`0 lt a lt 2 pi , sin^(-1)(sin a ) lt x^(2)-2x ` for all ` x in I` then ` a in `

Options

A. `(0,pi +1)`
B. `(pi+1,(3pi)/(2))`
C. `((3pi)/(2), 2 pi -1)`
D. `(2pi-1, 2 pi )`

Answer

To solve the inequality \( \sin^{-1}(\sin a) < x^2 - 2x \) for all \( x \in \mathbb{I} \) where \( 0 < a < 2\pi \), we will follow these steps:

### Step 1: Find the minimum value of the quadratic expression \( x^2 - 2x \)

The quadratic expression \( x^2 - 2x \) can be rewritten in vertex form. The vertex \( x \) of a quadratic \( ax^2 + bx + c \) is given by:

\[ x = -\frac{b}{2a} \]

Here, \( a = 1 \) and \( b = -2 \). Thus,

\[ x = -\frac{-2}{2 \cdot 1} = 1 \]

### Step 2: Calculate the minimum value of \( x^2 - 2x \) at \( x = 1 \)

Now, substitute \( x = 1 \) into the expression:

\[ x^2 - 2x = 1^2 - 2 \cdot 1 = 1 - 2 = -1 \]

### Step 3: Set up the inequality involving \( \sin^{-1}(\sin a) \)

From the previous steps, we have:

\[ \sin^{-1}(\sin a) < -1 \]

### Step 4: Analyze the behavior of \( \sin^{-1}(\sin a) \)

The function \( \sin^{-1}(\sin a) \) behaves differently depending on the value of \( a \):

1. For \( 0 \leq a \leq \frac{\pi}{2} \):
\[ \sin^{-1}(\sin a) = a \]

2. For \( \frac{\pi}{2} < a < \frac{3\pi}{2} \):
\[ \sin^{-1}(\sin a) = \pi - a \]

3. For \( \frac{3\pi}{2} < a < 2\pi \):
\[ \sin^{-1}(\sin a) = 2\pi - a \]

### Step 5: Determine the intervals for \( a \)

1. **Interval \( 0 < a \leq \frac{\pi}{2} \)**:
\[ a < -1 \quad \text{(not possible since \( a \) is positive)} \]

2. **Interval \( \frac{\pi}{2} < a < \frac{3\pi}{2} \)**:
\[ \pi - a < -1 \implies a > \pi + 1 \quad \text{(approximately \( 4.14 \))} \]
Since \( \frac{\pi}{2} \approx 1.57 \) and \( \frac{3\pi}{2} \approx 4.71 \), this interval is valid.

3. **Interval \( \frac{3\pi}{2} < a < 2\pi \)**:
\[ 2\pi - a < -1 \implies a > 2\pi + 1 \quad \text{(not possible since \( a < 2\pi \))} \]

### Step 6: Conclusion

From the analysis, the only valid interval for \( a \) is:

\[ a \in (\pi + 1, \frac{3\pi}{2}) \]

### Final Answer

Thus, the solution is:

\[ a \in (\pi + 1, \frac{3\pi}{2}) \]