Let `f:N->N` be defined by `f(x)=x^2+x+1,x in N`. Then `f(x)` is

Options

A. one-one and onto
B. many-one onto
C. one-one but not onto
D. none of these

Explanation

To analyze the function \( f: \mathbb{N} \to \mathbb{N} \) defined by \( f(x) = x^2 + x + 1 \), we will check if it is one-one (injective) and onto (surjective).

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

To determine if \( f \) is one-one, we need to show that if \( f(x_1) = f(x_2) \), then \( x_1 = x_2 \).

1. Assume \( f(x_1) = f(x_2) \).
2. This gives us the equation:
\[ x_1^2 + x_1 + 1 = x_2^2 + x_2 + 1 \]
3. Rearranging this, we get:
\[ x_1^2 - x_2^2 + x_1 - x_2 = 0 \]
4. Factoring the left-hand side, we have:
\[ (x_1 - x_2)(x_1 + x_2 + 1) = 0 \]
5. This implies either \( x_1 - x_2 = 0 \) (which means \( x_1 = x_2 \)) or \( x_1 + x_2 + 1 = 0 \). However, since \( x_1 \) and \( x_2 \) are natural numbers, \( x_1 + x_2 + 1 \) cannot be zero.
6. Therefore, \( x_1 = x_2 \).

Thus, \( f \) is one-one.

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

To check if \( f \) is onto, we need to see if for every \( y \in \mathbb{N} \), there exists an \( x \in \mathbb{N} \) such that \( f(x) = y \).

1. We set \( f(x) = y \):
\[ x^2 + x + 1 = y \]
2. Rearranging gives us:
\[ x^2 + x + (1 - y) = 0 \]
3. For this quadratic equation to have real solutions, the discriminant must be non-negative:
\[ D = b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot (1 - y) = 1 - 4 + 4y = 4y - 3 \]
4. For \( D \geq 0 \):
\[ 4y - 3 \geq 0 \implies y \geq \frac{3}{4} \]
5. Since \( y \) must be a natural number, the smallest value \( y \) can take is 1. However, we can check the values of \( f(x) \) for natural numbers:
- \( f(1) = 3 \)
- \( f(2) = 7 \)
- \( f(3) = 13 \)
- \( f(4) = 21 \)
- etc.

6. We observe that not all natural numbers can be expressed in the form \( f(x) \). For example, there is no \( x \) such that \( f(x) = 1 \) or \( f(x) = 2 \).

Thus, \( f \) is not onto.

### Conclusion

The function \( f(x) = x^2 + x + 1 \) is one-one but not onto.