Let f be a positive function. Let `I_(1) int_(1-k)^(k) x.f {x(1-x)} dx, I_(2)= int_(1-k)^(k) f{x(1-x)}dx` where `2k-1 gt 0`, then `I_(1)//I_(2)` is

Options

A. 2
B. k
C. `1//2`
D. 1

Answer

To solve the problem, we need to evaluate the ratios of the integrals \( I_1 \) and \( I_2 \) defined as follows:

\[ I_1 = \int_{1-k}^{k} x f(x)(1-x) \, dx \]
\[ I_2 = \int_{1-k}^{k} f(x)(1-x) \, dx \]

where \( 2k - 1 > 0 \).

### Step 1: Change of Variable for \( I_1 \)

We will use the property of definite integrals that states:

\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]

In our case, \( a = 1-k \) and \( b = k \), so \( a + b = (1-k) + k = 1 \). Thus, we can write:

\[ I_1 = \int_{1-k}^{k} x f(x)(1-x) \, dx = \int_{1-k}^{k} (1-x) f(1-x)(x) \, dx \]

### Step 2: Rewrite \( I_1 \)

Substituting \( 1-x \) for \( x \) in the integral:

\[ I_1 = \int_{1-k}^{k} (1-x) f(1-x) x \, dx \]

This gives us two expressions for \( I_1 \):

1. \( I_1 = \int_{1-k}^{k} x f(x)(1-x) \, dx \)
2. \( I_1 = \int_{1-k}^{k} (1-x) f(1-x) x \, dx \)

### Step 3: Combine the Integrals

Adding both expressions for \( I_1 \):

\[ 2I_1 = \int_{1-k}^{k} \left[ x f(x)(1-x) + (1-x) f(1-x) x \right] \, dx \]

Factoring out \( x \):

\[ 2I_1 = \int_{1-k}^{k} x \left[ f(x)(1-x) + f(1-x)(1-x) \right] \, dx \]

### Step 4: Relate \( I_1 \) and \( I_2 \)

Now, we can express \( I_2 \):

\[ I_2 = \int_{1-k}^{k} f(x)(1-x) \, dx \]

Using the symmetry property, we can also express \( I_2 \) in terms of \( f(1-x) \):

\[ I_2 = \int_{1-k}^{k} f(1-x)(1-x) \, dx \]

### Step 5: Establish the Ratio \( \frac{I_1}{I_2} \)

From the earlier steps, we have:

\[ 2I_1 = I_2 \implies I_1 = \frac{1}{2} I_2 \]

Thus, the ratio \( \frac{I_1}{I_2} \) is:

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

### Final Answer

The value of \( \frac{I_1}{I_2} \) is:

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