The function `f: R rarr R` defined as `f (x) = x^(2) + 1` 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 properties of the function \( f: \mathbb{R} \rightarrow \mathbb{R} \) defined by \( f(x) = x^2 + 1 \), we need to check if it is one-to-one (1-1) and onto.

### Step 1: Check if the function is one-to-one (1-1)
A function is one-to-one if different inputs produce different outputs. To check this, we can assume \( f(a) = f(b) \) and see if it implies \( a = b \).

Let:
\[ f(a) = f(b) \]
This implies:
\[ a^2 + 1 = b^2 + 1 \]
Subtracting 1 from both sides gives:
\[ a^2 = b^2 \]
Taking the square root of both sides yields:
\[ a = b \quad \text{or} \quad a = -b \]
Since \( a \) can equal \( -b \) (for example, \( f(1) = f(-1) = 2 \)), we conclude that the function is not one-to-one.

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

The function \( f(x) = x^2 + 1 \) always produces outputs that are greater than or equal to 1, since \( x^2 \geq 0 \) for all real \( x \). Therefore:
\[ f(x) \geq 1 \]
This means that the range of \( f \) is \( [1, \infty) \).

Since the codomain is \( \mathbb{R} \) (which includes negative numbers and numbers less than 1), and the range of \( f \) does not cover all of \( \mathbb{R} \), we conclude that the function is not onto.

### Conclusion
Since the function \( f(x) = x^2 + 1 \) is neither one-to-one nor onto, the correct answer is option 4: neither 1-1 nor onto.

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