Consider a real-valued function `f(x)= sqrt(sin^-1 x + 2) + sqrt(1 – sin^-1x)` then The domain of definition of `f(x)` is

Options

A. `[-1,1]`
B. `[sin 1,1]`
C. `[-1,sin1]`
D. `[-1,0]`

Answer

To find the domain of the function \( f(x) = \sqrt{\sin^{-1} x + 2} + \sqrt{1 - \sin^{-1} x} \), we need to ensure that both square root expressions are defined and non-negative.

### Step 1: Analyze the first square root \( \sqrt{\sin^{-1} x + 2} \)

For the expression \( \sqrt{\sin^{-1} x + 2} \) to be defined, we need:
\[ \sin^{-1} x + 2 \geq 0 \]
This simplifies to:
\[ \sin^{-1} x \geq -2 \]
Since the range of \( \sin^{-1} x \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), we can see that \( -2 \) is less than \( -\frac{\pi}{2} \). Therefore, this condition is satisfied for all \( x \) in the domain of \( \sin^{-1} x \), which is \( x \in [-1, 1] \).

### Step 2: Analyze the second square root \( \sqrt{1 - \sin^{-1} x} \)

For the expression \( \sqrt{1 - \sin^{-1} x} \) to be defined, we need:
\[ 1 - \sin^{-1} x \geq 0 \]
This simplifies to:
\[ \sin^{-1} x \leq 1 \]
The value \( \sin^{-1} x = 1 \) occurs when \( x = \sin(1) \). Therefore, the condition becomes:
\[ x \leq \sin(1) \]
Since \( \sin^{-1} x \) is defined for \( x \in [-1, 1] \), we need to find the intersection of this interval with \( [-1, \sin(1)] \).

### Step 3: Determine the intersection of intervals

The domain of \( \sin^{-1} x \) gives us \( x \in [-1, 1] \), and the condition from the second square root gives us \( x \in [-1, \sin(1)] \). Therefore, the intersection of these two intervals is:
\[ x \in [-1, \sin(1)] \]

### Conclusion

Thus, the domain of the function \( f(x) \) is:
\[ [-1, \sin(1)] \]