Let `f:RrarrR` be defined as `f(x)=2x`.

Options

A. f is one-one onto
B. f is many-one onto
C. f is one-one but not onto
D. f is neither one-one nor onto

Answer

To determine the nature of the function \( f: \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = 2x \), we need to check if it is one-one (injective) and onto (surjective).

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

A function \( f \) is one-one if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).

1. Assume \( f(x_1) = f(x_2) \).
\[ 2x_1 = 2x_2 \]
2. Divide both sides by 2:
\[ x_1 = x_2 \]

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

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

A function \( f \) is onto if for every \( y \in \mathbb{R} \), there exists an \( x \in \mathbb{R} \) such that \( f(x) = y \).

1. Let \( y \) be any real number.
2. We need to find \( x \) such that:
\[ f(x) = y \]
This means:
\[ 2x = y \]
3. Solving for \( x \):
\[ x = \frac{y}{2} \]

Since for every \( y \in \mathbb{R} \), we can find an \( x = \frac{y}{2} \) such that \( f(x) = y \), the function \( f \) is onto.

### Conclusion

Since \( f \) is both one-one and onto, we conclude that:

- The function \( f(x) = 2x \) is a bijection (one-one and onto).

### Final Answer

The function \( f(x) = 2x \) is one-one and onto.

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