Question

If \(f(x) = \frac{e^x}{1+e^x}, I_1 = \int_{-a}^a xg(x(1-x)) \, dx\) and \(I_2 = \int_{-a}^a g(x(1-x)) \, dx\), then the value of \(\frac{I_2}{I_1}\) is:

• A. -1
• B. -3
• C. 2
• D. 1

Hint:

For integrals with symmetric integrands, pay attention to how the variable behaves. In cases with even symmetry, terms involving odd powers of the variable will cancel out.

Answer 1

The Correct Option is C

Step 1: We are given the function \( f(x) = \frac{e^x}{1+e^x} \) and asked to find the ratio \(\frac{I_2}{I_1}\), where:

\[ I_1 = \int_{-a}^a x g(x(1-x)) \, dx \quad \text{and} \quad I_2 = \int_{-a}^a g(x(1-x)) \, dx. \]

Step 2: We analyze the integrands' symmetry. \( g(x(1-x)) \) is symmetric on \([-a, a]\), simplifying the evaluation of \( I_2 \).

Step 3: Considering the presence of \( x \) in \( I_1 \) and the symmetry of the integrand in \( I_2 \), the ratio \(\frac{I_2}{I_1}\) is determined to be 2.