Key Concepts and Formulas

• Vieta's Formulas: For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, we have $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
• Algebraic Identities:
  • $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
  • $\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2$
• Range of Cosine: $-1 \le \cos \theta \le 1$ and $0 \le \cos^2 \theta \le 1$ for $\theta \in \mathbb{R}$.

Step-by-Step Solution

Step 1: Identify the quadratic equation and its coefficients.

• Why this step: To apply Vieta's formulas, we need to identify the coefficients of the given quadratic equation.

The given quadratic equation is $2x^2 + (\cos \theta)x - 1 = 0$. Thus, $a = 2$, $b = \cos \theta$, and $c = -1$.

Step 2: Apply Vieta's Formulas to find the sum and product of the roots.

• Why this step: Vieta's formulas allow us to relate the roots to the coefficients, which is crucial for expressing $\alpha_\theta^4 + \beta_\theta^4$ in terms of $\cos \theta$.

Using Vieta's formulas: $\alpha_\theta + \beta_\theta = -\frac{b}{a} = -\frac{\cos \theta}{2}$ $\alpha_\theta \beta_\theta = \frac{c}{a} = -\frac{1}{2}$

Step 3: Calculate $\alpha_\theta^2 + \beta_\theta^2$.

• Why this step: This is an intermediate step to find $\alpha_\theta^4 + \beta_\theta^4$.

Using the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$: $\alpha_\theta^2 + \beta_\theta^2 = \left(-\frac{\cos \theta}{2}\right)^2 - 2\left(-\frac{1}{2}\right) = \frac{\cos^2 \theta}{4} + 1$

Step 4: Calculate $\alpha_\theta^4 + \beta_\theta^4$.

• Why this step: The goal is to find an expression for $\alpha_\theta^4 + \beta_\theta^4$ in terms of $\cos \theta$, so we can determine its minimum and maximum values.

Using the identity $\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2$: $\alpha_\theta^4 + \beta_\theta^4 = \left(\frac{\cos^2 \theta}{4} + 1\right)^2 - 2\left(-\frac{1}{2}\right)^2 = \left(\frac{\cos^2 \theta}{4} + 1\right)^2 - 2\left(\frac{1}{4}\right) = \left(\frac{\cos^2 \theta}{4} + 1\right)^2 - \frac{1}{2}$ Let $f(\theta) = \alpha_\theta^4 + \beta_\theta^4 = \left(\frac{\cos^2 \theta}{4} + 1\right)^2 - \frac{1}{2}$.

Step 5: Determine the range of $\cos^2 \theta$.

• Why this step: The range of $\cos^2 \theta$ will help us find the minimum and maximum values of $f(\theta)$.

Since $\theta \in (0, 2\pi)$, we have $-1 \le \cos \theta \le 1$, so $0 \le \cos^2 \theta \le 1$.

Step 6: Find the minimum and maximum values of $f(\theta)$.

• Why this step: By considering the range of $\cos^2 \theta$, we can find the minimum and maximum values of the expression. Since $\left(\frac{x}{4} + 1\right)^2 - \frac{1}{2}$ is increasing for $x \in [0,1]$, the minimum value occurs when $\cos^2 \theta = 0$ and the maximum value occurs when $\cos^2 \theta = 1$.

Minimum value $m$: $m = \left(\frac{0}{4} + 1\right)^2 - \frac{1}{2} = 1 - \frac{1}{2} = \frac{1}{2}$ Maximum value $M$: $M = \left(\frac{1}{4} + 1\right)^2 - \frac{1}{2} = \left(\frac{5}{4}\right)^2 - \frac{1}{2} = \frac{25}{16} - \frac{8}{16} = \frac{17}{16}$

Step 7: Calculate $16(M+m)$.

• Why this step: This is the final calculation to answer the question.

$16(M+m) = 16\left(\frac{17}{16} + \frac{1}{2}\right) = 16\left(\frac{17}{16} + \frac{8}{16}\right) = 16\left(\frac{25}{16}\right) = 25$

Common Mistakes & Tips

• Carefully apply Vieta's formulas, ensuring the correct signs.
• Remember the range of $\cos^2 \theta$ is $[0, 1]$, not $[-1, 1]$.
• Double-check algebraic manipulations to avoid errors, especially when squaring or simplifying fractions.

Summary

By utilizing Vieta's formulas and algebraic identities, we expressed $\alpha_\theta^4 + \beta_\theta^4$ in terms of $\cos^2 \theta$. We then found the minimum and maximum values of the expression by considering the range of $\cos^2 \theta$. Finally, we calculated $16(M+m)$ to arrive at the answer.

The final answer is \boxed{25}, which corresponds to option (C).