The inequality `(x^(2)-|x|-2)/(2|x|-x^(2)-2) gt 2` holds only if.

Options

A. `-1 lt x lt -(2)/(3)` only
B. only for `(2)/(3) lt x lt 1`
C. `-1 lt x lt 1`
D. `-1 lt x lt -(2)/(3)" or "(2)/(3) lt x lt 1`

Answer

To solve the inequality \(\frac{x^2 - |x| - 2}{2|x| - x^2 - 2} > 2\), we will follow a systematic approach:

### Step 1: Rearranging the Inequality
Start by moving the \(2\) from the right-hand side to the left-hand side:

\[ \frac{x^2 - |x| - 2}{2|x| - x^2 - 2} - 2 > 0 \]

### Step 2: Finding a Common Denominator
We can rewrite the left-hand side by finding a common denominator:

\[ \frac{x^2 - |x| - 2 - 2(2|x| - x^2 - 2)}{2|x| - x^2 - 2} > 0 \]

### Step 3: Simplifying the Numerator
Now simplify the numerator:

\[ x^2 - |x| - 2 - 4|x| + 2x^2 + 4 = 3x^2 - 5|x| + 2 \]

So the inequality becomes:

\[ \frac{3x^2 - 5|x| + 2}{2|x| - x^2 - 2} > 0 \]

### Step 4: Analyzing the Denominator
Next, we will analyze the denominator:

\[ 2|x| - x^2 - 2 \]

### Step 5: Finding Critical Points
To find the critical points, we need to set the numerator and denominator to zero.

1. **Numerator**: \(3x^2 - 5|x| + 2 = 0\)
2. **Denominator**: \(2|x| - x^2 - 2 = 0\)

### Step 6: Solving the Numerator
Let \(y = |x|\). Then we solve:

\[ 3y^2 - 5y + 2 = 0 \]

Using the quadratic formula:

\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 3 \cdot 2}}{2 \cdot 3} = \frac{5 \pm 1}{6} \]

This gives us:

\[ y = 1 \quad \text{and} \quad y = \frac{2}{3} \]

Thus, \( |x| = 1 \) or \( |x| = \frac{2}{3} \).

### Step 7: Solving the Denominator
Now solve for the denominator:

\[ 2|x| - x^2 - 2 = 0 \]

Rearranging gives us:

\[ x^2 - 2|x| + 2 = 0 \]

Using the quadratic formula again:

\[ |x| = 1 \pm i \]

Since this has no real solutions, we focus on the critical points from the numerator.

### Step 8: Testing Intervals
Now we will test the intervals defined by the critical points \( -1, -\frac{2}{3}, \frac{2}{3}, 1 \):

1. **Interval \( (-\infty, -1) \)**: Choose \( x = -2 \)
2. **Interval \( (-1, -\frac{2}{3}) \)**: Choose \( x = -0.8 \)
3. **Interval \( (-\frac{2}{3}, \frac{2}{3}) \)**: Choose \( x = 0 \)
4. **Interval \( (\frac{2}{3}, 1) \)**: Choose \( x = 0.8 \)
5. **Interval \( (1, \infty) \)**: Choose \( x = 2 \)

### Step 9: Conclusion
After testing these intervals, we find that the inequality holds true in the intervals:

\[ (-1, -\frac{2}{3}) \quad \text{and} \quad (\frac{2}{3}, 1) \]

Thus, the solution to the inequality is:

\[ x \in (-1, -\frac{2}{3}) \cup (\frac{2}{3}, 1) \]