2014 年高考数学（新课标卷 2理科）

怎么想到：关于虚轴对称会改变横坐标即实部的符号，而纵坐标即虚部保持不变。

为什么从这里入手：复数题先判断目标需要代数形式还是模与辐角。“怎么想到：关于虚轴对称会改变横坐标即实部的符号，而纵坐标即虚部保持不变。”与“识别结构并建立关系”直接相连，先利用共轭、模或实虚部关系，通常能避免把复数完全展开成冗长乘积。

详细展开：$\htmlData{tutor-start=0,tutor-end=1}{z}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{+}\mathrm{\htmlData{tutor-start=16,tutor-end=17}{i}}$对应点$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)}$，关于虚轴对称后为$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}$，所以$\htmlData{tutor-start=0,tutor-end=1}{z}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{+}\mathrm{\htmlData{tutor-start=17,tutor-end=18}{i}}$。相乘时展开并使用$\mathrm{\htmlData{tutor-start=8,tutor-end=9}{i}}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{1}$，交叉虚部$\htmlData{tutor-start=0,tutor-end=1}{2}\mathrm{\htmlData{tutor-start=9,tutor-end=10}{i}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{2}\mathrm{\htmlData{tutor-start=21,tutor-end=22}{i}}$抵消。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：乘积虚部为0，与两个点关于虚轴对称的结构相符；实部-5对应选项A。

怎么想到：和向量与差向量的模平方展开式只在数量积前的符号不同，作差能直接得到四倍数量积。

为什么从这里入手：空间关系只靠观察容易漏条件，而“怎么想到：和向量与差向量的模平方展开式只在数量积前的符号不同，作差能直接得到四倍数量积。”可以直接翻译成垂直、平行、数量积或体积公式。先识别结构并建立关系，是在建立可验证的几何链条，而不是凭图形猜结论。

详细展开：$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{|}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{|}\htmlData{tutor-start=11,tutor-end=12}{a}\htmlData{tutor-start=12,tutor-end=13}{|}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{|}\htmlData{tutor-start=19,tutor-end=20}{b}\htmlData{tutor-start=20,tutor-end=21}{|}^{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{+}\htmlData{tutor-start=26,tutor-end=27}{2}\htmlData{tutor-start=27,tutor-end=28}{a}\htmlData{tutor-start=28,tutor-end=34}{\cdot }\htmlData{tutor-start=34,tutor-end=35}{b}\htmlData{tutor-start=35,tutor-end=36}{=}\htmlData{tutor-start=36,tutor-end=37}{1}\htmlData{tutor-start=37,tutor-end=38}{0}$，$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{|}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{|}\htmlData{tutor-start=11,tutor-end=12}{a}\htmlData{tutor-start=12,tutor-end=13}{|}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{|}\htmlData{tutor-start=19,tutor-end=20}{b}\htmlData{tutor-start=20,tutor-end=21}{|}^{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{2}\htmlData{tutor-start=27,tutor-end=28}{a}\htmlData{tutor-start=28,tutor-end=34}{\cdot }\htmlData{tutor-start=34,tutor-end=35}{b}\htmlData{tutor-start=35,tutor-end=36}{=}\htmlData{tutor-start=36,tutor-end=37}{6}$。前式减后式，得到$\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=8}{\cdot }\htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{4}$，所以$\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=7}{\cdot }\htmlData{tutor-start=7,tutor-end=8}{b}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{1}$。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：所得数量积代回两式后，两个模平方之差为4，与已知$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{6}$完全一致，选择A。

怎么想到：已知AB、BC和面积，面积公式可求夹角B的正弦；正弦对应锐、钝两个角，必须用“钝角三角形”排除错误分支。

为什么从这里入手：三角题先把未知角转成特殊角、同一角或已知边角关系。“怎么想到：已知AB、BC和面积，面积公式可求夹角B的正弦；正弦对应锐、钝两个角，必须用“钝角三角形”排除错误分支。”指出了可用的象限、恒等式或几何关系，先识别结构并建立关系可以同时确定数值和正负号。

详细展开：$\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{A}\htmlData{tutor-start=14,tutor-end=15}{B}\htmlData{tutor-start=15,tutor-end=21}{\cdot }\htmlData{tutor-start=21,tutor-end=22}{B}\htmlData{tutor-start=22,tutor-end=23}{C}\sin \htmlData{tutor-start=28,tutor-end=29}{B}\htmlData{tutor-start=29,tutor-end=30}{=}\htmlData{tutor-start=30,tutor-end=31}{1}\htmlData{tutor-start=31,tutor-end=32}{/}\htmlData{tutor-start=32,tutor-end=33}{2}$给出$\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{2}}}{\htmlData{tutor-start=16,tutor-end=17}{2}}\sin \htmlData{tutor-start=23,tutor-end=24}{B}\htmlData{tutor-start=24,tutor-end=25}{=}\htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{/}\htmlData{tutor-start=27,tutor-end=28}{2}$，故$\sin \htmlData{tutor-start=5,tutor-end=6}{B}\htmlData{tutor-start=6,tutor-end=7}{=}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{/}\htmlData{tutor-start=16,tutor-end=17}{2}$，即$B=45^\circ$或$135^\circ$。若$B=45^\circ$，余弦定理给$\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}$，三边$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{2}}$构成直角三角形，不符合钝角条件；故$B=135^\circ$。于是$AC^{2}=1+2-2\sqrt{2}\cos135^\circ=5$。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：$\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{=}\sqrt{\htmlData{tutor-start=9,tutor-end=10}{5}}$为最大边，且$\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{5}\htmlData{tutor-start=8,tutor-end=9}{>}\htmlData{tutor-start=9,tutor-end=10}{A}\htmlData{tutor-start=10,tutor-end=11}{B}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{C}^{\htmlData{tutor-start=20,tutor-end=21}{2}}\htmlData{tutor-start=22,tutor-end=23}{=}\htmlData{tutor-start=23,tutor-end=24}{3}$，反向确认其对角为钝角，选择B。

怎么想到：题目给出单日优良概率和连续两日都优良概率，“已知某天优良”正是条件概率的分母事件。

为什么从这里入手：概率题的关键不是立刻代数，而是先说清随机试验与事件。“怎么想到：题目给出单日优良概率和连续两日都优良概率，“已知某天优良”正是条件概率的分母事件。”确定了基本事件或随机变量，所以先识别结构并建立关系，再依据独立、互斥、对立或期望线性性计算。

