If `(1)/sqrt(2) le x le 1 then sin^(-1) 2xsqrt(1-x^(2))` equals

Options

A. `2 sin^(-1)x`
B. `pi-2 sin^(-1)x`
C. `-pi--2sin^(-1)x`
D. none of these

Answer

To solve the problem, we need to find the value of \( \sin^{-1}(2x\sqrt{1-x^2}) \) given that \( \frac{1}{\sqrt{2}} \leq x \leq 1 \).

### Step-by-Step Solution:

1. **Identify the Range of x**:
We are given that \( \frac{1}{\sqrt{2}} \leq x \leq 1 \). This means \( x \) lies in the interval where \( x \) is between \( \frac{1}{\sqrt{2}} \) and \( 1 \).

2. **Use the Formula for Inverse Sine**:
We know the formula:
\[ 2 \sin^{-1}(x) = \sin^{-1}(2x\sqrt{1-x^2}) \quad \text{for } -\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}} \]
However, since we are in the range \( \frac{1}{\sqrt{2}} \leq x \leq 1 \), we need to use the modified version:
\[ 2 \sin^{-1}(x) = \pi - \sin^{-1}(2x\sqrt{1-x^2}) \]

3. **Rearranging the Formula**:
From the above formula, we can rearrange it to find \( \sin^{-1}(2x\sqrt{1-x^2}) \):
\[ \sin^{-1}(2x\sqrt{1-x^2}) = \pi - 2 \sin^{-1}(x) \]

4. **Conclusion**:
Thus, we conclude that:
\[ \sin^{-1}(2x\sqrt{1-x^2}) = \pi - 2 \sin^{-1}(x) \]

### Final Answer:
The expression \( \sin^{-1}(2x\sqrt{1-x^2}) \) equals \( \pi - 2 \sin^{-1}(x) \).