Let `f(x)=x/(1+x)` and let `g(x)=(rx)/(1-x)` , Let S be the set off all real numbers r such that `f(g(x))=g(f(x))` for infinitely many real number x. The number of elements in set S is

Options

A. 1
B. 2
C. 3
D. 5

Answer

To solve the problem, we need to find the values of \( r \) such that \( f(g(x)) = g(f(x)) \) for infinitely many real numbers \( x \).

1. **Define the functions**:
- We have \( f(x) = \frac{x}{1+x} \) and \( g(x) = \frac{rx}{1-x} \).

2. **Calculate \( f(g(x)) \)**:
- Substitute \( g(x) \) into \( f(x) \):
\[ f(g(x)) = f\left(\frac{rx}{1-x}\right) = \frac{\frac{rx}{1-x}}{1+\frac{rx}{1-x}}. \]
- Simplifying the denominator:
\[ 1 + \frac{rx}{1-x} = \frac{1-x + rx}{1-x} = \frac{1 - x + rx}{1-x}. \]
- Therefore,
\[ f(g(x)) = \frac{\frac{rx}{1-x}}{\frac{1 - x + rx}{1-x}} = \frac{rx}{1 - x + rx}. \]

3. **Calculate \( g(f(x)) \)**:
- Substitute \( f(x) \) into \( g(x) \):
\[ g(f(x)) = g\left(\frac{x}{1+x}\right) = \frac{r\left(\frac{x}{1+x}\right)}{1 - \frac{x}{1+x}}. \]
- Simplifying the denominator:
\[ 1 - \frac{x}{1+x} = \frac{1+x - x}{1+x} = \frac{1}{1+x}. \]
- Therefore,
\[ g(f(x)) = \frac{r\left(\frac{x}{1+x}\right)}{\frac{1}{1+x}} = r \cdot x. \]

4. **Set the two expressions equal**:
\[ \frac{rx}{1 - x + rx} = rx. \]

5. **Cross-multiply**:
\[ rx = (1 - x + rx) \cdot rx. \]
- This simplifies to:
\[ rx = rx - x^2r + r^2x^2. \]
- Rearranging gives:
\[ 0 = -x^2r + r^2x^2. \]
- Factoring out \( x^2 \):
\[ x^2(r^2 - r) = 0. \]

6. **Analyzing the equation**:
- For this equation to hold for infinitely many \( x \), we need:
\[ r^2 - r = 0. \]
- This factors to:
\[ r(r - 1) = 0. \]
- Thus, \( r = 0 \) or \( r = 1 \).

7. **Conclusion**:
- The set \( S \) consists of the values \( r = 0 \) and \( r = 1 \). Therefore, the number of elements in set \( S \) is \( 2 \).

### Final Answer:
The number of elements in set \( S \) is \( 2 \).