详细展开：设A表示某天优良，B表示随后一天优良。已知$\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{A}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{.}\htmlData{tutor-start=7,tutor-end=8}{7}\htmlData{tutor-start=8,tutor-end=9}{5}$，连续两天优良就是$\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=6}{\cap }\htmlData{tutor-start=6,tutor-end=7}{B}$，故$\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{A}\htmlData{tutor-start=3,tutor-end=8}{\cap }\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{0}\htmlData{tutor-start=12,tutor-end=13}{.}\htmlData{tutor-start=13,tutor-end=14}{6}$。根据条件概率定义，$\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=8}{\mid }\htmlData{tutor-start=8,tutor-end=9}{A}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{P}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{A}\htmlData{tutor-start=14,tutor-end=19}{\cap }\htmlData{tutor-start=19,tutor-end=20}{B}\htmlData{tutor-start=20,tutor-end=21}{)}\htmlData{tutor-start=21,tutor-end=22}{/}\htmlData{tutor-start=22,tutor-end=23}{P}\htmlData{tutor-start=23,tutor-end=24}{(}\htmlData{tutor-start=24,tutor-end=25}{A}\htmlData{tutor-start=25,tutor-end=26}{)}\htmlData{tutor-start=26,tutor-end=27}{=}\htmlData{tutor-start=27,tutor-end=28}{0}\htmlData{tutor-start=28,tutor-end=29}{.}\htmlData{tutor-start=29,tutor-end=30}{6}\htmlData{tutor-start=30,tutor-end=31}{/}\htmlData{tutor-start=31,tutor-end=32}{0}\htmlData{tutor-start=32,tutor-end=33}{.}\htmlData{tutor-start=33,tutor-end=34}{7}\htmlData{tutor-start=34,tutor-end=35}{5}\htmlData{tutor-start=35,tutor-end=36}{=}\htmlData{tutor-start=36,tutor-end=37}{0}\htmlData{tutor-start=37,tutor-end=38}{.}\htmlData{tutor-start=38,tutor-end=39}{8}$。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：反算$\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{A}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{P}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{B}\htmlData{tutor-start=7,tutor-end=12}{\mid }\htmlData{tutor-start=12,tutor-end=13}{A}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{0}\htmlData{tutor-start=16,tutor-end=17}{.}\htmlData{tutor-start=17,tutor-end=18}{7}\htmlData{tutor-start=18,tutor-end=19}{5}\htmlData{tutor-start=19,tutor-end=25}{\times}\htmlData{tutor-start=25,tutor-end=26}{0}\htmlData{tutor-start=26,tutor-end=27}{.}\htmlData{tutor-start=27,tutor-end=28}{8}\htmlData{tutor-start=28,tutor-end=29}{=}\htmlData{tutor-start=29,tutor-end=30}{0}\htmlData{tutor-start=30,tutor-end=31}{.}\htmlData{tutor-start=31,tutor-end=32}{6}$，恰好还原连续两天优良概率，选择A。

怎么想到：端面同心圆与两个阶梯轴视图共同说明零件是两段同轴实心圆柱；内圆是较细圆柱的投影，不是贯穿孔。

为什么从这里入手：目标是“识别结构并建立关系”，而“怎么想到：端面同心圆与两个阶梯轴视图共同说明零件是两段同轴实心圆柱；内圆是较细圆柱的投影，不是贯穿孔。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：毛坯体积为$\htmlData{tutor-start=0,tutor-end=1}{V}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=9}{\pi}\htmlData{tutor-start=9,tutor-end=14}{\cdot}\htmlData{tutor-start=14,tutor-end=15}{3}^{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=24}{\cdot}\htmlData{tutor-start=24,tutor-end=25}{6}\htmlData{tutor-start=25,tutor-end=26}{=}\htmlData{tutor-start=26,tutor-end=27}{5}\htmlData{tutor-start=27,tutor-end=28}{4}\htmlData{tutor-start=28,tutor-end=31}{\pi}$。原图网格显示前4 cm半径2、后2 cm半径3，零件体积为$\htmlData{tutor-start=0,tutor-end=1}{V}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=5}{\pi}\htmlData{tutor-start=5,tutor-end=10}{\cdot}\htmlData{tutor-start=10,tutor-end=11}{2}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=20}{\cdot}\htmlData{tutor-start=20,tutor-end=21}{4}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=25}{\pi}\htmlData{tutor-start=25,tutor-end=30}{\cdot}\htmlData{tutor-start=30,tutor-end=31}{3}^{\htmlData{tutor-start=33,tutor-end=34}{2}}\htmlData{tutor-start=35,tutor-end=40}{\cdot}\htmlData{tutor-start=40,tutor-end=41}{2}\htmlData{tutor-start=41,tutor-end=42}{=}\htmlData{tutor-start=42,tutor-end=43}{3}\htmlData{tutor-start=43,tutor-end=44}{4}\htmlData{tutor-start=44,tutor-end=47}{\pi}$。切去$\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=2}{4}\htmlData{tutor-start=2,tutor-end=5}{\pi}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{4}\htmlData{tutor-start=8,tutor-end=11}{\pi}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{0}\htmlData{tutor-start=14,tutor-end=17}{\pi}$，所求比为$\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{/}\htmlData{tutor-start=3,tutor-end=4}{5}\htmlData{tutor-start=4,tutor-end=5}{4}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{7}$。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：两段长度和为6，最大半径为3；保留比例$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{7}\htmlData{tutor-start=2,tutor-end=3}{/}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{7}$与切削比例$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{/}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{7}$之和为1，选择C。

怎么想到：判断是$\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=5}{\le }\htmlData{tutor-start=5,tutor-end=6}{t}$，所以k等于t时仍进入循环；列出每轮结束后的k、M、S最不易漏步。

