` f :[0,oo) to [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 determine the properties of the function \( f: [0, \infty) \to [0, \infty) \) defined by \( f(x) = \frac{x}{1+x} \), we need to check if it is one-to-one (injective) and onto (surjective).

### Step 1: Check if the function is one-to-one (1-1)

To check if the function is one-to-one, we need to see if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).

1. Assume \( f(x_1) = f(x_2) \):
\[ \frac{x_1}{1 + x_1} = \frac{x_2}{1 + x_2} \]

2. Cross-multiply:
\[ x_1(1 + x_2) = x_2(1 + x_1) \]

3. Expanding both sides gives:
\[ x_1 + x_1 x_2 = x_2 + x_1 x_2 \]

4. Subtract \( x_1 x_2 \) from both sides:
\[ x_1 = x_2 \]

Since \( f(x_1) = f(x_2) \) leads to \( x_1 = x_2 \), the function is one-to-one.

### Step 2: Check if the function is onto (onto)

To check if the function is onto, we need to see if for every \( y \in [0, \infty) \), there exists an \( x \in [0, \infty) \) such that \( f(x) = y \).

1. Set \( f(x) = y \):
\[ \frac{x}{1 + x} = y \]

2. Rearranging gives:
\[ x = y(1 + x) \]

3. This simplifies to:
\[ x - yx = y \quad \Rightarrow \quad x(1 - y) = y \]

4. Thus, we have:
\[ x = \frac{y}{1 - y} \]

5. For \( x \) to be in the domain \( [0, \infty) \), \( y \) must be less than 1 (since \( 1 - y \) must be positive). If \( y \geq 1 \), then \( x \) becomes negative, which is not in the domain.

Since there are values of \( y \) (specifically \( y \geq 1 \)) for which there is no corresponding \( x \) in the domain, the function is not onto.

### Conclusion

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

### Final Answer

The function is **1-1 but not onto**.

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