Let P and Q are two points on curve `y=log_((1)/(2))(x-(1)/(2))+log_(2) sqrt(4x^(2)-4x+1)` and P is also on `x^(2)+y^(2)=10`. Q lies inside the given circle such that its abscissa is integer. Find the smallest possible value of `vec(OP)*vec(OQ)` where 'O' being origin.

Answer

To solve the problem step by step, we will follow the given instructions and derive the required values systematically.

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

### Step 2: Simplify the curve equation
Using the properties of logarithms, we can rewrite the equation:
\[ y = -\log_{2}(x - \frac{1}{2}) + \log_{2}(\sqrt{4x^2 - 4x + 1}) \]
This can be combined into a single logarithm:
\[ y = \log_{2}\left(\frac{\sqrt{4x^2 - 4x + 1}}{x - \frac{1}{2}}\right) \]

### Step 3: Simplify the expression inside the logarithm
The expression \( \sqrt{4x^2 - 4x + 1} \) can be rewritten as:
\[ \sqrt{(2x - 1)^2} = |2x - 1| \]
Thus, we have:
\[ y = \log_{2}\left(\frac{|2x - 1|}{x - \frac{1}{2}}\right) \]

### Step 4: Determine the value of \( y \)
For \( x > \frac{1}{2} \), \( |2x - 1| = 2x - 1 \):
\[ y = \log_{2}\left(\frac{2x - 1}{x - \frac{1}{2}}\right) \]
This simplifies to:
\[ y = \log_{2}(2) = 1 \]
Thus, the points \( P \) and \( Q \) on the curve have coordinates \( (x, 1) \).

### Step 5: Find point \( P \) on the circle
The point \( P \) also lies on the circle defined by:
\[ x^2 + y^2 = 10 \]
Substituting \( y = 1 \):
\[ x^2 + 1^2 = 10 \]
\[ x^2 + 1 = 10 \]
\[ x^2 = 9 \]
Thus, \( x = 3 \) or \( x = -3 \). However, \( x = -3 \) is not valid since it makes the logarithm undefined. Therefore, we have:
\[ P = (3, 1) \]

### Step 6: Determine point \( Q \)
Point \( Q \) lies inside the circle and has an integer abscissa. The equation of the circle gives:
\[ x^2 + 1^2 < 10 \]
\[ x^2 < 9 \]
Thus, \( -3 < x < 3 \). The integer values for \( x \) are \( -2, -1, 0, 1, 2 \).

### Step 7: Valid values for \( Q \)
We need to ensure that the logarithm is defined for \( Q \):
- For \( x = -2 \): \( -2 - \frac{1}{2} < 0 \) (not valid)
- For \( x = -1 \): \( -1 - \frac{1}{2} < 0 \) (not valid)
- For \( x = 0 \): \( 0 - \frac{1}{2} < 0 \) (not valid)
- For \( x = 1 \): \( 1 - \frac{1}{2} > 0 \) (valid)
- For \( x = 2 \): \( 2 - \frac{1}{2} > 0 \) (valid)

Thus, the valid points for \( Q \) are \( (1, 1) \) and \( (2, 1) \).

### Step 8: Calculate dot products
The vector \( \vec{OP} = (3, 1) \) and for \( Q \):
- For \( Q = (1, 1) \):
\[ \vec{OQ} = (1, 1) \]
\[ \vec{OP} \cdot \vec{OQ} = 3 \cdot 1 + 1 \cdot 1 = 3 + 1 = 4 \]

- For \( Q = (2, 1) \):
\[ \vec{OQ} = (2, 1) \]
\[ \vec{OP} \cdot \vec{OQ} = 3 \cdot 2 + 1 \cdot 1 = 6 + 1 = 7 \]

### Step 9: Find the minimum value
The smallest possible value of \( \vec{OP} \cdot \vec{OQ} \) is:
\[ \text{Minimum value} = 4 \]

### Final Answer
The smallest possible value of \( \vec{OP} \cdot \vec{OQ} \) is \( \boxed{4} \).