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

Options

A. 2
B. `-3`
C. `-1`
D. 1

Answer

To solve the problem, we need to find the value of \( \frac{I_2}{I_1} \) where:

- \( I_1 = \int_{f(-a)}^{f(a)} x g(x(1-x)) \, dx \)
- \( I_2 = \int_{f(-a)}^{f(a)} g(x(1-x)) \, dx \)

Given the function \( f(x) = \frac{e^x}{1 + e^x} \), we first need to evaluate \( f(-a) \) and \( f(a) \).

### Step 1: Calculate \( f(-a) \) and \( f(a) \)

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

2. **Calculate \( f(-a) \)**:
\[ f(-a) = \frac{e^{-a}}{1 + e^{-a}} = \frac{1}{e^a + 1} \]

### Step 2: Use the property of definite integrals

We can use the property of definite integrals:
\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]

In our case, we replace \( x \) with \( f(a) + f(-a) - x \).

### Step 3: Evaluate \( I_1 \)

Using the property on \( I_1 \):
\[ I_1 = \int_{f(-a)}^{f(a)} x g(x(1-x)) \, dx \]
Substituting \( x \) with \( f(a) + f(-a) - x \):
\[ I_1 = \int_{f(-a)}^{f(a)} (f(a) + f(-a) - x) g(x(1-x)) \, dx \]

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

This can be expressed as:
\[ I_1 = \int_{f(-a)}^{f(a)} (f(a) + f(-a)) g(x(1-x)) \, dx - I_1 \]
Thus,
\[ 2I_1 = (f(a) + f(-a)) \int_{f(-a)}^{f(a)} g(x(1-x)) \, dx \]
This implies:
\[ I_1 = \frac{(f(a) + f(-a))}{2} I_2 \]

### Step 5: Calculate \( \frac{I_2}{I_1} \)

Now, we can find \( \frac{I_2}{I_1} \):
\[ \frac{I_2}{I_1} = \frac{2}{f(a) + f(-a)} \]

### Step 6: Evaluate \( f(a) + f(-a) \)

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

### Step 7: Final Calculation

Thus,
\[ \frac{I_2}{I_1} = \frac{2}{1} = 2 \]

### Conclusion

The final value of \( \frac{I_2}{I_1} \) is \( 2 \).