Let `f:ArarrB` is a function defined by `f(x)=(2x)/(1+x^(2))`. If the function f(x) is a bijective function, than the correct statement can be

Options

A. `A=B=[-1, 1]`
B. `A=B=[-2, 2]`
C. `A=[-1, 1], B=[-2, 2]`
D. `A=[-2, 2], B=[-1, 1]`

Answer

To determine the correct statement regarding the function \( f(x) = \frac{2x}{1+x^2} \) being a bijective function, we will analyze its injectivity and surjectivity step by step.

### Step 1: Check Injectivity
To check if the function is injective (one-to-one), we need to find the derivative \( f'(x) \) and analyze its sign.

1. **Find the derivative \( f'(x) \)**:
\[ f(x) = \frac{2x}{1+x^2} \]
Using the quotient rule:
\[ f'(x) = \frac{(1+x^2)(2) - (2x)(2x)}{(1+x^2)^2} \]
Simplifying the numerator:
\[ = \frac{2 + 2x^2 - 4x^2}{(1+x^2)^2} = \frac{2 - 2x^2}{(1+x^2)^2} = \frac{2(1-x^2)}{(1+x^2)^2} \]

2. **Determine where \( f'(x) = 0 \)**:
\[ 2(1-x^2) = 0 \Rightarrow 1 - x^2 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1 \]
The critical points are \( x = -1 \) and \( x = 1 \).

3. **Analyze the sign of \( f'(x) \)**:
- For \( x < -1 \): \( 1 - x^2 > 0 \) (increasing)
- For \( -1 < x < 1 \): \( 1 - x^2 > 0 \) (increasing)
- For \( x > 1 \): \( 1 - x^2 < 0 \) (decreasing)

Thus, \( f(x) \) is increasing on the intervals \( (-\infty, -1) \) and \( (-1, 1) \), and decreasing on \( (1, \infty) \). Therefore, \( f(x) \) is injective on the intervals \( (-\infty, -1) \cup (1, \infty) \).

### Step 2: Check Surjectivity
To check if the function is surjective (onto), we need to find the range of \( f(x) \).

1. **Evaluate the limits**:
- As \( x \to -1 \), \( f(-1) = \frac{2(-1)}{1+(-1)^2} = -1 \)
- As \( x \to 1 \), \( f(1) = \frac{2(1)}{1+(1)^2} = 1 \)

2. **Determine the range**:
- For \( x \) in \( (-1, 1) \), \( f(x) \) takes values from \( -1 \) to \( 1 \).
- For \( x < -1 \) and \( x > 1 \), \( f(x) \) approaches \( -1 \) and \( 1 \) respectively but does not include them.

Thus, the range of \( f(x) \) is \( (-1, 1) \).

### Conclusion
Since \( f(x) \) is injective on the intervals \( (-\infty, -1) \cup (1, \infty) \) and the range is \( (-1, 1) \), we conclude that \( f(x) \) is a bijective function on the interval \( (-1, 1) \).

### Final Answer
The correct statement is that \( f(x) \) is bijective on the interval \( (-1, 1) \).