If `f(x) = {{:(|x-1|([x]-x)",",x ne 1),(0",",x = 1):}`, then

Options

A. A. `f'(1^(+)) = 0`
B. B. `f'(1^(-)) = 0`
C. C. `f'(1^(-)) = - 1`
D. D. f(x) is differentiable at x = 1

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined as follows:

\[ f(x) = \begin{cases} |x - 1| \cdot ([x] - x) & \text{if } x \neq 1 \\ 0 & \text{if } x = 1 \end{cases} \]

Where \([x]\) denotes the greatest integer function (or floor function).

### Step 1: Analyze the function for \( x > 1 \)

When \( x > 1 \):
- The modulus \( |x - 1| = x - 1 \).
- The greatest integer function \([x]\) will be \( [x] = n \) where \( n \) is the largest integer less than or equal to \( x \). Thus, \([x] - x = n - x\).

Now, we can express \( f(x) \) as:
\[ f(x) = (x - 1)([x] - x) = (x - 1)(n - x) \]

Since \( n = [x] \) is an integer, we can write:
\[ f(x) = (x - 1)(n - x) = -(x - 1)(x - n) \]

### Step 2: Analyze the function for \( x < 1 \)

When \( x < 1 \):
- The modulus \( |x - 1| = -(x - 1) = 1 - x \).
- The greatest integer function \([x]\) will be \( [x] = -1 \) since \( x \) is less than 1.

Thus, we can express \( f(x) \) as:
\[ f(x) = (1 - x)(-1 - x) = -(1 - x)(x + 1) \]

This simplifies to:
\[ f(x) = -(x^2 + x - 1) \]

### Step 3: Evaluate the function at \( x = 1 \)

At \( x = 1 \):
\[ f(1) = 0 \]

### Step 4: Find the derivative \( f'(x) \)

#### Case 1: For \( x > 1 \)

Using the expression derived:
\[ f(x) = -(x - 1)(x - n) \]
To find \( f'(x) \), we differentiate:
\[ f'(x) = -[(x - n) + (x - 1)(-1)] = -[(x - n) - (x - 1)] = -[1 - n] \]

#### Case 2: For \( x < 1 \)

Using the expression derived:
\[ f(x) = -(x^2 + x - 1) \]
To find \( f'(x) \), we differentiate:
\[ f'(x) = -[2x + 1] \]

### Step 5: Evaluate \( f'(1^+) \) and \( f'(1^-) \)

1. **For \( x > 1 \)**:
- As \( x \) approaches 1 from the right, \( n = 1 \) (since \( [x] = 1 \)).
- Thus, \( f'(1^+) = -[1 - 1] = 0 \).

2. **For \( x < 1 \)**:
- As \( x \) approaches 1 from the left, we use \( f'(x) = -[2x + 1] \).
- Therefore, \( f'(1^-) = -[2(1) + 1] = -3 \).

### Step 6: Check differentiability at \( x = 1 \)

For \( f(x) \) to be differentiable at \( x = 1 \), we need:
\[ f'(1^+) = f'(1^-) \]
Since \( f'(1^+) = 0 \) and \( f'(1^-) = -3 \), they are not equal. Thus, \( f(x) \) is not differentiable at \( x = 1 \).

### Conclusion

The correct answer is option A, which states that \( f'(1^+) = 0 \).

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