The range of the function `sin^(-1)((x^(2))/(1+x^(2)))` is

Options

A. `(0,pi/2)`
B. `[0,pi/2)`
C. `[0,pi/2]`
D. none of these

Answer

To find the range of the function \( f(x) = \sin^{-1}\left(\frac{x^2}{1+x^2}\right) \), we will follow these steps:

### Step 1: Determine the range of the inner function
The first step is to analyze the expression \( \frac{x^2}{1+x^2} \).

- As \( x \) varies from \(-\infty\) to \(+\infty\), \( x^2 \) is always non-negative, meaning \( x^2 \geq 0 \).
- The denominator \( 1 + x^2 \) is always positive since \( x^2 \) is non-negative and \( 1 \) is positive.

Thus, \( \frac{x^2}{1+x^2} \) is always non-negative.

### Step 2: Find the limits of \( \frac{x^2}{1+x^2} \)
Now we need to find the minimum and maximum values of \( \frac{x^2}{1+x^2} \).

- **Minimum Value**: When \( x = 0 \),
\[ \frac{0^2}{1+0^2} = 0. \]

- **Maximum Value**: As \( x \) approaches \( \infty \),
\[ \frac{x^2}{1+x^2} \to \frac{x^2}{x^2} = 1. \]
However, it never actually reaches 1, so the maximum value is \( 1 \) (but not inclusive).

Thus, the range of \( \frac{x^2}{1+x^2} \) is \( [0, 1) \).

### Step 3: Apply the range to the inverse sine function
Next, we apply the range of the inner function to the inverse sine function \( \sin^{-1}(y) \).

- The range of \( \sin^{-1}(y) \) is from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \) for \( y \) in the interval \([-1, 1]\).
- Since \( \frac{x^2}{1+x^2} \) takes values from \( 0 \) to \( 1 \), we need to find the corresponding values of \( \sin^{-1}(y) \) for \( y \) in the interval \( [0, 1) \).

- **Minimum Value**: When \( y = 0 \),
\[ \sin^{-1}(0) = 0. \]

- **Maximum Value**: As \( y \) approaches \( 1 \) (but does not include it),
\[ \sin^{-1}(1) = \frac{\pi}{2}. \]

Thus, the range of \( f(x) = \sin^{-1}\left(\frac{x^2}{1+x^2}\right) \) is \( [0, \frac{\pi}{2}) \).

### Final Answer
The range of the function \( \sin^{-1}\left(\frac{x^2}{1+x^2}\right) \) is:
\[ [0, \frac{\pi}{2}). \]