The range of the function `f(x) = sin^(-1)[x^2+1/2]+cos^(-1)[x^2-1/2]`, where [ . ] denotes the greatest integer function.

Options

A. `{(pi)/2,pi}`
B. `{0,(pi)/2}`
C. `{pi}`
D. `(0,(pi)/2)`

Answer

To find the range of the function \( f(x) = \sin^{-1}\left[\lfloor x^2 + \frac{1}{2} \rfloor\right] + \cos^{-1}\left[\lfloor x^2 - \frac{1}{2} \rfloor\right] \), we will analyze the components of the function step by step.

### Step 1: Determine the values of \( x^2 \)

Since \( x^2 \) is always non-negative, we have:
\[ x^2 \geq 0 \]

### Step 2: Analyze the terms \( \lfloor x^2 + \frac{1}{2} \rfloor \) and \( \lfloor x^2 - \frac{1}{2} \rfloor \)

1. **For \( \lfloor x^2 + \frac{1}{2} \rfloor \)**:
- When \( x^2 = 0 \), \( \lfloor 0 + \frac{1}{2} \rfloor = \lfloor 0.5 \rfloor = 0 \)
- As \( x^2 \) increases, \( x^2 + \frac{1}{2} \) will take values from \( 0.5 \) to \( \infty \).
- Therefore, \( \lfloor x^2 + \frac{1}{2} \rfloor \) will take values starting from \( 0 \) and can go up to any integer \( n \) as \( x^2 \) increases.

2. **For \( \lfloor x^2 - \frac{1}{2} \rfloor \)**:
- When \( x^2 = 0 \), \( \lfloor 0 - \frac{1}{2} \rfloor = \lfloor -0.5 \rfloor = -1 \)
- As \( x^2 \) increases, \( x^2 - \frac{1}{2} \) will take values from \( -0.5 \) to \( \infty \).
- Therefore, \( \lfloor x^2 - \frac{1}{2} \rfloor \) will take values starting from \( -1 \) and can go up to any integer \( n \).

### Step 3: Determine the possible values of the floor functions

- \( \lfloor x^2 + \frac{1}{2} \rfloor \) can take values \( 0, 1, 2, \ldots \)
- \( \lfloor x^2 - \frac{1}{2} \rfloor \) can take values \( -1, 0, 1, 2, \ldots \)

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

1. **Case 1**: When \( \lfloor x^2 + \frac{1}{2} \rfloor = 0 \) and \( \lfloor x^2 - \frac{1}{2} \rfloor = -1 \):
\[ f(x) = \sin^{-1}(0) + \cos^{-1}(-1) = 0 + \pi = \pi \]

2. **Case 2**: When \( \lfloor x^2 + \frac{1}{2} \rfloor = 1 \) and \( \lfloor x^2 - \frac{1}{2} \rfloor = 0 \):
\[ f(x) = \sin^{-1}(1) + \cos^{-1}(0) = \frac{\pi}{2} + \frac{\pi}{2} = \pi \]

3. **Case 3**: When \( \lfloor x^2 + \frac{1}{2} \rfloor = 1 \) and \( \lfloor x^2 - \frac{1}{2} \rfloor = -1 \):
\[ f(x) = \sin^{-1}(1) + \cos^{-1}(-1) = \frac{\pi}{2} + \pi = \frac{3\pi}{2} \]

4. **Case 4**: When \( \lfloor x^2 + \frac{1}{2} \rfloor = 2 \) and \( \lfloor x^2 - \frac{1}{2} \rfloor = 0 \):
\[ f(x) = \sin^{-1}(2) + \cos^{-1}(0) \text{ (not valid since } \sin^{-1}(2) \text{ is undefined)} \]

### Conclusion

After analyzing all cases, we find that the only valid output of the function is \( \pi \). Thus, the range of the function \( f(x) \) is:
\[ \text{Range of } f(x) = \{ \pi \} \]