为什么从这里入手：目标是“识别结构并建立关系”，而“怎么想到：判断是$\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=5}{\le }\htmlData{tutor-start=5,tutor-end=6}{t}$，所以k等于t时仍进入循环；列出每轮结束后的k、M、S最不易漏步。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：初始$\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{S}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{k}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{1}$且$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{t}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{2}$。第一轮得到$\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{S}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{5}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{k}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{2}$；第二轮因$\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=4}{\le}\htmlData{tutor-start=4,tutor-end=5}{2}$仍执行，得到$\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{S}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{7}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{k}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{3}$。随后$\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=4}{\le}\htmlData{tutor-start=4,tutor-end=5}{2}$为假，程序输出S=7。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：循环恰执行k=1、2两轮，S从3增加2和2，最终7，选择D。

怎么想到：切线方程$\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{x}$直接给出斜率2，而曲线在某点的切线斜率就是该点导数值。

为什么从这里入手：函数的局部变化由导数控制。“怎么想到：切线方程$\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{x}$直接给出斜率2，而曲线在某点的切线斜率就是该点导数值。”给出了函数值、斜率或导数符号的入口，因此先识别结构并建立关系，就能把图象语言转换成方程或符号表。

详细展开：$\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{-}\ln\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{)}$的导数为$\htmlData{tutor-start=0,tutor-end=1}{y}'\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{a}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{)}$。在$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0}$处，$\htmlData{tutor-start=0,tutor-end=1}{y}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}$。令它等于切线斜率2，得到$\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{2}$，所以$\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}$。原曲线在x=0时y=0，确实经过切点。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：代入a=3后，点$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{)}$在曲线上且导数为2，因此切线唯一为$\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{x}$，选择D。

怎么想到：线性目标函数在多边形可行域的最值出现在顶点；边界直线两两联立比直接平移更便于确定数值。

为什么从这里入手：解析几何的目的，是把图形关系变成坐标等式。“怎么想到：线性目标函数在多边形可行域的最值出现在顶点；边界直线两两联立比直接平移更便于确定数值。”提供了点、斜率、距离或焦点条件，先识别结构并建立关系后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：三个边界$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{7}$、$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{y}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}$、$\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{y}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{5}$两两相交，得到$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{5}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{)}$、$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{3}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{4}\htmlData{tutor-start=4,tutor-end=5}{)}$、$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)}$，三点均满足全部约束。代入$\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{y}$，分别得到8、2、3，所以最大值为8。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：全部可行域顶点均已比较，目标直线沿增大方向最后经过$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{5}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{)}$，故选择B。

怎么想到：A、B都在同一条已知倾角的焦点弦上，三角形面积可用$\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{|}\htmlData{tutor-start=12,tutor-end=13}{A}\htmlData{tutor-start=13,tutor-end=14}{B}\htmlData{tutor-start=14,tutor-end=15}{|}\htmlData{tutor-start=15,tutor-end=16}{d}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{O}\htmlData{tutor-start=18,tutor-end=19}{,}\htmlData{tutor-start=19,tutor-end=20}{A}\htmlData{tutor-start=20,tutor-end=21}{B}\htmlData{tutor-start=21,tutor-end=22}{)}$；两部分都有现成公式。

为什么从这里入手：解析几何的目的，是把图形关系变成坐标等式。“怎么想到：A、B都在同一条已知倾角的焦点弦上，三角形面积可用$\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{|}\htmlData{tutor-start=12,tutor-end=13}{A}\htmlData{tutor-start=13,tutor-end=14}{B}\htmlData{tutor-start=14,tutor-end=15}{|}\htmlData{tutor-start=15,tutor-end=16}{d}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{O}\htmlData{tutor-start=18,tutor-end=19}{,}\htmlData{tutor-start=19,tutor-end=20}{A}\htmlData{tutor-start=20,tutor-end=21}{B}\htmlData{tutor-start=21,tutor-end=22}{)}$；两部分都有现成公式。”提供了点、斜率、距离或焦点条件，先识别结构并建立关系后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：$\htmlData{tutor-start=0,tutor-end=1}{y}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{x}$中$\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{p}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{3}$。倾角$30^\circ$的焦点弦长$|AB|=4p/\sin^{2}30^\circ=12$。焦点为$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{3}\htmlData{tutor-start=2,tutor-end=3}{/}\htmlData{tutor-start=3,tutor-end=4}{4}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)}$，弦方程$\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{4}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{/}\sqrt{\htmlData{tutor-start=16,tutor-end=17}{3}}$，即$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{-}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{3}}\htmlData{tutor-start=10,tutor-end=11}{y}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{3}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{4}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{0}$。原点到该直线距离为$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{3}\htmlData{tutor-start=2,tutor-end=3}{/}\htmlData{tutor-start=3,tutor-end=4}{4}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{/}\sqrt{\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{3}}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{3}\htmlData{tutor-start=18,tutor-end=19}{/}\htmlData{tutor-start=19,tutor-end=20}{8}$。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：底长12、高3/8都为正，面积为$\htmlData{tutor-start=0,tutor-end=1}{9}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{4}$；若直接联立求交点后作行列式也得到相同结果，选择D。

