answer:1. The given points are ( A(2,0) ) and ( B ) which lies on the ellipse defined by the equation: [ frac{x^{2}}{4} + frac{y^{2}}{3} = 1 ] with the conditions ( x > 0 ) and ( y > 0 ). 2. Assume ( B ) has coordinates ( B(2 cos theta, sqrt{3} sin theta) ) where ( theta in left(0, frac{pi}{2}right) ). 3. Point ( C ) is the foot of the perpendicular from ( B ) to the ( y )-axis, so the coordinates of ( C ) are ( C(0, sqrt{3} sin theta) ). 4. We need to find the area of quadrilateral ( OABC ). The area ( S_{OABC} ) can be divided into two parts: a right triangle ( triangle OBC ) and a right triangle ( triangle OAB ). 5. The area of ( triangle OBC ) is: [ text{Area}_{OBC} = frac{1}{2} times text{base} times text{height} = frac{1}{2} times (2 cos theta) times (sqrt{3} sin theta) = sqrt{3} cos theta sin theta ] 6. The area of ( triangle OAB ) is: [ text{Area}_{OAB} = frac{1}{2} times (text{base distance from } O text{ to } C - text{segment} AB) times text{height to } B ] [ = frac{1}{2} times (2 - 2 cos theta) times (sqrt{3} sin theta) = frac{sqrt{3}}{2} sin theta (2 - 2 cos theta) = sqrt{3} sin theta (1 - cos theta) ] 7. Therefore, the total area ( S_{OABC} ) is: [ S_{OABC} = (sqrt{3} cos theta sin theta) + (sqrt{3} sin theta (1 - cos theta)) ] [ = sqrt{3} cos theta sin theta + sqrt{3} sin theta - sqrt{3} cos theta sin theta ] [ = sqrt{3} sin theta + frac{sqrt{3}}{2} sin 2 theta ] 8. To find the maximum area, take the derivative of ( S_{OABC} ) with respect to ( theta ) and set it to zero: [ frac{d}{dtheta} left( sqrt{3} sin theta + frac{sqrt{3}}{2} sin 2 theta right) ] [ = sqrt{3} cos theta + frac{sqrt{3}}{2} cdot 2 cos 2 theta ] [ = sqrt{3} cos theta + sqrt{3} cos^2 theta - sqrt{3} sin^2 theta ] [ = sqrt{3} (cos^2 theta + cos theta - sin^2 theta) ] Using the identity ( cos^2 theta - sin^2 theta = cos 2 theta ), we have: [ = sqrt{3} (cos 2 theta + cos theta) = 0 ] 9. Solve the equation: [ 2 cos^2 theta + cos theta - 1 = 0 ] Factoring the quadratic equation: [ (2 cos theta - 1)(cos theta + 1) = 0 ] [ Rightarrow cos theta = frac{1}{2} text { or } cos theta = -1 ] Since ( theta in left(0, frac{pi}{2}right) ), we reject ( cos theta = -1 ); hence: [ cos theta = frac{1}{2} Rightarrow theta = frac{pi}{3} ] 10. Substitute ( theta = frac{pi}{3} ) into the area expression to verify: [ S_{OABC} = sqrt{3} sin left(frac{pi}{3}right) left( 1 - cos left( frac{pi}{3} right) right) + frac{sqrt{3}}{2} sin left( 2 times frac{pi}{3} right) ] [ = sqrt{3} cdot frac{sqrt{3}}{2} left( 1 - frac{1}{2} right) + frac{sqrt{3}}{2} (1 - cos^2 left(frac{pi}{3}right)) ] [ = frac{3}{2} left( frac{1}{2} right) + frac{sqrt{3}}{2} (sinleft( frac{2pi}{3} right)) ] Calculate specifically to confirm: [ = frac{3}{4} + frac{3}{4} = frac{9}{4} ] Conclusion: [ boxed{frac{9}{4}} ]

answer:Given z(1+i)=4-2i, we have z= dfrac {4-2i}{1+i}= dfrac {(4-2i)(1-i)}{(1+i)(1-i)}= dfrac {2-6i}{2}=1-3i, Therefore, |z|= sqrt {1^{2}+(-3)^{2}}= sqrt {10}. Hence, the correct choice is boxed{D}. This problem involves transforming the given equation and then simplifying it using the operations of multiplication and division in complex number algebraic form, followed by calculating the modulus of a complex number. It tests the operations of multiplication and division in complex number algebraic form and the method of finding the modulus of a complex number, making it a basic question.

