If `f(x) =[ sin ^(-1)(sin 2x )]` (where, [] denotes the greatest integer function ), then

Options

A. `int_(0)^(pi//2) f(x) dx=(pi)/(2) - sin ^(-1)(sin 1)`
B. `f(x)` is periodic with period `pi`
C. `underset(x rarr(pi)/(2))(lim) f(x)=-1`
D. None of thses

Answer

To solve the problem, we need to analyze the function \( f(x) = \lfloor \sin^{-1}(\sin(2x)) \rfloor \), where \( \lfloor \cdot \rfloor \) denotes the greatest integer function.

### Step 1: Understanding the function
The function \( \sin^{-1}(\sin(2x)) \) gives us the angle whose sine is \( \sin(2x) \). The output of \( \sin^{-1} \) is restricted to the interval \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).

### Step 2: Determine the range of \( 2x \)
Since \( 2x \) can take values from \( 0 \) to \( \pi \) as \( x \) varies from \( 0 \) to \( \frac{\pi}{2} \), we need to consider the periodic nature of the sine function.

### Step 3: Analyze the output of \( \sin^{-1}(\sin(2x)) \)
1. For \( 0 \leq 2x \leq \frac{\pi}{2} \) (which corresponds to \( 0 \leq x \leq \frac{\pi}{4} \)):
- Here, \( \sin^{-1}(\sin(2x)) = 2x \).

2. For \( \frac{\pi}{2} < 2x < \pi \) (which corresponds to \( \frac{\pi}{4} < x < \frac{\pi}{2} \)):
- In this range, \( \sin(2x) \) will be positive, but \( \sin^{-1} \) will yield \( 2x - \pi \) because \( \sin^{-1} \) takes the principal value in \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).

### Step 4: Applying the greatest integer function
- For \( 0 \leq x \leq \frac{\pi}{4} \):
\[ f(x) = \lfloor 2x \rfloor \]
Since \( 2x \) ranges from \( 0 \) to \( \frac{\pi}{2} \) (approximately \( 1.57 \)), \( \lfloor 2x \rfloor \) will be \( 0 \) for \( 0 \leq x < \frac{1}{2} \) and \( 1 \) for \( \frac{1}{2} \leq x < \frac{\pi}{4} \).

- For \( \frac{\pi}{4} < x < \frac{\pi}{2} \):
\[ f(x) = \lfloor 2x - \pi \rfloor \]
Here, \( 2x \) ranges from \( \frac{\pi}{2} \) to \( \pi \) (approximately \( 3.14 \)), so \( 2x - \pi \) will range from \( -\frac{\pi}{2} \) to \( 0 \). Thus, \( \lfloor 2x - \pi \rfloor \) will be \( -1 \) for \( \frac{\pi}{4} < x < \frac{\pi}{2} \).

### Conclusion
The function \( f(x) \) can be summarized as:
- \( f(x) = 0 \) for \( 0 \leq x < \frac{1}{2} \)
- \( f(x) = 1 \) for \( \frac{1}{2} \leq x < \frac{\pi}{4} \)
- \( f(x) = -1 \) for \( \frac{\pi}{4} < x < \frac{\pi}{2} \)