Question:

<p><strong>Step 1:</strong> Rewrite the equation as (1/2)|x| = x² - a, which means we need intersections of y = (1/2)|x| and y = x² - a.</p><p><strong>Step 2:</strong> Analyze y = (1/2)|x|: This is a V-shaped graph passing through origin with slopes ±1/2. For x ≥ 0, y = x/2; for x < 0, y = -x/2.</p><p><strong>Step 3:</strong> Analyze y = x² - a: This is a parabola opening upward with vertex at (0, -a). As 'a' increases, the parabola shifts downward; as 'a' decreases, it shifts upward.</p><p><strong>Step 4:</strong> For x ≥ 0, solve x/2 = x² - a, giving x² - x/2 - a = 0. This has solutions x = (1/4) ± √(1/16 + a) = (1 ± √(1 + 16a))/4.</p><p><strong>Step 5:</strong> For real solutions in x ≥ 0: we need 1 + 16a ≥ 0, so a ≥ -1/16. When this holds, we get at most 2 positive solutions (one from each ±).</p><p><strong>Step 6:</strong> By symmetry, for x < 0, we also get at most 2 negative solutions. However, we must check when the parabola is positioned to give maximum intersections.</p><p><strong>Step 7:</strong> The parabola y = x² - a passes through (0, -a). The V-shape y = (1/2)|x| has derivative ±1/2 at points away from origin. For maximum intersections (4 total: 2 on each side), the parabola vertex must be below the V-shape at x = 0, meaning -a < 0, so a > -1. Also, the parabola shouldn't be too low (otherwise no intersections), which occurs when a > -∞.</p><p><strong>Step 8:</strong> When a = -1, the parabola vertex is at (0, 1), which is exactly at the point where the V-shape has slope ±1/2. As a increases beyond -1, the vertex moves down and we maintain 4 intersection points. For a ≤ -1, we lose solutions.</p><p><strong>Step 9:</strong> Testing boundaries: When a = -1, the vertex is at (0,1) and tangency conditions give exactly 4 intersections. For a > -1, we continue to have 4 intersections (maximum). For a < -1, intersections reduce.</p><p><strong>∴ Answer:</strong> d</p>