Key Concepts and Formulas

• Roots of a Quadratic Equation: For a quadratic equation $ax^2 + bx + c = 0$, the sum of roots is $-\frac{b}{a}$ and the product of roots is $\frac{c}{a}$.
• Limit of Trigonometric Functions: The fundamental limit $\lim_{u \to 0} \frac{\sin u}{u} = 1$ is crucial for evaluating limits involving trigonometric functions.
• Half-Angle Identity: The identity $1 - \cos(2\theta) = 2 \sin^2(\theta)$ will be used to simplify the numerator of the limit expression.
• Algebraic Manipulation: Careful manipulation of algebraic expressions, especially when dealing with limits approaching a specific value, is essential.

Step-by-Step Solution

Step 1: Relate the roots of the given equation to the expression in the limit. We are given that $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + 1 = 0$. From Vieta's formulas, we have: $\alpha + \beta = -\frac{b}{a}$ $\alpha \beta = \frac{1}{a}$

Now, consider the expression $x^2 + bx + a$. We can rewrite this by substituting the relations from Vieta's formulas. Since $\frac{1}{a} = \alpha \beta$, we have $a = \frac{1}{\alpha \beta}$. Since $-\frac{b}{a} = \alpha + \beta$, we have $b = -a(\alpha + \beta) = -\frac{1}{\alpha \beta}(\alpha + \beta) = -\frac{\alpha + \beta}{\alpha \beta}$.

Let's consider a transformation. If we replace $x$ with $\frac{1}{x}$ in the original equation $ax^2 + bx + 1 = 0$, we get: $a\left(\frac{1}{x}\right)^2 + b\left(\frac{1}{x}\right) + 1 = 0$ $\frac{a}{x^2} + \frac{b}{x} + 1 = 0$ Multiplying by $x^2$ (assuming $x \neq 0$), we get: $a + bx + x^2 = 0$ So, $x^2 + bx + a = 0$. This means that the roots of the equation $x^2 + bx + a = 0$ are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$. Therefore, we can factor $x^2 + bx + a$ as $\left(x - \frac{1}{\alpha}\right)\left(x - \frac{1}{\beta}\right)$.

Step 2: Simplify the limit expression using trigonometric identities. The limit we need to evaluate is: $L = \lim_{x \rightarrow \frac{1}{\alpha}}\left(\frac{1-\cos \left(x^{2}+b x+a\right)}{2(1-\alpha x)^{2}}\right)^{\frac{1}{2}}$ Using the half-angle identity $1 - \cos(\theta) = 2 \sin^2\left(\frac{\theta}{2}\right)$, we can rewrite the numerator: $1 - \cos(x^2 + bx + a) = 2 \sin^2\left(\frac{x^2 + bx + a}{2}\right)$

Substituting this back into the limit expression: $L = \lim_{x \rightarrow \frac{1}{\alpha}}\left(\frac{2 \sin^2\left(\frac{x^2+b x+a}{2}\right)}{2(1-\alpha x)^{2}}\right)^{\frac{1}{2}}$ $L = \lim_{x \rightarrow \frac{1}{\alpha}}\left(\frac{\sin^2\left(\frac{x^2+b x+a}{2}\right)}{(1-\alpha x)^{2}}\right)^{\frac{1}{2}}$ $L = \lim_{x \rightarrow \frac{1}{\alpha}} \frac{\sin\left(\frac{x^2+b x+a}{2}\right)}{1-\alpha x}$ Since $x \to \frac{1}{\alpha}$, and $\alpha > 0$, $x$ will be positive. Also, $\alpha > \beta > 0$, so $\frac{1}{\alpha} < \frac{1}{\beta}$. As $x$ approaches $\frac{1}{\alpha}$, $x^2+bx+a$ approaches 0. For values of $x$ close to $\frac{1}{\alpha}$, $\frac{x^2+bx+a}{2}$ will be close to 0. For small values of $u$, $\sin(u) \approx u$. Therefore, we can remove the square root by considering the sign. As $x \to \frac{1}{\alpha}$, $1-\alpha x \to 0$. The sign of $1-\alpha x$ depends on whether $x$ approaches $\frac{1}{\alpha}$ from the left or right. However, the square root of a square is the absolute value. We will address the sign later if needed. For now, let's proceed with the absolute value implicitly handled by the limit.

