If `f(a) = log"" (1 + a)/(1-a)` then `f((2a)/(1+a^(2)))` is equal to

Options

A. 0
B. 1
C. `f(a)`
D. `2f(a)`

Answer

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

### Step-by-Step Solution:

1. **Substituting the Value into the Function**:
\[ f\left(\frac{2a}{1 + a^2}\right) = \log\left(\frac{1 + \frac{2a}{1 + a^2}}{1 - \frac{2a}{1 + a^2}}\right) \]

2. **Simplifying the Numerator**:
\[ 1 + \frac{2a}{1 + a^2} = \frac{(1 + a^2) + 2a}{1 + a^2} = \frac{1 + 2a + a^2}{1 + a^2} = \frac{(a + 1)^2}{1 + a^2} \]

3. **Simplifying the Denominator**:
\[ 1 - \frac{2a}{1 + a^2} = \frac{(1 + a^2) - 2a}{1 + a^2} = \frac{1 - 2a + a^2}{1 + a^2} = \frac{(1 - a)^2}{1 + a^2} \]

4. **Combining the Results**:
\[ f\left(\frac{2a}{1 + a^2}\right) = \log\left(\frac{\frac{(a + 1)^2}{1 + a^2}}{\frac{(1 - a)^2}{1 + a^2}}\right) = \log\left(\frac{(a + 1)^2}{(1 - a)^2}\right) \]

5. **Using Logarithmic Properties**:
\[ \log\left(\frac{(a + 1)^2}{(1 - a)^2}\right) = \log\left((a + 1)^2\right) - \log\left((1 - a)^2\right) = 2\log(a + 1) - 2\log(1 - a) \]
\[ = 2\left(\log(a + 1) - \log(1 - a)\right) = 2f(a) \]

### Final Result:
Thus, we have:
\[ f\left(\frac{2a}{1 + a^2}\right) = 2f(a) \]