Let `f(x)=[|x|]` where [.] denotes the greatest integer function, then `f'(-1)` is

Options

A. 0
B. 1
C. non-existent
D. none of these

Explanation

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

### Step 1: Understand the function \( f(x) \)
The function \( f(x) = [|x|] \) takes the absolute value of \( x \) and then applies the greatest integer function. This means:
- For \( x \) in the interval \( [-1, 0) \), \( |x| = -x \) and thus \( f(x) = [|x|] = [ -x ] = 1 \) (since \( -x \) is between 1 and 0).
- For \( x = -1 \), \( f(-1) = [|-1|] = [1] = 1 \).
- For \( x \) in the interval \( [0, 1) \), \( |x| = x \) and thus \( f(x) = [|x|] = [x] = 0 \).

### Step 2: Determine the left-hand derivative (LHD) at \( x = -1 \)
The left-hand derivative at \( x = -1 \) is given by:
\[ \text{LHD} = \lim_{x \to -1^-} \frac{f(x) - f(-1)}{x - (-1)} \]
Since \( f(-1) = 1 \) and for \( x \) approaching \(-1\) from the left, \( f(x) = 1 \):
\[ \text{LHD} = \lim_{x \to -1^-} \frac{1 - 1}{x + 1} = \lim_{x \to -1^-} \frac{0}{x + 1} = 0 \]

### Step 3: Determine the right-hand derivative (RHD) at \( x = -1 \)
The right-hand derivative at \( x = -1 \) is given by:
\[ \text{RHD} = \lim_{x \to -1^+} \frac{f(x) - f(-1)}{x - (-1)} \]
For \( x \) approaching \(-1\) from the right, \( f(x) = 0 \):
\[ \text{RHD} = \lim_{x \to -1^+} \frac{0 - 1}{x + 1} = \lim_{x \to -1^+} \frac{-1}{x + 1} \]
As \( x \) approaches \(-1\) from the right, \( x + 1 \) approaches \( 0 \) from the positive side, leading to:
\[ \text{RHD} = -\infty \]

### Step 4: Conclusion
Since the left-hand derivative (LHD) is \( 0 \) and the right-hand derivative (RHD) is \( -\infty \), we conclude that:
\[ f'(-1) \text{ does not exist} \]

### Final Answer
Thus, \( f'(-1) \) does not exist.