Let `alpha` and `beta` be the distinct roots of `ax^(2) + bx + c = 0` then `underset(x to alpha)(Lt) (1 - cos (ax^(2) + bx + c))/((x - alpha)^(2))` equal to

Options

A. `(1)/(2) (alpha - beta)^(2)`
B. `-(a^(2))/(2) (alpha - beta)^(2)`
C. 0
D. `(a^(2))/(2) (alpha - beta)^(2)`

Answer

To solve the limit problem given in the question, we will follow a systematic approach.

### Step-by-step Solution:

**Step 1: Identify the limit expression.**
We need to evaluate:
\[ \lim_{x \to \alpha} \frac{1 - \cos(ax^2 + bx + c)}{(x - \alpha)^2} \]

**Step 2: Substitute \(x = \alpha\).**
Since \(\alpha\) is a root of the equation \(ax^2 + bx + c = 0\), we have:
\[ a\alpha^2 + b\alpha + c = 0 \]
Thus, substituting \(x = \alpha\) gives:
\[ 1 - \cos(0) = 1 - 1 = 0 \]
This results in a \(0/0\) indeterminate form.

**Step 3: Apply L'Hôpital's Rule.**
Since we have a \(0/0\) form, we can apply L'Hôpital's Rule, which involves differentiating the numerator and the denominator with respect to \(x\).

**Step 4: Differentiate the numerator and denominator.**
The derivative of the numerator \(1 - \cos(ax^2 + bx + c)\) is:
\[ \sin(ax^2 + bx + c) \cdot (2ax + b) \]
The derivative of the denominator \((x - \alpha)^2\) is:
\[ 2(x - \alpha) \]
Thus, we have:
\[ \lim_{x \to \alpha} \frac{\sin(ax^2 + bx + c) \cdot (2ax + b)}{2(x - \alpha)} \]

**Step 5: Substitute \(x = \alpha\) again.**
Substituting \(x = \alpha\) in the new limit gives:
\[ \sin(0) \cdot (2a\alpha + b) = 0 \]
This again results in a \(0/0\) form.

**Step 6: Apply L'Hôpital's Rule again.**
We differentiate the numerator and denominator again. The numerator becomes:
\[ \cos(ax^2 + bx + c) \cdot (2ax + b)(2ax + b) + \sin(ax^2 + bx + c) \cdot (2a) \]
The denominator becomes:
\[ 2 \]
Now we evaluate:
\[ \lim_{x \to \alpha} \frac{\cos(0) \cdot (2a\alpha + b)^2 + 0}{2} = \frac{(2a\alpha + b)^2}{2} \]

**Step 7: Simplify the expression.**
We know that \(b = -a(\alpha + \beta)\) from Vieta's formulas, where \(\alpha\) and \(\beta\) are the roots of the quadratic equation. Thus:
\[ \frac{(2a\alpha - a(\alpha + \beta))^2}{2} = \frac{(a(\alpha - \beta))^2}{2} \]

### Final Result:
The limit evaluates to:
\[ \frac{a^2(\alpha - \beta)^2}{2} \]