If `f:RtoR` is a function which satisfies the condition `f(x)+f(y)=f((x+y)/(1+xy))` for all `x, y in R` except `xy=-1`, then range of `f(x)` is

Options

A. `[0,infty)`
B. `{0,1}`
C. `R`
D. `{0}`

Explanation

To find the range of the function \( f: \mathbb{R} \to \mathbb{R} \) that satisfies the condition

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

for all \( x, y \in \mathbb{R} \) except when \( xy = -1 \), we can follow these steps:

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

Let’s first substitute \( x = 1 \) and \( y = 1 \) into the functional equation:

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

This simplifies to:

\[ 2f(1) = f(1) \]

### Step 2: Solve for \( f(1) \)

From \( 2f(1) = f(1) \), we can rearrange this to:

\[ 2f(1) - f(1) = 0 \implies f(1) = 0 \]

### Step 3: Substitute \( y = 1 \)

Next, we substitute \( y = 1 \) into the original equation:

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

Since we found \( f(1) = 0 \), this simplifies to:

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

### Step 4: Analyze the function at \( x = 0 \)

Now, let’s substitute \( x = 0 \) and \( y = 0 \):

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

This gives us:

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

### Step 5: Substitute \( x = -1 \) and \( y = -1 \)

Now, let’s substitute \( x = -1 \) and \( y = -1 \):

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

The right side is undefined since it leads to division by zero. However, we can analyze the function by substituting \( x = -1 \) and \( y = 0 \):

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

This does not provide new information, but we can try \( x = -1 \) and \( y = 1 \):

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

Again, this leads to an undefined expression. However, we can conclude that \( f(-1) = 0 \) since \( f(1) = 0 \).

### Step 6: Generalization

From the previous steps, we have shown that:

- \( f(1) = 0 \)
- \( f(0) = 0 \)
- \( f(-1) = 0 \)

We can also substitute any \( y \) value in the equation and find that \( f(x) = 0 \) for all \( x \) in \( \mathbb{R} \).

### Conclusion

Thus, we conclude that the function \( f(x) = 0 \) for all \( x \in \mathbb{R} \).

### Final Answer

The range of \( f(x) \) is:

\[ \{0\} \]