The domain of definition of the function `f(x) = sqrt(3-2^(x) -2^(1-x)) + sqrt(sqrt(sin^(-1)x,)` is

Options

A. `[0, 1]`
B. ` (0, 1]`
C. `[0, 1]`
D. none of these

Explanation

To find the domain of the function \( f(x) = \sqrt{3 - 2^x - 2^{1-x}} + \sqrt{\sin^{-1}(x)} \), we need to ensure that both square root expressions are defined and non-negative.

### Step 1: Analyze the first square root \( \sqrt{3 - 2^x - 2^{1-x}} \)

For the expression under the first square root to be non-negative, we need:

\[ 3 - 2^x - 2^{1-x} \geq 0 \]

Rearranging gives:

\[ 2^x + 2^{1-x} \leq 3 \]

### Step 2: Simplify \( 2^{1-x} \)

We can rewrite \( 2^{1-x} \) as \( \frac{2}{2^x} \). Thus, we have:

\[ 2^x + \frac{2}{2^x} \leq 3 \]

Let \( y = 2^x \). The inequality becomes:

\[ y + \frac{2}{y} \leq 3 \]

### Step 3: Multiply through by \( y \) (assuming \( y > 0 \))

This gives:

\[ y^2 - 3y + 2 \leq 0 \]

### Step 4: Factor the quadratic

Factoring gives:

\[ (y - 1)(y - 2) \leq 0 \]

### Step 5: Determine the intervals

The critical points are \( y = 1 \) and \( y = 2 \). The intervals to test are \( (-\infty, 1] \), \( [1, 2] \), and \( [2, \infty) \). The inequality holds for:

\[ 1 \leq y \leq 2 \]

### Step 6: Convert back to \( x \)

Since \( y = 2^x \), we have:

\[ 1 \leq 2^x \leq 2 \]

Taking logarithms gives:

\[ 0 \leq x \leq 1 \]

### Step 7: Analyze the second square root \( \sqrt{\sin^{-1}(x)} \)

For this square root to be defined, we need:

\[ \sin^{-1}(x) \geq 0 \]

This implies:

\[ x \geq 0 \]

### Step 8: Combine the conditions

From the first square root, we have \( 0 \leq x \leq 1 \) and from the second square root, we have \( x \geq 0 \). Therefore, the combined domain is:

\[ 0 \leq x \leq 1 \]

### Final Answer
The domain of the function \( f(x) \) is:

\[ [0, 1] \]