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

Options

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

Explanation

To determine the set \( A \) for which the function \( f: \mathbb{R} - \{1, -1\} \to A \) defined by \( f(x) = \frac{x^2}{1 - x^2} \) is surjective, we will analyze the function step by step.

### Step 1: Analyze the function
The function is given by:
\[ f(x) = \frac{x^2}{1 - x^2} \]
We need to find the range of this function as \( x \) varies over \( \mathbb{R} - \{1, -1\} \).

### Step 2: Determine the behavior of \( f(x) \)
First, we can rewrite \( f(x) \) in terms of \( y \):
\[ y = f(x) = \frac{x^2}{1 - x^2} \]
Rearranging gives:
\[ y(1 - x^2) = x^2 \implies y - yx^2 = x^2 \implies y = x^2 + yx^2 \implies y = x^2(1 + y) \]
From this, we can express \( x^2 \) in terms of \( y \):
\[ x^2 = \frac{y}{1 + y} \]

### Step 3: Determine the range of \( y \)
Since \( x^2 \geq 0 \) for all \( x \), we have:
\[ \frac{y}{1 + y} \geq 0 \]
This implies that \( y \) must be non-negative. Therefore, we have:
\[ y \geq 0 \]

### Step 4: Determine the limits of \( y \)
Next, we analyze the limits of \( f(x) \) as \( x \) approaches the boundaries of its domain:
- As \( x \to 1 \) from the left, \( f(x) \to +\infty \).
- As \( x \to -1 \) from the right, \( f(x) \to +\infty \).
- As \( x \to 0 \), \( f(0) = 0 \).

Since \( f(x) \) is continuous and takes all values from \( 0 \) to \( +\infty \), we conclude that the range of \( f(x) \) is:
\[ [0, +\infty) \]

### Step 5: Conclusion about set \( A \)
For the function \( f \) to be surjective, the codomain \( A \) must equal the range of \( f(x) \). Therefore, we have:
\[ A = [0, +\infty) \]

### Final Answer
Thus, the set \( A \) is equal to:
\[ A = [0, +\infty) \]