answer:**Analysis** This problem mainly examines the knowledge of the scalar product formula of plane vectors, the operational properties of vector scalar products, the formula for the modulus, and the method of finding the area of a triangle. It is considered a medium-level question. **Solution** Since overrightarrow{OA} perp overrightarrow{OB}, we have overrightarrow{OA} cdot overrightarrow{OB} = (overrightarrow{a}-overrightarrow{b})cdot(overrightarrow{a}+overrightarrow{b})=0, Expanding and simplifying, we get overrightarrow{a}^2 - overrightarrow{b}^2 =0, which leads to |overrightarrow{a}|=|overrightarrow{b}|, |overrightarrow{a}|= sqrt{frac{1}{4}+frac{3}{4}} =1, |overrightarrow{OA}|=|overrightarrow{OB}|, that is |overrightarrow{a}-overrightarrow{b}|=|overrightarrow{a}+overrightarrow{b}|, This implies overrightarrow{a}^2 + overrightarrow{b}^2 -2 overrightarrow{a}cdot overrightarrow{b} = overrightarrow{a}^2 + overrightarrow{b}^2 +2 overrightarrow{a}cdot overrightarrow{b}, which means overrightarrow{a}cdot overrightarrow{b} =0, Thus, overrightarrow{a}, overrightarrow{b} are mutually perpendicular unit vectors. Therefore, |overrightarrow{OA}|=|overrightarrow{OB}|= sqrt{2}, and the area of triangle OAB, S= frac{1}{2} |overrightarrow{OA}| cdot |overrightarrow{OB}|=1. Hence, the answer is boxed{1}.

answer:Let's consider a tangent line to the parabola at point P(a,b). The distance from P to the line 2x-y-11=0 reaches its minimum when the tangent line at P is parallel to 2x-y-11=0. From y'=2x=2, we get x=1. Therefore, the point P(1,1). At this point, the distance d from P to the line is d= dfrac {|2-1-11|}{ sqrt {5}}=2 sqrt {5}. Hence, the minimum distance from P to the line is 2 sqrt {5}. Therefore, the correct answer is boxed{D}. To minimize the distance from P to the line, the tangent line at P must be parallel to 2x-y-11=0, which allows us to find the coordinates of P and thus the minimum distance. This problem examines the concept of extremal values in conic sections, as well as the idea of combining geometric and algebraic methods, and is considered to be of medium difficulty.

answer:First, simplify the complex number i(1 + i)i. i(1 + i)i = i(i + i^2) Recall that i^2 = -1. i(i + i^2) = i(i - 1) = i^2 - i = -1 - i Now, the complex number is in the standard form a + bi, where a is the real part and b is the imaginary part. Thus, the imaginary part of the complex number is -1. boxed{text{Imaginary part} = -1}

answer:1. **Understanding the Problem**: We need to determine the number of distinct integer solutions that satisfy the inequality ( |x| + |y| < 100 ). 2. **Working with the Equation**: - Consider the equation ( |x| + |y| = k ), where ( k ) is a non-negative integer. - For a fixed ( k ), the equation represents a diamond-shaped contour around the origin on the coordinate plane. 3. **Counting Solutions for Each ( k )**: - When ( k = 0 ): ( |x| + |y| = 0 implies x = 0, y = 0 ), so there is exactly **1 solution**. - When ( k = 1 ): ( |x| + |y| = 1 ) has points ((1,0), (-1,0), (0,1), (0,-1)). This totals ( 4 ) points, or **( 2^2 )** solutions. - For ( k geq 2 ): ( |x| + |y| = k ) can be visualized as a counting problem where points lie along a diamond-shaped contour. The number of integer solutions is given by the formula ( (k+1)^2 ). 4. **Summing Up the Solutions**: - To find the total number of solutions for ( |x| + |y| < 100 ), sum the solutions for ( k = 0 ) to ( k = 99 ): [ sum_{k=0}^{99} (k+1)^2 ] - The sum of the squares of the first ( 100 ) integers is given by the formula for the sum of squares: [ sum_{k=1}^{100} k^2 = frac{100 cdot 101 cdot 201}{6} ] 5. **Calculation**: - Using the formula: [ frac{100 cdot 101 cdot 201}{6} = 338350 ] # Conclusion: The total number of distinct integer solutions to the inequality ( |x| + |y| < 100 ) is thus: [ boxed{338350} ]