If `f(x)={x^2}-({x})^2, `where (x) denotes the fractional part of x, then

Options

A. `f (x) ` is continuous at ` x = 2 ` but not at ` x = - 2 `
B. `f (x) ` is continuous at ` x = - 2 ` but not at ` x = 2 `
C. ` f (x)` is continuous at ` x = 2 and x = - 2 `
D. `f (x) ` is discontinuous at ` x = 2 and x = - 2 `

Explanation

To solve the problem, we need to analyze the function \( f(x) = x^2 - (x)^{2} \), where \( (x) \) denotes the fractional part of \( x \). The fractional part of \( x \) is defined as \( (x) = x - \lfloor x \rfloor \), where \( \lfloor x \rfloor \) is the greatest integer less than or equal to \( x \).

We will check the continuity of the function at \( x = 2 \) and \( x = -2 \).

### Step 1: Evaluate \( f(2) \)

To find \( f(2) \):
\[ f(2) = 2^2 - (2)^2 = 4 - 0 = 4 \]

### Step 2: Calculate the Right-Hand Limit (RHL) as \( x \to 2^+ \)

For \( x \to 2^+ \), \( (x) = x - 2 \) because \( \lfloor x \rfloor = 2 \):
\[ f(x) = x^2 - (x - 2)^2 \]
Calculating \( (x - 2)^2 \):
\[ (x - 2)^2 = x^2 - 4x + 4 \]
Thus,
\[ f(x) = x^2 - (x^2 - 4x + 4) = 4x - 4 \]
Now, taking the limit as \( x \to 2^+ \):
\[ \lim_{x \to 2^+} f(x) = 4(2) - 4 = 8 - 4 = 4 \]

### Step 3: Calculate the Left-Hand Limit (LHL) as \( x \to 2^- \)

For \( x \to 2^- \), \( (x) = x - 2 \):
\[ f(x) = x^2 - (x - 2)^2 \]
Again, we calculate:
\[ f(x) = x^2 - (x - 2)^2 = x^2 - (x^2 - 4x + 4) = 4x - 4 \]
Now, taking the limit as \( x \to 2^- \):
\[ \lim_{x \to 2^-} f(x) = 4(2) - 4 = 8 - 4 = 4 \]

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

Since:
\[ f(2) = 4, \quad \lim_{x \to 2^+} f(x) = 4, \quad \lim_{x \to 2^-} f(x) = 4 \]
All values are equal, so \( f(x) \) is continuous at \( x = 2 \).

### Step 5: Evaluate \( f(-2) \)

To find \( f(-2) \):
\[ f(-2) = (-2)^2 - (-2)^2 = 4 - 0 = 4 \]

### Step 6: Calculate the Right-Hand Limit (RHL) as \( x \to -2^+ \)

For \( x \to -2^+ \), \( (x) = x + 2 \):
\[ f(x) = x^2 - (x + 2)^2 \]
Calculating \( (x + 2)^2 \):
\[ (x + 2)^2 = x^2 + 4x + 4 \]
Thus,
\[ f(x) = x^2 - (x^2 + 4x + 4) = -4x - 4 \]
Now, taking the limit as \( x \to -2^+ \):
\[ \lim_{x \to -2^+} f(x) = -4(-2) - 4 = 8 - 4 = 4 \]

### Step 7: Calculate the Left-Hand Limit (LHL) as \( x \to -2^- \)

For \( x \to -2^- \), \( (x) = x + 2 \):
\[ f(x) = x^2 - (x + 2)^2 \]
Again, we calculate:
\[ f(x) = x^2 - (x + 2)^2 = -4x - 4 \]
Now, taking the limit as \( x \to -2^- \):
\[ \lim_{x \to -2^-} f(x) = -4(-2) - 4 = 8 - 4 = 4 \]

### Step 8: Check Continuity at \( x = -2 \)

Since:
\[ f(-2) = 4, \quad \lim_{x \to -2^+} f(x) = 4, \quad \lim_{x \to -2^-} f(x) = 4 \]
All values are equal, so \( f(x) \) is continuous at \( x = -2 \).

### Conclusion

The function \( f(x) \) is continuous at both \( x = 2 \) and \( x = -2 \).

---