If the function `f: R -{1,-1} to A` definded by `f(x)=(x^(2))/(1-x^(2))`, is surjective, then A is equal to

Options

A. `R-(-1,0)`
B. `R-[-1,0)`
C. `R-{-1}`
D. `[0,oo)`

Explanation

To determine the set \( A \) such that the function \( f(x) = \frac{x^2}{1 - x^2} \) is surjective, we need to find the range of this function. Let's go through the steps to find the range.

### Step 1: Set up the equation
We start with the function:
\[ f(x) = \frac{x^2}{1 - x^2} \]
We want to find the values of \( y \) such that there exists an \( x \) in the domain \( \mathbb{R} \setminus \{1, -1\} \) for which \( f(x) = y \).

### Step 2: Rearranging the equation
Set \( f(x) = y \):
\[ y = \frac{x^2}{1 - x^2} \]
Rearranging gives:
\[ y(1 - x^2) = x^2 \]
\[ y - yx^2 = x^2 \]
\[ yx^2 + x^2 = y \]
Factoring out \( x^2 \):
\[ x^2(y + 1) = y \]
Thus, we can express \( x^2 \) as:
\[ x^2 = \frac{y}{y + 1} \]

### Step 3: Determine the conditions for \( x^2 \)
Since \( x^2 \) must be non-negative, we require:
\[ \frac{y}{y + 1} \geq 0 \]
This inequality holds when:
1. \( y \geq 0 \) (numerator is non-negative and denominator is positive).
2. \( y < -1 \) (numerator is negative and denominator is negative).

### Step 4: Analyze the intervals
Now we analyze the intervals:
- For \( y \geq 0 \): \( y + 1 > 0 \), hence \( \frac{y}{y + 1} \geq 0 \).
- For \( y < -1 \): \( y + 1 < 0 \), hence \( \frac{y}{y + 1} \geq 0 \).

Thus, the valid ranges for \( y \) are:
\[ y \in (-\infty, -1) \cup [0, \infty) \]

### Step 5: Conclusion
The range of the function \( f(x) \) is \( (-\infty, -1) \cup [0, \infty) \). Therefore, for the function to be surjective, the codomain \( A \) must equal the range of \( f(x) \).

Thus, we conclude:
\[ A = (-\infty, -1) \cup [0, \infty) \]