If f be a function from the set of natural numbers to the set of even natural numbers given by f(x) = 2x. Then, f is:

Options

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

Answer

To determine the nature of the function \( f(x) = 2x \) where \( f \) is a function from the set of natural numbers to the set of even natural numbers, we will check if the function is one-one (injective) and onto (surjective).

### Step 1: Check if \( f \) is one-one (injective)

A function is one-one if different inputs produce different outputs.

1. Assume \( f(a) = f(b) \) for some natural numbers \( a \) and \( b \).
2. This means \( 2a = 2b \).
3. Dividing both sides by 2 gives \( a = b \).

Since \( a \) must equal \( b \) if \( f(a) = f(b) \), the function is one-one.

### Step 2: Check if \( f \) is onto (surjective)

A function is onto if every element in the codomain has a pre-image in the domain.

1. The codomain of \( f \) is the set of even natural numbers: \( \{2, 4, 6, 8, \ldots\} \).
2. For any even natural number \( y \), we can express it as \( y = 2k \) where \( k \) is a natural number.
3. To find \( k \), we can rearrange this to \( k = \frac{y}{2} \).
4. Since \( k \) is a natural number for every even natural number \( y \), there exists a natural number \( x = k \) such that \( f(x) = y \).

Thus, every even natural number in the codomain has a corresponding natural number in the domain, confirming that \( f \) is onto.

### Conclusion

Since \( f \) is both one-one and onto, we conclude that \( f \) is a bijective function.

### Final Answer

The function \( f(x) = 2x \) is both one-one and onto, hence it is a **bijective function**.

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