answer:(a) Let's denote the function as f(x) = Asin(omega x + varphi) + B (where A > 0, omega > 0, and 0 < |varphi| < pi). According to pattern 1, we know the function has a period of 12, hence: frac{2pi}{omega} = 12, text{ so } omega = frac{pi}{6}. From pattern 2, we know that f(2) is the minimum, f(8) is the maximum, and f(8) - f(2) = 400, which implies the amplitude (A) is 200. Pattern 3 indicates that f(x) is monotonically increasing on the interval [2, 8] and f(2) = 100. Therefore, f(8) = 500 and we have: begin{cases} -A + B = 100 A + B = 500 end{cases} Solving this system of equations gives us: begin{cases} A = 200 B = 300 end{cases} Since f(2) is the minimum and f(8) is the maximum, we have: sinleft(frac{pi}{6} cdot 2 + varphiright) = -1 text{ and } sinleft(frac{pi}{6} cdot 8 + varphiright) = 1. Given that 0 < |varphi| < pi, we deduce that varphi = -frac{5pi}{6}. Thus: f(x) = 200sinleft(frac{pi}{6}x - frac{5pi}{6}right) + 300. (b) According to the condition, 200sinleft(frac{pi}{6}x - frac{5pi}{6}right) + 300 geq 400. Simplifying this expression gives us: sinleft(frac{pi}{6}x - frac{5pi}{6}right) geq frac{1}{2}, which leads to: 2kpi + frac{pi}{6} leq frac{pi}{6}x - frac{5pi}{6} leq 2kpi + frac{5pi}{6}, text{ where } k in mathbb{Z}. Solving for x gives us the inequality: 12k + 6 leq x leq 12k + 10, text{ where } k in mathbb{Z}. Considering x in mathbb{N^{*}} and 1 leq x leq 12 (since x represents the months of the year), we find that x = 6, 7, 8, 9, 10. Consequently: boxed{x = 6, 7, 8, 9, 10} Therefore, more than 400 portions of food need to be prepared in the months of June, July, August, September, and October.

answer:# Problem: Consider an infinite square grid on which we have placed dominoes (possibly infinitely many) such that each domino covers two adjacent squares and no two dominoes are touching, even at their corners. Prove that the uncovered part of the square grid can be filled seamlessly with 2 times 1 dominoes. 1. **Setup a Larger Grid:** We place a larger 2 times 2 grid on the original square lattice such that the edges of the larger grid coincide with the edges of the smaller squares in the original lattice. This creates cells in the pattern shown in the figure below:  2. **Analyze Domino Placement in 2 times 2 Cells:** Within each 2 times 2 cell of this larger grid, there are several possibilities for how the original dominoes might cover the smaller squares: - **Case 1:** None of the smaller squares inside the 2 times 2 cell are covered by a domino. - **Case 2:** One pair of adjacent smaller squares inside the 2 times 2 cell is covered by a domino. - **Case 3:** Two pairs of adjacent smaller squares inside the 2 times 2 cell are covered by separate dominoes. 3. **Handling Each Case**: - **Case 1 (0 dominoes in 2 times 2 cell):** All 4 smaller squares are empty. This can be straightforwardly covered by 2 dominoes. - **Case 2 (1 domino in 2 times 2 cell):** The single domino covers 2 of the smaller squares, with 2 smaller squares left uncovered. Since each 2 times 2 cell is adjacent to others, the domino covering part of one 2 times 2 cell necessarily ends up covering part of an adjacent 2 times 2 cell. In this scenario, we essentially need to consider two adjacent 2 times 2 cells together. Each of these paired larger cells will then have 6 uncovered smaller squares, which can also be covered by 3 additional dominoes. - **Case 3 (2 dominoes in 2 times 2 cell):** All 4 smaller squares are covered (2 squares per domino). This leaves no empty smaller squares within the 2 times 2 cell. 4. **Conclusion:** Each configuration within a 2 times 2 cell or pair of adjacent cells can be handled in such a way that the uncovered smaller squares are efficiently covered by additional 2 times 1 dominoes. Therefore, by iteratively applying this method across the entire grid, the entire plane can be seamlessly filled with 2 times 1 dominoes. boxed{}