To solve the inequality \(\frac{2x - 1}{2x^3 + 3x^2 + x} > 0\), we will follow these steps:
### Step 1: Factor the expression
We start by rewriting the inequality:
\[
\frac{2x - 1}{2x^3 + 3x^2 + x} > 0
\]
First, we can factor the denominator:
\[
2x^3 + 3x^2 + x = x(2x^2 + 3x + 1)
\]
So, the inequality becomes:
\[
\frac{2x - 1}{x(2x^2 + 3x + 1)} > 0
\]
### Step 2: Find the critical points
Next, we need to find the critical points where the expression is equal to zero or undefined. The critical points occur when the numerator or denominator is zero:
1. **Numerator:** \(2x - 1 = 0 \Rightarrow x = \frac{1}{2}\)
2. **Denominator:** \(x(2x^2 + 3x + 1) = 0\)
- \(x = 0\)
- To find the roots of \(2x^2 + 3x + 1 = 0\), we use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} = \frac{-3 \pm 1}{4}
\]
This gives us:
- \(x = -\frac{1}{2}\)
- \(x = -1\)
Thus, the critical points are \(x = -1, -\frac{1}{2}, 0, \frac{1}{2}\).
### Step 3: Test intervals
We will test the sign of the expression in the intervals defined by the critical points:
- Intervals: \((- \infty, -1)\), \((-1, -\frac{1}{2})\), \((- \frac{1}{2}, 0)\), \((0, \frac{1}{2})\), \((\frac{1}{2}, \infty)\)
1. **Interval \((- \infty, -1)\)**: Choose \(x = -2\)
\[
\frac{2(-2) - 1}{-2(2(-2)^2 + 3(-2) + 1)} = \frac{-4 - 1}{-2(8 - 6 + 1)} = \frac{-5}{-2 \cdot 3} > 0
\]
(Positive)
2. **Interval \((-1, -\frac{1}{2})\)**: Choose \(x = -0.75\)
\[
\frac{2(-0.75) - 1}{-0.75(2(-0.75)^2 + 3(-0.75) + 1)} = \frac{-1.5 - 1}{-0.75(1.125 - 2.25 + 1)} = \frac{-2.5}{-0.75(-0.125)} < 0
\]
(Negative)
3. **Interval \((- \frac{1}{2}, 0)\)**: Choose \(x = -0.25\)
\[
\frac{2(-0.25) - 1}{-0.25(2(-0.25)^2 + 3(-0.25) + 1)} = \frac{-0.5 - 1}{-0.25(0.125 - 0.75 + 1)} = \frac{-1.5}{-0.25(0.375)} > 0
\]
(Positive)
4. **Interval \((0, \frac{1}{2})\)**: Choose \(x = 0.25\)
\[
\frac{2(0.25) - 1}{0.25(2(0.25)^2 + 3(0.25) + 1)} = \frac{0.5 - 1}{0.25(0.125 + 0.75 + 1)} = \frac{-0.5}{0.25(1.875)} < 0
\]
(Negative)
5. **Interval \((\frac{1}{2}, \infty)\)**: Choose \(x = 1\)
\[
\frac{2(1) - 1}{1(2(1)^2 + 3(1) + 1)} = \frac{2 - 1}{1(2 + 3 + 1)} = \frac{1}{6} > 0
\]
(Positive)
### Step 4: Combine intervals
From the tests, the intervals where the expression is positive are:
- \((- \infty, -1)\)
- \((- \frac{1}{2}, 0)\)
- \((\frac{1}{2}, \infty)\)
Thus, the set \(S\) is:
\[
S = (-\infty, -1) \cup (-\frac{1}{2}, 0) \cup (\frac{1}{2}, \infty)
\]
### Step 5: Identify subset \(P\)
Now, we need to identify which intervals can belong to the subset \(P\) from the options given in the problem.
