If `f(x)=([x])/(|x|), x ne 0`, where [.] denotes the greatest integer function, then f'(1) is

Options

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

Explanation

To find \( f'(1) \) for the function \( f(x) = \frac{[x]}{|x|} \) where \( [x] \) denotes the greatest integer function and \( x \neq 0 \), we will follow these steps:

### Step 1: Analyze the function
The function is defined as:
\[ f(x) = \frac{[x]}{|x|} \]
We need to evaluate this function in different intervals to understand its behavior.

### Step 2: Determine the function values in different intervals
1. **For \( x \in (-\infty, 0) \)**:
- Here, \( |x| = -x \) and \( [x] \) is the greatest integer less than \( x \), which is negative.
- Thus, \( f(x) = \frac{[x]}{-x} \).

2. **For \( x \in (0, 1) \)**:
- Here, \( |x| = x \) and \( [x] = 0 \) since \( x \) is less than 1.
- Thus, \( f(x) = \frac{0}{x} = 0 \).

3. **For \( x \in [1, 2) \)**:
- Here, \( |x| = x \) and \( [x] = 1 \).
- Thus, \( f(x) = \frac{1}{x} \).

### Step 3: Check continuity at \( x = 1 \)
To check if \( f(x) \) is continuous at \( x = 1 \), we need to find the left-hand limit (LHL), right-hand limit (RHL), and the value of the function at that point.

- **LHL at \( x = 1 \)**:
\[ \text{LHL} = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} 0 = 0 \]

- **RHL at \( x = 1 \)**:
\[ \text{RHL} = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \frac{1}{x} = 1 \]

- **Value at \( x = 1 \)**:
\[ f(1) = \frac{1}{1} = 1 \]

### Step 4: Conclusion on continuity
Since:
\[ \text{LHL} = 0 \quad \text{and} \quad \text{RHL} = 1 \quad \text{and} \quad f(1) = 1 \]
we see that:
\[ \text{LHL} \neq \text{RHL} \]
Thus, \( f(x) \) is not continuous at \( x = 1 \).

### Step 5: Differentiability at \( x = 1 \)
A function must be continuous at a point to be differentiable there. Since \( f(x) \) is not continuous at \( x = 1 \), it follows that \( f'(1) \) does not exist.

### Final Answer
\[ f'(1) \text{ does not exist.} \]

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