Solution of inequality `|x+(1)/(x)|lt 4` is

Options

A. `(2-sqrt3,2+sqrt3)cup(-2-sqrt3,-2+sqrt3)`
B. `R-(2-sqrt3,2+sqrt3)`
C. `R-(-sqrt2-sqrt3,2+sqrt3)`
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: Set up the inequality
We start with the inequality:
\[ |x + \frac{1}{x}| < 4 \]
This means:
\[ -4 < x + \frac{1}{x} < 4 \]

### Step 2: Solve the first part of the inequality
**Case 1:**
\[ x + \frac{1}{x} < 4 \]
Rearranging gives:
\[ x + \frac{1}{x} - 4 < 0 \]
Multiplying through by \(x\) (noting that \(x \neq 0\)):
\[ x^2 - 4x + 1 < 0 \]

### Step 3: Find the roots of the quadratic
To find the roots of the quadratic \(x^2 - 4x + 1 = 0\), we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 1\), \(b = -4\), and \(c = 1\):
\[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3} \]
Thus, the roots are:
\[ x_1 = 2 - \sqrt{3}, \quad x_2 = 2 + \sqrt{3} \]

### Step 4: Test intervals for Case 1
We will test intervals defined by the roots \(2 - \sqrt{3}\) and \(2 + \sqrt{3}\):
- Choose a test point in the interval \((- \infty, 2 - \sqrt{3})\)
- Choose a test point in the interval \((2 - \sqrt{3}, 2 + \sqrt{3})\)
- Choose a test point in the interval \((2 + \sqrt{3}, \infty)\)

After testing, we find that the solution for this case is:
\[ x \in (2 - \sqrt{3}, 2 + \sqrt{3}) \]

### Step 5: Solve the second part of the inequality
**Case 2:**
\[ x + \frac{1}{x} > -4 \]
Rearranging gives:
\[ x + \frac{1}{x} + 4 > 0 \]
Multiplying through by \(x\) (noting that \(x \neq 0\)):
\[ x^2 + 4x + 1 > 0 \]

### Step 6: Find the roots of the quadratic for Case 2
Using the quadratic formula again:
\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-4 \pm \sqrt{16 - 4}}{2} = \frac{-4 \pm \sqrt{12}}{2} = \frac{-4 \pm 2\sqrt{3}}{2} = -2 \pm \sqrt{3} \]
Thus, the roots are:
\[ x_1 = -2 - \sqrt{3}, \quad x_2 = -2 + \sqrt{3} \]

### Step 7: Test intervals for Case 2
We will test intervals defined by the roots \(-2 - \sqrt{3}\) and \(-2 + \sqrt{3}\):
- Choose a test point in the interval \((- \infty, -2 - \sqrt{3})\)
- Choose a test point in the interval \((-2 - \sqrt{3}, -2 + \sqrt{3})\)
- Choose a test point in the interval \((-2 + \sqrt{3}, \infty)\)

After testing, we find that the solution for this case is:
\[ x \in (-\infty, -2 - \sqrt{3}) \cup (-2 + \sqrt{3}, \infty) \]

### Step 8: Combine the solutions
Now we need to find the intersection of the two cases:
- From Case 1: \(x \in (2 - \sqrt{3}, 2 + \sqrt{3})\)
- From Case 2: \(x \in (-\infty, -2 - \sqrt{3}) \cup (-2 + \sqrt{3}, \infty)\)

The final solution is:
\[ x \in (-2 - \sqrt{3}, -2 + \sqrt{3}) \cup (2 - \sqrt{3}, 2 + \sqrt{3}) \]