Domain of `f(x)=sin^(-1)[2-4x^2],` where [.] denotes the greatest integer function, is: `(a)[-(sqrt(3))/2,(sqrt(3))/2]-{0} ` (b) `[-(sqrt(3))/2,(sqrt(3))/2]` `(c)[-(sqrt(3))/2,(sqrt(3))/2)-{0}` (d) `(-(sqrt(3))/2,(sqrt(3))/2)-0`

Answer

To find the domain of the function \( f(x) = \sin^{-1}[2 - 4x^2] \), where \([.]\) denotes the greatest integer function, we need to ensure that the expression inside the sine inverse function is valid. The sine inverse function, \(\sin^{-1}(x)\), is defined for \(x\) in the interval \([-1, 1]\).

### Step-by-step Solution:

1. **Set the Inequality**:
We need to find the values of \(x\) such that:
\[ -1 \leq 2 - 4x^2 \leq 1 \]

2. **Solve the Left Inequality**:
Start with the left part of the inequality:
\[ 2 - 4x^2 \geq -1 \]
Rearranging gives:
\[ 4x^2 \leq 3 \quad \Rightarrow \quad x^2 \leq \frac{3}{4} \]

3. **Solve the Right Inequality**:
Now, consider the right part of the inequality:
\[ 2 - 4x^2 \leq 1 \]
Rearranging gives:
\[ 4x^2 \geq 1 \quad \Rightarrow \quad x^2 \geq \frac{1}{4} \]

4. **Combine the Results**:
We now have two inequalities:
\[ \frac{1}{4} \leq x^2 \leq \frac{3}{4} \]
Taking square roots gives:
\[ \sqrt{\frac{1}{4}} \leq |x| \leq \sqrt{\frac{3}{4}} \quad \Rightarrow \quad \frac{1}{2} \leq |x| \leq \frac{\sqrt{3}}{2} \]

5. **Express in Interval Notation**:
This means:
\[ -\frac{\sqrt{3}}{2} \leq x \leq -\frac{1}{2} \quad \text{or} \quad \frac{1}{2} \leq x \leq \frac{\sqrt{3}}{2} \]
In interval notation, this can be expressed as:
\[ x \in \left[-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right] \cup \left[\frac{1}{2}, \frac{\sqrt{3}}{2}\right] \]

6. **Remove the Point where \(x = 0\)**:
Since the function involves the greatest integer function, we need to exclude \(x = 0\) from the domain:
\[ \text{Domain} = \left[-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right] \cup \left[\frac{1}{2}, \frac{\sqrt{3}}{2}\right] - \{0\} \]

### Final Domain:
Thus, the domain of \( f(x) = \sin^{-1}[2 - 4x^2] \) is:
\[ \left[-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right] \cup \left[\frac{1}{2}, \frac{\sqrt{3}}{2}\right] - \{0\} \]

### Answer:
The correct option is (c) \([- \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}] - \{0\}\).