The product of all possible values of α, for which lim x→0 1 - cos(αx) cos((α+1)x) cos((α+2)x)/ sin²((α+1)x) = 2, is :

Mathematics · Limits, Continuity and Differentiability

JEE Main 2026 — 5 April, Morning Shift — Question 40

The product of all possible values of α, for which $lim_{x→0} \frac{1 - cos(αx) cos((α+1)x) cos((α+2)x)} { sin²((α+1)x)} = 2,$ is :

Option A:
$-2$
Option B:
$1$
Option C:
$-1$
Correct
Option D:
$\frac{5}{4}$

Answer: C

Step-by-step solution

$\lim _{x \rightarrow 0} \frac{1-\cos \alpha x \cdot \cos ((\alpha+1) x) \cos ((\alpha+2) x)}{(\alpha+1)^{2} x^{2}}$ By using L.H. Rule we get $\Rightarrow \frac{1}{2}(\alpha+1)^{2}\left[\alpha^{2}+(\alpha+1)^{2}+(\alpha+2)^{2}\right]=2$ $\alpha^{2}+(\alpha+1)^{2}+(\alpha+2)^{2}=4(\alpha+1)^{2}$ $\alpha^{2}+(\alpha+2)^{2}=3(\alpha+1)^{2}$ $\Rightarrow \alpha^{2}+2 \alpha-1=0$ Product $=-1$

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Exam: JEE Main 2026
Paper: 5 April, Morning Shift
Subject: Mathematics
Chapter: Limits, Continuity and Differentiability
Topic: Evaluation of Limit of Functions