If the function `f:R rarrA` defined as `f(x)=tan^(-1)((2x^(3))/(1+x^(6)))` is a surjective function, then the set A is equal to

Options

A. `[-(pi)/(2), (pi)/(2)]`
B. `[-(pi)/(4), (pi)/(4)]`
C. `(-(pi)/(2), (pi)/(2))`
D. `[0, (pi)/(4)]`

Answer

To determine the set \( A \) for which the function \( f: \mathbb{R} \to A \) defined by

\[ f(x) = \tan^{-1}\left(\frac{2x^3}{1+x^6}\right) \]

is a surjective function, we need to analyze the range of \( f(x) \).

### Step 1: Understanding the Function

The function \( f(x) \) is defined as the arctangent of a rational function. The output of the arctangent function, \( \tan^{-1}(y) \), is restricted to the interval \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \) for all real \( y \). However, we need to find the specific range of \( f(x) \).

### Step 2: Analyzing the Argument of the Arctangent

We need to analyze the expression \( \frac{2x^3}{1+x^6} \):

1. **Behavior as \( x \to 0 \)**:
\[ f(0) = \tan^{-1}\left(\frac{2 \cdot 0^3}{1 + 0^6}\right) = \tan^{-1}(0) = 0 \]

2. **Behavior as \( x \to \infty \)**:
\[ \text{As } x \to \infty, \frac{2x^3}{1+x^6} \to 0 \text{ (since } x^6 \text{ dominates)} \]
Thus, \( f(x) \to 0 \).

3. **Behavior as \( x \to -\infty \)**:
\[ \text{As } x \to -\infty, \frac{2x^3}{1+x^6} \to 0 \text{ (similar reasoning)} \]
Thus, \( f(x) \to 0 \).

### Step 3: Finding Critical Points

Next, we find the critical points by taking the derivative of \( f(x) \) and setting it to zero:

\[ f'(x) = \frac{d}{dx} \left( \tan^{-1}\left(\frac{2x^3}{1+x^6}\right) \right) \]

Using the chain rule and quotient rule, we can analyze the behavior of \( f(x) \) to find local maxima and minima.

### Step 4: Evaluating Specific Values

To determine the range, we can evaluate \( f(x) \) at specific points:

- **At \( x = 1 \)**:
\[ f(1) = \tan^{-1}\left(\frac{2 \cdot 1^3}{1 + 1^6}\right) = \tan^{-1}\left(\frac{2}{2}\right) = \tan^{-1}(1) = \frac{\pi}{4} \]

- **At \( x = -1 \)**:
\[ f(-1) = \tan^{-1}\left(\frac{2 \cdot (-1)^3}{1 + (-1)^6}\right) = \tan^{-1}\left(\frac{-2}{2}\right) = \tan^{-1}(-1) = -\frac{\pi}{4} \]

### Step 5: Conclusion on the Range

From the analysis, we find that:

- As \( x \) approaches \( \pm \infty \), \( f(x) \) approaches \( 0 \).
- The maximum value of \( f(x) \) is \( \frac{\pi}{4} \) and the minimum value is \( -\frac{\pi}{4} \).

Thus, the range of \( f(x) \) is:

\[ \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \]

Since \( f(x) \) is surjective, the set \( A \) must equal the range of \( f(x) \).

### Final Answer

The set \( A \) is:

\[ A = \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \]