<p>Let \(f(x) = x^2 - 2x - 3\); \(x \geq 1\) and \(g(x) = 1 + \sqrt{x + 4}\); \(x \geq -4\). Then the number of real solutions of equation \(f(x) = g(x)\) is/are:</p>
Step-by-Step Solution
Key Concept: We need to find the intersection points of f(x) and g(x) within their common domain [1, ∞). This requires solving the equation algebraically and verifying solutions satisfy domain constraints.
<p><strong>Step 1: Identify the domain.</strong> We need x ≥ 1 (from f) and x ≥ -4 (from g), so the common domain is x ≥ 1.</p><p><strong>Step 2: Set up the equation.</strong> We solve f(x) = g(x):<br>x² - 2x - 3 = 1 + √(x + 4)</p><p><strong>Step 3: Isolate the square root.</strong><br>x² - 2x - 3 - 1 = √(x + 4)<br>x² - 2x - 4 = √(x + 4)</p><p><strong>Step 4: Square both sides.</strong> For this to be valid, we need x² - 2x - 4 ≥ 0.<br>(x² - 2x - 4)² = x + 4<br>x⁴ - 4x³ - 8x² + 4x² + 16x + 16 = x + 4<br>x⁴ - 4x³ - 4x² + 16x + 16 = x + 4<br>x⁴ - 4x³ - 4x² + 15x + 12 = 0</p><p><strong>Step 5: Test x = 1 (left endpoint).</strong><br>f(1) = 1 - 2 - 3 = -4<br>g(1) = 1 + √5 ≈ 1 + 2.236 ≈ 3.236<br>f(1) ≠ g(1), so x = 1 is not a solution.</p><p><strong>Step 6: Analyze behavior.</strong> For x ≥ 1:<br>• f(x) = x² - 2x - 3 is increasing (since f'(x) = 2x - 2 > 0 for x > 1)<br>• g(x) = 1 + √(x + 4) is also increasing but at a slower rate<br>• At x = 1: f(1) = -4 < g(1) ≈ 3.236<br>• As x → ∞: f(x) grows as x² while g(x) grows as √x, so eventually f(x) > g(x)</p><p><strong>Step 7: Verify the intersection condition.</strong> Check the constraint x² - 2x - 4 ≥ 0 at the solution. For x ≥ 1, find where x² - 2x - 4 = 0: x = (2 ± √(4+16))/2 = (2 ± √20)/2 = 1 ± √5. Since √5 ≈ 2.236, we get x ≈ 3.236 or x ≈ -1.236. So x² - 2x - 4 ≥ 0 for x ≥ 1 + √5 ≈ 3.236.</p><p><strong>Step 8: Determine the number of solutions.</strong> Since f(1) < g(1), f is increasing faster than g (eventually), and both are continuous, there is exactly one intersection point in the domain x ≥ 1.</p><p><strong>∴ Answer:</strong> B</p>
Correct Answer: B