If `|(x^(2)+mx+1)/(x^(2)+x+1)|lt3` for all real x, then

Options

A. `m lt -1`
B. `-1 lt m lt 6`
C. `-1 lt m lt 5`
D. `m gt 6`

Answer

To solve the inequality \( \left| \frac{x^2 + mx + 1}{x^2 + x + 1} \right| < 3 \) for all real \( x \), we will follow these steps:

### Step 1: Rewrite the Inequality
We start by rewriting the inequality without the absolute value:
\[ -3 < \frac{x^2 + mx + 1}{x^2 + x + 1} < 3 \]

### Step 2: Break it into Two Inequalities
This gives us two inequalities to solve:
1. \( \frac{x^2 + mx + 1}{x^2 + x + 1} < 3 \)
2. \( \frac{x^2 + mx + 1}{x^2 + x + 1} > -3 \)

### Step 3: Solve the First Inequality
Starting with the first inequality:
\[ \frac{x^2 + mx + 1}{x^2 + x + 1} < 3 \]
Multiply both sides by \( x^2 + x + 1 \) (which is always positive for all real \( x \)):
\[ x^2 + mx + 1 < 3(x^2 + x + 1) \]
This simplifies to:
\[ x^2 + mx + 1 < 3x^2 + 3x + 3 \]
Rearranging gives:
\[ -2x^2 + (m - 3)x - 2 < 0 \]

### Step 4: Analyze the Quadratic
For the quadratic \( -2x^2 + (m - 3)x - 2 \) to be less than zero for all \( x \), its discriminant must be less than zero:
\[ D = (m - 3)^2 - 4(-2)(-2) < 0 \]
This simplifies to:
\[ (m - 3)^2 - 16 < 0 \]
\[ (m - 3)^2 < 16 \]
Taking the square root gives:
\[ -4 < m - 3 < 4 \]
Adding 3 to all parts results in:
\[ -1 < m < 7 \]

### Step 5: Solve the Second Inequality
Now we solve the second inequality:
\[ \frac{x^2 + mx + 1}{x^2 + x + 1} > -3 \]
Again, multiply both sides by \( x^2 + x + 1 \):
\[ x^2 + mx + 1 > -3(x^2 + x + 1) \]
This simplifies to:
\[ x^2 + mx + 1 > -3x^2 - 3x - 3 \]
Rearranging gives:
\[ 4x^2 + (m + 3)x + 4 > 0 \]

### Step 6: Analyze the Second Quadratic
For the quadratic \( 4x^2 + (m + 3)x + 4 \) to be greater than zero for all \( x \), its discriminant must also be less than zero:
\[ D = (m + 3)^2 - 4(4)(4) < 0 \]
This simplifies to:
\[ (m + 3)^2 - 64 < 0 \]
Taking the square root gives:
\[ -8 < m + 3 < 8 \]
Subtracting 3 from all parts results in:
\[ -11 < m < 5 \]

### Step 7: Combine the Results
Now we have two ranges for \( m \):
1. From the first inequality: \( -1 < m < 7 \)
2. From the second inequality: \( -11 < m < 5 \)

The common range is:
\[ -1 < m < 5 \]

### Conclusion
Thus, the final answer is:
\[ m \in (-1, 5) \]