The function `f(x) = {x} sin (pi [x])`, where `[.]` denotes the greatest integer function and `{.}` is the fraction part function , is discontinuous at

Options

A. all `x`
B. all integer points
C. no `x`
D. `x` which is not an integer

Answer

To determine the points of discontinuity of the function \( f(x) = \{x\} \sin(\pi [x]) \), where \([x]\) is the greatest integer function and \(\{x\}\) is the fractional part function, we will analyze the components of the function step by step.

### Step 1: Understand the components of the function
The function consists of two parts:
- The fractional part function \(\{x\} = x - [x]\), which gives the non-integer part of \(x\).
- The greatest integer function \([x]\), which gives the largest integer less than or equal to \(x\).

### Step 2: Analyze the sine function
The sine function \(\sin(\pi [x])\) can be simplified based on the value of \([x]\):
- Since \([x]\) is an integer, we can denote it as \(n\), where \(n = [x]\).
- Therefore, \(\sin(\pi [x]) = \sin(n \pi)\), which equals \(0\) for all integers \(n\).

### Step 3: Substitute back into the function
Substituting this back into the function, we have:
\[ f(x) = \{x\} \sin(\pi [x]) = \{x\} \cdot 0 = 0 \]
Thus, for all \(x\), \(f(x) = 0\).

### Step 4: Determine points of discontinuity
Since \(f(x) = 0\) for all \(x\), the function is constant everywhere. A constant function is continuous everywhere in its domain.

### Conclusion
The function \(f(x) = \{x\} \sin(\pi [x])\) is continuous for all \(x\) in the real numbers, and therefore, there are no points of discontinuity.

### Final Answer
The function is discontinuous at **no points**.

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