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. `oo`

Explanation

To find \( f'(1) \) for the function \( f(x) = \frac{[x]}{|x|} \) where \( x \neq 0 \) and \([x]\) denotes the greatest integer function, we will analyze the function in the intervals around \( x = 1 \).

### Step 1: Define the function in intervals
The greatest integer function \([x]\) takes the value of the largest integer less than or equal to \( x \). Therefore, we can define \( f(x) \) in different intervals:

1. For \( 0 < x < 1 \):
\[ [x] = 0 \implies f(x) = \frac{0}{|x|} = 0 \]

2. For \( 1 \leq x < 2 \):
\[ [x] = 1 \implies f(x) = \frac{1}{|x|} = \frac{1}{x} \]

3. For \( x < 0 \):
\[ [x] = -1 \implies f(x) = \frac{-1}{|x|} = -\frac{1}{-x} = \frac{1}{x} \]

Thus, we can summarize the function as:
\[ f(x) = \begin{cases} 0 & \text{for } 0 < x < 1 \\ \frac{1}{x} & \text{for } 1 \leq x < 2 \\ \frac{1}{x} & \text{for } x < 0 \end{cases} \]

### Step 2: Check continuity at \( x = 1 \)
To determine if \( f(x) \) is differentiable at \( x = 1 \), we first check its continuity at this point.

- **Left-hand limit** as \( x \to 1^- \):
\[ \lim_{x \to 1^-} f(x) = 0 \]

- **Right-hand limit** as \( x \to 1^+ \):
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \frac{1}{x} = 1 \]

Since the left-hand limit \( (0) \) is not equal to the right-hand limit \( (1) \), we conclude that:
\[ \lim_{x \to 1} f(x) \text{ does not exist.} \]

### Step 3: Conclusion about differentiability
Since \( f(x) \) is not continuous at \( x = 1 \), it cannot be differentiable at that point. Therefore, we conclude that:
\[ f'(1) \text{ does not exist.} \]

### Final Answer
Thus, the value of \( f'(1) \) is:
\[ \text{non-existent} \]

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