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

Options

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

Answer

To find the range of the function \( f(x) = \sin^{-1}\left(\frac{x^2}{1+x^2}\right) \), we will first analyze the inner function \( g(x) = \frac{x^2}{1+x^2} \).

### Step 1: Analyze the function \( g(x) = \frac{x^2}{1+x^2} \)

1. **Domain**: The function \( g(x) \) is defined for all real numbers \( x \).
2. **Behavior as \( x \to \infty \)**:
\[ g(x) = \frac{x^2}{1+x^2} \approx \frac{x^2}{x^2} = 1 \quad \text{as } x \to \infty \]
Thus, \( g(x) \) approaches 1 but never actually reaches it.
3. **Behavior as \( x \to 0 \)**:
\[ g(0) = \frac{0^2}{1+0^2} = 0 \]
4. **Behavior for negative \( x \)**: Since \( x^2 \) is always non-negative, \( g(x) \) is also non-negative for all real \( x \).

### Step 2: Determine the range of \( g(x) \)

From the analysis:
- As \( x \) varies from \( -\infty \) to \( \infty \), \( g(x) \) takes values from \( 0 \) to \( 1 \) (not including \( 1 \)).
- Therefore, the range of \( g(x) \) is:
\[ [0, 1) \]

### Step 3: Find the range of \( f(x) = \sin^{-1}(g(x)) \)

1. **Apply the range of \( g(x) \)**:
Since \( g(x) \) takes values in the interval \( [0, 1) \), we can substitute this into the inverse sine function.
2. **Calculate the range of \( f(x) \)**:
\[ f(x) = \sin^{-1}(y) \quad \text{where } y \in [0, 1) \]
- The minimum value occurs when \( y = 0 \):
\[ \sin^{-1}(0) = 0 \]
- The maximum value occurs as \( y \) approaches \( 1 \):
\[ \sin^{-1}(1) = \frac{\pi}{2} \quad \text{(but never actually reaches this value)} \]

### Conclusion

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