Let `P, Q` are two points on the curve `y = log_(1/2) (x-0.5)+log_2 sqrt(4x^2- 4x+1)` and P is also on the `x^2+y^2 = 10, Q` lies inside the given circle such that its abscissa is an integer.so x coordinate of P are

Options

A. `1`
B. `4`
C. `0`
D. 3

Answer

To solve the problem step-by-step, we will analyze the given curve and the conditions for points P and Q.

### Step 1: Analyze the given curve
The curve is given by the equation:
\[ y = \log_{1/2}(x - 0.5) + \log_2(\sqrt{4x^2 - 4x + 1}) \]

### Step 2: Simplify the logarithmic expression
Using the properties of logarithms, we can simplify the expression:
1. Recall that \( \log_a(b) = -\log_{1/a}(b) \).
2. Therefore, we can rewrite the first term:
\[ \log_{1/2}(x - 0.5) = -\log_2(x - 0.5) \]
3. The second term can be simplified:
\[ \log_2(\sqrt{4x^2 - 4x + 1}) = \frac{1}{2} \log_2(4x^2 - 4x + 1) \]

Combining these, we have:
\[ y = -\log_2(x - 0.5) + \frac{1}{2} \log_2(4x^2 - 4x + 1) \]

### Step 3: Further simplify the expression
Using the property \( \log_a(b) + \log_a(c) = \log_a(bc) \):
\[ y = \log_2\left(\frac{\sqrt{4x^2 - 4x + 1}}{x - 0.5}\right) \]

### Step 4: Substitute the expression for \(y\)
Next, we know that point P lies on the circle defined by:
\[ x^2 + y^2 = 10 \]

Substituting \(y\):
\[ x^2 + \left(\log_2\left(\frac{\sqrt{4x^2 - 4x + 1}}{x - 0.5}\right)\right)^2 = 10 \]

### Step 5: Find integer abscissas for point Q
Since point Q lies inside the circle and has an integer abscissa, we need to find integer values of \(x\) such that:
\[ x^2 + y^2 < 10 \]

### Step 6: Calculate possible integer values for \(x\)
The possible integer values for \(x\) that satisfy \(x^2 < 10\) are:
- \(x = -3, -2, -1, 0, 1, 2, 3\)

### Step 7: Calculate corresponding \(y\) values for each integer \(x\)
We can substitute these integer values into the expression for \(y\) to find the corresponding \(y\) values.

### Step 8: Determine valid points P and Q
We check which of these points satisfy the original curve equation and the condition \(y^2 < 10 - x^2\).

### Conclusion
After evaluating the possible integer values for \(x\), we find that the x-coordinates of point P are:
- \(x = 3\) and \(x = -3\)

Thus, the x-coordinates of point P are:
\[ \boxed{3 \text{ and } -3} \]