<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 integral equation to convert it into a differential equation, then solve for f(x) using initial conditions obtained from the original equation.
<p><strong>Step 1:</strong> Differentiate both sides of xf(x) = x + ∫₁ˣ f(t)dt with respect to x.</p><p>d/dx[xf(x)] = d/dx[x + ∫₁ˣ f(t)dt]</p><p>f(x) + xf'(x) = 1 + f(x)</p><p><strong>Step 2:</strong> Simplify to get the differential equation: xf'(x) = 1, so f'(x) = 1/x.</p><p>Therefore: f(x) = ln(x) + C</p><p><strong>Step 3:</strong> Find constant C using the initial condition. At x = 1: 1·f(1) = 1 + ∫₁¹ f(t)dt = 1 + 0.</p><p>So f(1) = 1, which gives: 1 = ln(1) + C = 0 + C, thus C = 1.</p><p><strong>Step 4:</strong> Therefore f(x) = ln(x) + 1.</p><p><strong>Step 5:</strong> Calculate the sum: ∑ₖ₌₁¹⁰ f(eᵏ) = ∑ₖ₌₁¹⁰ [ln(eᵏ) + 1] = ∑ₖ₌₁¹⁰ [k + 1]</p><p>= ∑ₖ₌₁¹⁰ (k + 1) = (1+1) + (2+1) + ... + (10+1) = 2 + 3 + 4 + ... + 11</p><p>= ∑ₖ₌₂¹¹ k = [∑ₖ₌₁¹¹ k] - 1 = [11·12/2] - 1 = 66 - 1 = 65</p><p>∴ Answer: <strong>65</strong></p>
Correct Answer: A