Question

Let \[ S = \{ \sin^2 2\theta : (\sin^4 \theta + \cos^4 \theta)x^2 + (\sin 2\theta)x + (\sin^6 \theta + \cos^6 \theta) = 0 \, \text{has real roots} \}. \] If \( \alpha \) and \( \beta \) are the smallest and largest elements of the set \( S \), respectively, then \[ 3 \big((\alpha - 2)^2 + (\beta - 1)^2 \big) \] equals:

Answer 1

Correct Answer: 4

To determine the values of \( \alpha \) and \( \beta \), we analyze the discriminant \( D \) of the given quadratic equation:

\[ D = (\sin 2\theta)^2 - 4 \left( 1 - \frac{\sin^2 2\theta}{2} \right) \left( 1 - \frac{3}{4}\sin^2 2\theta \right). \]

Expanding this, we get:

\[ D = (\sin 2\theta)^2 - 4 \left( 1 - \frac{5}{4}\sin^2 2\theta + \frac{3}{8}\sin^4 2\theta \right). \]

Simplifying further,

\[ D = -\frac{3}{2}\sin^4 2\theta + 6\sin^2 2\theta - 4 > 0. \]

This inequality leads us to solve for \( \sin^2 2\theta \):

\[ 3\sin^4 2\theta - 12\sin^2 2\theta + 8 < 0. \]

Solving this inequality, we get:

\[ \sin^2 2\theta = \frac{12 \pm \sqrt{144 - 12.8}}{6} = \frac{12 \pm 4\sqrt{3}}{6} = 2 \pm \frac{2\sqrt{3}}{3}. \]

Thus, we have:

\[ \sin^2 2\theta = 2 \pm \frac{2}{\sqrt{3}}, \quad \text{but } \sin^2 2\theta \in [0, 1]. \]

Therefore,

\[ \alpha = 2 - \frac{2}{\sqrt{3}}, \quad \beta = 1. \]

Now, we calculate \( 3((\alpha - 2)^2 + (\beta - 1)^2) \):

\[ (\alpha - 2)^2 = \frac{4}{3}, \quad (\beta - 1)^2 = 0. \] \[ 3((\alpha - 2)^2 + (\beta - 1)^2) = 3 \times \frac{4}{3} = 4. \]