If `1/sqrt(2) le x le 1` and `sin^(-1)(2xsqrt(1-x^(2))) = A + B sin^(-1)x`, then (A,B)=

Options

A. `(pi,2)`
B. (0,2)
C. `(pi,-2)`
D. (0,-2)

Answer

To solve the problem, we need to analyze the equation given:

\[ \sin^{-1}(2x\sqrt{1-x^2}) = A + B\sin^{-1}(x) \]

where \( \frac{1}{\sqrt{2}} \leq x \leq 1 \).

### Step 1: Use the double angle formula for sine

We know that:

\[ \sin(2\theta) = 2\sin(\theta)\cos(\theta) \]

If we let \( \theta = \sin^{-1}(x) \), then we have:

\[ \sin(2\theta) = 2x\sqrt{1-x^2} \]

Thus, we can rewrite the left side of our equation:

\[ \sin^{-1}(2x\sqrt{1-x^2}) = 2\sin^{-1}(x) \]

### Step 2: Determine the range of \( x \)

Given that \( \frac{1}{\sqrt{2}} \leq x \leq 1 \), we need to consider the implications for \( 2\sin^{-1}(x) \):

- When \( x = \frac{1}{\sqrt{2}} \), \( \sin^{-1}(x) = \frac{\pi}{4} \), thus \( 2\sin^{-1}(x) = \frac{\pi}{2} \).
- When \( x = 1 \), \( \sin^{-1}(x) = \frac{\pi}{2} \), thus \( 2\sin^{-1}(x) = \pi \).

### Step 3: Relate the expressions

Now, we can equate:

\[ \sin^{-1}(2x\sqrt{1-x^2}) = 2\sin^{-1}(x) \]

to the form:

\[ A + B\sin^{-1}(x) \]

From this, we can see that:

\[ A + B\sin^{-1}(x) = 2\sin^{-1}(x) \]

### Step 4: Identify coefficients \( A \) and \( B \)

From the equation \( A + B\sin^{-1}(x) = 2\sin^{-1}(x) \), we can identify:

- \( B = 2 \)
- \( A = 0 \)

### Step 5: Conclusion

Thus, we conclude that the values of \( A \) and \( B \) are:

\[ (A, B) = (0, 2) \]