Let `S = {(x,y): (y(3x-1))/(x(3x-2))<0}, S'= {(x,y) in AxxB: -1 leqAleq1, -1 leqBleq1},` then the area of the region enclosed by all points in `SnnS'` is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the area of the region enclosed by the sets \( S \) and \( S' \).

### Step 1: Analyze the set \( S \)

The set \( S \) is defined as:
\[ S = \left\{ (x,y) : \frac{y(3x-1)}{x(3x-2)} < 0 \right\} \]

To understand this inequality, we need to determine when the fraction is negative. This occurs when the numerator and denominator have opposite signs.

### Step 2: Determine the critical points

1. **Numerator**: \( y(3x-1) \)
- \( y = 0 \) or \( 3x - 1 = 0 \) gives \( x = \frac{1}{3} \).

2. **Denominator**: \( x(3x-2) \)
- \( x = 0 \) or \( 3x - 2 = 0 \) gives \( x = \frac{2}{3} \).

Thus, the critical points are \( x = 0, \frac{1}{3}, \frac{2}{3} \).

### Step 3: Test intervals

We will test the sign of the expression in the intervals defined by the critical points:

- **Interval \( (-\infty, 0) \)**:
- Choose \( x = -1 \): \( \frac{y(3(-1)-1)}{-1(3(-1)-2)} = \frac{y(-4)}{(-1)(-5)} = \frac{-4y}{5} < 0 \) if \( y > 0 \).

- **Interval \( (0, \frac{1}{3}) \)**:
- Choose \( x = \frac{1}{6} \): \( \frac{y(3(\frac{1}{6})-1)}{\frac{1}{6}(3(\frac{1}{6})-2)} = \frac{y(\frac{1}{2}-1)}{\frac{1}{6}(\frac{1}{2}-2)} = \frac{y(-\frac{1}{2})}{\frac{1}{6}(-\frac{3}{2})} > 0 \) if \( y < 0 \).

- **Interval \( (\frac{1}{3}, \frac{2}{3}) \)**:
- Choose \( x = \frac{1}{2} \): \( \frac{y(3(\frac{1}{2})-1)}{\frac{1}{2}(3(\frac{1}{2})-2)} = \frac{y(0)}{\frac{1}{2}(-\frac{1}{2})} = 0 \).

- **Interval \( (\frac{2}{3}, \infty) \)**:
- Choose \( x = 1 \): \( \frac{y(3(1)-1)}{1(3(1)-2)} = \frac{y(2)}{1} > 0 \) if \( y > 0 \).

### Step 4: Determine the regions for \( S \)

From the tests, we find:
- \( y > 0 \): \( x \in (-\infty, 0) \cup \left(\frac{2}{3}, \infty\right) \)
- \( y < 0 \): \( x \in \left(0, \frac{1}{3}\right) \cup \left(\frac{1}{3}, \frac{2}{3}\right) \)

### Step 5: Analyze the set \( S' \)

The set \( S' \) is defined as:
\[ S' = \{ (x,y) : -1 \leq x \leq 1, -1 \leq y \leq 1 \} \]
This represents a square region in the Cartesian plane.

### Step 6: Find the intersection \( S \cap S' \)

Now we find the area of the intersection \( S \cap S' \):
- For \( y > 0 \), the valid \( x \) values are \( (-1, 0) \) and \( (2/3, 1) \).
- For \( y < 0 \), the valid \( x \) values are \( (0, 1/3) \) and \( (1/3, 2/3) \).

### Step 7: Calculate the areas

1. **Area for \( y > 0 \)**:
- Area of rectangle from \( (-1, 0) \) to \( (0, 1) \):
\[ \text{Area} = 1 \times 1 = 1 \]
- Area of rectangle from \( (2/3, 1) \) to \( (1, 1) \):
\[ \text{Area} = \left(1 - \frac{2}{3}\right) \times 1 = \frac{1}{3} \]

2. **Area for \( y < 0 \)**:
- Area of rectangle from \( (0, -1) \) to \( (1/3, 0) \):
\[ \text{Area} = \frac{1}{3} \times 1 = \frac{1}{3} \]
- Area of rectangle from \( (1/3, -1) \) to \( (2/3, 0) \):
\[ \text{Area} = \frac{1}{3} \times 1 = \frac{1}{3} \]

### Step 8: Total area

Adding all the areas together:
\[ \text{Total Area} = 1 + \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1 + 1 = 2 \]

### Final Answer

The area of the region enclosed by all points in \( S \cap S' \) is:
\[ \boxed{2} \]