The function f : `[0,oo)to[0,oo)` defined by `f(x)=(2x)/(1+2x)` is

Options

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

Explanation

To determine the type of the function \( f: [0, \infty) \to [0, \infty) \) defined by \( f(x) = \frac{2x}{1 + 2x} \), we will analyze its properties step by step.

### Step 1: Determine if the function is one-to-one (injective)
To check if the function is one-to-one, we need to see if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).

Assume \( f(x_1) = f(x_2) \):
\[ \frac{2x_1}{1 + 2x_1} = \frac{2x_2}{1 + 2x_2} \]

Cross-multiplying gives:
\[ 2x_1(1 + 2x_2) = 2x_2(1 + 2x_1) \]
Expanding both sides:
\[ 2x_1 + 4x_1x_2 = 2x_2 + 4x_1x_2 \]
Subtracting \( 4x_1x_2 \) from both sides:
\[ 2x_1 = 2x_2 \]
Dividing by 2:
\[ x_1 = x_2 \]
Thus, \( f \) is one-to-one.

### Step 2: Determine if the function is onto (surjective)
Next, we need to check if every value in the codomain \( [0, \infty) \) can be achieved by \( f(x) \).

To find the range of \( f(x) \), we can analyze its behavior as \( x \) approaches the boundaries of its domain:
- As \( x \to 0 \):
\[ f(0) = \frac{2(0)}{1 + 2(0)} = 0 \]
- As \( x \to \infty \):
\[ f(x) = \frac{2x}{1 + 2x} \to 1 \text{ (since the highest degree terms dominate)} \]

Now, to find the range more rigorously, we can set \( y = f(x) \):
\[ y = \frac{2x}{1 + 2x} \]
Rearranging gives:
\[ y(1 + 2x) = 2x \implies y + 2xy = 2x \implies 2xy - 2x = -y \implies x(2y - 2) = -y \implies x = \frac{-y}{2y - 2} \]
This expression is valid as long as \( 2y - 2 \neq 0 \), which gives \( y \neq 1 \).

Thus, the function can take values from \( 0 \) to \( 1 \) (not including \( 1 \)).

### Conclusion
The function \( f(x) = \frac{2x}{1 + 2x} \) is:
- One-to-one (injective)
- Not onto (surjective) since it does not cover the entire codomain \( [0, \infty) \) (it only covers \( [0, 1) \))

### Final Answer
The function \( f(x) = \frac{2x}{1 + 2x} \) is one-to-one but not onto.

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