The function `f(x)=[x]+1/2,x!inI` is a/an (wher [.] denotes greatest integer function)

Options

A. Even
B. odd
C. neither even nor odd
D. even as well as odd

Explanation

To determine whether the function \( f(x) = [x] + \frac{1}{2} \) (where \([x]\) denotes the greatest integer function) is even, odd, or neither, we can follow these steps:

### Step 1: Understand the Function
The function \( f(x) \) is defined as:
- \( f(x) = [x] + \frac{1}{2} \) for \( x \notin \mathbb{Z} \) (i.e., \( x \) is not an integer).

### Step 2: Check for Evenness
A function \( f(x) \) is even if \( f(-x) = f(x) \) for all \( x \).

### Step 3: Calculate \( f(-x) \)
Let’s calculate \( f(-x) \):
- If \( x \) is not an integer, then \( -x \) is also not an integer.
- Therefore, \( f(-x) = [-x] + \frac{1}{2} \).

### Step 4: Relate \([-x]\) to \([x]\)
The greatest integer function has the property that:
- \([-x] = -[x] - 1\) when \( x \) is not an integer.

Thus, we can write:
\[ f(-x) = [-x] + \frac{1}{2} = (-[x] - 1) + \frac{1}{2} = -[x] - \frac{1}{2}. \]

### Step 5: Compare \( f(-x) \) and \( f(x) \)
Now we need to compare \( f(-x) \) and \( f(x) \):
- From our earlier steps, we have:
\[ f(x) = [x] + \frac{1}{2}. \]
\[ f(-x) = -[x] - \frac{1}{2}. \]

### Step 6: Check if \( f(-x) = -f(x) \)
Now, we check if \( f(-x) = -f(x) \):
\[ -f(x) = -([x] + \frac{1}{2}) = -[x] - \frac{1}{2}. \]
This shows that:
\[ f(-x) = -f(x). \]

### Conclusion
Since \( f(-x) = -f(x) \), the function \( f(x) \) is an **odd function**.

### Final Answer
The function \( f(x) = [x] + \frac{1}{2} \) is an **odd function**.

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