<p>The real number <i>x</i> when added to its inverse gives the minimum value of the sum at <i>x</i> equal to</p>
Step-by-Step Solution
Key Concept: For a function f(x) = x + 1/x where x > 0, the minimum occurs when the first derivative equals zero. By AM-GM inequality, x + 1/x ≥ 2√(x·1/x) = 2, with equality when x = 1/x, giving x = 1.
<p><strong>Step 1:</strong> Let f(x) = x + 1/x. We need to find the value of x that minimizes this sum (assuming x > 0).</p><p><strong>Step 2:</strong> Using AM-GM Inequality: For positive numbers a and b, (a+b)/2 ≥ √(ab). Thus: (x + 1/x)/2 ≥ √(x · 1/x) = 1, which gives x + 1/x ≥ 2.</p><p><strong>Step 3:</strong> Equality holds in AM-GM when x = 1/x, solving: x² = 1 → x = 1 (for x > 0).</p><p><strong>Step 4:</strong> Verify using calculus: f'(x) = 1 - 1/x² = 0 → x² = 1 → x = 1. Since f''(x) = 2/x³ > 0 at x = 1, this is a minimum.</p><p><strong>Step 5:</strong> The minimum value is f(1) = 1 + 1/1 = 2, occurring at x = 1.</p><p>∴ Answer: B (x = 1)</p>
Correct Answer: B