<p>If \(f(x)\) and \(g(x)\) are both continuous functions then the value of \[\displaystyle\int_{\ln \lambda}^{\ln(1/\lambda)} \dfrac{f\!\left(\dfrac{x^2}{4}\right)(f(x) - f(-x))}{g\!\left(\dfrac{x^2}{4}\right)(g(x) + g(-x))} \, dx\] is equal to:</p>
Step-by-Step Solution
Key Concept: The integrand contains both odd and even function components: (f(x) - f(-x)) is odd while (g(x) + g(-x)) is even. When an odd function is divided by an even function and the limits are symmetric about zero (−ln λ to ln λ), the integral equals zero.
<p><strong>Step 1:</strong> Rewrite the limits. Note that ∫₍ₗₙ λ₎^(ln(1/λ)) = ∫₍ₗₙ λ₎^(-ln λ) by using ln(1/λ) = −ln λ. Reverse to get: −∫₍₋ₗₙ λ₎^(ln λ)</p><p><strong>Step 2:</strong> Analyze the integrand's parity. Let h(x) = f(x²/4)(f(x) − f(−x)) / [g(x²/4)(g(x) + g(−x))]</p><p><strong>Step 3:</strong> Check h(−x): The numerator has f(−x)² (even) times [f(−x) − f(x)] = −[f(x) − f(−x)] (odd), giving odd behavior. The denominator has g(−x)² (even) times [g(−x) + g(x)] = [g(x) + g(−x)] (even), staying even. Therefore h(−x) = −h(x), making h(x) an <strong>odd function</strong>.</p><p><strong>Step 4:</strong> Apply the odd function property. For any odd function integrated over a symmetric interval [−a, a], the integral equals zero.</p><p>∴ Answer: <strong>0</strong> (Option D)</p>
Correct Answer: D