If the function `f:R rarr A` defined as `f(x)=sin^(-1)((x)/(1+x^(2)))` is a surjective function, then the set A is

Options

A. `[-(pi)/(6),(pi)/(6)]`
B. `[-(pi)/(2),(pi)/(2)]`
C. `[-(pi)/(3),(pi)/(6)]`
D. `[0,(pi)/(3)]`

Answer

To determine the set \( A \) for which the function \( f: \mathbb{R} \to A \) defined by \( f(x) = \sin^{-1}\left(\frac{x}{1+x^2}\right) \) is surjective, we need to find the range of the function \( f(x) \).

### Step 1: Understand the function
The function \( f(x) = \sin^{-1}\left(\frac{x}{1+x^2}\right) \) takes inputs from the real numbers \( \mathbb{R} \) and outputs values in the range of the inverse sine function, which is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).

### Step 2: Determine the range of \( \frac{x}{1+x^2} \)
To find the range of \( f(x) \), we first need to analyze the expression \( \frac{x}{1+x^2} \).

1. **Find the critical points**: We can differentiate \( g(x) = \frac{x}{1+x^2} \) and set the derivative to zero to find the maximum and minimum values.
\[ g'(x) = \frac{(1+x^2)(1) - x(2x)}{(1+x^2)^2} = \frac{1+x^2 - 2x^2}{(1+x^2)^2} = \frac{1-x^2}{(1+x^2)^2} \]
Setting \( g'(x) = 0 \) gives:
\[ 1 - x^2 = 0 \implies x = \pm 1 \]

2. **Evaluate \( g(x) \) at critical points and endpoints**:
- \( g(1) = \frac{1}{1+1^2} = \frac{1}{2} \)
- \( g(-1) = \frac{-1}{1+(-1)^2} = -\frac{1}{2} \)
- As \( x \to \infty \) or \( x \to -\infty \), \( g(x) \to 0 \).

### Step 3: Determine the maximum and minimum values
From the evaluations:
- The maximum value of \( g(x) \) is \( \frac{1}{2} \) (at \( x = 1 \)).
- The minimum value of \( g(x) \) is \( -\frac{1}{2} \) (at \( x = -1 \)).

Thus, the range of \( g(x) = \frac{x}{1+x^2} \) is \( \left[-\frac{1}{2}, \frac{1}{2}\right] \).

### Step 4: Find the range of \( f(x) \)
Since \( f(x) = \sin^{-1}(g(x)) \) and \( g(x) \) takes values in \( \left[-\frac{1}{2}, \frac{1}{2}\right] \), we find:
\[ f(x) \text{ takes values in } \left[\sin^{-1}\left(-\frac{1}{2}\right), \sin^{-1}\left(\frac{1}{2}\right)\right] \]
Calculating these:
- \( \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \)
- \( \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \)

### Step 5: Conclusion
Thus, the range of \( f(x) \) is \( \left[-\frac{\pi}{6}, \frac{\pi}{6}\right] \). Since \( f \) is surjective, the set \( A \) must be equal to this range.

Therefore, the set \( A \) is:
\[ \boxed{\left[-\frac{\pi}{6}, \frac{\pi}{6}\right]} \]