<p>If \(x, y \in \mathbb{R}\) and \(2x^2 + 6xy + 5y^2 = 1\), then</p>
Step-by-Step Solution
Key Concept: Treat the constraint as a quadratic in x and use the discriminant condition (Δ ≥ 0) to find the range of y, then analyze which expressions remain bounded or achieve specific values.
<p><strong>Step 1:</strong> Rewrite 2x² + 6xy + 5y² = 1 as a quadratic in x:</p><p>2x² + 6xy + (5y² - 1) = 0</p><p><strong>Step 2:</strong> For real x, the discriminant must be non-negative:</p><p>Δ = (6y)² - 4(2)(5y² - 1) ≥ 0</p><p>36y² - 40y² + 8 ≥ 0</p><p>-4y² + 8 ≥ 0</p><p>y² ≤ 2</p><p>∴ |y| ≤ √2, so <strong>-√2 ≤ y ≤ √2</strong></p><p><strong>Step 3:</strong> From the original equation: 2x² = 1 - 6xy - 5y²</p><p>Since 2x² ≥ 0: 1 - 6xy - 5y² ≥ 0</p><p>This gives us bounds on expressions involving x and y.</p><p><strong>Step 4:</strong> For extrema analysis, rewrite as:</p><p>2x² + 6xy + 5y² = 1 is an ellipse. The range of y is [-√2, √2], and x ranges accordingly based on each y value.</p><p>Check boundary cases and critical points to determine which options (A, C) are achievable or always satisfied.</p><p>∴ Answer: A,C</p>
Correct Answer: A,C