqrt{n^4+16}}-\frac{8n}{(n^2+4)\sqrt{n^4+16}}+\ldots+\frac{n}{\sqrt{n^4+n^4}}-\frac{2n\cdot n^2}{(n^2+n^2)\sqrt{n^4+n^4}}\right) be πk\frac{\pi}{k}, using only the principal values of the inverse trigonometric functions. Then k2k^2 is equal to ________.

Official answer

32

NTA final key.

Chapter
Integrals

Same idea in other shifts

Asked 9× in all
  1. 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
  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.