Let `f: R to R` be defined as f(x)=3x. Choose the correct answer

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 solve the problem, we need to analyze the function \( f: \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = 3x \).

### Step 1: Determine if the function is one-to-one (injective)
A function is one-to-one if \( f(x_1) = f(x_2) \) implies that \( x_1 = x_2 \).

1. Assume \( f(x_1) = f(x_2) \).
2. This gives us \( 3x_1 = 3x_2 \).
3. Dividing both sides by 3, we get \( x_1 = x_2 \).

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

### Step 2: Determine if the function is onto (surjective)
A function is onto if for every \( y \) in the codomain \( \mathbb{R} \), there exists an \( x \) in the domain \( \mathbb{R} \) such that \( f(x) = y \).

1. Let \( y \) be any real number.
2. We need to find \( x \) such that \( f(x) = y \).
3. This gives us the equation \( 3x = y \).
4. Solving for \( x \), we get \( x = \frac{y}{3} \).

Since for every \( y \) in \( \mathbb{R} \), we can find an \( x = \frac{y}{3} \) in \( \mathbb{R} \) such that \( f(x) = y \), the function is onto.

### Conclusion
Since the function \( f(x) = 3x \) is both one-to-one and onto, it is a bijective function.

### Final Answer
The correct answer is that the function is both one-to-one and onto.

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