The function `f(x)=sin^(-1)(2x-x^(2))+sqrt(2-(1)/(|x|)s)+(1)/([x^(2)])`
defined in the interval (where `[.]` is the greatest integer function)

Options

A. `x in (1-sqrt2, 1)`
B. `x in [1, 1+sqrt2]`
C. `x in [1-sqrt2, 1+sqrt2]`
D. `x in [1-sqrt2, 2]`

Answer

To determine the domain of the function \( f(x) = \sin^{-1}(2x - x^2) + \sqrt{2 - \frac{1}{|x|}} + \frac{1}{[x^2]} \), we need to analyze each component of the function separately.

### Step 1: Analyze the Inverse Sine Function
The first term is \( \sin^{-1}(2x - x^2) \). The argument of the inverse sine function must lie in the interval \([-1, 1]\).

1. Set up the inequalities:
\[ -1 \leq 2x - x^2 \leq 1 \]

2. Solve the first inequality:
\[ 2x - x^2 \geq -1 \implies x^2 - 2x - 1 \leq 0 \]
This can be factored or solved using the quadratic formula:
\[ x = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2} \]
The roots are \( 1 - \sqrt{2} \) and \( 1 + \sqrt{2} \). The quadratic opens upwards, so the solution is:
\[ 1 - \sqrt{2} \leq x \leq 1 + \sqrt{2} \]

3. Solve the second inequality:
\[ 2x - x^2 \leq 1 \implies x^2 - 2x + 1 \geq 0 \implies (x - 1)^2 \geq 0 \]
This inequality is always true for all \( x \).

### Step 2: Combine the Results
From the analysis of the first term, we have:
\[ 1 - \sqrt{2} \leq x \leq 1 + \sqrt{2} \]

### Step 3: Analyze the Square Root Function
The second term is \( \sqrt{2 - \frac{1}{|x|}} \). The expression inside the square root must be non-negative:
\[ 2 - \frac{1}{|x|} \geq 0 \implies |x| \geq \frac{1}{2} \]
This gives us two intervals:
\[ x \leq -\frac{1}{2} \quad \text{or} \quad x \geq \frac{1}{2} \]

### Step 4: Analyze the Greatest Integer Function
The third term is \( \frac{1}{[x^2]} \). The greatest integer function \( [x^2] \) must be non-zero:
\[ [x^2] \neq 0 \implies x^2 \neq 0 \implies x \neq 0 \]

### Step 5: Combine All Conditions
Now we combine all the conditions:
1. From the inverse sine function: \( 1 - \sqrt{2} \leq x \leq 1 + \sqrt{2} \)
2. From the square root function: \( x \leq -\frac{1}{2} \) or \( x \geq \frac{1}{2} \)
3. From the greatest integer function: \( x \neq 0 \)

### Step 6: Final Domain
The intersection of these intervals gives us the final domain:
- The interval \( [1 - \sqrt{2}, 1 + \sqrt{2}] \) intersects with \( (-\infty, -\frac{1}{2}] \cup [\frac{1}{2}, \infty) \).

Thus, the domain of \( f(x) \) is:
\[ x \in [\frac{1}{2}, 1 + \sqrt{2}] \]