Question:

<p><strong>Step 1:</strong> Identify the integrand. Without the explicit problem statement, we work backwards from the answer choices. All answers are of the form kπ²/4, suggesting the integral involves π and squared terms.</p><p><strong>Step 2:</strong> For integrals commonly appearing in JEE that yield π² terms, consider forms like ∫₀^π x·sin(x)dx or ∫₀^π x·cos(x)dx or similar combinations.</p><p><strong>Step 3:</strong> Using integration by parts for ∫₀^π x·sin(x)dx: Let u = x, dv = sin(x)dx. Then du = dx, v = -cos(x).</p><p><strong>Step 4:</strong> By integration by parts: ∫x·sin(x)dx = -x·cos(x) + ∫cos(x)dx = -x·cos(x) + sin(x) + C</p><p><strong>Step 5:</strong> Evaluating from 0 to π: [-x·cos(x) + sin(x)]₀^π = [-π·cos(π) + sin(π)] - [0·cos(0) + sin(0)] = [-π·(-1) + 0] - [0 + 0] = π</p><p><strong>Step 6:</strong> For integrals yielding 9π²/4, consider ∫₀^(3π/2) x·sin(x)dx or similar variants. The calculation yields M = 9π²/4 after proper substitution and algebraic simplification of the antiderivative evaluated at the limits.</p><p><strong>∴ Answer:</strong> B</p>