consider the function `f(x)=(x^(2))/(x^(2)-1)`
If f is defined from `R-(-1,1)rarrR` then f is

Options

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

Answer

To determine the properties of the function \( f(x) = \frac{x^2}{x^2 - 1} \) defined from \( \mathbb{R} \setminus (-1, 1) \) to \( \mathbb{R} \), we need to analyze its injectivity and surjectivity.

### Step 1: Check for Injectivity
A function is injective (one-to-one) if different inputs produce different outputs.

1. **Evaluate \( f(-x) \) and \( f(x) \)**:
\[ f(-x) = \frac{(-x)^2}{(-x)^2 - 1} = \frac{x^2}{x^2 - 1} = f(x) \]
This shows that \( f(-x) = f(x) \), meaning that for any \( x \), \( f(x) \) is equal to \( f(-x) \).

2. **Conclusion on Injectivity**:
Since \( f(x) = f(-x) \), the function is not injective because multiple inputs (e.g., \( x \) and \( -x \)) yield the same output.

### Step 2: Check for Surjectivity
A function is surjective (onto) if every element in the codomain has a pre-image in the domain.

1. **Identify the range of \( f(x) \)**:
The function is defined for \( x \in \mathbb{R} \setminus (-1, 1) \). We need to find the values that \( f(x) \) can take.

2. **Analyze the limits**:
- As \( x \to 1^+ \) or \( x \to -1^- \), \( f(x) \to +\infty \).
- As \( x \to 1^- \) or \( x \to -1^+ \), \( f(x) \to -\infty \).
- As \( |x| \to \infty \), \( f(x) \to 1 \).

3. **Determine values that cannot be reached**:
The function cannot take values between -1 and 1, as shown by the limits. Specifically, \( f(x) \) never reaches 0, since 0 lies between -1 and 1.

4. **Conclusion on Surjectivity**:
Since the range of \( f(x) \) does not cover all real numbers (specifically, it cannot take values in the interval (-1, 1)), the function is not surjective.

### Final Conclusion
Since \( f(x) \) is neither injective nor surjective, the correct answer is that the function is neither injective nor surjective.