A point `P(x,y)` moves such that `[x+y+1]=[x]`. Where [.] denotes greatest intetger function and `x in (0,2)`, then the area represented by all the possible position of P , is

Options

A. `sqrt(2)`
B. `2sqrt(2)`
C. `4sqrt(2)`
D. 2

Answer

To solve the problem, we need to find the area represented by all possible positions of the point \( P(x, y) \) under the condition given by the greatest integer function. The condition is:

\[ [x + y + 1] = [x] \]

where \( [.] \) denotes the greatest integer function, and \( x \) is in the interval \( (0, 2) \).

### Step-by-Step Solution:

1. **Understanding the Condition**:
We start with the equation:
\[ [x + y + 1] = [x] \]
This implies that the greatest integer of \( x + y + 1 \) is equal to the greatest integer of \( x \).

2. **Rearranging the Equation**:
We can rewrite the equation as:
\[ [x + y] = [x] - 1 \]
This means that \( x + y \) lies in the range of integers defined by \( [x] - 1 \).

3. **Finding Intervals for \( x \)**:
Since \( x \) is in the interval \( (0, 2) \), we will consider two cases based on the value of \( x \):
- Case 1: \( 0 < x < 1 \)
- Case 2: \( 1 \leq x < 2 \)

4. **Case 1: \( 0 < x < 1 \)**:
Here, \( [x] = 0 \). Thus:
\[ [x + y] = -1 \]
This implies:
\[ -1 \leq x + y < 0 \]
Therefore:
\[ -1 - x < y < -x \]

5. **Case 2: \( 1 \leq x < 2 \)**:
In this case, \( [x] = 1 \). Thus:
\[ [x + y] = 0 \]
This implies:
\[ 0 \leq x + y < 1 \]
Therefore:
\[ -x < y < 1 - x \]

6. **Graphing the Regions**:
We need to graph the inequalities derived from both cases.

- For Case 1, the lines are:
- \( y = -x - 1 \) (which intersects the y-axis at -1)
- \( y = -x \) (which intersects the y-axis at 0)
- For Case 2, the lines are:
- \( y = -x \) (which intersects the y-axis at 0)
- \( y = 1 - x \) (which intersects the y-axis at 1)

The points of intersection for these lines will help us define the bounded region.

7. **Finding Points of Intersection**:
- For Case 1, the lines intersect at:
- \( (0, 0) \) and \( (1, 0) \)
- For Case 2, the lines intersect at:
- \( (1, 0) \) and \( (1, 0) \) (which is the same point)
- \( (0, 1) \)

8. **Calculating the Area**:
The area of the bounded region can be calculated as follows:
- The area consists of two triangles:
- Triangle from Case 1 with vertices at \( (0, -1), (0, 0), (1, 0) \)
- Triangle from Case 2 with vertices at \( (1, 0), (1, -1), (0, 1) \)

The area of each triangle is:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
- For the triangle in Case 1:
- Base = 1, Height = 1
- Area = \( \frac{1}{2} \times 1 \times 1 = \frac{1}{2} \)
- For the triangle in Case 2:
- Base = 1, Height = 1
- Area = \( \frac{1}{2} \times 1 \times 1 = \frac{1}{2} \)

Total Area = Area from Case 1 + Area from Case 2 = \( \frac{1}{2} + \frac{1}{2} = 1 \).

### Final Area Calculation:
The total area represented by all possible positions of \( P(x, y) \) is:

\[ \text{Total Area} = 1 \]