Find the possible values of `sin^(-1) (1 - x) + cos^(-1) sqrt(x -2)`

Options

A. 0
B. `{-1,1}
C. 2
D. -1/2

Explanation

To solve the problem of finding the possible values of \( \sin^{-1}(1 - x) + \cos^{-1}(\sqrt{x - 2}) \), we will follow these steps:

### Step 1: Determine the domain of \( \sin^{-1}(1 - x) \)

The function \( \sin^{-1}(y) \) is defined for \( y \) in the interval \([-1, 1]\). Therefore, we need to find the values of \( x \) such that:

\[ -1 \leq 1 - x \leq 1 \]

From the left inequality:

\[ -1 \leq 1 - x \implies x \leq 2 \]

From the right inequality:

\[ 1 - x \leq 1 \implies x \geq 0 \]

Thus, we have:

\[ 0 \leq x \leq 2 \]

### Step 2: Determine the domain of \( \cos^{-1}(\sqrt{x - 2}) \)

The function \( \cos^{-1}(y) \) is defined for \( y \) in the interval \([0, 1]\). Therefore, we need to find the values of \( x \) such that:

\[ 0 \leq \sqrt{x - 2} \leq 1 \]

From the left inequality:

\[ 0 \leq \sqrt{x - 2} \implies x - 2 \geq 0 \implies x \geq 2 \]

From the right inequality:

\[ \sqrt{x - 2} \leq 1 \implies x - 2 \leq 1 \implies x \leq 3 \]

Thus, we have:

\[ 2 \leq x \leq 3 \]

### Step 3: Combine the domains

Now we combine the two intervals we found:

1. From \( \sin^{-1}(1 - x) \): \( 0 \leq x \leq 2 \)
2. From \( \cos^{-1}(\sqrt{x - 2}) \): \( 2 \leq x \leq 3 \)

The intersection of these intervals is:

\[ x = 2 \]

### Step 4: Evaluate the expression at \( x = 2 \)

Now we substitute \( x = 2 \) into the expression:

\[ \sin^{-1}(1 - 2) + \cos^{-1}(\sqrt{2 - 2}) \]

This simplifies to:

\[ \sin^{-1}(-1) + \cos^{-1}(0) \]

We know that:

\[ \sin^{-1}(-1) = -\frac{\pi}{2} \quad \text{and} \quad \cos^{-1}(0) = \frac{\pi}{2} \]

Thus, we have:

\[ -\frac{\pi}{2} + \frac{\pi}{2} = 0 \]

### Conclusion

The possible value of \( \sin^{-1}(1 - x) + \cos^{-1}(\sqrt{x - 2}) \) is:

\[ \boxed{0} \]