`f:N-{1}rightarrowNrightarrow{1}`
f(x)= largest prime factor of n then f is-

Options

A. one one onto
B. one one into
C. many one onto
D. many one into

Answer

To solve the problem, we need to analyze the function \( f: \mathbb{N} - \{1\} \rightarrow \mathbb{N} \) defined as \( f(n) = \) largest prime factor of \( n \). We need to determine whether this function is one-one (injective), onto (surjective), many-one, or into.

### Step-by-Step Solution:

1. **Understanding the Function**:
The function \( f(n) \) gives the largest prime factor of \( n \). For example:
- \( f(8) = 2 \) (largest prime factor of 8 is 2)
- \( f(16) = 2 \) (largest prime factor of 16 is also 2)
- \( f(6) = 3 \) (largest prime factor of 6 is 3)
- \( f(9) = 3 \) (largest prime factor of 9 is also 3)

2. **Checking if the Function is One-One**:
A function is one-one (injective) if different inputs map to different outputs.
- From the examples above, \( f(8) = 2 \) and \( f(16) = 2 \), which means two different inputs (8 and 16) give the same output (2).
- Similarly, \( f(6) = 3 \) and \( f(9) = 3 \).
- Therefore, the function is **not one-one**.

3. **Checking if the Function is Onto**:
A function is onto (surjective) if every element in the codomain is mapped by some element in the domain.
- The codomain is \( \mathbb{N} \), which includes all natural numbers.
- The largest prime factor of any natural number \( n \) is a prime number. However, not every natural number is a prime number.
- For example, there is no \( n \) such that \( f(n) = 4 \) because 4 is not a prime number.
- Therefore, the function is **not onto**.

4. **Checking if the Function is Many-One**:
A function is many-one if two or more different inputs can map to the same output.
- As shown in the previous examples, multiple inputs (like 8 and 16) can yield the same output (2).
- Thus, the function is **many-one**.

5. **Checking if the Function is Into**:
A function is into if the range of the function is a proper subset of the codomain.
- Since the largest prime factor can only be a prime number, and not all natural numbers are prime, the range of \( f(n) \) (which consists of prime numbers) is indeed a proper subset of \( \mathbb{N} \).
- Therefore, the function is **into**.

### Conclusion:
The function \( f(n) = \) largest prime factor of \( n \) is **many-one** and **into**.