怎么想到：直三棱柱有三条互相垂直的方向，且各长度相等，建立坐标系后中点坐标和异面直线夹角都可一次算出。

为什么从这里入手：解析几何的目的，是把图形关系变成坐标等式。“怎么想到：直三棱柱有三条互相垂直的方向，且各长度相等，建立坐标系后中点坐标和异面直线夹角都可一次算出。”提供了点、斜率、距离或焦点条件，先识别结构并建立关系后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：设公共长度为1，取$\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{,}\htmlData{tutor-start=15,tutor-end=16}{0}\htmlData{tutor-start=16,tutor-end=17}{,}\htmlData{tutor-start=17,tutor-end=18}{0}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{,}\htmlData{tutor-start=20,tutor-end=21}{B}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{0}\htmlData{tutor-start=24,tutor-end=25}{,}\htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{,}\htmlData{tutor-start=27,tutor-end=28}{0}\htmlData{tutor-start=28,tutor-end=29}{)}$，侧棱沿z轴。则$\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{0}\htmlData{tutor-start=10,tutor-end=11}{,}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{)}\htmlData{tutor-start=13,tutor-end=14}{,}\htmlData{tutor-start=14,tutor-end=15}{B}_{\htmlData{tutor-start=17,tutor-end=18}{1}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{0}\htmlData{tutor-start=22,tutor-end=23}{,}\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{,}\htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{)}\htmlData{tutor-start=27,tutor-end=28}{,}\htmlData{tutor-start=28,tutor-end=29}{C}_{\htmlData{tutor-start=31,tutor-end=32}{1}}\htmlData{tutor-start=33,tutor-end=34}{=}\htmlData{tutor-start=34,tutor-end=35}{(}\htmlData{tutor-start=35,tutor-end=36}{0}\htmlData{tutor-start=36,tutor-end=37}{,}\htmlData{tutor-start=37,tutor-end=38}{0}\htmlData{tutor-start=38,tutor-end=39}{,}\htmlData{tutor-start=39,tutor-end=40}{1}\htmlData{tutor-start=40,tutor-end=41}{)}$，$\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{,}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{)}\htmlData{tutor-start=13,tutor-end=14}{,}\htmlData{tutor-start=14,tutor-end=15}{N}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{/}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{,}\htmlData{tutor-start=21,tutor-end=22}{0}\htmlData{tutor-start=22,tutor-end=23}{,}\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{)}$。所以$\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{M}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=23}{/}\htmlData{tutor-start=23,tutor-end=24}{2}\htmlData{tutor-start=24,tutor-end=25}{,}\htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{/}\htmlData{tutor-start=28,tutor-end=29}{2}\htmlData{tutor-start=29,tutor-end=30}{,}\htmlData{tutor-start=30,tutor-end=31}{1}\htmlData{tutor-start=31,tutor-end=32}{)}$，$\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{N}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=24}{/}\htmlData{tutor-start=24,tutor-end=25}{2}\htmlData{tutor-start=25,tutor-end=26}{,}\htmlData{tutor-start=26,tutor-end=27}{0}\htmlData{tutor-start=27,tutor-end=28}{,}\htmlData{tutor-start=28,tutor-end=29}{1}\htmlData{tutor-start=29,tutor-end=30}{)}$。点积为$\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{4}$，两模分别为$\sqrt{\htmlData{tutor-start=6,tutor-end=7}{6}}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{2}$、$\sqrt{\htmlData{tutor-start=6,tutor-end=7}{5}}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{2}$。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：异面直线所成角取锐角，故点积取绝对值；结果介于0与1之间，选择C。

怎么想到：所有极值点的函数值平方都等于3，因此只需让极值点横坐标的绝对值尽可能小，再检验最小值是否小于$\htmlData{tutor-start=0,tutor-end=1}{m}^{\htmlData{tutor-start=3,tutor-end=4}{2}}$。

为什么从这里入手：函数的局部变化由导数控制。“怎么想到：所有极值点的函数值平方都等于3，因此只需让极值点横坐标的绝对值尽可能小，再检验最小值是否小于$\htmlData{tutor-start=0,tutor-end=1}{m}^{\htmlData{tutor-start=3,tutor-end=4}{2}}$。”给出了函数值、斜率或导数符号的入口，因此先识别结构并建立关系，就能把图象语言转换成方程或符号表。

详细展开：极值点满足$\cos\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=9}{\pi }\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{/}\htmlData{tutor-start=11,tutor-end=12}{m}\htmlData{tutor-start=12,tutor-end=13}{)}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{0}$，故$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{m}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{k}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{)}$，且$\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{f}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{]}^{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{3}$。在所有整数k中，$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{k}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{|}$最小为1/2，所以$\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{[}\htmlData{tutor-start=11,tutor-end=12}{f}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{x}_{\htmlData{tutor-start=16,tutor-end=17}{0}}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{]}^{\htmlData{tutor-start=22,tutor-end=23}{2}}$的最小可能值为$\htmlData{tutor-start=0,tutor-end=1}{m}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{3}$。存在所需极值点当且仅当$\htmlData{tutor-start=0,tutor-end=1}{m}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{3}\htmlData{tutor-start=9,tutor-end=10}{<}\htmlData{tutor-start=10,tutor-end=11}{m}^{\htmlData{tutor-start=13,tutor-end=14}{2}}$，即$\htmlData{tutor-start=0,tutor-end=1}{m}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{>}\htmlData{tutor-start=6,tutor-end=7}{4}$。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：当$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{m}\htmlData{tutor-start=2,tutor-end=3}{|}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{2}$时只能取等号而题设要求严格小于，所以端点不取；$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{m}\htmlData{tutor-start=2,tutor-end=3}{|}\htmlData{tutor-start=3,tutor-end=4}{>}\htmlData{tutor-start=4,tutor-end=5}{2}$时取$\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0}$即可实现，选择C。

怎么想到：$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{a}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{0}}$中选择7个x、3个a才产生$\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{7}}$，对应组合数$\binom{\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{0}}{\htmlData{tutor-start=11,tutor-end=12}{7}}$。

为什么从这里入手：目标是“识别结构并建立关系”，而“怎么想到：$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{a}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{0}}$中选择7个x、3个a才产生$\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{7}}$，对应组合数$\binom{\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{0}}{\htmlData{tutor-start=11,tutor-end=12}{7}}$。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：展开式中$\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{7}}$项为$\binom{\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{0}}{\htmlData{tutor-start=11,tutor-end=12}{7}}\htmlData{tutor-start=13,tutor-end=14}{x}^{\htmlData{tutor-start=16,tutor-end=17}{7}}\htmlData{tutor-start=18,tutor-end=19}{a}^{\htmlData{tutor-start=21,tutor-end=22}{3}}$，所以系数是$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{a}^{\htmlData{tutor-start=6,tutor-end=7}{3}}$。由题设$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{a}^{\htmlData{tutor-start=6,tutor-end=7}{3}}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{5}$，得到$\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{8}$，实数解$\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{2}$。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：代回后系数$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{)}^{\htmlData{tutor-start=10,tutor-end=11}{3}}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{0}\htmlData{tutor-start=16,tutor-end=17}{/}\htmlData{tutor-start=17,tutor-end=18}{8}\htmlData{tutor-start=18,tutor-end=19}{=}\htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{5}$，精确符合题设。

怎么想到：第二项是$\htmlData{tutor-start=0,tutor-end=1}{2}\sin\htmlData{tutor-start=5,tutor-end=12}{\varphi}\cos\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{x}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=26}{\varphi}\htmlData{tutor-start=26,tutor-end=27}{)}$，恰好适用$\htmlData{tutor-start=0,tutor-end=1}{2}\sin \htmlData{tutor-start=6,tutor-end=7}{A}\cos \htmlData{tutor-start=12,tutor-end=13}{B}$公式，并会出现第一项$\sin\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=15}{\varphi}\htmlData{tutor-start=15,tutor-end=16}{)}$。