Step 3: Substitute the factored form of $x^2+bx+a$ and apply the limit. We know that $x^2 + bx + a = \left(x - \frac{1}{\alpha}\right)\left(x - \frac{1}{\beta}\right)$. So, the expression inside the limit becomes: $\frac{\sin\left(\frac{\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)}{2}\right)}{1-\alpha x}$ We can rewrite $1 - \alpha x$ as $-\alpha\left(x - \frac{1}{\alpha}\right)$. The expression is now: $\frac{\sin\left(\frac{\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)}{2}\right)}{-\alpha\left(x - \frac{1}{\alpha}\right)}$ Let $u = \frac{\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)}{2}$. As $x \rightarrow \frac{1}{\alpha}$, $u \rightarrow 0$. We want to use the limit $\lim_{u \to 0} \frac{\sin u}{u} = 1$. To do this, we need to manipulate the denominator to match the argument of the sine function. $\frac{\sin\left(\frac{\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)}{2}\right)}{\frac{\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)}{2}} \times \frac{\frac{\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)}{2}}{-\alpha\left(x - \frac{1}{\alpha}\right)}$ As $x \rightarrow \frac{1}{\alpha}$, the first part of the expression tends to 1: $\lim_{x \rightarrow \frac{1}{\alpha}} \frac{\sin\left(\frac{\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)}{2}\right)}{\frac{\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)}{2}} = 1$ Now, we evaluate the limit of the second part: $\lim_{x \rightarrow \frac{1}{\alpha}} \frac{\frac{\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)}{2}}{-\alpha\left(x - \frac{1}{\alpha}\right)}$ We can cancel out the $\left(x - \frac{1}{\alpha}\right)$ term, since $x \neq \frac{1}{\alpha}$ as $x \to \frac{1}{\alpha}$: $\lim_{x \rightarrow \frac{1}{\alpha}} \frac{\frac{1}{2}\left(x-\frac{1}{\beta}\right)}{-\alpha}$ Now, substitute $x = \frac{1}{\alpha}$: $\frac{\frac{1}{2}\left(\frac{1}{\alpha}-\frac{1}{\beta}\right)}{-\alpha} = \frac{1}{2\alpha} \left(\frac{1}{\alpha}-\frac{1}{\beta}\right) \times (-1) = -\frac{1}{2\alpha}\left(\frac{1}{\alpha}-\frac{1}{\beta}\right) = \frac{1}{2\alpha}\left(\frac{1}{\beta}-\frac{1}{\alpha}\right)$

Step 4: Consider the square root and the sign. The original limit was: $L = \lim_{x \rightarrow \frac{1}{\alpha}}\left(\frac{1-\cos \left(x^{2}+b x+a\right)}{2(1-\alpha x)^{2}}\right)^{\frac{1}{2}}$ This is equivalent to: $L = \frac{1}{\sqrt{2}} \lim_{x \rightarrow \frac{1}{\alpha}} \frac{\left| \sin\left(\frac{x^2+b x+a}{2}\right) \right|}{|1-\alpha x|}$ As $x \to \frac{1}{\alpha}$, $\frac{x^2+bx+a}{2} \to 0$. For $x$ close to $\frac{1}{\alpha}$, $\sin\left(\frac{x^2+b x+a}{2}\right)$ has the same sign as its argument. The argument is $\frac{1}{2}\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right)$. Since $\alpha > \beta > 0$, we have $\frac{1}{\alpha} < \frac{1}{\beta}$. As $x \to \frac{1}{\alpha}$, if $x > \frac{1}{\alpha}$, then $x - \frac{1}{\alpha} > 0$. Since $x$ is close to $\frac{1}{\alpha}$, $x - \frac{1}{\beta} < 0$. So, $\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right) < 0$, and thus $\frac{x^2+bx+a}{2} < 0$. Therefore, $\sin\left(\frac{x^2+b x+a}{2}\right) < 0$. Also, $1-\alpha x = -\alpha\left(x - \frac{1}{\alpha}\right)$. If $x > \frac{1}{\alpha}$, then $x - \frac{1}{\alpha} > 0$, so $1-\alpha x < 0$. Thus, $\frac{\sin\left(\frac{x^2+b x+a}{2}\right)}{1-\alpha x} = \frac{(-)}{(-)} > 0$. If $x < \frac{1}{\alpha}$, then $x - \frac{1}{\alpha} < 0$. Since $x$ is close to $\frac{1}{\alpha}$, $x - \frac{1}{\beta} < 0$. So, $\left(x-\frac{1}{\alpha}\right)\left(x-\frac{1}{\beta}\right) > 0$, and thus $\frac{x^2+bx+a}{2} > 0$. Therefore, $\sin\left(\frac{x^2+b x+a}{2}\right) > 0$. Also, $1-\alpha x = -\alpha\left(x - \frac{1}{\alpha}\right)$. If $x < \frac{1}{\alpha}$, then $x - \frac{1}{\alpha} < 0$, so $1-\alpha x > 0$. Thus, $\frac{\sin\left(\frac{x^2+b x+a}{2}\right)}{1-\alpha x} = \frac{(+)}{(+)} > 0$. In both cases, the ratio is positive. So the absolute value signs do not change the sign of the overall expression. The limit evaluated in Step 3 is indeed the correct value of the expression inside the square root.

