The function `f (x) = log((1+x)/(1-x))` satisfies the equation

Options

A. f(x+2) – 2f (x+1) + f (x) =0
B. `f((2x)/(1+x^2))=2f (x)`
C. `f(p) f(q) = f(p+q)`
D. ` f(p) + f(q) = f((p+q)/(1+pq))`

Answer

To solve the problem, we need to verify which of the given options satisfies the equation for the function \( f(x) = \log\left(\frac{1+x}{1-x}\right) \).

### Step-by-Step Solution:

1. **Check Option 1: \( f(x+2) = f(x) + 2 \)**

Substitute \( x+2 \) into the function:
\[ f(x+2) = \log\left(\frac{1+(x+2)}{1-(x+2)}\right) = \log\left(\frac{1+x+2}{1-x-2}\right) = \log\left(\frac{x+3}{-x-1}\right) \]
Now, calculate \( f(x) + 2 \):
\[ f(x) + 2 = \log\left(\frac{1+x}{1-x}\right) + 2 = \log\left(\frac{1+x}{1-x}\right) + \log(e^2) = \log\left(\frac{(1+x)e^2}{1-x}\right) \]
Since \( f(x+2) \) does not equal \( f(x) + 2 \), **Option 1 is incorrect**.

2. **Check Option 2: \( f\left(\frac{2x}{1+x^2}\right) = 2f(x) \)**

Substitute \( \frac{2x}{1+x^2} \) into the function:
\[ f\left(\frac{2x}{1+x^2}\right) = \log\left(\frac{1+\frac{2x}{1+x^2}}{1-\frac{2x}{1+x^2}}\right) \]
Simplifying the numerator:
\[ 1 + \frac{2x}{1+x^2} = \frac{1+x^2+2x}{1+x^2} = \frac{(x+1)^2}{1+x^2} \]
Simplifying the denominator:
\[ 1 - \frac{2x}{1+x^2} = \frac{1+x^2-2x}{1+x^2} = \frac{(1-x)^2}{1+x^2} \]
Thus,
\[ f\left(\frac{2x}{1+x^2}\right) = \log\left(\frac{(x+1)^2}{(1-x)^2}\right) = 2\log\left(\frac{x+1}{1-x}\right) = 2f(x) \]
Therefore, **Option 2 is correct**.

3. **Check Option 3: \( f(p)f(q) = f(p+q) \)**

Calculate \( f(p)f(q) \):
\[ f(p) = \log\left(\frac{1+p}{1-p}\right), \quad f(q) = \log\left(\frac{1+q}{1-q}\right) \]
Thus,
\[ f(p)f(q) = \log\left(\frac{1+p}{1-p}\right) + \log\left(\frac{1+q}{1-q}\right) = \log\left(\frac{(1+p)(1+q)}{(1-p)(1-q)}\right) \]
Now, calculate \( f(p+q) \):
\[ f(p+q) = \log\left(\frac{1+(p+q)}{1-(p+q)}\right) \]
Since \( f(p)f(q) \neq f(p+q) \), **Option 3 is incorrect**.

4. **Check Option 4: \( f(p) + f(q) = f(p+q) \)**

From the previous calculations:
\[ f(p) + f(q) = \log\left(\frac{1+p}{1-p}\right) + \log\left(\frac{1+q}{1-q}\right) = \log\left(\frac{(1+p)(1+q)}{(1-p)(1-q)}\right) \]
This can be simplified to:
\[ = \log\left(\frac{(1+p)(1+q)}{(1-p)(1-q)}\right) = f(p+q) \]
Therefore, **Option 4 is correct**.

### Conclusion:
The correct options are **Option 2 and Option 4**.