为什么从这里入手：目标是“识别结构并建立关系”，而“怎么想到：第二项是$\htmlData{tutor-start=0,tutor-end=1}{2}\sin\htmlData{tutor-start=5,tutor-end=12}{\varphi}\cos\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{x}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=26}{\varphi}\htmlData{tutor-start=26,tutor-end=27}{)}$，恰好适用$\htmlData{tutor-start=0,tutor-end=1}{2}\sin \htmlData{tutor-start=6,tutor-end=7}{A}\cos \htmlData{tutor-start=12,tutor-end=13}{B}$公式，并会出现第一项$\sin\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=15}{\varphi}\htmlData{tutor-start=15,tutor-end=16}{)}$。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：$\htmlData{tutor-start=0,tutor-end=1}{2}\sin\htmlData{tutor-start=5,tutor-end=12}{\varphi}\cos\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{x}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=26}{\varphi}\htmlData{tutor-start=26,tutor-end=27}{)}\htmlData{tutor-start=27,tutor-end=28}{=}\sin\htmlData{tutor-start=32,tutor-end=33}{(}\htmlData{tutor-start=33,tutor-end=34}{x}\htmlData{tutor-start=34,tutor-end=35}{+}\htmlData{tutor-start=35,tutor-end=36}{2}\htmlData{tutor-start=36,tutor-end=43}{\varphi}\htmlData{tutor-start=43,tutor-end=44}{)}\htmlData{tutor-start=44,tutor-end=45}{+}\sin\htmlData{tutor-start=49,tutor-end=50}{[}\htmlData{tutor-start=50,tutor-end=51}{-}\htmlData{tutor-start=51,tutor-end=52}{x}\htmlData{tutor-start=52,tutor-end=53}{]}\htmlData{tutor-start=53,tutor-end=54}{=}\sin\htmlData{tutor-start=58,tutor-end=59}{(}\htmlData{tutor-start=59,tutor-end=60}{x}\htmlData{tutor-start=60,tutor-end=61}{+}\htmlData{tutor-start=61,tutor-end=62}{2}\htmlData{tutor-start=62,tutor-end=69}{\varphi}\htmlData{tutor-start=69,tutor-end=70}{)}\htmlData{tutor-start=70,tutor-end=71}{-}\sin \htmlData{tutor-start=76,tutor-end=77}{x}$。因此原式中$\sin\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=15}{\varphi}\htmlData{tutor-start=15,tutor-end=16}{)}$相消，只剩$\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\sin \htmlData{tutor-start=10,tutor-end=11}{x}$，其最大值为1。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：取$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=5}{\pi}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{k}\htmlData{tutor-start=10,tutor-end=13}{\pi}$可达到1，且结果与参数$\htmlData{tutor-start=0,tutor-end=7}{\varphi}$无关，最大值确定为1。

怎么想到：偶函数的函数值只取决于$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{u}\htmlData{tutor-start=2,tutor-end=3}{|}$；在非负半轴单调递减且$\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{0}$，所以正值恰对应$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{u}\htmlData{tutor-start=2,tutor-end=3}{|}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{2}$。

为什么从这里入手：函数的局部变化由导数控制。“怎么想到：偶函数的函数值只取决于$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{u}\htmlData{tutor-start=2,tutor-end=3}{|}$；在非负半轴单调递减且$\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{0}$，所以正值恰对应$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{u}\htmlData{tutor-start=2,tutor-end=3}{|}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{2}$。”给出了函数值、斜率或导数符号的入口，因此先识别结构并建立关系，就能把图象语言转换成方程或符号表。

详细展开：令$\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}$。因为f为偶函数，$\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{u}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{f}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{|}\htmlData{tutor-start=8,tutor-end=9}{u}\htmlData{tutor-start=9,tutor-end=10}{|}\htmlData{tutor-start=10,tutor-end=11}{)}$。f在$\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=10}{\infty}\htmlData{tutor-start=10,tutor-end=11}{)}$单调递减且$\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{0}$，所以$\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=4}{\le}\htmlData{tutor-start=4,tutor-end=5}{|}\htmlData{tutor-start=5,tutor-end=6}{u}\htmlData{tutor-start=6,tutor-end=7}{|}\htmlData{tutor-start=7,tutor-end=8}{<}\htmlData{tutor-start=8,tutor-end=9}{2}$时$\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{|}\htmlData{tutor-start=3,tutor-end=4}{u}\htmlData{tutor-start=4,tutor-end=5}{|}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{>}\htmlData{tutor-start=7,tutor-end=8}{0}$，而$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{u}\htmlData{tutor-start=2,tutor-end=3}{|}\htmlData{tutor-start=3,tutor-end=4}{>}\htmlData{tutor-start=4,tutor-end=5}{2}$时$\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{|}\htmlData{tutor-start=3,tutor-end=4}{u}\htmlData{tutor-start=4,tutor-end=5}{|}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{<}\htmlData{tutor-start=7,tutor-end=8}{0}$。于是$\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{>}\htmlData{tutor-start=7,tutor-end=8}{0}$等价于$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{|}\htmlData{tutor-start=5,tutor-end=6}{<}\htmlData{tutor-start=6,tutor-end=7}{2}$。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：端点$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{3}$都使$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{|}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}$、函数值等于0，不满足严格大于0，所以最终是开区间。

怎么想到：从圆外点M看圆，射线相对MO的最大偏角在相切时取得；只需让这个最大角不小于45度。

为什么从这里入手：最值与范围题要先找到变量受限方式以及等号何时成立。“怎么想到：从圆外点M看圆，射线相对MO的最大偏角在相切时取得；只需让这个最大角不小于45度。”给出了可行域或关键界，先识别结构并建立关系能把‘可能有多大’转成单调性、基本不等式或边界比较。

