The range of the function `f(x)=sin^(-1)[x^(2)-(1)/(3)]-cos^(-1)[x^(2)+(2)/()]` is (where, [x] represents the greatest integer value of x)

Options

A. `[-pi, 0]`
B. `{-pi, 0}`
C. `{0, pi}`
D. `{0, pi, -pi}`

Answer

To find the range of the function \( f(x) = \sin^{-1}\left[x^2 - \frac{1}{3}\right] - \cos^{-1}\left[x^2 + 2\right] \), where \([x]\) represents the greatest integer value of \(x\), we will follow these steps:

### Step 1: Define the function components
We start with the function:
\[ f(x) = \sin^{-1}\left[x^2 - \frac{1}{3}\right] - \cos^{-1}\left[x^2 + 2\right] \]
Here, we need to analyze the arguments of the inverse sine and cosine functions.

### Step 2: Determine the range of \(x^2\)
Since \([x]\) is the greatest integer less than or equal to \(x\), let \(n = [x]\). Then \(n \leq x < n + 1\), which implies:
\[ n^2 \leq x^2 < (n + 1)^2 \]
Thus, the possible values for \(x^2\) are in the range:
\[ [n^2, (n + 1)^2) \]

### Step 3: Analyze the argument of \(\sin^{-1}\)
The argument of \(\sin^{-1}\) is:
\[ x^2 - \frac{1}{3} \]
For \(x^2\) in the range \([n^2, (n + 1)^2)\), we have:
\[ n^2 - \frac{1}{3} \leq x^2 - \frac{1}{3} < (n + 1)^2 - \frac{1}{3} \]
This means:
\[ \left[n^2 - \frac{1}{3}, (n + 1)^2 - \frac{1}{3}\right) \]

### Step 4: Analyze the argument of \(\cos^{-1}\)
The argument of \(\cos^{-1}\) is:
\[ x^2 + 2 \]
For \(x^2\) in the range \([n^2, (n + 1)^2)\), we have:
\[ n^2 + 2 \leq x^2 + 2 < (n + 1)^2 + 2 \]
This means:
\[ [n^2 + 2, (n + 1)^2 + 2) \]

### Step 5: Determine the range of \(\sin^{-1}\) and \(\cos^{-1}\)
The function \(\sin^{-1}(k)\) is defined for \(k \in [-1, 1]\) and the range is \([- \frac{\pi}{2}, \frac{\pi}{2}]\).
The function \(\cos^{-1}(k)\) is defined for \(k \in [-1, 1]\) and the range is \([0, \pi]\).

### Step 6: Find the valid range for \(k\)
We need to ensure that the arguments of \(\sin^{-1}\) and \(\cos^{-1}\) fall within their respective domains:
1. For \(\sin^{-1}\):
\[ -1 \leq x^2 - \frac{1}{3} \leq 1 \]
This gives:
\[ -\frac{2}{3} \leq x^2 \leq \frac{4}{3} \]
Since \(x^2\) is always non-negative, we have:
\[ 0 \leq x^2 \leq \frac{4}{3} \]

2. For \(\cos^{-1}\):
\[ -1 \leq x^2 + 2 \leq 1 \]
This gives:
\[ -3 \leq x^2 \leq -1 \]
This condition is not possible since \(x^2\) cannot be negative.

### Step 7: Conclusion
Since the argument for \(\cos^{-1}\) does not yield valid values, we need to focus on the valid outputs from \(\sin^{-1}\):
- The valid range from \(\sin^{-1}\) is from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).

Thus, the overall range of the function \(f(x)\) is:
\[ \text{Range of } f(x) = \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \]