<p>The equation \(|x+1|\,|x-1| = a^2 - 2a - 3\) can have real solution in x if a belongs to</p>
<p>(a) \([1-\sqrt{5}, -1]\)</p>
<p>(b) \([-1, 3]\)</p>
<p>(c) \([3, 1+\sqrt{5}]\)</p>
<p>(d) \([1+\sqrt{3}, 1+\sqrt{5}]\)</p>
Step-by-Step Solution
Key Concept: The left side |x+1||x-1| = |x²-1| always has minimum value 0, so we need a² - 2a - 3 ≥ 0. Additionally, |x²-1| can achieve any non-negative value, making the range constraint a² - 2a - 3 ≥ 0 the only requirement for real solutions to exist.
<p><strong>Step 1:</strong> Analyze the left side: |x+1||x-1| = |x+1||x-1| = |(x+1)(x-1)| = |x² - 1|</p><p><strong>Step 2:</strong> Since |x² - 1| ≥ 0 for all real x, the minimum value is 0 (achieved when x = ±1).</p><p><strong>Step 3:</strong> As |x| → ∞, we have |x² - 1| → ∞, so the left side can take any value in [0, ∞).</p><p><strong>Step 4:</strong> For real solutions in x to exist, the RHS must lie in the range [0, ∞): a² - 2a - 3 ≥ 0</p><p><strong>Step 5:</strong> Factor: (a - 3)(a + 1) ≥ 0</p><p><strong>Step 6:</strong> Solving the inequality: a ≤ -1 or a ≥ 3</p><p>∴ Answer: A (or equivalent: a ∈ (-∞, -1] ∪ [3, ∞))</p>
Correct Answer: A