详细展开：$\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{M}\htmlData{tutor-start=2,tutor-end=3}{=}\sqrt{\htmlData{tutor-start=9,tutor-end=10}{x}_{\htmlData{tutor-start=12,tutor-end=13}{0}}^{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{1}}$。作切线MT，直角三角形OMT中$\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{T}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}$，最大偏角$\htmlData{tutor-start=0,tutor-end=6}{\alpha}$满足$\sin\htmlData{tutor-start=4,tutor-end=10}{\alpha}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{/}\htmlData{tutor-start=13,tutor-end=14}{O}\htmlData{tutor-start=14,tutor-end=15}{M}$。存在$\angle OMN=45^\circ$等价于$\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{M}\htmlData{tutor-start=2,tutor-end=5}{\le}\sqrt{\htmlData{tutor-start=11,tutor-end=12}{2}}$，于是$\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=14}{\le}\htmlData{tutor-start=14,tutor-end=15}{2}$，即$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{x}_{\htmlData{tutor-start=4,tutor-end=5}{0}}\htmlData{tutor-start=6,tutor-end=7}{|}\htmlData{tutor-start=7,tutor-end=10}{\le}\htmlData{tutor-start=10,tutor-end=11}{1}$。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：端点对应恰好相切，角45度可取；区间外连最大偏角都不足45度，因此范围为闭区间。

怎么想到：线性递推$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{3}\htmlData{tutor-start=9,tutor-end=10}{a}_{\htmlData{tutor-start=12,tutor-end=13}{n}}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{1}$要变成纯倍数关系，可令$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{c}$并要求常数项也乘3，即$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{3}\htmlData{tutor-start=5,tutor-end=6}{c}$。

为什么从这里入手：数列题要先识别相邻项之间保持不变的结构。“怎么想到：线性递推$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{3}\htmlData{tutor-start=9,tutor-end=10}{a}_{\htmlData{tutor-start=12,tutor-end=13}{n}}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{1}$要变成纯倍数关系，可令$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{c}$并要求常数项也乘3，即$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{3}\htmlData{tutor-start=5,tutor-end=6}{c}$。”提示了差、比或可累加关系，先识别结构并建立关系能把第 $\htmlData{tutor-start=0,tutor-end=1}{n}$ 项问题改写成已经会处理的等差、等比或裂项模型。

详细展开：由$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{3}\htmlData{tutor-start=5,tutor-end=6}{c}$得$\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{2}$。于是$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{3}\htmlData{tutor-start=13,tutor-end=14}{a}_{\htmlData{tutor-start=16,tutor-end=17}{n}}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{3}\htmlData{tutor-start=20,tutor-end=21}{/}\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{=}\htmlData{tutor-start=23,tutor-end=24}{3}\htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{a}_{\htmlData{tutor-start=28,tutor-end=29}{n}}\htmlData{tutor-start=30,tutor-end=31}{+}\htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{/}\htmlData{tutor-start=33,tutor-end=34}{2}\htmlData{tutor-start=34,tutor-end=35}{)}$，所以$\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=13}{\}}$是公比3的等比数列。首项$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{3}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{2}$，故$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{3}\htmlData{tutor-start=12,tutor-end=13}{/}\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{)}\htmlData{tutor-start=15,tutor-end=16}{3}^{\htmlData{tutor-start=18,tutor-end=19}{n}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1}}\htmlData{tutor-start=22,tutor-end=23}{=}\htmlData{tutor-start=23,tutor-end=24}{3}^{\htmlData{tutor-start=26,tutor-end=27}{n}}\htmlData{tutor-start=28,tutor-end=29}{/}\htmlData{tutor-start=29,tutor-end=30}{2}$，从而$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{3}^{\htmlData{tutor-start=10,tutor-end=11}{n}}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{)}\htmlData{tutor-start=15,tutor-end=16}{/}\htmlData{tutor-start=16,tutor-end=17}{2}$。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：代入n=1得$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}$；并且$\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{a}_{\htmlData{tutor-start=4,tutor-end=5}{n}}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{[}\htmlData{tutor-start=10,tutor-end=11}{3}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{3}^{\htmlData{tutor-start=15,tutor-end=16}{n}}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{)}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{]}\htmlData{tutor-start=23,tutor-end=24}{/}\htmlData{tutor-start=24,tutor-end=25}{2}\htmlData{tutor-start=25,tutor-end=26}{=}\htmlData{tutor-start=26,tutor-end=27}{(}\htmlData{tutor-start=27,tutor-end=28}{3}^{\htmlData{tutor-start=30,tutor-end=31}{n}\htmlData{tutor-start=31,tutor-end=32}{+}\htmlData{tutor-start=32,tutor-end=33}{1}}\htmlData{tutor-start=34,tutor-end=35}{-}\htmlData{tutor-start=35,tutor-end=36}{1}\htmlData{tutor-start=36,tutor-end=37}{)}\htmlData{tutor-start=37,tutor-end=38}{/}\htmlData{tutor-start=38,tutor-end=39}{2}\htmlData{tutor-start=39,tutor-end=40}{=}\htmlData{tutor-start=40,tutor-end=41}{a}_{\htmlData{tutor-start=43,tutor-end=44}{n}\htmlData{tutor-start=44,tutor-end=45}{+}\htmlData{tutor-start=45,tutor-end=46}{1}}$，通项满足初值和递推。

怎么想到：通项含$\htmlData{tutor-start=0,tutor-end=1}{3}^{\htmlData{tutor-start=3,tutor-end=4}{k}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}$，从第二项起可把它与$\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=6}{\cdot}\htmlData{tutor-start=6,tutor-end=7}{3}^{\htmlData{tutor-start=9,tutor-end=10}{k}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}}$比较，使倒数被一个公比1/3的等比数列控制。

为什么从这里入手：数列题要先识别相邻项之间保持不变的结构。“怎么想到：通项含$\htmlData{tutor-start=0,tutor-end=1}{3}^{\htmlData{tutor-start=3,tutor-end=4}{k}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}$，从第二项起可把它与$\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=6}{\cdot}\htmlData{tutor-start=6,tutor-end=7}{3}^{\htmlData{tutor-start=9,tutor-end=10}{k}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}}$比较，使倒数被一个公比1/3的等比数列控制。”提示了差、比或可累加关系，先识别结构并建立关系能把第 $\htmlData{tutor-start=0,tutor-end=1}{n}$ 项问题改写成已经会处理的等差、等比或裂项模型。

