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.

Options

A. `(1, 2)`
B. `(2, 4)`
C. `(3, 1)`
D. `(3, 5)`

Answer

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

### Step 1: Understand the 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 Curve Equation
We can simplify the equation:
1. Rewrite the first term using the change of base formula:
\[ \log_{1/2}(x - 0.5) = -\log_2(x - 0.5) \]
2. The second term can be simplified:
\[ \log_2(\sqrt{4x^2 - 4x + 1}) = \frac{1}{2} \log_2(4x^2 - 4x + 1) \]

Thus, the equation becomes:
\[ y = -\log_2(x - 0.5) + \frac{1}{2} \log_2(4x^2 - 4x + 1) \]

### Step 3: Further Simplification
The expression \(4x^2 - 4x + 1\) can be factored:
\[ 4x^2 - 4x + 1 = (2x - 1)^2 \]
So we have:
\[ \log_2(4x^2 - 4x + 1) = \log_2((2x - 1)^2) = 2\log_2(2x - 1) \]
Thus:
\[ y = -\log_2(x - 0.5) + \log_2(2x - 1) \]

### Step 4: Combine the Logs
Using properties of logarithms:
\[ y = \log_2\left(\frac{2x - 1}{x - 0.5}\right) \]

### Step 5: Find Point P
Point P lies on the circle defined by:
\[ x^2 + y^2 = 10 \]
Substituting \(y\) from the curve into the circle equation:
\[ x^2 + \left(\log_2\left(\frac{2x - 1}{x - 0.5}\right)\right)^2 = 10 \]

### Step 6: Find Integer Abscissa for Point Q
Point Q is inside the circle and has an integer abscissa. We need to find integer values of \(x\) such that:
1. \(x^2 + y^2 < 10\)
2. \(y\) must be calculated from the curve equation.

### Step 7: Calculate Possible Values
To find \(P\):
1. Set \(y = 1\) (as derived from the simplification).
2. Substitute \(y = 1\) into the circle equation:
\[ x^2 + 1^2 = 10 \]
\[ x^2 + 1 = 10 \]
\[ x^2 = 9 \]
\[ x = 3 \text{ or } -3 \]

However, \(x = -3\) leads to a negative argument in the logarithm, so:
\[ x = 3 \]
Thus, point \(P\) is:
\[ P(3, 1) \]

### Step 8: Find Q
Now, we need to find integer values for \(x\) such that \(x^2 + y^2 < 10\). The possible integer values for \(x\) are \(-3, -2, -1, 0, 1, 2, 3\).

### Step 9: Check Each Integer
For \(x = -2, -1, 0, 1, 2\):
- Calculate \(y\) using the curve equation.
- Check if \(x^2 + y^2 < 10\).

### Conclusion
After checking, we find that \(Q\) can be any point with integer abscissa that satisfies the conditions.