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

Options

A. `2f(x)`
B. `2f(x^(2))`
C. `(f(x))^(2)`
D. `-2f(x)`

Explanation

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

### Step-by-Step Solution:

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

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

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

4. **Combine the results**:
\[ f\left(\frac{2x}{1+x^2}\right) = \log_e\left(\frac{\frac{(1-x)^2}{1+x^2}}{\frac{(1+x)^2}{1+x^2}}\right) = \log_e\left(\frac{(1-x)^2}{(1+x)^2}\right) \]

5. **Use properties of logarithms**:
\[ f\left(\frac{2x}{1+x^2}\right) = \log_e\left((1-x)^2\right) - \log_e\left((1+x)^2\right) = 2\log_e(1-x) - 2\log_e(1+x) \]

6. **Factor out the common term**:
\[ f\left(\frac{2x}{1+x^2}\right) = 2\left(\log_e(1-x) - \log_e(1+x)\right) \]

7. **Recognize the function**:
\[ \log_e(1-x) - \log_e(1+x) = \log_e\left(\frac{1-x}{1+x}\right) = f(x) \]

8. **Final result**:
\[ f\left(\frac{2x}{1+x^2}\right) = 2f(x) \]

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