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 involving trigonometric functions that evaluates to 2. The limit depends on a parameter $\alpha$. We need to find all possible values of $\alpha$ that satisfy this condition and then find their product.
Step 2: Key Formula or Approach:
The limit is of the indeterminate form 0/0 as $x \to 0$. We can use standard limits and Taylor series expansions for small $x$.
The key Taylor expansions are:
1. $\cos(u) \approx 1 - \frac{u^2}{2}$ for small $u$.
2. $\sin(u) \approx u$ for small $u$.
We will apply these approximations to the numerator and denominator of the given limit expression.
Step 3: Detailed Explanation:
Let the given limit be $L$. \[ L = \lim_{x \to 0} \frac{1-\cos(\alpha x)\cos((\alpha+1)x)\cos((\alpha+2)x)}{\sin^2((\alpha+1)x)} \] For the limit to exist and be non-zero, the denominator must not be identically zero, which means $\alpha+1 \neq 0$.
Let's approximate the terms for small $x$:
$\cos(\alpha x) \approx 1 - \frac{(\alpha x)^2}{2}$
$\cos((\alpha+1)x) \approx 1 - \frac{((\alpha+1)x)^2}{2}$
$\cos((\alpha+2)x) \approx 1 - \frac{((\alpha+2)x)^2}{2}$
$\sin((\alpha+1)x) \approx (\alpha+1)x$
The denominator becomes:
$\sin^2((\alpha+1)x) \approx ((\alpha+1)x)^2 = (\alpha+1)^2 x^2$.
The numerator becomes:
$N(x) = 1 - \left(1 - \frac{\alpha^2 x^2}{2}\right)\left(1 - \frac{(\alpha+1)^2 x^2}{2}\right)\left(1 - \frac{(\alpha+2)^2 x^2}{2}\right)$
When expanding the product of the three terms in the parenthesis, we are interested in terms up to $x^2$, as higher order terms will vanish when divided by $x^2$ in the limit.
The product is approximately $1 - \left(\frac{\alpha^2 x^2}{2} + \frac{(\alpha+1)^2 x^2}{2} + \frac{(\alpha+2)^2 x^2}{2}\right) + O(x^4)$.
So, the numerator simplifies to:
$N(x) \approx 1 - \left(1 - \frac{\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2}{2} x^2\right)$
$N(x) \approx \frac{\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2}{2} x^2$
Now, substitute these approximations back into the limit:
\[ L = \lim_{x \to 0} \frac{\frac{\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2}{2} x^2}{(\alpha+1)^2 x^2} \] Cancel out the $x^2$ term:
\[ L = \frac{\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2}{2(\alpha+1)^2} \] We are given that this limit is equal to 2.
\[ \frac{\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2}{2(\alpha+1)^2} = 2 \] Expand the terms in the numerator:
$\alpha^2 + (\alpha^2+2\alpha+1) + (\alpha^2+4\alpha+4) = 4(\alpha+1)^2$
$3\alpha^2 + 6\alpha + 5 = 4(\alpha^2+2\alpha+1)$
$3\alpha^2 + 6\alpha + 5 = 4\alpha^2 + 8\alpha + 4$
Rearrange to form a standard quadratic equation:
$0 = (4\alpha^2 - 3\alpha^2) + (8\alpha - 6\alpha) + (4 - 5)$
$0 = \alpha^2 + 2\alpha - 1$
This is a quadratic equation for $\alpha$. The product of all possible values of $\alpha$ is the product of the roots of this equation. For a quadratic equation $ax^2+bx+c=0$, the product of roots is $c/a$.
Product of roots = $\frac{-1}{1} = -1$.
The roots are $\alpha = -1 \pm \sqrt{2}$, neither of which is $-1$, so our initial assumption $\alpha+1 \neq 0$ is valid for these solutions.
Step 4: Final Answer:
The product of all possible values of $\alpha$ is -1.