A point where function `f(x) = [sin [x]]` is not continuous in `(0,2pi)` [.] denotes the greatest integer ` le x,` is

Options

A. `(3,0)`
B. `(2,0)`
C. `(1,0)`
D. none of these

Answer

To determine the points where the function \( f(x) = \lfloor \sin(\lfloor x \rfloor) \rfloor \) is not continuous in the interval \( (0, 2\pi) \), we will follow these steps:

### Step 1: Understand the Function
The function involves the greatest integer function (floor function) applied to the sine of the greatest integer function of \( x \). The greatest integer function \( \lfloor x \rfloor \) returns the largest integer less than or equal to \( x \).

### Step 2: Identify the Intervals
In the interval \( (0, 2\pi) \) (approximately \( 0 \) to \( 6.28 \)), we can break it down into integer intervals:
- \( 0 \leq x < 1 \) (where \( \lfloor x \rfloor = 0 \))
- \( 1 \leq x < 2 \) (where \( \lfloor x \rfloor = 1 \))
- \( 2 \leq x < 3 \) (where \( \lfloor x \rfloor = 2 \))
- \( 3 \leq x < 4 \) (where \( \lfloor x \rfloor = 3 \))
- \( 4 \leq x < 5 \) (where \( \lfloor x \rfloor = 4 \))
- \( 5 \leq x < 6 \) (where \( \lfloor x \rfloor = 5 \))
- \( 6 \leq x < 2\pi \) (where \( \lfloor x \rfloor = 6 \))

### Step 3: Evaluate \( f(x) \) in Each Interval
Now we calculate \( f(x) \) in each interval:
- For \( 0 \leq x < 1 \): \( f(x) = \lfloor \sin(0) \rfloor = \lfloor 0 \rfloor = 0 \)
- For \( 1 \leq x < 2 \): \( f(x) = \lfloor \sin(1) \rfloor \) (approximately \( 0.84 \)), so \( f(x) = 0 \)
- For \( 2 \leq x < 3 \): \( f(x) = \lfloor \sin(2) \rfloor \) (approximately \( 0.91 \)), so \( f(x) = 0 \)
- For \( 3 \leq x < 4 \): \( f(x) = \lfloor \sin(3) \rfloor \) (approximately \( 0.14 \)), so \( f(x) = 0 \)
- For \( 4 \leq x < 5 \): \( f(x) = \lfloor \sin(4) \rfloor \) (approximately \( -0.76 \)), so \( f(x) = -1 \)
- For \( 5 \leq x < 6 \): \( f(x) = \lfloor \sin(5) \rfloor \) (approximately \( -0.99 \)), so \( f(x) = -1 \)
- For \( 6 \leq x < 2\pi \): \( f(x) = \lfloor \sin(6) \rfloor \) (approximately \( -0.28 \)), so \( f(x) = -1 \)

### Step 4: Check Continuity at the Transition Points
We need to check the continuity at the transition points where \( x \) changes from one interval to another:
- At \( x = 1 \): \( f(1^-) = 0 \) and \( f(1^+) = 0 \) (continuous)
- At \( x = 2 \): \( f(2^-) = 0 \) and \( f(2^+) = 0 \) (continuous)
- At \( x = 3 \): \( f(3^-) = 0 \) and \( f(3^+) = 0 \) (continuous)
- At \( x = 4 \): \( f(4^-) = 0 \) and \( f(4^+) = -1 \) (discontinuous)
- At \( x = 5 \): \( f(5^-) = -1 \) and \( f(5^+) = -1 \) (continuous)
- At \( x = 6 \): \( f(6^-) = -1 \) and \( f(6^+) = -1 \) (continuous)

### Conclusion
The function \( f(x) \) is discontinuous at \( x = 4 \).