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Friday, 9 October

JEE Main 2023 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 13 April 2023, Shift 2, Maths Q9

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

If lim⁡x→0eax−cos⁡(bx)−cxe−cx21−cos⁡(2x)=17\lim_{x\to0}\dfrac{e^{ax}-\cos(bx)-\dfrac{cxe^{-cx}}{2}}{1-\cos(2x)}=17, then 5a2+b25a^2+b^2 is equal to
  1. (1)6868Official answer
  2. (2)6464
  3. (3)7272
  4. (4)7676

Official answer

Option 1

NTA final key.

Chapter
Limits and Derivatives
Topic
L'Hopital's Rule
Idea tested
Choosing Series Expansion over L'Hopital

Same idea in other shifts

Asked 6× in all
  1. 6 Apr 2026, Shift 1 · Q17The value of lim⁡x→0(x2sin⁡2xx2−sin⁡2x)\lim_{x\to 0}\left(\frac{x^2\sin^2 x}{x^2-\sin^2 x}\right) is:MediumSingle correct
  2. 8 Apr 2025, Shift 2 · Q16Given below are two statements: Statement I :…MediumTwo statements
  3. 8 Apr 2024, Shift 1 · Q26The value of lim⁡x→02(1−cos⁡xcos⁡2xcos⁡3x3⋯cos⁡10x10x2)\lim_{x\to0}2\left(\frac{1-\cos x\sqrt{\cos2x}\sqrt[3]{\cos3x}\cdots\sqrt[10]{\cos10x}}{x^2}\right) is ____________.MediumNumerical value
  4. 31 Jan 2024, Shift 2 · Q25If lim⁡x→0ax2ex−blog⁡e(1+x)+cxe−xx2sin⁡x=1\lim_{x\to0}\frac{ax^2e^x-b\log_e(1+x)+cxe^{-x}}{x^2\sin x}=1, then 16(a2+b2+c2)16(a^2+b^2+c^2) is equal to ________.HardNumerical value
  5. 18 Mar 2021, Shift 1 · Q72If lim⁡x→0sin⁡−1x−tan⁡−1x3x3\lim_{x\to0}\dfrac{\sin^{-1}x-\tan^{-1}x}{3x^3} is equal to L, then the value of (6L+1)(6L+1) is :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 42 of the 174 JEE Main shifts held from 2017 to 2026.