Quadratic Equations
Number of Solutions
Grade 11
Question:
<p>Let \(f(x) = x^2 - 2x - 3\), then \(\lambda = |f(|x|)|\) has:</p>
<p>exactly one solution, if \(\lambda < 0\)</p>
<p>exactly two solutions, if \(\lambda = \{0\} \cup (4,\infty)\)</p>
<p>exactly three solutions, if \(\lambda = 3\)</p>
<p>exactly four solutions, if \(\lambda = \{4\} \cup (0,3)\)</p>
Step-by-Step Solution
Key Concept: Analyzing |f(|x|)| requires understanding that |x| creates even symmetry, and then |f(·)| reflects negative portions above the x-axis. You must track where f(|x|) changes sign to correctly identify the structure of |f(|x|)|.
<p><strong>Step 1:</strong> Find f(|x|). Since f(x) = x² - 2x - 3, we have f(|x|) = |x|² - 2|x| - 3 = x² - 2|x| - 3 (since |x|² = x²).</p><p><strong>Step 2:</strong> Identify where f(|x|) = 0: x² - 2|x| - 3 = 0. Let u = |x| ≥ 0, then u² - 2u - 3 = 0 → (u-3)(u+1) = 0 → u = 3 (u = -1 rejected). So f(|x|) = 0 at x = ±3.</p><p><strong>Step 3:</strong> Determine the sign of f(|x|). For |x| < 3: f(|x|) < 0; for |x| > 3: f(|x|) > 0; at |x| = 3: f(|x|) = 0.</p><p><strong>Step 4:</strong> Apply absolute value: λ = |f(|x|)| = f(|x|) when |x| ≥ 3, and λ = -f(|x|) when |x| < 3.</p><p><strong>Step 5:</strong> Analyze critical points. At x = 0: λ = |-3| = 3 (local minimum on (-3,3)). At x = ±1: λ = |1 - 2 - 3| = 4 (local maxima). At x = ±3: λ = 0 (global minima). The function has 5 turning points (even symmetry about y-axis).</p><p>∴ λ = |f(|x|)| has minimum value 0, achieves local maximum 4 at x = ±1, and has exactly 5 local extrema.</p>
Correct Answer: B,D