The complete set of solution of Inequality `(x^2-5 x+6 sgn(x))/(x sgn(x-1)+1) >= 0` is (where sgn respresent signum

Options

A. `(-oo,-1]uu[0,2]uu[3,oo)`
B. `(-oo,0]uu[2,oo)`
C. `(-oo,2]uu[3,oo)`
D. `(-oo,-1]uu[0,oo)`

Answer

To solve the inequality

\[ \frac{x^2 - 5x + 6 \cdot \text{sgn}(x)}{x \cdot \text{sgn}(x - 1) + 1} \geq 0, \]

we will analyze the expression step by step.

### Step 1: Understand the Signum Function
The signum function, denoted as \(\text{sgn}(x)\), is defined as follows:
- \(\text{sgn}(x) = 1\) if \(x > 0\)
- \(\text{sgn}(x) = 0\) if \(x = 0\)
- \(\text{sgn}(x) = -1\) if \(x < 0\)

### Step 2: Analyze the Cases Based on the Signum Function
We will consider different cases based on the value of \(x\).

#### Case 1: \(x > 1\)
In this case, \(\text{sgn}(x) = 1\) and \(\text{sgn}(x - 1) = 1\). The inequality simplifies to:

\[ \frac{x^2 - 5x + 6}{x + 1} \geq 0. \]

Factoring the numerator:

\[ x^2 - 5x + 6 = (x - 2)(x - 3). \]

Thus, the inequality becomes:

\[ \frac{(x - 2)(x - 3)}{(x + 1)} \geq 0. \]

The critical points are \(x = 2\), \(x = 3\), and \(x = -1\). We will test the intervals determined by these points:

- For \(x > 3\): Positive
- For \(2 < x < 3\): Negative
- For \(1 < x < 2\): Positive
- For \(x < -1\): Not applicable since \(x > 1\)

The solution in this case is:

\[ x \in (1, 2) \cup (3, \infty). \]

#### Case 2: \(x = 1\)
Substituting \(x = 1\) into the inequality:

\[ \frac{1^2 - 5 \cdot 1 + 6 \cdot 0}{1 \cdot 0 + 1} = \frac{2}{1} = 2 \geq 0. \]

Thus, \(x = 1\) is a solution.

#### Case 3: \(0 < x < 1\)
Here, \(\text{sgn}(x) = 1\) and \(\text{sgn}(x - 1) = -1\). The inequality becomes:

\[ \frac{x^2 - 5x + 6}{-x + 1} \geq 0. \]

This simplifies to:

\[ \frac{(x - 2)(x - 3)}{1 - x} \geq 0. \]

The critical points are \(x = 2\), \(x = 3\), and \(x = 1\). Testing the intervals:

- For \(x < 1\): Positive
- For \(1 < x < 2\): Negative
- For \(2 < x < 3\): Negative

Thus, the solution in this case is:

\[ x \in (0, 1). \]

#### Case 4: \(x = 0\)
Substituting \(x = 0\):

\[ \frac{0^2 - 5 \cdot 0 + 6 \cdot 0}{0 \cdot -1 + 1} = \frac{0}{1} = 0. \]

Thus, \(x = 0\) is also a solution.

#### Case 5: \(x < 0\)
Here, \(\text{sgn}(x) = -1\) and \(\text{sgn}(x - 1) = -1\). The inequality becomes:

\[ \frac{- (x^2 - 5x + 6)}{-x + 1} \geq 0. \]

This simplifies to:

\[ \frac{(x - 2)(x - 3)}{1 - x} \leq 0. \]

The critical points are \(x = 2\), \(x = 3\), and \(x = 1\). Testing the intervals:

- For \(x < -1\): Positive
- For \(-1 < x < 0\): Negative

Thus, the solution in this case is:

\[ x \in (-\infty, -1). \]

### Step 3: Combine All Solutions
Combining all the solutions from the cases we analyzed:

1. From Case 1: \( (1, 2) \cup (3, \infty) \)
2. From Case 2: \( \{1\} \)
3. From Case 3: \( (0, 1) \)
4. From Case 4: \( \{0\} \)
5. From Case 5: \( (-\infty, -1) \)

The complete solution set is:

\[ (-\infty, -1) \cup [0, 2] \cup [3, \infty). \]

### Final Answer
The complete set of solutions of the inequality is:

\[ (-\infty, -1) \cup [0, 2] \cup [3, \infty). \]