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: The function \( f(x) = \frac{e^x}{1+e^x} \) is given, and we are tasked with evaluating 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: To evaluate the integrals, we first look at the symmetry of the integrands. The function \( g(x(1-x)) \) is symmetric in the interval \([-a, a]\), and thus, the integral \( I_2 \) becomes straightforward.

Step 3: Given that \( x \) appears in \( I_1 \), and the symmetry of the integrand in \( I_2 \) cancels out the effect of \( x \), the value of \(\frac{I_2}{I_1}\) simplifies to 2.