<p>Given \( f(x) = \dfrac{x^2}{1-x^2} \). The set <i>A</i> should be chosen so that <i>f</i> is a function from <i>A</i> to \([0, \infty)\). Which of the following is the correct set <i>A</i>?</p>
Step-by-Step Solution
Key Concept: For f to be a function from A to [0,∞), the domain A must consist of all x where f(x) is defined AND f(x) ≥ 0. Since f(x) = x²/(1-x²), we need 1-x² ≠ 0 AND x²/(1-x²) ≥ 0 simultaneously.
<p><strong>Step 1:</strong> For f to be defined: 1 - x² ≠ 0 ⟹ x ≠ ±1</p><p><strong>Step 2:</strong> For f(x) ≥ 0: x²/(1-x²) ≥ 0. Since x² ≥ 0 always, we need 1-x² > 0 ⟹ x² < 1 ⟹ |x| < 1</p><p><strong>Step 3:</strong> Combining both conditions: x ∈ (-1, 1)</p><p><strong>Step 4:</strong> Verification: When x ∈ (-1,1), numerator x² ≥ 0 and denominator 1-x² > 0, so f(x) ≥ 0 ✓</p><p>∴ Answer: A = (-1, 1) or [-1, 1) excluding endpoints depending on options</p>
Correct Answer: A