If `f(x) = [x^(-2) [x^(2)]]`, (where `[*]` denotes the greatest integer function) `x ne 0`, then incorrect statement

Options

A. f(x) is continuous everywhere
B. f(x) is discontinuous at `x = sqrt(2)`
C. f(x) is non-differentiable at x = 1
D. f(x) is discontinuous at infinitely many points

Answer

To solve the problem, we need to analyze the function \( f(x) = \left[ x^{-2} \left[ x^2 \right] \right] \), where \([ \cdot ]\) denotes the greatest integer function (also known as the floor function). We will determine the continuity and differentiability of this function at various points.

### Step 1: Understanding the Function
The function can be rewritten as:
\[ f(x) = \left[ \frac{\left[ x^2 \right]}{x^2} \right] \]
where \( x \neq 0 \).

### Step 2: Analyzing for Integer Values of \( x \)
1. **If \( x \) is an integer**:
- Then \( x^2 \) is also an integer.
- Therefore, \(\left[ x^2 \right] = x^2\).
- Thus, \( f(x) = \left[ \frac{x^2}{x^2} \right] = \left[ 1 \right] = 1 \).
- So, for integer values, \( f(x) = 1 \).

### Step 3: Analyzing for Non-Integer Values of \( x \)
2. **If \( x \) is not an integer**:
- Then \( x^2 \) is not an integer, and we have \(\left[ x^2 \right] < x^2\).
- Therefore, \(\left[ x^2 \right] = k\) where \( k \) is the greatest integer less than \( x^2 \).
- Thus, \( f(x) = \left[ \frac{k}{x^2} \right] \).
- Since \( k < x^2 \), it follows that \( \frac{k}{x^2} < 1 \).
- Hence, \( f(x) = \left[ \frac{k}{x^2} \right] = 0 \) for non-integer \( x \).

### Step 4: Summary of Function Values
- For integer \( x \): \( f(x) = 1 \)
- For non-integer \( x \): \( f(x) = 0 \)

### Step 5: Continuity and Discontinuity
3. **Continuity**:
- The function is discontinuous at every integer point because the limit from the left (approaching an integer) gives 0, while the value at the integer point is 1.
- Therefore, \( f(x) \) is discontinuous at all integers.

4. **Specific Points**:
- At \( x = \sqrt{2} \), which is not an integer, the function is continuous in a small neighborhood around this point since it takes the value 0 for all non-integer values.
- At \( x = 1 \), the function is not differentiable because the left-hand limit and right-hand limit do not match (0 from the left and 1 at the point).

### Conclusion
Now we can evaluate the statements:
- **Option A**: \( f(x) \) is continuous everywhere. **(Incorrect)**
- **Option B**: \( f(x) \) is discontinuous at \( x = \sqrt{2} \). **(Incorrect)**
- **Option C**: \( f(x) \) is non-differentiable at \( x = 1 \). **(Correct)**
- **Option D**: \( f(x) \) is discontinuous at infinitely many points. **(Correct)**

Thus, the incorrect statements are **A** and **B**.