If `S` is the set of all real `x` such that `(2x-1)/(2x^3+3x^2+x)` is positive, then `S` contains

Options

A. `(-oo,-3/2)`
B. `(-3/2,1/4)`
C. `(-1/4,1/2)`
D. None of these

Answer

To solve the problem of finding the set \( S \) of all real \( x \) such that

\[ \frac{2x - 1}{2x^3 + 3x^2 + x} > 0, \]

we will follow these steps:

### Step 1: Identify where the expression is positive
The expression is positive when both the numerator and denominator are either both positive or both negative.

### Step 2: Analyze the numerator
Set the numerator \( 2x - 1 > 0 \):

\[ 2x - 1 > 0 \implies 2x > 1 \implies x > \frac{1}{2}. \]

### Step 3: Analyze the denominator
Set the denominator \( 2x^3 + 3x^2 + x \):

First, factor out \( x \):

\[ x(2x^2 + 3x + 1). \]

Now, we need to find the roots of the quadratic \( 2x^2 + 3x + 1 = 0 \) using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} = \frac{-3 \pm \sqrt{9 - 8}}{4} = \frac{-3 \pm 1}{4}. \]

Calculating the roots:

1. \( x = \frac{-2}{4} = -\frac{1}{2} \)
2. \( x = \frac{-4}{4} = -1 \)

Thus, the denominator can be factored as:

\[ x(2x + 1)(x + 1). \]

### Step 4: Determine the intervals
The critical points from the numerator and denominator are:

- From the numerator: \( x = \frac{1}{2} \)
- From the denominator: \( x = 0, -\frac{1}{2}, -1 \)

### Step 5: Test intervals
The critical points divide the number line into intervals. We will test the sign of the expression in each interval:

1. \( (-\infty, -1) \)
2. \( (-1, -\frac{1}{2}) \)
3. \( (-\frac{1}{2}, 0) \)
4. \( (0, \frac{1}{2}) \)
5. \( (\frac{1}{2}, \infty) \)

**Testing each interval:**

- For \( x < -1 \) (e.g., \( x = -2 \)):
\[ \frac{2(-2) - 1}{2(-2)^3 + 3(-2)^2 + (-2)} = \frac{-4 - 1}{-16 + 12 - 2} = \frac{-5}{-6} > 0. \]

- For \( -1 < x < -\frac{1}{2} \) (e.g., \( x = -0.75 \)):
\[ \frac{2(-0.75) - 1}{2(-0.75)^3 + 3(-0.75)^2 + (-0.75)} = \frac{-1.5 - 1}{-0.84375 + 1.6875 - 0.75} < 0. \]

- For \( -\frac{1}{2} < x < 0 \) (e.g., \( x = -0.25 \)):
\[ \frac{2(-0.25) - 1}{2(-0.25)^3 + 3(-0.25)^2 + (-0.25)} < 0. \]

- For \( 0 < x < \frac{1}{2} \) (e.g., \( x = 0.25 \)):
\[ \frac{2(0.25) - 1}{2(0.25)^3 + 3(0.25)^2 + (0.25)} < 0. \]

- For \( x > \frac{1}{2} \) (e.g., \( x = 1 \)):
\[ \frac{2(1) - 1}{2(1)^3 + 3(1)^2 + (1)} = \frac{2 - 1}{2 + 3 + 1} = \frac{1}{6} > 0. \]

### Step 6: Compile the intervals where the expression is positive
From our tests, the expression is positive in the intervals:

1. \( (-\infty, -1) \)
2. \( (\frac{1}{2}, \infty) \)

### Conclusion
Thus, the set \( S \) contains:

\[ S = (-\infty, -1) \cup \left( \frac{1}{2}, \infty \right). \]