The function `f(x) = {{:(x^(2)[1/x^(2)], x ne 0, ),(0, x=0,):}` where [x] represents the greatest integer less than or equal to x

Options

A. continous at x=1
B. continous at x=-1
C. continous at x=0
D. continous at x=2

Answer

To determine the continuity of the function \( f(x) = \begin{cases} x^2 \cdot \lfloor \frac{1}{x^2} \rfloor & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \), we need to check the continuity at the points \( x = 1, -1, 0, \) and \( 2 \).

### Step 1: Check continuity at \( x = 1 \)

1. **Calculate \( f(1) \)**:
\[ f(1) = 1^2 \cdot \lfloor \frac{1}{1^2} \rfloor = 1 \cdot \lfloor 1 \rfloor = 1 \]

2. **Calculate the right-hand limit (RHL) as \( x \to 1^+ \)**:
\[ \text{For } x > 1, \quad \lfloor \frac{1}{x^2} \rfloor = 0 \quad \text{(since } \frac{1}{x^2} < 1\text{)} \]
\[ \Rightarrow \text{RHL} = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} x^2 \cdot 0 = 0 \]

3. **Calculate the left-hand limit (LHL) as \( x \to 1^- \)**:
\[ \text{For } x < 1, \quad \lfloor \frac{1}{x^2} \rfloor = 1 \quad \text{(since } \frac{1}{x^2} \geq 1\text{)} \]
\[ \Rightarrow \text{LHL} = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x^2 \cdot 1 = 1 \]

4. **Conclusion for continuity at \( x = 1 \)**:
\[ \text{RHL} \neq \text{LHL} \Rightarrow f(x) \text{ is not continuous at } x = 1. \]

### Step 2: Check continuity at \( x = -1 \)

1. **Calculate \( f(-1) \)**:
\[ f(-1) = (-1)^2 \cdot \lfloor \frac{1}{(-1)^2} \rfloor = 1 \cdot \lfloor 1 \rfloor = 1 \]

2. **Calculate the right-hand limit (RHL) as \( x \to -1^+ \)**:
\[ \text{For } x > -1, \quad \lfloor \frac{1}{x^2} \rfloor = 1 \]
\[ \Rightarrow \text{RHL} = \lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} x^2 \cdot 1 = 1 \]

3. **Calculate the left-hand limit (LHL) as \( x \to -1^- \)**:
\[ \text{For } x < -1, \quad \lfloor \frac{1}{x^2} \rfloor = 0 \]
\[ \Rightarrow \text{LHL} = \lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} x^2 \cdot 0 = 0 \]

4. **Conclusion for continuity at \( x = -1 \)**:
\[ \text{RHL} \neq \text{LHL} \Rightarrow f(x) \text{ is not continuous at } x = -1. \]

### Step 3: Check continuity at \( x = 0 \)

1. **Calculate \( f(0) \)**:
\[ f(0) = 0 \]

2. **Calculate the right-hand limit (RHL) as \( x \to 0^+ \)**:
\[ \text{For } x > 0, \quad \lfloor \frac{1}{x^2} \rfloor \to \infty \text{ as } x \to 0^+ \]
\[ \Rightarrow \text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x^2 \cdot \lfloor \frac{1}{x^2} \rfloor = \infty \]

3. **Calculate the left-hand limit (LHL) as \( x \to 0^- \)**:
\[ \text{For } x < 0, \quad \lfloor \frac{1}{x^2} \rfloor \to \infty \text{ as } x \to 0^- \]
\[ \Rightarrow \text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x^2 \cdot \lfloor \frac{1}{x^2} \rfloor = \infty \]

4. **Conclusion for continuity at \( x = 0 \)**:
\[ \text{RHL} \neq f(0) \Rightarrow f(x) \text{ is not continuous at } x = 0. \]

### Step 4: Check continuity at \( x = 2 \)

1. **Calculate \( f(2) \)**:
\[ f(2) = 2^2 \cdot \lfloor \frac{1}{2^2} \rfloor = 4 \cdot \lfloor \frac{1}{4} \rfloor = 4 \cdot 0 = 0 \]

2. **Calculate the right-hand limit (RHL) as \( x \to 2^+ \)**:
\[ \text{For } x > 2, \quad \lfloor \frac{1}{x^2} \rfloor = 0 \]
\[ \Rightarrow \text{RHL} = \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} x^2 \cdot 0 = 0 \]

3. **Calculate the left-hand limit (LHL) as \( x \to 2^- \)**:
\[ \text{For } x < 2, \quad \lfloor \frac{1}{x^2} \rfloor = 0 \]
\[ \Rightarrow \text{LHL} = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} x^2 \cdot 0 = 0 \]

4. **Conclusion for continuity at \( x = 2 \)**:
\[ \text{RHL} = \text{LHL} = f(2) = 0 \Rightarrow f(x) \text{ is continuous at } x = 2. \]

### Final Conclusion
The function \( f(x) \) is continuous only at \( x = 2 \) and not continuous at \( x = 1, -1, \) and \( 0 \).