Re: [中學] 102學測數學多選11,12

看板Math作者JohnMash (Paul)時間11年前 (2013/01/30 11:16)推噓0(0推 0噓 2→)

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※ 引述《JohnMash (Paul)》之銘言： ※ 引述《weilai81241 (Hungry Bird)》之銘言： : http://mag.udn.com/html/campus/e102/e102_02a.pdf : 整份考卷就這兩題完全不知道該如何下手... : 請版上各位大大幫忙了... Problem 11 Consider ellipse E : x^2/a^2 + y^2/b^2 = 1 and a^2-b^2=c^2 Point F1 : (c,0) Consider the graph of ellipse E above x-axis, E_+ Let an arbitrary point P on E_+. Denote d(F1,P) the distance between F1 and P. Then d(F1,P) is strictly increasing function for P from (a,0) to (-a,0). ^^^^^^^^^^^^^^^^^^^ --------------------------------------------------------------- 簡單證明一下嚴格遞增這回事首先證明 P 在 (a,0) 到 (0,b) 的範圍內 d(F1,P) 是嚴格遞增如若不然則存在相異點 Q,R 使得 d(F1,Q)=d(F1,R) 又因為 △F1 F2 Q 和 △F1 F2 R 周長相等故得到 △F1 F2 Q 和 △F1 F2 R 全等但兩三角形的面積卻不等(等底不等高)因而矛盾又因為 F1+F2 是常數故可得 P 在 (a,0) 到 (-a,0) 的範圍內 d(F1,P) 是嚴格遞增 ------------------------------------------------------------------- However, for the four points Q1,Q2,Q3,Q4, on a square with center F1, we have d(F1,Q1)=d(F1,Q2)=d(F1,Q3)=d(F1,Q4) hence, we have only one of {Qi} to take above the x-axis. For example, you take Q1, then the only other possible point is below the x-axis. But this line segment constructed from the points must be parallel to the axis, and then leads to a contradiction. Hence, the correct answer is no point or one point only. (我認為正確答案僅有1,5兩個選項) ------------------------------------------- Problem 12 Plot an arithmetic sequence for b_n. They are all on a line B and y-intercept positive Plot a geometric sequence for a_n. It is an oscillating and convergent curve A. a_9 and a_{10} must be one point above x-axis and the other below x-axis. You can easily see from the graph that line B must have negative common difference because of its positive y-intercept. However, because curve A is convergent, we can have a_8 > b_8 or a_8 < b_8 or a_8 = b_8 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 27.147.57.77

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