If `|(x^(2) +6)/(5x)| ge 1`, then x belongs to

Options

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

Answer

To solve the inequality \(\left|\frac{x^2 + 6}{5x}\right| \geq 1\), we will break it down step by step.

### Step 1: Rewrite the Inequality
We start with the given inequality:
\[ \left|\frac{x^2 + 6}{5x}\right| \geq 1 \]
This can be rewritten as:
\[ \frac{|x^2 + 6|}{|5x|} \geq 1 \]
This implies:
\[ |x^2 + 6| \geq |5x| \]

### Step 2: Analyze the Absolute Values
Since \(x^2 + 6\) is always positive (as \(x^2\) is non-negative and 6 is positive), we have:
\[ x^2 + 6 \geq 5|x| \]

### Step 3: Split into Cases
We will consider two cases based on the sign of \(x\).

#### Case 1: \(x \geq 0\)
In this case, \(|5x| = 5x\). The inequality becomes:
\[ x^2 + 6 \geq 5x \]
Rearranging gives:
\[ x^2 - 5x + 6 \geq 0 \]
Now we can factor this quadratic:
\[ (x - 2)(x - 3) \geq 0 \]

### Step 4: Find the Critical Points
The critical points are \(x = 2\) and \(x = 3\). We will analyze the sign of the expression \((x - 2)(x - 3)\) on the intervals:
- \( (-\infty, 2) \)
- \( (2, 3) \)
- \( (3, \infty) \)

Using a sign chart:
- For \(x < 2\), both factors are negative, so the product is positive.
- For \(2 < x < 3\), the first factor is positive and the second is negative, so the product is negative.
- For \(x > 3\), both factors are positive, so the product is positive.

Thus, the solution for this case is:
\[ x \in (-\infty, 2] \cup [3, \infty) \]

#### Case 2: \(x < 0\)
In this case, \(|5x| = -5x\). The inequality becomes:
\[ x^2 + 6 \geq -5x \]
Rearranging gives:
\[ x^2 + 5x + 6 \geq 0 \]
Factoring gives:
\[ (x + 2)(x + 3) \geq 0 \]

### Step 5: Find the Critical Points
The critical points are \(x = -2\) and \(x = -3\). We analyze the sign of \((x + 2)(x + 3)\):
- For \(x < -3\), both factors are negative, so the product is positive.
- For \(-3 < x < -2\), the first factor is negative and the second is positive, so the product is negative.
- For \(x > -2\), both factors are positive, so the product is positive.

Thus, the solution for this case is:
\[ x \in (-\infty, -3] \cup [-2, \infty) \]

### Step 6: Combine the Solutions
Now we combine the solutions from both cases:
1. From Case 1: \(x \in (-\infty, 2] \cup [3, \infty)\)
2. From Case 2: \(x \in (-\infty, -3] \cup [-2, \infty)\)

The final solution is:
\[ x \in (-\infty, -3] \cup [-2, 0) \cup (0, 2] \cup [3, \infty) \]

### Final Answer
Thus, the solution set is:
\[ x \in (-\infty, -3] \cup [-2, 0) \cup (0, 2] \cup [3, \infty) \]