<p>If \(f(x)\) is a differentiable function defined for all positive real numbers such that \(xf(x) = x + \displaystyle\int_1^x f(t)\,dt\), then the value of \(\displaystyle\sum_{k=1}^{10} f(e^k)\) is:</p>
Step-by-Step Solution
Key Concept: Differentiate both sides of the functional equation xf(x) = x + ∫₁ˣ f(t)dt to find a differential equation for f(x), then solve it using the initial condition obtained by substituting x = 1.
<p><strong>Step 1:</strong> Start with xf(x) = x + ∫₁ˣ f(t)dt. Find f(1) by substituting x = 1: 1·f(1) = 1 + 0, so f(1) = 1.</p><p><strong>Step 2:</strong> Differentiate both sides with respect to x: f(x) + xf'(x) = 1 + f(x), which simplifies to xf'(x) = 1, giving f'(x) = 1/x.</p><p><strong>Step 3:</strong> Integrate: f(x) = ln(x) + C. Using f(1) = 1: 1 = ln(1) + C = 0 + C, so C = 1. Therefore f(x) = ln(x) + 1.</p><p><strong>Step 4:</strong> Verify: xf(x) = x(ln(x) + 1) = x·ln(x) + x, and x + ∫₁ˣ (ln(t) + 1)dt = x + [t·ln(t) - t + t]₁ˣ = x + [x·ln(x) - x + x - (0 - 1 + 1)] = x + x·ln(x). ✓</p><p><strong>Step 5:</strong> Calculate f(eᵏ) = ln(eᵏ) + 1 = k + 1.</p><p><strong>Step 6:</strong> ∑ₖ₌₁¹⁰ f(eᵏ) = ∑ₖ₌₁¹⁰ (k + 1) = ∑ₖ₌₂¹¹ k = (2 + 3 + ... + 11) = (1 + 2 + ... + 11) - 1 = 66 - 1 = 65.</p><p>∴ Answer: B (65)</p>
Correct Answer: B