The point ([P + 1], [P]) (where, [x] is the greatest integer function) lying inside the region bounded by the circle `x^(2)+y^(2)-2x-15`=0 and `x^(2)+ y^(2) -2x - 7=0` , then

Options

A. `P in [-1,2) - { 0,1}`
B. `P in [-1 ,0) cup ( 0,1 ) cup (1,2]`
C. `P in ( -1,2) `
D. none of these

Answer

To solve the problem, we need to analyze the conditions given by the two circles and the point defined by the greatest integer function.

### Step-by-Step Solution:

1. **Identify the Circles**:
The equations of the circles are:
- Circle 1: \( x^2 + y^2 - 2x - 15 = 0 \)
- Circle 2: \( x^2 + y^2 - 2x - 7 = 0 \)

2. **Rewrite the Circle Equations**:
We can rewrite the equations in standard form:
- Circle 1:
\[ (x - 1)^2 + y^2 = 16 \quad \text{(Center: (1, 0), Radius: 4)} \]
- Circle 2:
\[ (x - 1)^2 + y^2 = 8 \quad \text{(Center: (1, 0), Radius: \(2\sqrt{2}\))} \]

3. **Determine the Region**:
The region bounded by these circles is the annular region between the two circles. The point \(([P + 1], [P])\) must lie inside the first circle and outside the second circle.

4. **Point Inside the First Circle**:
For the point \(([P + 1], [P])\) to lie inside Circle 1:
\[ (P + 1 - 1)^2 + (P - 0)^2 < 16 \]
Simplifying this:
\[ P^2 + P^2 < 16 \implies 2P^2 < 16 \implies P^2 < 8 \implies -\sqrt{8} < P < \sqrt{8} \]

5. **Point Outside the Second Circle**:
For the point \(([P + 1], [P])\) to lie outside Circle 2:
\[ (P + 1 - 1)^2 + (P - 0)^2 > 8 \]
Simplifying this:
\[ P^2 + P^2 > 8 \implies 2P^2 > 8 \implies P^2 > 4 \implies P < -2 \text{ or } P > 2 \]

6. **Combine the Conditions**:
We now have two inequalities:
- From Circle 1: \( -\sqrt{8} < P < \sqrt{8} \) (approximately \(-2.83 < P < 2.83\))
- From Circle 2: \( P < -2 \text{ or } P > 2 \)

The valid ranges for \(P\) that satisfy both conditions are:
- For \(P < -2\): This is valid as it lies within \(-\sqrt{8}\).
- For \(P > 2\): This is valid as it lies within \(\sqrt{8}\).

Thus, we have:
- \(P < -2\) or \(P > 2\)

7. **Determine Integer Values**:
Since \([P]\) is the greatest integer function, we consider integer values:
- For \(P < -2\), possible integer values are \(-3, -4, -5, \ldots\)
- For \(P > 2\), possible integer values are \(3, 4, 5, \ldots\)

### Final Answer:
The possible integer values of \(P\) are \(P \in (-\infty, -2) \cup (2, \infty)\).