If set A `={x|(x^2(x-5)(2x-1))/((5x+1)(x+2)) lt 0 }` and
Set `B={x|(3x+1)/(6x^3+x^2-x) gt 0 } " then " A cap B` does not contain

Options

A. (1,4)
B. (5,11)
C. `(-3/2,(-1)/2)`
D. None of these

Explanation

To solve the problem, we need to determine the sets A and B based on the given inequalities and then find their intersection \( A \cap B \). Finally, we will identify which of the provided options is not contained in \( A \cap B \).

### Step 1: Define Set A
Set A is defined as:
\[ A = \{ x \mid \frac{x^2 (x - 5)(2x - 1)}{(5x + 1)(x + 2)} < 0 \} \]
To find the values of \( x \) for which this expression is negative, we identify the points where the numerator and denominator are zero.

**Numerator:**
- \( x^2 = 0 \) gives \( x = 0 \)
- \( x - 5 = 0 \) gives \( x = 5 \)
- \( 2x - 1 = 0 \) gives \( x = \frac{1}{2} \)

**Denominator:**
- \( 5x + 1 = 0 \) gives \( x = -\frac{1}{5} \)
- \( x + 2 = 0 \) gives \( x = -2 \)

### Step 2: Identify Critical Points
The critical points are:
- \( x = -2 \)
- \( x = -\frac{1}{5} \)
- \( x = 0 \)
- \( x = \frac{1}{2} \)
- \( x = 5 \)

### Step 3: Test Intervals
We will test the sign of the expression in the intervals defined by these critical points:
1. \( (-\infty, -2) \)
2. \( (-2, -\frac{1}{5}) \)
3. \( (-\frac{1}{5}, 0) \)
4. \( (0, \frac{1}{2}) \)
5. \( (\frac{1}{2}, 5) \)
6. \( (5, \infty) \)

After testing these intervals, we find that:
- \( A = (-2, -\frac{1}{5}) \cup (\frac{1}{2}, 5) \)

### Step 4: Define Set B
Set B is defined as:
\[ B = \{ x \mid \frac{3x + 1}{6x^3 + x^2 - x} > 0 \} \]
First, we factor the denominator:
\[ 6x^3 + x^2 - x = x(6x^2 + x - 1) \]
Factoring \( 6x^2 + x - 1 \) gives:
\[ 6x^2 + x - 1 = (3x - 1)(2x + 1) \]
Thus,
\[ B = \{ x \mid \frac{3x + 1}{x(3x - 1)(2x + 1)} > 0 \} \]

### Step 5: Identify Critical Points for Set B
The critical points are:
- \( x = -\frac{1}{3} \) (numerator zero)
- \( x = 0 \) (denominator zero)
- \( x = \frac{1}{3} \) (denominator zero)
- \( x = -\frac{1}{2} \) (denominator zero)

### Step 6: Test Intervals for Set B
We test the sign of the expression in the intervals defined by these critical points:
1. \( (-\infty, -\frac{1}{2}) \)
2. \( (-\frac{1}{2}, -\frac{1}{3}) \)
3. \( (-\frac{1}{3}, 0) \)
4. \( (0, \frac{1}{3}) \)
5. \( (\frac{1}{3}, \infty) \)

After testing these intervals, we find that:
- \( B = (-\infty, -\frac{1}{2}) \cup (-\frac{1}{3}, 0) \cup (\frac{1}{3}, \infty) \)

### Step 7: Find Intersection \( A \cap B \)
Now we find the intersection of sets A and B:
- From A: \( (-2, -\frac{1}{5}) \cup (\frac{1}{2}, 5) \)
- From B: \( (-\infty, -\frac{1}{2}) \cup (-\frac{1}{3}, 0) \cup (\frac{1}{3}, \infty) \)

The intersection \( A \cap B \) is:
- \( (-2, -\frac{1}{2}) \) (from \( (-2, -\frac{1}{5}) \) and \( (-\infty, -\frac{1}{2}) \))

### Step 8: Analyze Options
We need to check which of the provided options is not contained in \( A \cap B \):
1. \( (1, 4) \)
2. \( (5, 11) \)
3. \( (-\frac{3}{2}, -\frac{1}{2}) \)
4. \( (-2, -1) \)

**Conclusion:**
The only interval that is not contained in \( A \cap B \) is \( (5, 11) \).