<p>Let \(f\) be a positive function. Let \(I_1 = \int_{1-k}^{k} x f\{x(1-x)\}dx\) and \(I_2 = \int_{1-k}^{k} f\{x(1-x)\}dx\), where \(2k-1 > 0\). Then \(I_1/I_2\) is equal to</p>
Step-by-Step Solution
Key Concept: Use the property that f{x(1-x)} is symmetric about x = 1/2, so the integral of x·f{x(1-x)} can be split using this symmetry to show I₁ = (1/2)I₂.
<p><strong>Step 1:</strong> Recognize the symmetry. Note that the limits satisfy: (1-k) + k = 1, so the interval [1-k, k] is symmetric about x = 1/2.</p><p><strong>Step 2:</strong> Observe that g(x) = f{x(1-x)} satisfies g(1-x) = f{(1-x)(1-(1-x))} = f{(1-x)x} = g(x). Thus f{x(1-x)} is symmetric about x = 1/2.</p><p><strong>Step 3:</strong> For I₁, split the integrand using substitution. Let u = 1-x in the second half:</p><p>I₁ = ∫₁₋ₖᵏ x·f{x(1-x)}dx</p><p><strong>Step 4:</strong> Using the substitution property and symmetry about x = 1/2, write x = (1/2) + [(x - 1/2)]. The odd part (x - 1/2)·f{x(1-x)} integrates to zero over the symmetric interval, leaving:</p><p>I₁ = ∫₁₋ₖᵏ (1/2)·f{x(1-x)}dx = (1/2)I₂</p><p><strong>Step 5:</strong> Therefore, I₁/I₂ = <strong>1/2</strong></p><p>∴ Answer: C</p>
Correct Answer: C