<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: Analyze the composite function λ = |f(|x|)| by first understanding f(x)'s behavior, then applying |x| substitution, and finally taking the absolute value of the result. This triple composition creates multiple symmetries that determine domain, range, and other properties.
<p><strong>Step 1: Analyze f(x) = x² - 2x - 3</strong></p><p>f(x) = (x-1)² - 4, with vertex at (1, -4). Roots: x = 3, -1. Domain: ℝ, Range: [-4, ∞)</p><p><strong>Step 2: Apply |x| substitution to get f(|x|)</strong></p><p>f(|x|) = |x|² - 2|x| - 3 = x² - 2|x| - 3</p><p>This function is EVEN since f(|-x|) = f(|x|). It's symmetric about the y-axis.</p><p><strong>Step 3: Determine properties of f(|x|)</strong></p><p>• f(|0|) = -3 (minimum value of f(|x|))</p><p>• f(|x|) → ∞ as x → ±∞</p><p>• f(|x|) ranges from [-3, ∞)</p><p><strong>Step 4: Apply outer absolute value to get λ = |f(|x|)|</strong></p><p>• Since minimum of f(|x|) is -3, we have f(|x|) ∈ [-3, ∞)</p><p>• |f(|x|)| = 0 when f(|x|) = 0, i.e., x² - 2|x| - 3 = 0</p><p>• This gives |x|² - 2|x| - 3 = 0 → (|x| - 3)(|x| + 1) = 0 → |x| = 3</p><p>• So λ = 0 when x = ±3</p><p>• Minimum of λ occurs when f(|x|) is closest to 0. Since f(|x|) ≥ -3, we have λ = |f(|x|)| ∈ [0, ∞)</p><p><strong>Step 5: Key properties of λ = |f(|x|)|</strong></p><p><strong>B)</strong> λ is an EVEN function ✓ (proven in Step 2: |f(|-x|)| = |f(|x|)|)</p><p><strong>C)</strong> λ has MINIMUM value 0 ✓ (occurs at x = ±3)</p><p><strong>D)</strong> λ is CONTINUOUS everywhere ✓ (composition of continuous functions)</p><p>∴ Answer: <strong>B, C, D</strong></p>
Correct Answer: B,C,D