Question

If \( \lim_{x \to 0} \frac{1 - \cos(\alpha x)\cos((\alpha + 1)x)\cos((\alpha + 2)x)}{\sin^2((\alpha + 1)x)} = 2 \), then the product of all possible values of \( \alpha \) is:

• A. 1
• B. -1
• C. 2
• D. -2

Hint:

For limits involving \( 1 - \cos(ax)\cos(bx)\dots \), the expansion shortcut \( \frac{a^2 + b^2 + \dots}{2} \) is much faster than applying L'Hôpital's Rule multiple times.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This limit is of the form \( 0/0 \). For small \( x \), we can use the Taylor expansion \( \cos(kx) \approx 1 - \frac{(kx)^2}{2} \) or the identity \( 1 - \cos A \cos B \cos C \approx \frac{A^2 + B^2 + C^2}{2} x^2 \).

Step 2: Key Formula or Approach:
The numerator \( 1 - \prod \cos(k_i x) \) as \( x \to 0 \) is equivalent to \( \frac{\sum k_i^2}{2} x^2 \). The denominator \( \sin^2((\alpha+1)x) \) is equivalent to \( (\alpha+1)^2 x^2 \).

Step 3: Detailed Explanation:
1. Equate the limit to 2: \[ \frac{\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2}{2(\alpha+1)^2} = 2 \] 2. Simplify the equation: \[ \alpha^2 + \alpha^2 + 2\alpha + 1 + \alpha^2 + 4\alpha + 4 = 4(\alpha^2 + 2\alpha + 1) \] \[ 3\alpha^2 + 6\alpha + 5 = 4\alpha^2 + 8\alpha + 4 \] 3. Rearrange into a quadratic: \[ \alpha^2 + 2\alpha - 1 = 0 \] 4. The product of roots (values of \( \alpha \)) is given by \( c/a \): Product = \( -1 / 1 = -1 \).

Step 4: Final Answer:
The product of all possible values of \( \alpha \) is -1.