详细展开：$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}$。当$\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{2}$时，$\htmlData{tutor-start=0,tutor-end=1}{3}^{\htmlData{tutor-start=3,tutor-end=4}{k}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{>}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=14}{\cdot}\htmlData{tutor-start=14,tutor-end=15}{3}^{\htmlData{tutor-start=17,tutor-end=18}{k}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{1}}$，所以$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{k}}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{3}^{\htmlData{tutor-start=14,tutor-end=15}{k}}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{<}\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{/}\htmlData{tutor-start=22,tutor-end=23}{3}^{\htmlData{tutor-start=25,tutor-end=26}{k}\htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{1}}$。因此有限和小于$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\sum_{\htmlData{tutor-start=8,tutor-end=9}{k}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{2}}^{\htmlData{tutor-start=14,tutor-end=15}{n}}\htmlData{tutor-start=16,tutor-end=17}{3}^{\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{k}\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{)}}$，再把有限正项和放大为无穷等比级数，得到$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\sum_{\htmlData{tutor-start=8,tutor-end=9}{j}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{1}}^{\htmlData{tutor-start=14,tutor-end=20}{\infty}}\htmlData{tutor-start=21,tutor-end=22}{(}\htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=24}{/}\htmlData{tutor-start=24,tutor-end=25}{3}\htmlData{tutor-start=25,tutor-end=26}{)}^{\htmlData{tutor-start=28,tutor-end=29}{j}}\htmlData{tutor-start=30,tutor-end=31}{=}\htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{+}\htmlData{tutor-start=33,tutor-end=34}{1}\htmlData{tutor-start=34,tutor-end=35}{/}\htmlData{tutor-start=35,tutor-end=36}{2}\htmlData{tutor-start=36,tutor-end=37}{=}\htmlData{tutor-start=37,tutor-end=38}{3}\htmlData{tutor-start=38,tutor-end=39}{/}\htmlData{tutor-start=39,tutor-end=40}{2}$。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：从k=2起第一处比较就是严格不等式，所以即使把有限和扩充为无穷和，最终仍严格小于3/2，满足题目要求。

怎么想到：E是PD中点，矩形对角线交点O又是BD中点；在三角形PDB中连接两中点即可得到PB的平行线。

为什么从这里入手：目标是“识别结构并建立关系”，而“怎么想到：E是PD中点，矩形对角线交点O又是BD中点；在三角形PDB中连接两中点即可得到PB的平行线。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：令$\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{A}\htmlData{tutor-start=3,tutor-end=4}{C}\htmlData{tutor-start=4,tutor-end=9}{\cap }\htmlData{tutor-start=9,tutor-end=10}{B}\htmlData{tutor-start=10,tutor-end=11}{D}$。矩形对角线互相平分，所以O是BD中点。E为PD中点，故在三角形PDB中$\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=12}{\parallel }\htmlData{tutor-start=12,tutor-end=13}{P}\htmlData{tutor-start=13,tutor-end=14}{B}$。E、O都属于平面AEC，所以$\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=9}{\subset}$平面AEC；PB不在该平面内，由线面平行判定定理得$\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=11}{\parallel}$平面AEC。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：构造线EO位于目标平面且不与PB重合，判定定理的全部条件均满足。

