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

Options

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

Answer

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 first square root to be defined, the expression inside must be greater than or equal to zero:
\[ 3 - 2^x - 2^{1-x} \geq 0 \]

### Step 2: Simplify the inequality

We can rewrite \( 2^{1-x} \) as \( \frac{2}{2^x} \):
\[ 3 - 2^x - \frac{2}{2^x} \geq 0 \]
Multiplying through by \( 2^x \) (which is positive for all real \( x \)):
\[ 3 \cdot 2^x - (2^x)^2 - 2 \geq 0 \]
Let \( y = 2^x \):
\[ 3y - y^2 - 2 \geq 0 \]
Rearranging gives:
\[ -y^2 + 3y - 2 \geq 0 \]
Factoring the quadratic:
\[ -(y - 1)(y - 2) \geq 0 \]
This implies:
\[ (y - 1)(y - 2) \leq 0 \]

### Step 3: Determine the intervals for \( y \)

The roots of the equation are \( y = 1 \) and \( y = 2 \). The intervals to check are:
- \( (-\infty, 1) \)
- \( [1, 2] \)
- \( (2, \infty) \)

The product \( (y - 1)(y - 2) \) is:
- Positive in \( (-\infty, 1) \)
- Zero at \( y = 1 \)
- Negative in \( (1, 2) \)
- Zero at \( y = 2 \)
- Positive in \( (2, \infty) \)

Thus, the inequality \( (y - 1)(y - 2) \leq 0 \) holds for:
\[ 1 \leq y \leq 2 \]
Converting back to \( x \):
\[ 1 \leq 2^x \leq 2 \]
Taking logarithms:
\[ 0 \leq x \leq 1 \]

### Step 4: Analyze the second square root \( \sqrt{\sin^{-1}(x)} \)

The domain of \( \sin^{-1}(x) \) is \( -1 \leq x \leq 1 \). For \( \sqrt{\sin^{-1}(x)} \) to be defined, we need:
\[ \sin^{-1}(x) \geq 0 \implies x \geq 0 \]
Thus, the second square root is defined for:
\[ 0 \leq x \leq 1 \]

### Step 5: Combine the domains

The domain of \( f(x) \) is the intersection of the two conditions:
1. From the first square root: \( 0 \leq x \leq 1 \)
2. From the second square root: \( 0 \leq x \leq 1 \)

Thus, the domain of \( f(x) \) is:
\[ \boxed{[0, 1]} \]