<p>If \(\int_{a}^{b} x f(x)\, dx = \frac{a+b}{2} \int_{a}^{b} f(x)\, dx\), then \(f(a+b-x)\) equals:</p>
Step-by-Step Solution
Key Concept: Use the property that if an integral equation holds for all intervals [a,b], substitute x → (a+b-x) to create a second equation, then solve the system of two equations to find f(a+b-x) in terms of f(x).
<p><strong>Step 1:</strong> Given condition: $\int_{a}^{b} x f(x)\, dx = \frac{a+b}{2} \int_{a}^{b} f(x)\, dx$ ... (1)</p><p><strong>Step 2:</strong> Substitute $x \to (a+b-x)$ in equation (1). Note that $dx \to -dx$, and limits reverse from $[a,b]$ to $[b,a]$:</p><p>$\int_{b}^{a} (a+b-x) f(a+b-x)\, dx = \frac{a+b}{2} \int_{b}^{a} f(a+b-x)\, dx$</p><p><strong>Step 3:</strong> Reverse limits (multiply by -1):</p><p>$\int_{a}^{b} (a+b-x) f(a+b-x)\, dx = \frac{a+b}{2} \int_{a}^{b} f(a+b-x)\, dx$ ... (2)</p><p><strong>Step 4:</strong> Expand the LHS of (2):</p><p>$(a+b)\int_{a}^{b} f(a+b-x)\, dx - \int_{a}^{b} x f(a+b-x)\, dx = \frac{a+b}{2} \int_{a}^{b} f(a+b-x)\, dx$</p><p><strong>Step 5:</strong> Rearranging:</p><p>$\int_{a}^{b} x f(a+b-x)\, dx = \frac{a+b}{2} \int_{a}^{b} f(a+b-x)\, dx$ ... (3)</p><p><strong>Step 6:</strong> Comparing equations (1) and (3): both have the form $\int_{a}^{b} x g(x)\, dx = \frac{a+b}{2} \int_{a}^{b} g(x)\, dx$. This is satisfied when $g(x) = f(x) = f(a+b-x)$ (by uniqueness for the given functional form).</p><p>∴ <strong>Answer:</strong> $f(a+b-x) = f(x)$</p>
Correct Answer: A