The function `f:R rarr R, f(x)=x^(2)` is

Options

A. injective but not surjective
B. surjective but not injective
C. injective as well as surjective
D. neither injective nor surjective.

Answer

To determine the properties of the function \( f: \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = x^2 \), we will analyze whether the function is injective (one-to-one) and surjective (onto).

### Step-by-step Solution:

1. **Understand the function**: The function given is \( f(x) = x^2 \). This is a quadratic function.

**Hint**: Identify the type of function and its general properties.

2. **Determine the range of the function**: The output of \( f(x) = x^2 \) is always non-negative since squaring any real number (positive or negative) results in a non-negative number. Therefore, the range of \( f \) is \( [0, \infty) \).

**Hint**: Consider the values that the function can take as \( x \) varies over all real numbers.

3. **Check if the function is injective**: A function is injective if different inputs map to different outputs. For \( f(x) = x^2 \):
- For \( x = 2 \), \( f(2) = 4 \).
- For \( x = -2 \), \( f(-2) = 4 \).
- Since both \( 2 \) and \( -2 \) yield the same output, the function is not injective.

**Hint**: Test specific values to see if different inputs can produce the same output.

4. **Check if the function is surjective**: A function is surjective if every element in the codomain (in this case, \( \mathbb{R} \)) has a pre-image in the domain. The codomain is \( \mathbb{R} \), but the range of \( f(x) = x^2 \) is \( [0, \infty) \). Since there are no negative outputs, not every real number has a corresponding input.

**Hint**: Compare the range of the function with the codomain to determine if every element in the codomain is covered.

5. **Conclusion**: Since \( f(x) = x^2 \) is neither injective nor surjective, we conclude that the function is neither one-to-one nor onto.

**Hint**: Summarize your findings based on the definitions of injective and surjective functions.

### Final Answer:
The function \( f(x) = x^2 \) is neither injective nor surjective.