<p>If the expression \([mx - 1 + (1/x)]\) is non-negative for all positive real \(x\), then the minimum value of \(m\) must be</p>
Step-by-Step Solution
Key Concept: For f(x) = mx - 1 + 1/x to be non-negative for all x > 0, we need min(f(x)) ≥ 0. Find the critical point using calculus and apply the non-negativity condition at the minimum.
<p><strong>Step 1:</strong> Let f(x) = mx - 1 + 1/x for x > 0. For f(x) ≥ 0 for all positive x, we need min f(x) ≥ 0.</p><p><strong>Step 2:</strong> Find critical points: f'(x) = m - 1/x² = 0 ⟹ x² = 1/m ⟹ x = 1/√m (valid only if m > 0).</p><p><strong>Step 3:</strong> Check second derivative: f''(x) = 2/x³ > 0 for x > 0, confirming this is a minimum.</p><p><strong>Step 4:</strong> At x = 1/√m, the minimum value is: f(1/√m) = m(1/√m) - 1 + √m = √m - 1 + √m = 2√m - 1.</p><p><strong>Step 5:</strong> For non-negativity: 2√m - 1 ≥ 0 ⟹ √m ≥ 1/2 ⟹ m ≥ 1/4.</p><p><strong>Alternative (AM-GM):</strong> For x > 0, by AM-GM: mx + 1/x ≥ 2√(m · 1/x · x) requires careful balancing. At equality in weighted AM-GM, mx = 1/x gives the minimum, so 2√m - 1 ≥ 0.</p><p>∴ Minimum value of m = <strong>1/4</strong></p>
Correct Answer: C