Given `f (x) =log {(1 + x)//(1-x)}`and `g(x) = (3x+x^3)//(1+3x^2).` Then f (g(x)] equals

Options

A. `-f(x)`
B. `3[f(x)]`
C. `[f(x)]^3`
D. None of these

Answer

To solve the problem, we need to evaluate \( f(g(x)) \) where \( f(x) = \log\left(\frac{1+x}{1-x}\right) \) and \( g(x) = \frac{3x + x^3}{1 + 3x^2} \).

### Step-by-Step Solution:

1. **Substitute \( g(x) \) into \( f(x) \)**:
\[ f(g(x)) = f\left(\frac{3x + x^3}{1 + 3x^2}\right) \]
This means we will replace \( x \) in \( f(x) \) with \( g(x) \).

2. **Write out \( f(g(x)) \)**:
\[ f(g(x)) = \log\left(\frac{1 + \frac{3x + x^3}{1 + 3x^2}}{1 - \frac{3x + x^3}{1 + 3x^2}}\right) \]

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

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

5. **Combine the results**:
\[ f(g(x)) = \log\left(\frac{\frac{1 + 3x + 3x^2 + x^3}{1 + 3x^2}}{\frac{1 - 3x + 3x^2 - x^3}{1 + 3x^2}}\right) \]
The \( 1 + 3x^2 \) cancels out:
\[ f(g(x)) = \log\left(\frac{1 + 3x + 3x^2 + x^3}{1 - 3x + 3x^2 - x^3}\right) \]

6. **Factor the numerator and denominator**:
The numerator can be factored as:
\[ 1 + 3x + 3x^2 + x^3 = (x + 1)^3 \]
The denominator can be factored as:
\[ 1 - 3x + 3x^2 - x^3 = (1 - x)^3 \]

7. **Final expression**:
\[ f(g(x)) = \log\left(\frac{(x + 1)^3}{(1 - x)^3}\right) \]
This can be simplified using properties of logarithms:
\[ f(g(x)) = 3 \log\left(\frac{x + 1}{1 - x}\right) \]

8. **Recognize the function**:
Since \( f(x) = \log\left(\frac{1+x}{1-x}\right) \), we can express our final result as:
\[ f(g(x)) = 3f(x) \]

### Conclusion:
Thus, the final answer is:
\[ f(g(x)) = 3f(x) \]