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 if the function \( f(x) = \frac{2x}{1+x^2} \) is bijective, we need to check if it is both injective (one-to-one) and surjective (onto).

### Step 1: Find the derivative of \( f(x) \)

To check for injectivity, we first find the derivative \( f'(x) \):

\[ f'(x) = \frac{d}{dx} \left( \frac{2x}{1+x^2} \right) \]

Using the quotient rule, where \( u = 2x \) and \( v = 1+x^2 \):

\[ f'(x) = \frac{v \cdot u' - u \cdot v'}{v^2} \]

Calculating \( u' \) and \( v' \):

- \( u' = 2 \)
- \( v' = 2x \)

Now substituting back into the quotient rule:

\[ f'(x) = \frac{(1+x^2)(2) - (2x)(2x)}{(1+x^2)^2} \]

This simplifies to:

\[ f'(x) = \frac{2 + 2x^2 - 4x^2}{(1+x^2)^2} = \frac{2 - 2x^2}{(1+x^2)^2} \]

Factoring out the common term:

\[ f'(x) = \frac{2(1 - x^2)}{(1+x^2)^2} \]

### Step 2: Analyze the sign of \( f'(x) \)

The critical points occur when \( f'(x) = 0 \):

\[ 2(1 - x^2) = 0 \implies 1 - x^2 = 0 \implies x^2 = 1 \implies x = \pm 1 \]

Now we check the intervals determined by these critical points:

- For \( x < -1 \): \( 1 - x^2 < 0 \) → \( f'(x) < 0 \) (decreasing)
- For \( -1 < x < 1 \): \( 1 - x^2 > 0 \) → \( f'(x) > 0 \) (increasing)
- For \( x > 1 \): \( 1 - x^2 < 0 \) → \( f'(x) < 0 \) (decreasing)

### Step 3: Determine the behavior of \( f(x) \)

Since \( f(x) \) is increasing on the interval \( (-1, 1) \) and decreasing on \( (-\infty, -1) \) and \( (1, \infty) \), it is injective on the interval \( (-1, 1) \).

### Step 4: Find the range of \( f(x) \)

Now we need to find the range of \( f(x) \) when \( x \) is in the interval \( (-1, 1) \):

- \( f(-1) = \frac{2(-1)}{1+(-1)^2} = -1 \)
- \( f(1) = \frac{2(1)}{1+(1)^2} = 1 \)

Thus, as \( x \) varies from \(-1\) to \(1\), \( f(x) \) takes all values from \(-1\) to \(1\).

### Conclusion

Since \( f(x) \) is injective and the range of \( f(x) \) is \((-1, 1)\), we conclude that \( f(x) \) is bijective on the interval \((-1, 1)\).

### Correct Statement

The correct statement is that the function \( f(x) \) is bijective when its domain is restricted to \( (-1, 1) \) and its range is \( (-1, 1) \).