The function `f : (0, oo) rarr [0, oo), f(x) = (x)/(1+x)` is

Options

A. One-one and onto
B. One-one but not onto
C. Onto but not one-one
D. Neither one-one nor onto

Answer

To analyze the function \( f: (0, \infty) \rightarrow [0, \infty) \) defined by \( f(x) = \frac{x}{1+x} \), we need to determine whether it is a one-to-one function (injective) and find its range.

### Step-by-Step Solution:

1. **Check if the function is one-to-one (injective)**:
- We need to show that if \( f(x_1) = f(x_2) \), then \( x_1 = x_2 \).
- Start with the equation:
\[ f(x_1) = f(x_2) \implies \frac{x_1}{1+x_1} = \frac{x_2}{1+x_2} \]
- Cross-multiply to eliminate the fractions:
\[ x_1(1 + x_2) = x_2(1 + x_1) \]
- Expanding both sides:
\[ x_1 + x_1 x_2 = x_2 + x_1 x_2 \]
- Subtract \( x_1 x_2 \) from both sides:
\[ x_1 = x_2 \]
- Since \( x_1 = x_2 \), the function is one-to-one.

2. **Find the range of the function**:
- Set \( f(x) = y \):
\[ y = \frac{x}{1+x} \]
- Rearranging gives:
\[ y(1 + x) = x \implies y + xy = x \implies y = x - xy \]
- Factor out \( x \):
\[ y = x(1 - y) \implies x = \frac{y}{1 - y} \]
- The expression \( \frac{y}{1 - y} \) is valid as long as \( y \neq 1 \).
- Since \( x \) must be positive, we need \( \frac{y}{1 - y} > 0 \). This implies:
- \( y > 0 \) and \( 1 - y > 0 \) which means \( y < 1 \).
- Therefore, the range of \( f(x) \) is \( [0, 1) \).

3. **Conclusion**:
- The function \( f(x) = \frac{x}{1+x} \) is one-to-one (injective) and its range is \( [0, 1) \).
- Since the range does not cover all of \( [0, \infty) \), the function is not onto (surjective).

### Final Answer:
The function \( f(x) = \frac{x}{1+x} \) is one-to-one but not onto.

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