If `f(x)=e^(sin(x-[x])cospix)`, where [x] denotes the greatest integer function, then f(x) is

Options

A. non-periodic
B. periodic with no fundamental period
C. periodic with period 2
D. periodic with period `pi`

Answer

To solve the problem, we need to analyze the function \( f(x) = e^{\sin(x - [x]) \cos(\pi x)} \), where \([x]\) denotes the greatest integer function (or floor function).

### Step 1: Understand the components of the function
The function consists of two main components:
1. \( \sin(x - [x]) \)
2. \( \cos(\pi x) \)

### Step 2: Analyze \( \sin(x - [x]) \)
The term \( x - [x] \) represents the fractional part of \( x \), denoted as \( \{x\} \). The fractional part \( \{x\} \) is periodic with a period of 1 because:
\[ \{x + 1\} = \{x\} \]
Thus, \( \sin(x - [x]) = \sin(\{x\}) \) is also periodic with a period of 1.

### Step 3: Analyze \( \cos(\pi x) \)
The function \( \cos(\pi x) \) has a known period. The period of \( \cos(kx) \) is given by:
\[ \text{Period} = \frac{2\pi}{|k|} \]
For \( k = \pi \), the period becomes:
\[ \text{Period} = \frac{2\pi}{\pi} = 2 \]
Thus, \( \cos(\pi x) \) is periodic with a period of 2.

### Step 4: Determine the overall period of \( f(x) \)
To find the overall period of \( f(x) \), we need to find the least common multiple (LCM) of the individual periods:
- The period of \( \sin(\{x\}) \) is 1.
- The period of \( \cos(\pi x) \) is 2.

The LCM of 1 and 2 is:
\[ \text{LCM}(1, 2) = 2 \]

### Conclusion
Since the LCM of the periods is 2, the function \( f(x) \) is periodic with a period of 2.

Thus, the final answer is:
**f(x) is periodic with period 2.**