answer:The median score for player A is 28, and the median score for player B is 36; therefore, the sum of the median scores of players A and B in these games is boxed{64}. Hence, the correct answer is C.

answer:First, we find the derivative of f(x) with respect to x to determine the conditions for f(x) to have an extremum. Taking the derivative, we get: f'(x) = x - 2a + frac{b}{x}. Since we are given that f(1) = frac{1}{2} and f'(1) = 0, we can set up the following equations: f(1) = frac{1}{2} Rightarrow frac{1}{2}(1)^2 - 2a(1) + bln(1) + 2a^2 = frac{1}{2}, f'(1) = 0 Rightarrow 1 - 2a + frac{b}{1} = 0. The second equation simplifies to b = 2a - 1. Substituting this into the first equation, we have: frac{1}{2} - 2a + 2a^2 = frac{1}{2}, 2a^2 - 2a = 0, a(a - 1) = 0. This gives us two possible values for a, either a = 0 or a = 1. If a = 0, then from b = 2a - 1 we get b = -1. If a = 1, then b = 2a - 1 would give us b = 1. However, we are given that f(x) achieves an extremum at x = 1. We need to determine which of these solutions is valid. For a = 1, the derivative f'(x) = x - 2 + frac{1}{x} does not change sign around x = 1, thus it does not yield an extremum at x = 1. Therefore, the only valid solution is a = 0 and b = -1. Combining these, we have a + b = 0 + (-1) = -1. Therefore, the answer is boxed{-1}.

answer:1. We are given a cube and inside this cube, two regular tetrahedra are inscribed such that four vertices of the cube are vertices of one tetrahedron, and the remaining four vertices of the cube are vertices of the other tetrahedron. 2. Let's denote the side length of the cube as ( a ). 3. Each of the tetrahedra shares four of the cube's vertices such that the arrangement of the tetrahedra will form a regular octahedron in their intersection. 4. To determine the volume of the regular octahedron that results from the overlapping part of the two tetrahedra, we recognize the properties of these geometric shapes. 5. The formula for the volume of a cubic shape with side length ( a ) is: [ V_{text{cube}} = a^3 ] 6. The tetrahedra inscribed in the cube each occupy a symmetric section of the cube. When considering their overlap, they intersect in such a way that the resulting shape is a regular octahedron. 7. The volume of this regular octahedron can be derived as follows. Consider that the volume of one tetrahedron inscribed within the cube is 1/3 the volume of a pyramid with the same base and height: [ V_{text{tetrahedron}} = frac{1}{6} a^3 ] 8. The overlapping volume of the two tetrahedra forming a regular octahedron is exactly half the volume of one tetrahedron. Hence, the volume of the shared part (the octahedron) is given by: [ V_{text{octahedron}} = frac{1}{6} a^3 ] 9. To find the fraction of the cube’s volume that the octahedron occupies, we calculate: [ text{Fraction} = frac{V_{text{octahedron}}}{V_{text{cube}}} = frac{frac{1}{6} a^3}{a^3} = frac{1}{6} ] 10. Therefore, the volume of the shared region, which is the octahedron, constitutes one-sixth of the overall volume of the cube. # Conclusion: [ boxed{frac{1}{6}} ]

answer:**Analysis** This question examines the domain of the tangent function, which is a basic topic. **Solution** Given frac{pi}{4}-xneq kpi+ frac{pi}{2}, kinmathbb{Z}, we can derive xneq -kpi- frac{pi}{4}, kinmathbb{Z}, thus, the domain of the function y=tan left( frac{pi}{4}-x right) is {x|xneq kpi + frac{3pi}{4} , kinmathbb{Z}, xinmathbb{R}}. Therefore, the correct answer is boxed{D}.