Statement I `sin x=a , ` where `-1 lt a lt 0 `, then for ` x in [0,npi]` has `2(n-1) ` solution ` AA n in N` .
Statement II sin x takes value a exactly two times when we take one complete rotation covering all the quadrants starting from x=0 .

Options

A. Statement I is true , Statement II is true , Statement II is a correct explanation for Statement I.
B. Statement I is true , Statement II is true , Statement II is not a correct explanation for Statement II.
C. Statement I is true , Statement II is false
D. Statement I is false , Statement II true .

Answer

To solve the problem, we need to analyze both statements provided in the question regarding the equation \( \sin x = a \) where \( -1 < a < 0 \).

### Step 1: Analyze Statement I

Given the equation \( \sin x = a \) with \( -1 < a < 0 \), we need to determine how many solutions exist for \( x \) in the interval \( [0, n\pi] \) for \( n \in \mathbb{N} \).

1. **Understanding the sine function**: The sine function is negative in the third and fourth quadrants. Therefore, for \( \sin x = a \) where \( a \) is negative, \( x \) must lie in these quadrants.

2. **Finding solutions in one complete rotation**: In one complete rotation (from \( 0 \) to \( 2\pi \)), \( \sin x = a \) will have exactly two solutions:
- One solution in the third quadrant.
- One solution in the fourth quadrant.

3. **Extending to \( n\pi \)**: For each additional \( \pi \) added (i.e., moving from \( 2\pi \) to \( 3\pi \), \( 3\pi \) to \( 4\pi \), etc.), we will again find two solutions in each interval of \( [2k\pi, (2k+1)\pi] \) and \( [(2k+1)\pi, (2k+2)\pi] \) for \( k = 0, 1, 2, \ldots, n-1 \).

4. **Counting the total solutions**: Therefore, for \( n \) complete rotations, the total number of solutions is:
\[ 2n \]
However, since the statement claims \( 2(n-1) \), we need to check this.

5. **Final conclusion for Statement I**: The correct number of solutions for \( x \) in \( [0, n\pi] \) is \( 2n \), not \( 2(n-1) \). Thus, Statement I is **incorrect**.

### Step 2: Analyze Statement II

1. **Revisiting the sine function**: As established, \( \sin x = a \) where \( a < 0 \) has two solutions in the interval \( [0, 2\pi] \).

2. **Generalizing for \( n \)**: For any \( n \), the sine function will still yield two solutions for each complete rotation of \( 2\pi \). Therefore, in the interval \( [0, n\pi] \), the sine function will take the value \( a \) exactly \( 2n \) times.

3. **Conclusion for Statement II**: Since the sine function indeed takes the value \( a \) exactly two times for each complete rotation, Statement II is **correct**.

### Final Conclusion

- **Statement I**: Incorrect.
- **Statement II**: Correct.