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

Options

A. `[0,sqrt3]`
B. `[1,sqrt2]`
C. `[1,sqrt6]`
D. `[sqrt3,sqrt6]`

Answer

To find the range of the function \( f(x) = \sqrt{\sin^{-1} x + 2} + \sqrt{1 - \sin^{-1} x} \), we will follow these steps:

### Step 1: Determine the Domain of the Function
The function involves the inverse sine function, \( \sin^{-1} x \), which has a domain of \( x \in [-1, 1] \).

1. **For \( \sqrt{\sin^{-1} x + 2} \)**:
\[ \sin^{-1} x + 2 \geq 0 \implies \sin^{-1} x \geq -2 \]
Since \( \sin^{-1} x \) ranges from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \), this condition is always satisfied for \( x \in [-1, 1] \).

2. **For \( \sqrt{1 - \sin^{-1} x} \)**:
\[ 1 - \sin^{-1} x \geq 0 \implies \sin^{-1} x \leq 1 \]
This condition is satisfied for \( x \in [-1, 1] \) since \( \sin^{-1} x \) reaches a maximum of \( \frac{\pi}{2} \) when \( x = 1 \).

Thus, the domain of \( f(x) \) is \( x \in [-1, 1] \).

### Step 2: Find Critical Points
We need to find the critical points of \( f(x) \) by differentiating it.

1. Differentiate \( f(x) \):
\[ f'(x) = \frac{1}{2\sqrt{\sin^{-1} x + 2}} \cdot \frac{1}{\sqrt{1 - x^2}} + \frac{1}{2\sqrt{1 - \sin^{-1} x}} \cdot \left(-\frac{1}{\sqrt{1 - x^2}}\right) \]

2. Set \( f'(x) = 0 \) to find critical points:
\[ \frac{1}{2\sqrt{\sin^{-1} x + 2}} - \frac{1}{2\sqrt{1 - \sin^{-1} x}} = 0 \]
This leads to:
\[ \sqrt{\sin^{-1} x + 2} = \sqrt{1 - \sin^{-1} x} \]
Squaring both sides gives:
\[ \sin^{-1} x + 2 = 1 - \sin^{-1} x \]
Rearranging yields:
\[ 2\sin^{-1} x = -1 \implies \sin^{-1} x = -\frac{1}{2} \]
Thus, \( x = \sin\left(-\frac{1}{2}\right) = -\frac{\sqrt{3}}{2} \).

### Step 3: Evaluate \( f(x) \) at Critical Points and Endpoints
1. **At \( x = -1 \)**:
\[ f(-1) = \sqrt{\sin^{-1}(-1) + 2} + \sqrt{1 - \sin^{-1}(-1)} = \sqrt{-\frac{\pi}{2} + 2} + \sqrt{1 + \frac{\pi}{2} \]
This simplifies to:
\[ f(-1) = \sqrt{2 - \frac{\pi}{2}} + \sqrt{1 + \frac{\pi}{2}} \]

2. **At \( x = 1 \)**:
\[ f(1) = \sqrt{\sin^{-1}(1) + 2} + \sqrt{1 - \sin^{-1}(1)} = \sqrt{\frac{\pi}{2} + 2} + \sqrt{1 - \frac{\pi}{2}} \]

3. **At \( x = -\frac{\sqrt{3}}{2} \)**:
\[ f\left(-\frac{\sqrt{3}}{2}\right) = \sqrt{-\frac{1}{2} + 2} + \sqrt{1 + \frac{1}{2}} = \sqrt{\frac{3}{2}} + \sqrt{\frac{3}{2}} = 2\sqrt{\frac{3}{2}} = \sqrt{6} \]

### Step 4: Determine the Range
Now we compare the values:
- \( f(-1) \)
- \( f(1) \)
- \( f\left(-\frac{\sqrt{3}}{2}\right) = \sqrt{6} \)

The minimum value occurs at \( f(-1) \) and the maximum at \( f(1) \).

### Conclusion
The range of \( f(x) \) is:
\[ \text{Range of } f(x) = [\sqrt{3}, \sqrt{6}] \]