The limit is: $L = \sqrt{\lim_{x \rightarrow \frac{1}{\alpha}}\left(\frac{\sin\left(\frac{x^2+b x+a}{2}\right)}{1-\alpha x}\right)^2}$ Since the expression inside the square root is positive as $x \to \frac{1}{\alpha}$, the limit of the square root is the square root of the limit. $L = \left| \frac{1}{2\alpha}\left(\frac{1}{\beta}-\frac{1}{\alpha}\right) \right|$ Since $\alpha > \beta > 0$, we have $\frac{1}{\alpha} < \frac{1}{\beta}$, so $\frac{1}{\beta} - \frac{1}{\alpha} > 0$. Also, $2\alpha > 0$. Thus, the absolute value is not needed: $L = \frac{1}{2\alpha}\left(\frac{1}{\beta}-\frac{1}{\alpha}\right)$

Step 5: Equate the given limit expression with the calculated limit and solve for k. We are given that: $\lim_{x \rightarrow \frac{1}{\alpha}}\left(\frac{1-\cos \left(x^{2}+b x+a\right)}{2(1-\alpha x)^{2}}\right)^{\frac{1}{2}}=\frac{1}{k}\left(\frac{1}{\beta}-\frac{1}{\alpha}\right)$ We have calculated the left-hand side to be: $\frac{1}{2\alpha}\left(\frac{1}{\beta}-\frac{1}{\alpha}\right)$ Equating the two expressions: $\frac{1}{2\alpha}\left(\frac{1}{\beta}-\frac{1}{\alpha}\right) = \frac{1}{k}\left(\frac{1}{\beta}-\frac{1}{\alpha}\right)$ Since $\alpha > \beta > 0$, $\frac{1}{\beta} - \frac{1}{\alpha} \neq 0$. We can divide both sides by $\left(\frac{1}{\beta}-\frac{1}{\alpha}\right)$: $\frac{1}{2\alpha} = \frac{1}{k}$ Therefore, $k = 2\alpha$.

Common Mistakes & Tips

• Sign Errors with Square Roots: Be careful when taking the square root of a squared expression. The result is the absolute value. Analyze the sign of the expression inside the limit as $x$ approaches the given value to determine if the absolute value is necessary.
• Algebraic Simplification: Ensure that all algebraic manipulations, especially when dealing with fractions and terms approaching zero, are done correctly. Factorization and cancellation of terms require careful attention.
• Misapplication of Limit Formulas: The limit $\lim_{u \to 0} \frac{\sin u}{u} = 1$ is fundamental. Make sure the argument of the sine function in the numerator matches the denominator, or can be made to match through algebraic manipulation.

Summary The problem involves evaluating a limit of a function involving a quadratic expression and a trigonometric term. We first established the relationship between the roots of the given quadratic equation and the terms appearing in the limit expression. By using the half-angle identity for cosine and the fundamental limit involving sine, we simplified the expression. After carefully handling the square root and ensuring the correct sign, we equated the calculated limit with the given expression to solve for the unknown constant $k$. The value of $k$ was found to be $2\alpha$.

The final answer is \boxed{2 \alpha}.