The solution set of the inequation `|x+(1)/(x)| lt 4`, is

Options

A. `(2-sqrt(3), 2 +sqrt(3)) cup (-2-sqrt(3), -2 + sqrt(3))`
B. `R-(2-sqrt(3), 2+sqrt(3))`
C. `R-(-2-sqrt(3), -2 + sqrt(3))`
D. none of these

Answer

To solve the inequality \( |x + \frac{1}{x}| < 4 \), we will break it down into two cases based on the properties of absolute values.

### Step 1: Remove the absolute value
The inequality \( |A| < B \) can be rewritten as \( -B < A < B \). Here, we have:
\[ -4 < x + \frac{1}{x} < 4 \]

### Step 2: Solve the first part of the inequality
Let's solve the left part of the inequality:
\[ x + \frac{1}{x} > -4 \]
Multiplying through by \( x \) (assuming \( x > 0 \) and \( x < 0 \) separately later), we get:
\[ x^2 + 1 > -4x \]
Rearranging gives:
\[ x^2 + 4x + 1 > 0 \]
Now, we will find the roots of the quadratic equation \( x^2 + 4x + 1 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 - 4}}{2} = \frac{-4 \pm \sqrt{12}}{2} = \frac{-4 \pm 2\sqrt{3}}{2} = -2 \pm \sqrt{3} \]
Thus, the roots are \( -2 - \sqrt{3} \) and \( -2 + \sqrt{3} \).

### Step 3: Analyze the intervals
The quadratic \( x^2 + 4x + 1 \) opens upwards (as the coefficient of \( x^2 \) is positive). Therefore, the expression is positive outside the roots:
\[ x < -2 - \sqrt{3} \quad \text{or} \quad x > -2 + \sqrt{3} \]

### Step 4: Solve the second part of the inequality
Now, we solve the right part of the inequality:
\[ x + \frac{1}{x} < 4 \]
Again, multiplying through by \( x \) (considering \( x > 0 \) and \( x < 0 \)):
\[ x^2 - 4x + 1 < 0 \]
Finding the roots of \( x^2 - 4x + 1 = 0 \):
\[ x = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3} \]
Thus, the roots are \( 2 - \sqrt{3} \) and \( 2 + \sqrt{3} \).

### Step 5: Analyze the intervals
The quadratic \( x^2 - 4x + 1 \) opens upwards, so it is negative between the roots:
\[ 2 - \sqrt{3} < x < 2 + \sqrt{3} \]

### Step 6: Find the intersection of the intervals
Now we need to find the intersection of the intervals from both parts:
1. From \( x + \frac{1}{x} > -4 \): \( x < -2 - \sqrt{3} \) or \( x > -2 + \sqrt{3} \)
2. From \( x + \frac{1}{x} < 4 \): \( 2 - \sqrt{3} < x < 2 + \sqrt{3} \)

The relevant intervals for \( x > -2 + \sqrt{3} \) and \( 2 - \sqrt{3} < x < 2 + \sqrt{3} \) gives us:
\[ (-2 + \sqrt{3}, 2 + \sqrt{3}) \]

### Final Answer
Thus, the solution set of the inequality \( |x + \frac{1}{x}| < 4 \) is:
\[ x \in (2 - \sqrt{3}, 2 + \sqrt{3}) \]