Key Concepts and Formulas

• Relationship between roots and coefficients of a quadratic equation: For a quadratic equation $a{x^2} + bx + c = 0$ with distinct roots $\alpha$ and $\beta$, we have $\alpha + \beta = -\frac{b}{a}$ and $\alpha \beta = \frac{c}{a}$. Also, the quadratic can be factored as $a{x^2} + bx + c = a(x - \alpha)(x - \beta)$.
• Trigonometric Identity: The double angle identity for cosine can be rewritten as $1 - \cos(2\theta) = 2\sin^2(\theta)$.
• Standard Limit: The fundamental trigonometric limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. This implies $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin^2 \theta}{\theta^2} = 1$.

Step-by-Step Solution

Let the given limit be denoted by $L$. $L = \mathop {\lim }\limits_{x \to \alpha } {{1 - \cos \left( {a{x^2} + bx + c} \right)} \over {{{\left( {x - \alpha } \right)}^2}}}$

Step 1: Rewrite the quadratic expression in terms of its roots. Since $\alpha$ and $\beta$ are the roots of $a{x^2} + bx + c = 0$, we can write $a{x^2} + bx + c = a(x - \alpha)(x - \beta)$. Substituting this into the limit expression, we get: $L = \mathop {\lim }\limits_{x \to \alpha } {{1 - \cos \left( {a\left( {x - \alpha } \right)\left( {x - \beta } \right)} \right)} \over {{{\left( {x - \alpha } \right)}^2}}}$

Step 2: Apply the trigonometric identity $1 - \cos(2\theta) = 2\sin^2(\theta)$. Let $\theta = \frac{a\left( {x - \alpha } \right)\left( {x - \beta } \right)}{2}$. Then $2\theta = a\left( {x - \alpha } \right)\left( {x - \beta } \right)$. Applying the identity, the numerator becomes $2\sin^2\left( \frac{a\left( {x - \alpha } \right)\left( {x - \beta } \right)}{2} \right)$. $L = \mathop {\lim }\limits_{x \to \alpha } {{2{{\sin }^2}\left( {a{{\left( {x - \alpha } \right)\left( {x - \beta } \right)} \over 2}} \right)} \over {{{\left( {x - \alpha } \right)}^2}}}$

Step 3: Manipulate the expression to use the standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. We have $\sin^2$ in the numerator. To use the standard limit, we need to divide by the square of the argument of the sine function, which is $\left( \frac{a\left( {x - \alpha } \right)\left( {x - \beta } \right)}{2} \right)^2 = \frac{a^2\left( {x - \alpha } \right)^2\left( {x - \beta } \right)^2}{4}$. We multiply and divide by this term, but it's more efficient to directly multiply and divide by the term needed for the limit. We need $\left(x - \alpha\right)^2$ in the denominator. Let's introduce the square of the entire argument of the sine function in the denominator and then balance it. $L = \mathop {\lim }\limits_{x \to \alpha } 2 \times \frac{{{{\sin }^2}\left( {a{{\left( {x - \alpha } \right)\left( {x - \beta } \right)} \over 2}} \right)}}{{{{\left( {x - \alpha } \right)}^2}}}$ We want to have $\left( \frac{a\left( {x - \alpha } \right)\left( {x - \beta } \right)}{2} \right)^2$ in the denominator of the $\sin^2$ term. L = \mathop {\lim }\limits_{x \to \alpha } 2 \times \frac{{{{\sin }^2}\left( {a{{\left( {x - \alpha } \right)\left( {x - \beta } \right)} \over 2}} \right)}}{{\left( \frac{a\left( {x - \alpha } \right)\left( {x - \beta } \right)}{2} \right)^2}} \times \frac{{{{\left( {a{{\left( {x - \alpha } \right)\left( {x - \beta } \right)} \over 2}} \right)^2}}}{{{{\left( {x - \alpha } \right)}^2}}} L = \mathop {\lim }\limits_{x \to \alpha } 2 \times \frac{{{{\sin }^2}\left( {a{{\left( {x - \alpha } \right)\left( {x - \beta } \right)} \over 2}} \right)}}{{\left( \frac{a\left( {x - \alpha } \right)\left( {x - \beta } \right)}{2} \right)^2}} \times \frac{{\frac{{{a^2}{{\left( {x - \alpha } \right)}^2}{{\left( {x - \beta } \right)}^2}}}}{4}}{{{{\left( {x - \alpha } \right)}^2}}}

Step 4: Simplify the expression and evaluate the limit. As $x \to \alpha$, the argument of the sine function, $\frac{a\left( {x - \alpha } \right)\left( {x - \beta } \right)}{2}$, approaches $\frac{a(0)(\alpha - \beta)}{2} = 0$. Therefore, $\mathop {\lim }\limits_{x \to \alpha } \frac{{{{\sin }^2}\left( {a{{\left( {x - \alpha } \right)\left( {x - \beta } \right)} \over 2}} \right)}}{{\left( \frac{a\left( {x - \alpha } \right)\left( {x - \beta } \right)}{2} \right)^2}} = 1$. Now, let's simplify the remaining part of the expression: \frac{2 \times \frac{{{a^2}{{\left( {x - \alpha } \right)}^2}{{\left( {x - \beta } \right)}^2}}}}{4} \times \frac{1}{{{{\left( {x - \alpha } \right)}^2}}} = \frac{{{{a^2}{{\left( {x - \alpha } \right)}^2}{{\left( {x - \beta } \right)}^2}}}}{2} \times \frac{1}{{{{\left( {x - \alpha } \right)}^2}}} $= \frac{{{a^2}{{\left( {x - \beta } \right)}^2}}}{2}$ Now, we take the limit as $x \to \alpha$: $L = \mathop {\lim }\limits_{x \to \alpha } \frac{{{a^2}{{\left( {x - \beta } \right)}^2}}}{2} = \frac{{{a^2}{{\left( {\alpha - \beta } \right)}^2}}}{2}$

Common Mistakes & Tips

• Incorrectly applying the standard limit: Ensure that the argument of the sine function and the denominator are squared identically before applying $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.
• Algebraic errors: Be careful with squaring terms and simplifying fractions, especially when introducing terms to match the standard limit form.
• Forgetting the factor 'a': The coefficient 'a' of the quadratic plays a crucial role in the final answer. Make sure it's carried through the calculations correctly.

Summary

The problem requires evaluating a limit involving a trigonometric function of a quadratic expression. By first rewriting the quadratic in terms of its roots and then applying the trigonometric identity $1 - \cos(2\theta) = 2\sin^2(\theta)$, we transform the limit into a form where the standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ can be used. Careful algebraic manipulation is needed to isolate the standard limit and evaluate the remaining terms. The final result depends on the coefficient 'a' and the difference between the roots squared.

The final answer is $\boxed{{\frac{{{a^2}{{\left( {\alpha - \beta } \right)}^2}}}{2}}}$. This corresponds to option (A).