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. (a)`sqrt(2)`
B. (b)`2sqrt(2)`
C. (c)`4sqrt(2)`
D. (d)2

Answer

To solve the problem step by step, we need to analyze the condition given for the point \( P(x, y) \) and find the area represented by all possible positions of \( P \).

### Step 1: Understand the condition
The condition given is:
\[ [x + y + 1] = [x] \]
where \([.]\) denotes the greatest integer function (floor function). This means that \(x + y + 1\) and \(x\) must lie in the same integer interval.

### Step 2: Analyze the greatest integer function
Since \(x \in (0, 2)\), we can break down the intervals for \(x\):
1. For \(0 < x < 1\), we have \([x] = 0\).
2. For \(1 \leq x < 2\), we have \([x] = 1\).

### Step 3: Case 1: \(0 < x < 1\)
In this case, since \([x] = 0\), we have:
\[ [x + y + 1] = 0 \implies 0 \leq x + y + 1 < 1 \]
This simplifies to:
\[ -1 < x + y < 0 \implies -1 - x < y < -x \]
For \(x\) in the interval \((0, 1)\), we can express the bounds for \(y\):
- Lower bound: \(y = -1 - x\)
- Upper bound: \(y = -x\)

### Step 4: Case 2: \(1 \leq x < 2\)
In this case, since \([x] = 1\), we have:
\[ [x + y + 1] = 1 \implies 1 \leq x + y + 1 < 2 \]
This simplifies to:
\[ 0 \leq x + y < 1 \implies -x \leq y < 1 - x \]
For \(x\) in the interval \((1, 2)\), we can express the bounds for \(y\):
- Lower bound: \(y = -x\)
- Upper bound: \(y = 1 - x\)

### Step 5: Determine the area bounded by the lines
Now we need to find the area represented by these inequalities in the \(xy\)-plane.

1. **For \(0 < x < 1\)**:
- The lines are \(y = -1 - x\) and \(y = -x\).
- At \(x = 0\), \(y = -1\) and at \(x = 1\), \(y = -2\).
- The area forms a triangle with vertices at \((0, -1)\), \((1, -2)\), and \((1, -1)\).

2. **For \(1 \leq x < 2\)**:
- The lines are \(y = -x\) and \(y = 1 - x\).
- At \(x = 1\), \(y = 0\) and at \(x = 2\), \(y = -1\).
- The area forms a triangle with vertices at \((1, 0)\), \((2, -1)\), and \((2, 0)\).

### Step 6: Calculate the area of both triangles
1. **Area of the first triangle**:
- Base = 1 (from \(x = 0\) to \(x = 1\))
- Height = 1 (from \(y = -1\) to \(y = -2\))
- Area = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}\)

2. **Area of the second triangle**:
- Base = 1 (from \(x = 1\) to \(x = 2\))
- Height = 1 (from \(y = 0\) to \(y = -1\))
- Area = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}\)

### Step 7: Total area
Total area = Area of first triangle + Area of second triangle = \(\frac{1}{2} + \frac{1}{2} = 1\).

### Final Answer
The area represented by all possible positions of \(P\) is \(1\).