Question:

Step 1: Identify the integral and simplify the limits. We need to evaluate: $$I = \int_{\ln\lambda}^{\ln(1/\lambda)} \frac{f\!\left(\frac{x^2}{4}\right)(f(x) - f(-x))}{g\!\left(\frac{x^2}{4}\right)(g(x) + g(-x))}\,dx$$ First, observe that $\ln(1/\lambda) = -\ln\lambda$. Therefore, the integral becomes: $$I = \int_{\ln\lambda}^{-\ln\lambda} \frac{f\!\left(\frac{x^2}{4}\right)(f(x) - f(-x))}{g\!\left(\frac{x^2}{4}\right)(g(x) + g(-x))}\,dx$$ This can be rewritten with symmetric limits about zero: $$I = \int_{-\ln\lambda}^{\ln\lambda} \frac{f\!\left(\frac{x^2}{4}\right)(f(x) - f(-x))}{g\!\left(\frac{x^2}{4}\right)(g(x) + g(-x))}\,dx$$ Step 2: Define the integrand as a function and check its parity. Let us define: $$h(x) = \frac{f\!\left(\frac{x^2}{4}\right)(f(x) - f(-x))}{g\!\left(\frac{x^2}{4}\right)(g(x) + g(-x))}$$ Now we check whether $h(x)$ is an odd or even function by computing $h(-x)$: $$h(-x) = \frac{f\!\left(\frac{(-x)^2}{4}\right)(f(-x) - f(-(-x)))}{g\!\left(\frac{(-x)^2}{4}\right)(g(-x) + g(-(-x)))}$$ Step 3: Simplify $h(-x)$ using properties of the argument. Since $(-x)^2 = x^2$, we have: $$h(-x) = \frac{f\!\left(\frac{x^2}{4}\right)(f(-x) - f(x))}{g\!\left(\frac{x^2}{4}\right)(g(-x) + g(x))}$$ Factoring out $-1$ from the numerator: $$h(-x) = \frac{f\!\left(\frac{x^2}{4}\right) \cdot (-(f(x) - f(-x)))}{g\!\left(\frac{x^2}{4}\right)(g(x) + g(-x))}$$ $$h(-x) = -\frac{f\!\left(\frac{x^2}{4}\right)(f(x) - f(-x))}{g\!\left(\frac{x^2}{4}\right)(g(x) + g(-x))} = -h(x)$$ Step 4: Apply the property of odd functions over symmetric intervals. Since $h(x)$ is an odd function (satisfying $h(-x) = -h(x)$) and the integral is evaluated over a symmetric interval $[-\ln\lambda, \ln\lambda]$ about the origin, we use the fundamental property: $$\int_{-a}^{a} h(x)\,dx = 0 \quad \text{for any odd function } h(x)$$ Therefore: $$I = 0$$ The answer is **Option 4: $0$** <div class="key-concept"><strong>Key Concept:</strong> Property of definite integrals: integral of an odd function over a symmetric interval is zero</div> <div class="trap-box"><strong>Trap:</strong> Students may not recognize that $\ln(1/\lambda) = -\ln\lambda$ making the interval symmetric, or may not check that the integrand is an odd function.</div>