The value of x satisfying `|(x^(2)-1)/(x^(2)+x+1)| lt 1` , are :

Options

A. `x in(-2,oo)`
B. `x in (-oo,-(1)/(2)) cup( 0 ,oo) `
C. `x in (-2,-(1)/(2)) cup (0 , oo ) `
D. None of these

Answer

To solve the inequality \( \left| \frac{x^2 - 1}{x^2 + x + 1} \right| < 1 \), we can break it down into two inequalities:

1. \( \frac{x^2 - 1}{x^2 + x + 1} < 1 \)
2. \( \frac{x^2 - 1}{x^2 + x + 1} > -1 \)

### Step 1: Solve the first inequality \( \frac{x^2 - 1}{x^2 + x + 1} < 1 \)

To solve this, we can rearrange it:

\[ \frac{x^2 - 1}{x^2 + x + 1} - 1 < 0 \]

This simplifies to:

\[ \frac{x^2 - 1 - (x^2 + x + 1)}{x^2 + x + 1} < 0 \]

\[ \frac{x^2 - 1 - x^2 - x - 1}{x^2 + x + 1} < 0 \]

\[ \frac{-x - 2}{x^2 + x + 1} < 0 \]

### Step 2: Analyze the denominator \( x^2 + x + 1 \)

The quadratic \( x^2 + x + 1 \) has a discriminant \( b^2 - 4ac = 1 - 4 = -3 \), which is less than 0. Therefore, \( x^2 + x + 1 \) is always positive for all \( x \).

### Step 3: Solve the numerator \( -x - 2 < 0 \)

This implies:

\[ -x - 2 < 0 \implies -x < 2 \implies x > -2 \]

### Step 4: Combine with the denominator condition

Since the denominator is always positive, we conclude from the first inequality that:

\[ x > -2 \]

### Step 5: Solve the second inequality \( \frac{x^2 - 1}{x^2 + x + 1} > -1 \)

Rearranging gives:

\[ \frac{x^2 - 1}{x^2 + x + 1} + 1 > 0 \]

This simplifies to:

\[ \frac{x^2 - 1 + (x^2 + x + 1)}{x^2 + x + 1} > 0 \]

\[ \frac{2x^2 + x}{x^2 + x + 1} > 0 \]

### Step 6: Analyze the numerator \( 2x^2 + x \)

Factoring gives:

\[ x(2x + 1) > 0 \]

### Step 7: Find the critical points

The critical points are:

1. \( x = 0 \)
2. \( 2x + 1 = 0 \implies x = -\frac{1}{2} \)

### Step 8: Test intervals

We test the intervals determined by the critical points \( -2, -\frac{1}{2}, 0 \):

- For \( x < -2 \): Choose \( x = -3 \) → \( (-3)(-6) > 0 \) (True)
- For \( -2 < x < -\frac{1}{2} \): Choose \( x = -1 \) → \( (-1)(-1) > 0 \) (True)
- For \( -\frac{1}{2} < x < 0 \): Choose \( x = -0.25 \) → \( (-0.25)(0.5) < 0 \) (False)
- For \( x > 0 \): Choose \( x = 1 \) → \( (1)(3) > 0 \) (True)

### Step 9: Combine results

From the two inequalities, we have:

1. From \( \frac{-x - 2}{x^2 + x + 1} < 0 \): \( x > -2 \)
2. From \( \frac{2x^2 + x}{x^2 + x + 1} > 0 \): \( x < -\frac{1}{2} \) or \( x > 0 \)

### Final solution

Combining these gives us:

\[ x \in (-2, -\frac{1}{2}) \cup (0, \infty) \]