Let `f(x) = |x - 1|([x] - [-x])`, then which of the following statement(s) is/are correct. (where [.] denotes greatest integer function.)

Options

A. f(x) is continuous at x = 1
B. f(x) is derivable at x = 1
C. f(X) is non-derivable at x = 1
D. f(x) is discontinuous at x = 1

Answer

To solve the problem, we need to analyze the function \( f(x) = |x - 1|([x] - [-x]) \), where \([.]\) denotes the greatest integer function.

### Step-by-Step Solution

1. **Understanding the Function**:
The function can be broken down into two parts:
- \( |x - 1| \): This is the absolute value function, which is continuous everywhere.
- \([x] - [-x]\): This involves the greatest integer function.

2. **Evaluating \([x] - [-x]\)**:
The expression \([x] - [-x]\) can be simplified based on the value of \(x\):
- For \(x \in [n, n+1)\) where \(n\) is an integer, \([x] = n\) and \([-x] = -n - 1\) if \(x\) is not an integer, or \([-x] = -n\) if \(x\) is an integer.
- Thus, \([x] - [-x] = n - (-n - 1) = 2n + 1\) if \(x\) is not an integer and \([x] - [-x] = n - (-n) = 0\) if \(x\) is an integer.

3. **Finding \(f(x)\) for Different Intervals**:
- For \(x\) in the interval \((n, n+1)\) (not including integers):
\[ f(x) = |x - 1| \cdot (2n + 1) \]
- For \(x = n\) (integer):
\[ f(n) = |n - 1| \cdot 0 = 0 \]

4. **Checking Continuity at \(x = 1\)**:
- Calculate \(f(1)\):
\[ f(1) = 0 \]
- Calculate the right-hand limit as \(x\) approaches 1 from the right (\(1 + h\)):
\[ f(1 + h) = |(1 + h) - 1| \cdot (2 \cdot 1 + 1) = h \cdot 3 = 3h \]
As \(h \to 0\), \(f(1 + h) \to 0\).
- Calculate the left-hand limit as \(x\) approaches 1 from the left (\(1 - h\)):
\[ f(1 - h) = |(1 - h) - 1| \cdot (2 \cdot 1 + 1) = h \cdot 3 = 3h \]
As \(h \to 0\), \(f(1 - h) \to 0\).

Since both limits equal \(f(1)\), \(f(x)\) is continuous at \(x = 1\).

5. **Checking Differentiability at \(x = 1\)**:
- Right-hand derivative:
\[ f'(1^+) = \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0} \frac{3h - 0}{h} = 3 \]
- Left-hand derivative:
\[ f'(1^-) = \lim_{h \to 0} \frac{f(1 - h) - f(1)}{-h} = \lim_{h \to 0} \frac{3h - 0}{-h} = -3 \]

Since the right-hand and left-hand derivatives are not equal, \(f(x)\) is not differentiable at \(x = 1\).

### Conclusion
- \(f(x)\) is continuous at \(x = 1\).
- \(f(x)\) is not differentiable at \(x = 1\).

### Correct Statements
Based on the analysis:
- The correct statements are:
- \(f(x)\) is continuous at \(x = 1\).
- \(f(x)\) is not differentiable at \(x = 1\).