Suppose [x] = greatest integer `le x`
Let `f(x) = sin^(-1)[x^(2) + 1/2]-cos^(-1)[x^(2)-1/2]`, Then range of f is:

Options

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

Answer

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

### Step 1: Understanding the Greatest Integer Function
The greatest integer function, denoted as \([x]\), gives the largest integer less than or equal to \(x\). Therefore, we need to evaluate \([x^2 + \frac{1}{2}]\) and \([x^2 - \frac{1}{2}]\).

### Step 2: Determine the Range of \(x^2\)
Since \(x^2\) is always non-negative, we have:
\[ 0 \leq x^2 < \infty \]
Thus, we can analyze the expressions \(x^2 + \frac{1}{2}\) and \(x^2 - \frac{1}{2}\):
- For \(x^2 + \frac{1}{2}\), as \(x^2\) varies from \(0\) to \(\infty\), \(x^2 + \frac{1}{2}\) varies from \(\frac{1}{2}\) to \(\infty\).
- For \(x^2 - \frac{1}{2}\), as \(x^2\) varies from \(0\) to \(\infty\), \(x^2 - \frac{1}{2}\) varies from \(-\frac{1}{2}\) to \(\infty\).

### Step 3: Evaluate the Greatest Integer Functions
Now we evaluate:
1. \([x^2 + \frac{1}{2}]\)
- For \(0 \leq x^2 < 1\), \([x^2 + \frac{1}{2}] = 0\).
- For \(1 \leq x^2 < 2\), \([x^2 + \frac{1}{2}] = 1\).
- For \(x^2 \geq 2\), \([x^2 + \frac{1}{2}] \geq 1\).

2. \([x^2 - \frac{1}{2}]\)
- For \(0 \leq x^2 < \frac{1}{2}\), \([x^2 - \frac{1}{2}] = -1\).
- For \(\frac{1}{2} \leq x^2 < \frac{3}{2}\), \([x^2 - \frac{1}{2}] = 0\).
- For \(x^2 \geq \frac{3}{2}\), \([x^2 - \frac{1}{2}] \geq 0\).

### Step 4: Analyze the Function \(f(x)\)
Now we can analyze \(f(x)\) based on the values of \([x^2 + \frac{1}{2}]\) and \([x^2 - \frac{1}{2}]\):
1. **Case 1**: \(0 \leq x^2 < 1\)
- \(f(x) = \sin^{-1}(0) - \cos^{-1}(-1) = 0 - \pi = -\pi\).

2. **Case 2**: \(1 \leq x^2 < 2\)
- \(f(x) = \sin^{-1}(1) - \cos^{-1}(0) = \frac{\pi}{2} - \frac{\pi}{2} = 0\).

3. **Case 3**: \(x^2 \geq 2\)
- \(f(x) = \sin^{-1}(n) - \cos^{-1}(m)\) where \(n\) and \(m\) are integers greater than or equal to \(1\). The values of \(f(x)\) will vary but will always yield results between \(-\pi\) and \(0\).

### Step 5: Conclusion
The function \(f(x)\) takes values from \(-\pi\) to \(0\). Therefore, the range of \(f\) is:
\[ \text{Range of } f = [-\pi, 0] \]