f(x) is one-one if- (A) x \in (0,\infty) (B) x \in (-\infty,0) (C) x \in (1,\infty) (D) x \in (-\infty,1) - \{0\}
Step-by-Step Solution
Key Concept: A function is one-one (injective) if different inputs produce different outputs, i.e., f(x₁) = f(x₂) ⟹ x₁ = x₂. We need to find where f(x) = x + 1/x is strictly monotonic or check injectivity by examining f'(x).
<p><strong>Step 1:</strong> Assume f(x) = x + 1/x (the standard function for such problems). For f to be one-one, we need: f(x₁) = f(x₂) ⟹ x₁ = x₂.</p><p><strong>Step 2:</strong> Check using calculus. Find f'(x) = 1 - 1/x² = (x² - 1)/x².</p><p><strong>Step 3:</strong> Analyze the sign of f'(x):</p><ul><li>When x ∈ (0, 1): x² < 1, so f'(x) < 0 (strictly decreasing)</li><li>When x ∈ (1, ∞): x² > 1, so f'(x) > 0 (strictly increasing)</li><li>When x ∈ (-∞, -1): x² > 1, so f'(x) > 0 (strictly increasing)</li><li>When x ∈ (-1, 0): x² < 1, so f'(x) < 0 (strictly decreasing)</li></ul></p><p><strong>Step 4:</strong> A strictly monotonic function is always one-one. Among the given options, x ∈ (1, ∞) is where f'(x) > 0 throughout, making f strictly increasing and hence one-one.</p><p><strong>Step 5:</strong> Verify option (C) x ∈ (1, ∞): For all x in this interval, f is strictly increasing, so f(x₁) = f(x₂) ⟹ x₁ = x₂. ✓</p><p><strong>∴ Answer:</strong> C</p>
Correct Answer: C