If `(f(x))^(2)xxf((1-x)/(1+x))=64x AAx in D_(f),` then
The domain of `f(x)` is

Options

A. `[0,oo)`
B. `R-{1}`
C. `(-oo,oo)`
D. None of these

Explanation

To solve the problem, we start with the equation given:

\[ (f(x))^2 \cdot f\left(\frac{1-x}{1+x}\right) = 64x \]

### Step 1: Substitute \( x \) with \( \frac{1-x}{1+x} \)

We will replace \( x \) in the original equation with \( \frac{1-x}{1+x} \):

\[ f\left(\frac{1-x}{1+x}\right)^2 \cdot f\left(\frac{1 - \frac{1-x}{1+x}}{1 + \frac{1-x}{1+x}}\right) = 64 \cdot \frac{1-x}{1+x} \]

### Step 2: Simplify the second term

Now, we need to simplify the second term in the equation:

\[ \frac{1 - \frac{1-x}{1+x}}{1 + \frac{1-x}{1+x}} = \frac{\frac{(1+x) - (1-x)}{1+x}}{\frac{(1+x) + (1-x)}{1+x}} = \frac{\frac{2x}{1+x}}{\frac{2}{1+x}} = x \]

### Step 3: Rewrite the equation

Now we can rewrite the equation as:

\[ f\left(\frac{1-x}{1+x}\right)^2 \cdot f(x) = 64 \cdot \frac{1-x}{1+x} \]

### Step 4: Divide the two equations

Now we have two equations:

1. \( (f(x))^2 \cdot f\left(\frac{1-x}{1+x}\right) = 64x \)
2. \( f\left(\frac{1-x}{1+x}\right)^2 \cdot f(x) = 64 \cdot \frac{1-x}{1+x} \)

We can divide the first equation by the second:

\[ \frac{(f(x))^2 \cdot f\left(\frac{1-x}{1+x}\right)}{f\left(\frac{1-x}{1+x}\right)^2 \cdot f(x)} = \frac{64x}{64 \cdot \frac{1-x}{1+x}} \]

This simplifies to:

\[ \frac{f(x)}{f\left(\frac{1-x}{1+x}\right)} = \frac{x(1+x)}{1-x} \]

### Step 5: Find the form of \( f(x) \)

From the above equation, we can express \( f(x) \) in terms of \( x \):

\[ f(x) = k \cdot \frac{x(1+x)}{1-x} \]

where \( k \) is a constant.

### Step 6: Determine the domain of \( f(x) \)

The function \( f(x) \) will be undefined when the denominator is zero:

\[ 1 - x = 0 \implies x = 1 \]

Thus, the domain of \( f(x) \) is all real numbers except \( x = 1 \).

### Final Answer

The domain of \( f(x) \) is:

\[ D_f = \mathbb{R} \setminus \{1\} \]