Let be a real function satisfying `f(x)+f(y)=f((x+y)/(1-xy))`for all ` x ,y in R ` and ` xy ne1`.
Then f(x) is

Options

A. a periodic function with period ` pi//2`
B. an odd function
C. an even function
D. none of these

Explanation

To solve the problem, we need to analyze the functional equation given:

\[ f(x) + f(y) = f\left(\frac{x+y}{1-xy}\right) \]

for all \( x, y \in \mathbb{R} \) where \( xy \neq 1 \).

### Step 1: Substitute \( x = 0 \) and \( y = 0 \)

Let's start by substituting \( x = 0 \) and \( y = 0 \) into the functional equation:

\[ f(0) + f(0) = f\left(\frac{0 + 0}{1 - 0 \cdot 0}\right) \]

This simplifies to:

\[ 2f(0) = f(0) \]

### Step 2: Solve for \( f(0) \)

From the equation \( 2f(0) = f(0) \), we can rearrange it to find:

\[ 2f(0) - f(0) = 0 \implies f(0) = 0 \]

### Step 3: Substitute \( y = -x \)

Next, let's substitute \( y = -x \) into the original equation:

\[ f(x) + f(-x) = f\left(\frac{x + (-x)}{1 - x(-x)}\right) \]

This simplifies to:

\[ f(x) + f(-x) = f\left(\frac{0}{1 + x^2}\right) = f(0) \]

Since we found \( f(0) = 0 \), we have:

\[ f(x) + f(-x) = 0 \]

### Step 4: Conclude that \( f(x) \) is an odd function

The equation \( f(x) + f(-x) = 0 \) implies that:

\[ f(-x) = -f(x) \]

This is the definition of an odd function. Therefore, we conclude that \( f(x) \) is an odd function.

### Final Conclusion

Thus, the function \( f(x) \) is an odd function.

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