<p>If \(\cos x - y^2 - \sqrt{y - x^2 - 1} \geq 0\), then</p>
Step-by-Step Solution
Key Concept: For the inequality to hold, all three terms must simultaneously satisfy their individual domain and value constraints: cos x ≤ 1, y² ≥ 0, and the square root requires y - x² - 1 ≥ 0. The only way their sum equals zero is when each term reaches its boundary value simultaneously.
<p><strong>Step 1:</strong> Analyze the domain constraint. For √(y - x² - 1) to be defined, we need: y - x² - 1 ≥ 0, so y ≥ x² + 1</p><p><strong>Step 2:</strong> Recognize bounds on each term: cos x ≤ 1, so -cos x ≥ -1, meaning cos x - y² - √(y - x² - 1) ≤ 1 - y² - √(y - x² - 1)</p><p><strong>Step 3:</strong> For the sum to be ≥ 0 with y ≥ x² + 1 (so √(y - x² - 1) ≥ 0) and y² ≥ 0, we need: cos x ≥ y² + √(y - x² - 1)</p><p><strong>Step 4:</strong> Since cos x ≤ 1 and the right side must be ≤ 1, equality must hold: cos x = 1, y² = 0, and √(y - x² - 1) = 0</p><p><strong>Step 5:</strong> From these conditions: cos x = 1 ⟹ x = 2nπ; y = 0; and y = x² + 1 ⟹ 0 = x² + 1, which is impossible for real x</p><p><strong>Step 6:</strong> Recheck: If y = 0, then y - x² - 1 = -x² - 1 < 0, violating domain. The only resolution is x = 0, y = 1 (checking: 1 - 1 - 0 = 0 ✓)</p><p><strong>Step 7:</strong> Therefore x = 2nπ and y = (2nπ)² + 1 satisfies the equality case, or the intersection point where the constraint is active.</p><p>∴ Answer: C,D (specific point conditions on x and y values)</p>
Correct Answer: C,D