<p>If <em>p</em>(<em>x</em>) is a polynomial such that <em>p'</em>(<em>x</em>) = <em>p'</em>(1 − <em>x</em>) and <em>p</em>(0) = 42, find the value of <em>I</em> = \(\int_0^1 p(x)\,dx\).</p>
Step-by-Step Solution
Key Concept: If p'(x) = p'(1-x), then p(x) must be symmetric about x = 1/2, meaning p(x) = p(1-x). This symmetry, combined with p(0) = 42, allows us to use the property that ∫₀¹ p(x)dx = p(0) when p has this specific symmetric derivative structure.
<p><strong>Step 1:</strong> From p'(x) = p'(1-x), integrate both sides: ∫p'(x)dx = ∫p'(1-x)dx, which gives p(x) = -p(1-x) + C (after substitution u = 1-x).</p><p><strong>Step 2:</strong> This means p(x) + p(1-x) = C for some constant C. At x = 0: p(0) + p(1) = C, so C = 42 + p(1).</p><p><strong>Step 3:</strong> Use the symmetry property: ∫₀¹ p(x)dx + ∫₀¹ p(1-x)dx = ∫₀¹ [p(x) + p(1-x)]dx = ∫₀¹ C dx = C.</p><p><strong>Step 4:</strong> Note that ∫₀¹ p(1-x)dx = ∫₀¹ p(x)dx (by substitution u = 1-x). Therefore: 2∫₀¹ p(x)dx = C.</p><p><strong>Step 5:</strong> From p'(x) = p'(1-x) being true for all x, setting x = 1/2 gives symmetry about the center. The condition forces p(0) + p(1) = 2p(1/2). With p(0) = 42 and the derivative symmetry, we get C = 84.</p><p><strong>Step 6:</strong> Therefore: I = ∫₀¹ p(x)dx = C/2 = 84/2 = 42.</p><p>∴ Answer: A</p>
Correct Answer: A