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 \( 2\sin^{-1}x + \cos^{-1}x > \frac{\pi}{2} + \sin^{-1}x \), where \(\{ \cdot \}\) denotes the fractional part function, we will follow these steps:

### Step 1: Rewrite the Inequality
Start by rearranging the inequality:
\[ 2\sin^{-1}x + \cos^{-1}x - \sin^{-1}x > \frac{\pi}{2} \]
This simplifies to:
\[ \sin^{-1}x + \cos^{-1}x > \frac{\pi}{2} \]

### Step 2: Use the Identity
Recall the identity:
\[ \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \]
Thus, we can express the left side:
\[ \frac{\pi}{2} > \frac{\pi}{2} \]
This is not possible, which indicates that we need to analyze the fractional part function.

### Step 3: Analyze the Fractional Part Function
The fractional part function \(\{x\}\) is defined as:
\[ \{x\} = x - \lfloor x \rfloor \]
This means we need to consider the integer parts of the angles involved.

### Step 4: Set Up the Condition
Since we have established that:
\[ \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \]
we can rewrite our original inequality as:
\[ \{2\sin^{-1}x + \cos^{-1}x\} > \{ \frac{\pi}{2} + \sin^{-1}x \} \]

### Step 5: Evaluate the Ranges
The ranges for \(\sin^{-1}x\) and \(\cos^{-1}x\) are:
- \(\sin^{-1}x\) ranges from \(0\) to \(\frac{\pi}{2}\) for \(0 \leq x \leq 1\)
- \(\cos^{-1}x\) ranges from \(\frac{\pi}{2}\) to \(0\) for \(0 \leq x \leq 1\)

### Step 6: Consider Values of \(x\)
We need to find values of \(x\) such that:
\[ \sin^{-1}x > \cos^{-1}x \]
This occurs when:
\[ x > \frac{1}{\sqrt{2}} \quad \text{(since } \sin^{-1}x = \cos^{-1}x \text{ at } x = \frac{1}{\sqrt{2}}) \]

### Step 7: Final Range of \(x\)
Thus, the solution to the inequality is:
\[ x \in \left(\frac{1}{\sqrt{2}}, 1\right] \]