State whether the following function is one-one onto or bijective:
`f:RrarrR` defined by `f(x)=1+x^(2)`.

Answer

To determine whether the function \( f: \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = 1 + x^2 \) is one-one, onto, or bijective, we will analyze the function step by step.

### Step 1: Check if the function is one-one (injective)
A function is one-one if different inputs produce different outputs. To check this, we can assume that \( f(a) = f(b) \) for some \( a, b \in \mathbb{R} \) and see if it leads to \( a = b \).

1. Start with the equation:
\[ f(a) = f(b) \]
This means:
\[ 1 + a^2 = 1 + b^2 \]
Simplifying gives:
\[ a^2 = b^2 \]
This implies:
\[ a = b \quad \text{or} \quad a = -b \]

2. Since \( a \) can be equal to \( b \) or \( -b \), we see that the function can produce the same output for different inputs (e.g., \( f(1) = f(-1) = 2 \)). Thus, the function is **not one-one**.

### Step 2: Check if the function is onto (surjective)
A function is onto if every element in the codomain (in this case, \( \mathbb{R} \)) has a pre-image in the domain.

1. The range of \( f(x) = 1 + x^2 \) can be analyzed:
- The minimum value occurs when \( x = 0 \):
\[ f(0) = 1 + 0^2 = 1 \]
- As \( x \) increases or decreases, \( f(x) \) increases without bound (as \( x^2 \) approaches infinity).

2. Therefore, the range of \( f \) is:
\[ [1, \infty) \]
Since the codomain is \( \mathbb{R} \) (which includes all real numbers), and the range \( [1, \infty) \) does not cover all real numbers (it does not include values less than 1), the function is **not onto**.

### Conclusion
Since the function \( f(x) = 1 + x^2 \) is neither one-one nor onto, it is not bijective.

### Final Answer
The function \( f(x) = 1 + x^2 \) is neither one-one nor onto.

---