<p>If the function \(f : R - \{1, -1\} \to A\) defined by \(f(x) = \dfrac{x^2}{1 - x^2}\), is surjective, then <i>A</i> is equal to:</p>
Step-by-Step Solution
Key Concept: For a function to be surjective onto set A, the range of f must equal A exactly. Find the range by analyzing all possible output values that f(x) can achieve as x varies over its domain R - {1, -1}.
<p><strong>Step 1:</strong> Let y = f(x) = x²/(1 - x²). Solve for x in terms of y.</p><p>y(1 - x²) = x²</p><p>y - yx² = x²</p><p>y = x² + yx² = x²(1 + y)</p><p>x² = y/(1 + y)</p><p><strong>Step 2:</strong> For x to be real, we need x² ≥ 0, so y/(1 + y) ≥ 0.</p><p>This inequality holds when: y ∈ [-∞, -1) ∪ [0, ∞)</p><p><strong>Step 3:</strong> Also, y ≠ -1 because if y = -1, then x²(1 + y) = 0, giving x = 0, but f(0) = 0 ≠ -1. Also, the denominator (1 + y) cannot be zero.</p><p><strong>Step 4:</strong> Check boundary: as x → 1⁻ or x → 1⁺, we have x² → 1, so f(x) = x²/(1-x²) → -∞. As x → ∞, f(x) → -1⁻. As x = 0, f(0) = 0. For x small, f(x) ≈ x² ≥ 0.</p><p>Therefore, the range is [0, ∞) ∪ (-∞, -1).</p><p>∴ Answer: A = [0, ∞) ∪ (-∞, -1) or equivalently ℝ - (-1, 0)</p>
Correct Answer: C