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 \( y = 2 \sin^{-1}(x^2 + \frac{1}{2}) + \cos^{-1}(x^2 - \frac{1}{2}) \), we will analyze the components of the function step by step.

### Step 1: Determine the domain of the inverse sine and cosine functions.

The function \( \sin^{-1}(u) \) is defined for \( -1 \leq u \leq 1 \), and \( \cos^{-1}(v) \) is also defined for \( -1 \leq v \leq 1 \).

- For \( \sin^{-1}(x^2 + \frac{1}{2}) \):
\[ -1 \leq x^2 + \frac{1}{2} \leq 1 \]
This simplifies to:
\[ -\frac{1}{2} \leq x^2 \leq \frac{1}{2} \]
Since \( x^2 \) is always non-negative, we have:
\[ 0 \leq x^2 \leq \frac{1}{2} \]
Therefore, \( x \) can take values in the range:
\[ -\sqrt{\frac{1}{2}} \leq x \leq \sqrt{\frac{1}{2}} \]

- For \( \cos^{-1}(x^2 - \frac{1}{2}) \):
\[ -1 \leq x^2 - \frac{1}{2} \leq 1 \]
This simplifies to:
\[ -\frac{1}{2} \leq x^2 \leq \frac{3}{2} \]
Again, since \( x^2 \) is non-negative, we have:
\[ 0 \leq x^2 \leq \frac{3}{2} \]
Therefore, \( x \) can take values in the range:
\[ -\sqrt{\frac{3}{2}} \leq x \leq \sqrt{\frac{3}{2}} \]

### Step 2: Analyze the function values.

Now we will evaluate the function at the boundaries of the determined ranges.

1. **When \( x^2 = 0 \)**:
\[ y = 2 \sin^{-1}(0 + \frac{1}{2}) + \cos^{-1}(0 - \frac{1}{2}) = 2 \sin^{-1}(\frac{1}{2}) + \cos^{-1}(-\frac{1}{2}) \]
We know \( \sin^{-1}(\frac{1}{2}) = \frac{\pi}{6} \) and \( \cos^{-1}(-\frac{1}{2}) = \frac{2\pi}{3} \):
\[ y = 2 \cdot \frac{\pi}{6} + \frac{2\pi}{3} = \frac{\pi}{3} + \frac{2\pi}{3} = \pi \]

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

### Step 3: Determine the range of the function.

The values of \( y \) at the boundaries are \( \pi \) and \( \frac{3\pi}{2} \). Since the function is continuous within the domain, the range of \( y \) is:
\[ \{ \pi, \frac{3\pi}{2} \} \]

### Step 4: Apply the greatest integer function.

The greatest integer function \( [y] \) will take the values:
- For \( y = \pi \), \( [\pi] = 3 \) (since \( \pi \approx 3.14 \))
- For \( y = \frac{3\pi}{2} \), \( \left[\frac{3\pi}{2}\right] = 4 \) (since \( \frac{3\pi}{2} \approx 4.71 \))

Thus, the range of the function \( y \) in terms of the greatest integer function is:
\[ \{3, 4\} \]

### Final Answer:
The range of the function \( y = 2 \sin^{-1}(x^2 + \frac{1}{2}) + \cos^{-1}(x^2 - \frac{1}{2}) \) is \( \{3, 4\} \).