If `f(x)=(e^x)/(1+e^x), I_1=int_(f(-a))^(f(a))xg(x(1-x)dx` and `I_2=int_(f(-a))^(f(a)) g(x(1-x))dx`, then the value of `(I_2)/(I_1)` is (a) `-1` (b) `-2` (c) 2 (d) 1

Options

A. `-1`
B. `-2`
C. `2`
D. `1`

Explanation

To solve the problem, we need to evaluate the integrals \( I_1 \) and \( I_2 \) and find the ratio \( \frac{I_2}{I_1} \).

### Step 1: Define the function and calculate \( f(a) \) and \( f(-a) \)

Given:
\[ f(x) = \frac{e^x}{1 + e^x} \]

Calculating \( f(a) \):
\[ f(a) = \frac{e^a}{1 + e^a} \]

Calculating \( f(-a) \):
\[ f(-a) = \frac{e^{-a}}{1 + e^{-a}} = \frac{1}{e^a + 1} \]

### Step 2: Find the sum \( f(a) + f(-a) \)

Now, we will find \( f(a) + f(-a) \):
\[ f(a) + f(-a) = \frac{e^a}{1 + e^a} + \frac{1}{e^a + 1} = \frac{e^a + 1}{1 + e^a} = 1 \]

### Step 3: Express \( f(-a) \) in terms of \( f(a) \)

Let:
\[ f(a) = \alpha \quad \text{and} \quad f(-a) = 1 - \alpha \]

### Step 4: Evaluate \( I_1 \)

The integral \( I_1 \) is defined as:
\[ I_1 = \int_{f(-a)}^{f(a)} x g(x(1-x)) \, dx \]

Using the property of definite integrals, we can change the variable:
\[ I_1 = \int_{\alpha}^{1 - \alpha} x g(x(1-x)) \, dx \]

### Step 5: Evaluate \( I_2 \)

The integral \( I_2 \) is defined as:
\[ I_2 = \int_{f(-a)}^{f(a)} g(x(1-x)) \, dx \]

### Step 6: Use the property of integrals

Using the substitution \( x = 1 - t \):
\[ I_1 = \int_{\alpha}^{1 - \alpha} (1 - t) g((1 - t)t) \, dt \]

This leads to:
\[ I_1 = \int_{\alpha}^{1 - \alpha} (1 - t) g(t(1-t)) \, dt \]

### Step 7: Relate \( I_1 \) and \( I_2 \)

Now, we can express \( I_2 \) in terms of \( I_1 \):
\[ I_2 = \int_{\alpha}^{1 - \alpha} g(x(1-x)) \, dx \]

From the properties of integrals, we find that:
\[ 2I_1 = I_2 \]

### Step 8: Calculate the ratio \( \frac{I_2}{I_1} \)

Thus, we have:
\[ \frac{I_2}{I_1} = 2 \]

### Final Answer

The value of \( \frac{I_2}{I_1} \) is:
\[ \boxed{2} \]