If f(x) = ` log_(e) ((1-x)/(1+x))` , then `f((2x)/(1 + x^(2)))` is equal to :

Options

A. `[f(x)]^(2)`
B. 2f(x)
C. 4f(x)
D. None of these

Answer

To solve the problem, we need to find \( f\left(\frac{2x}{1 + x^2}\right) \) given that \( f(x) = \log_e\left(\frac{1 - x}{1 + x}\right) \).

### Step-by-Step Solution:

1. **Substitute into the function**:
We start by substituting \( \frac{2x}{1 + x^2} \) into the function \( f(x) \):
\[ f\left(\frac{2x}{1 + x^2}\right) = \log_e\left(\frac{1 - \frac{2x}{1 + x^2}}{1 + \frac{2x}{1 + x^2}}\right) \]

2. **Simplify the numerator**:
The numerator becomes:
\[ 1 - \frac{2x}{1 + x^2} = \frac{(1 + x^2) - 2x}{1 + x^2} = \frac{1 + x^2 - 2x}{1 + x^2} \]

3. **Simplify the denominator**:
The denominator becomes:
\[ 1 + \frac{2x}{1 + x^2} = \frac{(1 + x^2) + 2x}{1 + x^2} = \frac{1 + x^2 + 2x}{1 + x^2} \]

4. **Combine the fractions**:
Now we can combine the numerator and denominator:
\[ f\left(\frac{2x}{1 + x^2}\right) = \log_e\left(\frac{\frac{1 + x^2 - 2x}{1 + x^2}}{\frac{1 + x^2 + 2x}{1 + x^2}}\right) = \log_e\left(\frac{1 + x^2 - 2x}{1 + x^2 + 2x}\right) \]

5. **Factor the expressions**:
Notice that:
\[ 1 + x^2 - 2x = (1 - x)^2 \quad \text{and} \quad 1 + x^2 + 2x = (1 + x)^2 \]
Thus, we can rewrite the logarithm:
\[ f\left(\frac{2x}{1 + x^2}\right) = \log_e\left(\frac{(1 - x)^2}{(1 + x)^2}\right) \]

6. **Use the properties of logarithms**:
Using the property of logarithms \( \log(a^b) = b \log(a) \):
\[ f\left(\frac{2x}{1 + x^2}\right) = 2 \log_e\left(\frac{1 - x}{1 + x}\right) \]

7. **Recognize the original function**:
Since \( f(x) = \log_e\left(\frac{1 - x}{1 + x}\right) \), we can express our result as:
\[ f\left(\frac{2x}{1 + x^2}\right) = 2 f(x) \]

### Final Answer:
Thus, the final result is:
\[ f\left(\frac{2x}{1 + x^2}\right) = 2 f(x) \]