<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 functions (f(x) - f(-x)) in numerator and even functions (g(x) + g(-x)) in denominator. Recognize that the limits are symmetric about zero: ln(1/λ) = -ln(λ), making the integration domain [-ln(λ), ln(λ)]. The product of an even function with an odd function yields an odd function, whose integral over symmetric limits is zero.
<p><strong>Step 1:</strong> Identify the parity of components. Note that f(x) - f(-x) is odd (difference of f and its reflection), and f(x/2²) is even. So the numerator f(x²/4)·[f(x) - f(-x)] is even × odd = odd.</p><p><strong>Step 2:</strong> Similarly, g(x) + g(-x) is the even part of g(x), so the denominator g(x²/4)·[g(x) + g(-x)] is even × even = even.</p><p><strong>Step 3:</strong> Therefore the integrand = odd/even = odd function.</p><p><strong>Step 4:</strong> Observe the limits: from ln(λ) to ln(1/λ) = from ln(λ) to -ln(λ). These are symmetric about zero.</p><p><strong>Step 5:</strong> By the property of odd functions: ∫₍₋ₐ₎ᵃ [odd function] dx = 0.</p><p><strong>Step 6:</strong> Therefore, ∫_{ln(λ)}^{ln(1/λ)} = ∫_{-ln(λ)}^{ln(λ)} [odd function] dx = 0.</p><p>∴ Answer: <strong>D (0)</strong></p>
Correct Answer: D