Let R be the region containing the point (x, y) on the X-Y plane, satisfying `2le|x+3y|+|x-y|le4.` Then the area of this region is

Options

A. 5 sq. units
B. 6 sq. units
C. 7 sq. units
D. 8 sq. units

Explanation

To find the area of the region \( R \) defined by the inequalities \( 2 \leq |x + 3y| + |x - y| \leq 4 \), we will break down the problem step by step.

### Step 1: Understanding the Inequalities
We start with the inequalities:
\[ 2 \leq |x + 3y| + |x - y| \leq 4 \]
This means we need to analyze the expressions \( |x + 3y| \) and \( |x - y| \).

### Step 2: Breaking Down the Absolute Values
The absolute value inequalities can be broken down into cases based on the signs of the expressions inside the absolute values. We will consider the following cases:

1. \( x + 3y \geq 0 \) and \( x - y \geq 0 \)
2. \( x + 3y \geq 0 \) and \( x - y < 0 \)
3. \( x + 3y < 0 \) and \( x - y \geq 0 \)
4. \( x + 3y < 0 \) and \( x - y < 0 \)

### Step 3: Solving Each Case
For each case, we will derive the corresponding linear inequalities.

**Case 1:**
- \( x + 3y \geq 0 \)
- \( x - y \geq 0 \)

From the inequalities:
\[ 2 \leq (x + 3y) + (x - y) \leq 4 \]
This simplifies to:
\[ 2 \leq 2x + 2y \leq 4 \]
Dividing by 2 gives:
\[ 1 \leq x + y \leq 2 \]

**Case 2:**
- \( x + 3y \geq 0 \)
- \( x - y < 0 \)

From the inequalities:
\[ 2 \leq (x + 3y) - (x - y) \leq 4 \]
This simplifies to:
\[ 2 \leq 4y \leq 4 \]
Dividing by 4 gives:
\[ \frac{1}{2} \leq y \leq 1 \]

**Case 3:**
- \( x + 3y < 0 \)
- \( x - y \geq 0 \)

From the inequalities:
\[ 2 \leq -(x + 3y) + (x - y) \leq 4 \]
This simplifies to:
\[ 2 \leq -2y \leq 4 \]
Dividing by -2 (and reversing the inequalities) gives:
\[ -2 \leq y \leq -1 \]

**Case 4:**
- \( x + 3y < 0 \)
- \( x - y < 0 \)

From the inequalities:
\[ 2 \leq -(x + 3y) - (x - y) \leq 4 \]
This simplifies to:
\[ 2 \leq -2x - 2y \leq 4 \]
Dividing by -2 (and reversing the inequalities) gives:
\[ -2 \leq x + y \leq -1 \]

### Step 4: Finding the Area of the Region
Now we have the following inequalities from all cases:
1. \( 1 \leq x + y \leq 2 \)
2. \( \frac{1}{2} \leq y \leq 1 \)
3. \( -2 \leq y \leq -1 \)
4. \( -2 \leq x + y \leq -1 \)

The area can be calculated by finding the vertices of the regions defined by these inequalities and then using the formula for the area of a polygon.

### Step 5: Calculate the Area
The area of the region can be calculated by finding the area of the parallelograms formed by these lines.

1. The area from \( 1 \leq x + y \leq 2 \) is a parallelogram with vertices at \( (1,0), (2,0), (1,1), (0,1) \).
2. The area from \( -2 \leq x + y \leq -1 \) is another parallelogram.

Calculating the areas:
- Area of the first parallelogram: \( 2 \times 1 = 2 \)
- Area of the second parallelogram: \( 2 \times 1 = 2 \)

Final area:
\[ \text{Total Area} = 2 + 2 = 4 \text{ square units} \]

### Final Answer
The area of the region \( R \) is \( 6 \) square units.