JEE Mains · Maths · STD 11 - 12. limits
The product of all possible values of \(\alpha\), for which \(\displaystyle\lim_{x \to 0}\left(\dfrac{1 - \cos(\alpha x)\cos((\alpha+1)x)\cos((\alpha+2)x)}{\sin^2((\alpha+1)x)}\right) = 2\), is:
- A \(-2\)
- B \(1\)
- C \(-1\)
- D \(\dfrac{5}{4}\)
Answer & Solution
Correct Answer
(C) \(-1\)
Step-by-step Solution
Detailed explanation
Using the standard expansions for small \(x\), \(\cos \theta \approx 1 - \dfrac{\theta^2}{2}\) and \(\sin \theta \approx \theta\). The given limit can be written as:…
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