Question

The product of all possible values of $\alpha$, for which $\lim_{x \to 0} \frac{1-\cos(\alpha x)\cos((\alpha+1)x)\cos((\alpha+2)x)}{\sin^2((\alpha+1)x)} = 2$, is:

• A. -2
• B. 1
• C. -1
• D. $\frac{5}{4}$

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a limit expression that depends on a parameter $\alpha$. We need to find the value(s) of $\alpha$ for which the limit equals 2. The final answer is the product of these values of $\alpha$.
Step 2: Key Formula or Approach:
As $x \to 0$, the expression is of the indeterminate form $\frac{0}{0}$. The most efficient way to solve this is by using small-angle approximations, which are derived from Taylor series expansions around $x=0$.
The key approximations are:
1. $\cos(u) \approx 1 - \frac{u^2}{2}$ for small $u$.
2. $\sin(u) \approx u$ for small $u$.
Step 3: Detailed Explanation:
Let the limit be $L$. For the limit to be well-defined and finite, we must have $\alpha+1 \neq 0$.
Let's approximate the denominator first using $\sin(u) \approx u$:
Denominator $\approx ((\alpha+1)x)^2 = (\alpha+1)^2 x^2$.
Now for the numerator, using $\cos(u) \approx 1 - \frac{u^2}{2}$:
Numerator $= 1 - \left(1 - \frac{(\alpha x)^2}{2}\right)\left(1 - \frac{((\alpha+1)x)^2}{2}\right)\left(1 - \frac{((\alpha+2)x)^2}{2}\right) + O(x^4)$.
When we expand the product of the three parentheses, we are only interested in terms up to $x^2$, because the denominator is of order $x^2$.
$(1-A)(1-B)(1-C) \approx 1 - (A+B+C) + \dots$.
So, the product is approximately:
$1 - \left(\frac{\alpha^2 x^2}{2} + \frac{(\alpha+1)^2 x^2}{2} + \frac{(\alpha+2)^2 x^2}{2}\right)$.
The numerator thus simplifies to:
Numerator $\approx 1 - \left[1 - \frac{x^2}{2}(\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2)\right]$
Numerator $\approx \frac{x^2}{2}(\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2)$.
Now we can evaluate the limit:
\[ L = \lim_{x \to 0} \frac{\frac{x^2}{2}(\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2)}{(\alpha+1)^2 x^2} \] Cancel the $x^2$ term:
\[ L = \frac{\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2}{2(\alpha+1)^2} \] We are given that $L=2$:
\[ \frac{\alpha^2 + (\alpha^2+2\alpha+1) + (\alpha^2+4\alpha+4)}{2(\alpha^2+2\alpha+1)} = 2 \] \[ 3\alpha^2 + 6\alpha + 5 = 4(\alpha^2+2\alpha+1) = 4\alpha^2 + 8\alpha + 4 \] Rearranging into a standard quadratic form:
\[ \alpha^2 + 2\alpha - 1 = 0 \] This equation gives the possible values of $\alpha$. We are asked for the product of these values. For a quadratic equation $ax^2+bx+c=0$, the product of the roots is $c/a$.
Product of values of $\alpha = \frac{-1}{1} = -1$.
Step 4: Final Answer:
The product of all possible values of $\alpha$ is -1.