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 \([.]\) denotes the greatest integer function) is even, odd, or neither, we will follow these steps:

### Step 1: Understand the Function
The function is defined as:
\[ f(x) = [x] + \frac{1}{2} \]
where \([x]\) is the greatest integer less than or equal to \(x\). This function is defined for \(x\) not belonging to the set of integers.

### Step 2: Check if the Function is Even
A function \(f(x)\) is even if:
\[ f(-x) = f(x) \quad \text{for all } x \]
Let's calculate \(f(-x)\):
- For \(x\) not an integer, \(-x\) will also not be an integer.
- Thus, we have:
\[ f(-x) = [-x] + \frac{1}{2} \]
Using the property of the greatest integer function, we know that:
\[ [-x] = -[x] - 1 \quad \text{(since } -x \text{ is the negative of } x\text{)} \]
Therefore:
\[ f(-x) = -[x] - 1 + \frac{1}{2} = -[x] - \frac{1}{2} \]
Now, we need to compare \(f(-x)\) with \(f(x)\):
\[ f(x) = [x] + \frac{1}{2} \]
Clearly, \(f(-x) \neq f(x)\), so the function is not even.

### Step 3: Check if the Function is Odd
A function \(f(x)\) is odd if:
\[ f(-x) = -f(x) \quad \text{for all } x \]
From our previous calculations:
\[ f(-x) = -[x] - \frac{1}{2} \]
And:
\[ -f(x) = -([x] + \frac{1}{2}) = -[x] - \frac{1}{2} \]
Since \(f(-x) = -f(x)\), the function satisfies the condition for being odd.

### Conclusion
Thus, the function \(f(x) = [x] + \frac{1}{2}\) is an **odd function**.