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

Options

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

Answer

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

### Step 1: Analyze the components of the function

1. **Identify the input for the greatest integer function (GIF)**:
- The term \( \lfloor x^2 + \frac{1}{2} \rfloor \) can take integer values based on the value of \( x^2 \).
- The term \( \lfloor x^2 - \frac{1}{2} \rfloor \) also takes integer values.

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

- Since \( x^2 \) is always non-negative, we can analyze the range of \( x^2 \):
- \( x^2 \geq 0 \)

### Step 3: Calculate the values of \( \lfloor x^2 + \frac{1}{2} \rfloor \)

- For \( x^2 \) in the range \( [0, 1) \):
- \( \lfloor x^2 + \frac{1}{2} \rfloor = \lfloor 0 + \frac{1}{2} \rfloor = 0 \)

- For \( x^2 \) in the range \( [1, 2) \):
- \( \lfloor x^2 + \frac{1}{2} \rfloor = \lfloor 1 + \frac{1}{2} \rfloor = 1 \)

- For \( x^2 \) in the range \( [2, 3) \):
- \( \lfloor x^2 + \frac{1}{2} \rfloor = \lfloor 2 + \frac{1}{2} \rfloor = 2 \)

### Step 4: Calculate the values of \( \lfloor x^2 - \frac{1}{2} \rfloor \)

- For \( x^2 \) in the range \( [0, 1) \):
- \( \lfloor x^2 - \frac{1}{2} \rfloor = \lfloor 0 - \frac{1}{2} \rfloor = -1 \)

- For \( x^2 \) in the range \( [1, 2) \):
- \( \lfloor x^2 - \frac{1}{2} \rfloor = \lfloor 1 - \frac{1}{2} \rfloor = 0 \)

- For \( x^2 \) in the range \( [2, 3) \):
- \( \lfloor x^2 - \frac{1}{2} \rfloor = \lfloor 2 - \frac{1}{2} \rfloor = 1 \)

### Step 5: Evaluate the function for the possible values

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

2. **When \( \lfloor x^2 + \frac{1}{2} \rfloor = 1 \) and \( \lfloor x^2 - \frac{1}{2} \rfloor = 0 \)**:
\[ y = 2 \sin^{-1}(1) + \cos^{-1}(0) = 2 \cdot \frac{\pi}{2} + \frac{\pi}{2} = \pi + \frac{\pi}{2} = \frac{3\pi}{2} \]

3. **When \( \lfloor x^2 + \frac{1}{2} \rfloor = 2 \) and \( \lfloor x^2 - \frac{1}{2} \rfloor = 1 \)**:
\[ y = 2 \sin^{-1}(2) + \cos^{-1}(1) \]
- Note: \( \sin^{-1}(2) \) is undefined; hence this case does not contribute to the range.

### Step 6: Conclusion on the range

The only valid outputs from the function are \( \pi \) and \( \frac{3\pi}{2} \). Therefore, the range of the function is:
\[ \{ \pi, \frac{3\pi}{2} \} \]