JEE Main 2026 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 5 April 2026, Shift 1, Maths Q17

Question 17 of 75 in this shift, Maths question 17 of 25, Section A.

The product of all possible values of

\alpha

, for which

\lim_{x \to 0}\left(\frac{1 - \cos(\alpha x)\cos((\alpha+1)x)\cos((\alpha+2)x)}{\sin^2((\alpha+1)x)}\right) = 2

, is:

(1)-2
(2)1
(3)-1Official answer
(4) $\frac{5}{4}$

Official answer

Option 3

NTA final key.

Chapter: Limits and Derivatives
Topic: L'Hopital's Rule
Idea tested: Simplifying with Standard Limits Before L'Hopital

Same idea in other shifts

Asked 6× in all

5 Apr 2026, Shift 2 · Q17Let $f(x)=\lim_{y\to 0}\frac{(1-\cos(xy))\tan(xy)}{y^3}$ . Then the number of solutions of the equation $f(x)=\sin x$ , $x \in \mathbf{R}$ …MediumSingle correct
7 Apr 2025, Shift 1 · Q16…MediumSingle correct
7 Apr 2025, Shift 2 · Q23For $t>-1$ , let $\alpha_t$ and $\beta_t$ be the roots of the equation…HardNumerical value
31 Jan 2024, Shift 1 · Q6 $\lim_{x\to0}\frac{e^{2|\sin x|}-2|\sin x|-1}{x^2}$ MediumSingle correct
8 Apr 2023, Shift 2 · Q9If $\alpha > \beta > 0$ are the roots of the equation $ax^2 + bx + 1 = 0$ , and…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 42 of the 174 JEE Main shifts held from 2017 to 2026.