If `2sin^(-1)x+{cos^(-1)x} gt (pi)/(2)+{sin^(-1)x}`, then `x in `: (where `{*}` denotes fractional part function)

Options

A. (cos 1, 1]
B. [sin 1, 1]
C. (sin 1, 1]
D. none of these

Answer

To solve the inequality \( 2\sin^{-1}x + \{ \cos^{-1}x \} > \frac{\pi}{2} + \{ \sin^{-1}x \} \), we will follow these steps:

### Step 1: Rewrite the fractional parts
We know that the fractional part function can be expressed as:
\[ \{ y \} = y - \lfloor y \rfloor \]
Thus, we can rewrite the inequality as:
\[ 2\sin^{-1}x + \left( \cos^{-1}x - \lfloor \cos^{-1}x \rfloor \right) > \frac{\pi}{2} + \left( \sin^{-1}x - \lfloor \sin^{-1}x \rfloor \right) \]

### Step 2: Simplify the inequality
Rearranging the terms gives us:
\[ 2\sin^{-1}x + \cos^{-1}x - \frac{\pi}{2} > \lfloor \cos^{-1}x \rfloor + \lfloor \sin^{-1}x \rfloor + \sin^{-1}x \]
Using the identity \( \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \), we can substitute \( \cos^{-1}x \):
\[ 2\sin^{-1}x + \left( \frac{\pi}{2} - \sin^{-1}x \right) - \frac{\pi}{2} > \lfloor \cos^{-1}x \rfloor + \lfloor \sin^{-1}x \rfloor + \sin^{-1}x \]
This simplifies to:
\[ \sin^{-1}x > \lfloor \cos^{-1}x \rfloor + \lfloor \sin^{-1}x \rfloor \]

### Step 3: Analyze the ranges
- The range of \( \sin^{-1}x \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).
- The range of \( \cos^{-1}x \) is \( [0, \pi] \).

The greatest integer values for \( \sin^{-1}x \) can be \( -1, 0, 1 \) (since \( \frac{\pi}{2} \approx 1.57 \)).
For \( \cos^{-1}x \), the greatest integer values can be \( 0, 1, 2, 3 \) (since \( \pi \approx 3.14 \)).

### Step 4: Set up the inequality
From the inequality \( \sin^{-1}x > \lfloor \cos^{-1}x \rfloor + \lfloor \sin^{-1}x \rfloor \), we need to find the values of \( x \) such that this holds true.

### Step 5: Solve for \( x \)
1. **Case 1**: If \( \lfloor \sin^{-1}x \rfloor = 0 \) and \( \lfloor \cos^{-1}x \rfloor = 0 \):
\[ \sin^{-1}x > 0 \implies x > 0 \]
2. **Case 2**: If \( \lfloor \sin^{-1}x \rfloor = 1 \) and \( \lfloor \cos^{-1}x \rfloor = 0 \):
\[ \sin^{-1}x > 1 \implies x > \sin(1) \]
3. **Case 3**: If \( \lfloor \sin^{-1}x \rfloor = 1 \) and \( \lfloor \cos^{-1}x \rfloor = 1 \):
\[ \sin^{-1}x > 2 \implies \text{Not possible since } \sin^{-1}x \leq \frac{\pi}{2} \]

### Conclusion
The valid range for \( x \) is:
\[ x \in (\sin(1), 1] \]