A parabola of the form <span class="math-inline">y = ax^2 + bx + c</span> with <span class="math-inline">a > 0</span> intersects the graph of <span class="math-inline">f(x) = \frac{1}{x^2 - 4}</span>. Number of possible distinct intersection(s) of these graph is
Step-by-Step Solution
Key Concept: We need to find all possible numbers of intersections between a upward-opening parabola y = ax² + bx + c (a > 0) and the rational function f(x) = 1/(x² - 4). The intersection points satisfy ax² + bx + c = 1/(x² - 4), which leads to a polynomial equation. By analyzing the behavior of both functions and considering various parabola configurations, we determine which intersection counts are achievable.
<p><strong>Step 1: Set up the intersection equation</strong></p><p>For intersections, we solve: ax² + bx + c = 1/(x² - 4)</p><p>Multiplying both sides by (x² - 4): (ax² + bx + c)(x² - 4) = 1</p><p>This expands to: ax⁴ + bx³ + (c - 4a)x² - 4bx - 4c = 1</p><p>Or: ax⁴ + bx³ + (c - 4a)x² - 4bx - (4c + 1) = 0</p><p>This is a polynomial of degree 4, so at most 4 real solutions exist.</p><p><strong>Step 2: Analyze the rational function f(x) = 1/(x² - 4)</strong></p><p>• Vertical asymptotes at x = ±2</p><p>• For |x| > 2: f(x) > 0 and f(x) → 0 as |x| → ∞</p><p>• For |x| < 2: f(x) < 0</p><p>• f(x) is even, symmetric about y-axis</p><p><strong>Step 3: Check if 0 intersections is possible</strong></p><p>If the parabola y = ax² + bx + c has its vertex far above the x-axis and opens upward, staying entirely in the region where f(x) < 0 (between x = -2 and x = 2) or entirely above where f(x) > 0 (outside x = ±2), then 0 intersections is possible. ✓ But this is NOT in the answer choices.</p><p><strong>Step 4: Check if 2 intersections is possible</strong></p><p>A parabola can intersect one branch of the hyperbola on each side of the y-axis (utilizing symmetry). For example, if the parabola dips below into the regions where f(x) < 0 and also extends into regions where f(x) > 0, we get 2 intersections. This is achievable. ✓</p><p><strong>Step 5: Check if 3 intersections is possible</strong></p><p>Consider a parabola with vertex on the negative y-axis (c < 0). The parabola can intersect: one branch in the left region (x < -2), pass through the middle region (−2 < x < 2), and intersect the right branch (x > 2) twice. This gives 3 intersections. This is achievable. ✓</p><p><strong>Step 6: Check if 4 intersections is possible</strong></p><p>A parabola opening upward with appropriate placement can intersect each of the two branches of f(x) = 1/(x² - 4) at two points each. For instance, if the parabola passes through both regions |x| > 2 appropriately, we can have 2 intersections on the left (x < -2) and 2 on the right (x > -2). This is achievable. ✓</p><p><strong>Step 7: Verify which options are achievable</strong></p><p>• 0 intersections: Possible but not in the multiple correct answers</p><p>• 2 intersections: POSSIBLE ✓</p><p>• 3 intersections: POSSIBLE ✓</p><p>• 4 intersections: POSSIBLE ✓</p><p><strong>∴ Answer: B,C,D</strong></p>
Correct Answer: B,C,D