怎么想到：PA垂直底面，矩形三条方向天然正交，坐标法可把二面角转成两个平面法向量夹角，同时E坐标直接给出体积高。

为什么从这里入手：解析几何的目的，是把图形关系变成坐标等式。“怎么想到：PA垂直底面，矩形三条方向天然正交，坐标法可把二面角转成两个平面法向量夹角，同时E坐标直接给出体积高。”提供了点、斜率、距离或焦点条件，先识别结构并建立关系后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：设$\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{D}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{0}\htmlData{tutor-start=14,tutor-end=15}{,}\sqrt{\htmlData{tutor-start=21,tutor-end=22}{3}}\htmlData{tutor-start=23,tutor-end=24}{,}\htmlData{tutor-start=24,tutor-end=25}{0}\htmlData{tutor-start=25,tutor-end=26}{)}\htmlData{tutor-start=26,tutor-end=27}{,}\htmlData{tutor-start=27,tutor-end=28}{B}\htmlData{tutor-start=28,tutor-end=29}{=}\htmlData{tutor-start=29,tutor-end=30}{(}\htmlData{tutor-start=30,tutor-end=31}{s}\htmlData{tutor-start=31,tutor-end=32}{,}\htmlData{tutor-start=32,tutor-end=33}{0}\htmlData{tutor-start=33,tutor-end=34}{,}\htmlData{tutor-start=34,tutor-end=35}{0}\htmlData{tutor-start=35,tutor-end=36}{)}\htmlData{tutor-start=36,tutor-end=37}{,}\htmlData{tutor-start=37,tutor-end=38}{C}\htmlData{tutor-start=38,tutor-end=39}{=}\htmlData{tutor-start=39,tutor-end=40}{(}\htmlData{tutor-start=40,tutor-end=41}{s}\htmlData{tutor-start=41,tutor-end=42}{,}\sqrt{\htmlData{tutor-start=48,tutor-end=49}{3}}\htmlData{tutor-start=50,tutor-end=51}{,}\htmlData{tutor-start=51,tutor-end=52}{0}\htmlData{tutor-start=52,tutor-end=53}{)}\htmlData{tutor-start=53,tutor-end=54}{,}\htmlData{tutor-start=54,tutor-end=55}{P}\htmlData{tutor-start=55,tutor-end=56}{=}\htmlData{tutor-start=56,tutor-end=57}{(}\htmlData{tutor-start=57,tutor-end=58}{0}\htmlData{tutor-start=58,tutor-end=59}{,}\htmlData{tutor-start=59,tutor-end=60}{0}\htmlData{tutor-start=60,tutor-end=61}{,}\htmlData{tutor-start=61,tutor-end=62}{1}\htmlData{tutor-start=62,tutor-end=63}{)}$，则$\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{,}\sqrt{\htmlData{tutor-start=11,tutor-end=12}{3}}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{,}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{/}\htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{)}$。平面DAE为$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0}$，法向量$\htmlData{tutor-start=0,tutor-end=1}{n}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{0}\htmlData{tutor-start=10,tutor-end=11}{,}\htmlData{tutor-start=11,tutor-end=12}{0}\htmlData{tutor-start=12,tutor-end=13}{)}$。由$\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{E}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{,}\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}\htmlData{tutor-start=14,tutor-end=15}{/}\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{,}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{/}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{)}$、$\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{s}\htmlData{tutor-start=5,tutor-end=6}{,}\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}\htmlData{tutor-start=14,tutor-end=15}{,}\htmlData{tutor-start=15,tutor-end=16}{0}\htmlData{tutor-start=16,tutor-end=17}{)}$得平面CAE法向量可取$\htmlData{tutor-start=0,tutor-end=1}{n}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{-}\sqrt{\htmlData{tutor-start=14,tutor-end=15}{3}}\htmlData{tutor-start=16,tutor-end=17}{,}\htmlData{tutor-start=17,tutor-end=18}{s}\htmlData{tutor-start=18,tutor-end=19}{,}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{s}\sqrt{\htmlData{tutor-start=27,tutor-end=28}{3}}\htmlData{tutor-start=29,tutor-end=30}{)}$。二面角60度给$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{n}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=12}{\cdot }\htmlData{tutor-start=12,tutor-end=13}{n}_{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{|}\htmlData{tutor-start=18,tutor-end=19}{/}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{|}\htmlData{tutor-start=21,tutor-end=22}{n}_{\htmlData{tutor-start=24,tutor-end=25}{1}}\htmlData{tutor-start=26,tutor-end=27}{|}\htmlData{tutor-start=27,tutor-end=28}{|}\htmlData{tutor-start=28,tutor-end=29}{n}_{\htmlData{tutor-start=31,tutor-end=32}{2}}\htmlData{tutor-start=33,tutor-end=34}{|}\htmlData{tutor-start=34,tutor-end=35}{)}\htmlData{tutor-start=35,tutor-end=36}{=}\sqrt{\htmlData{tutor-start=42,tutor-end=43}{3}}\htmlData{tutor-start=44,tutor-end=45}{/}\sqrt{\htmlData{tutor-start=51,tutor-end=52}{3}\htmlData{tutor-start=52,tutor-end=53}{+}\htmlData{tutor-start=53,tutor-end=54}{4}\htmlData{tutor-start=54,tutor-end=55}{s}^{\htmlData{tutor-start=57,tutor-end=58}{2}}}\htmlData{tutor-start=60,tutor-end=61}{=}\htmlData{tutor-start=61,tutor-end=62}{1}\htmlData{tutor-start=62,tutor-end=63}{/}\htmlData{tutor-start=63,tutor-end=64}{2}$，解得$\htmlData{tutor-start=0,tutor-end=1}{s}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{2}$。底面三角形ACD面积为$\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{A}\htmlData{tutor-start=12,tutor-end=13}{D}\htmlData{tutor-start=13,tutor-end=19}{\cdot }\htmlData{tutor-start=19,tutor-end=20}{D}\htmlData{tutor-start=20,tutor-end=21}{C}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{3}\sqrt{\htmlData{tutor-start=29,tutor-end=30}{3}}\htmlData{tutor-start=31,tutor-end=32}{/}\htmlData{tutor-start=32,tutor-end=33}{4}$，E到z=0的高为1/2。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：所得$\htmlData{tutor-start=0,tutor-end=1}{s}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0}$且回代法向量夹角余弦恰为1/2；E是PD中点，所以其到底面的高度确为PA的一半，体积核验完整。

怎么想到：题目已给斜率和截距公式，应先算$\bar{\htmlData{tutor-start=5,tutor-end=6}{t}}\htmlData{tutor-start=7,tutor-end=8}{,}\bar{\htmlData{tutor-start=13,tutor-end=14}{y}}$，再把七组数据围绕均值列表相乘，避免直接展开大数。

为什么从这里入手：统计量都来自明确的样本计数或加权。“怎么想到：题目已给斜率和截距公式，应先算$\bar{\htmlData{tutor-start=5,tutor-end=6}{t}}\htmlData{tutor-start=7,tutor-end=8}{,}\bar{\htmlData{tutor-start=13,tutor-end=14}{y}}$，再把七组数据围绕均值列表相乘，避免直接展开大数。”给出了数据与权重，先识别结构并建立关系可以避免把频数、频率和概率混为一谈，并能用总数或概率和反查计算。

详细展开：$\htmlData{tutor-start=0,tutor-end=7}{\bar{t}}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{+}\cdots\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{7}\htmlData{tutor-start=19,tutor-end=20}{)}\htmlData{tutor-start=20,tutor-end=21}{/}\htmlData{tutor-start=21,tutor-end=22}{7}\htmlData{tutor-start=22,tutor-end=23}{=}\htmlData{tutor-start=23,tutor-end=24}{4}$，$\bar{\htmlData{tutor-start=5,tutor-end=6}{y}}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{.}\htmlData{tutor-start=11,tutor-end=12}{9}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{3}\htmlData{tutor-start=14,tutor-end=15}{.}\htmlData{tutor-start=15,tutor-end=16}{3}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{3}\htmlData{tutor-start=18,tutor-end=19}{.}\htmlData{tutor-start=19,tutor-end=20}{6}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{4}\htmlData{tutor-start=22,tutor-end=23}{.}\htmlData{tutor-start=23,tutor-end=24}{4}\htmlData{tutor-start=24,tutor-end=25}{+}\htmlData{tutor-start=25,tutor-end=26}{4}\htmlData{tutor-start=26,tutor-end=27}{.}\htmlData{tutor-start=27,tutor-end=28}{8}\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{5}\htmlData{tutor-start=30,tutor-end=31}{.}\htmlData{tutor-start=31,tutor-end=32}{2}\htmlData{tutor-start=32,tutor-end=33}{+}\htmlData{tutor-start=33,tutor-end=34}{5}\htmlData{tutor-start=34,tutor-end=35}{.}\htmlData{tutor-start=35,tutor-end=36}{9}\htmlData{tutor-start=36,tutor-end=37}{)}\htmlData{tutor-start=37,tutor-end=38}{/}\htmlData{tutor-start=38,tutor-end=39}{7}\htmlData{tutor-start=39,tutor-end=40}{=}\htmlData{tutor-start=40,tutor-end=41}{4}\htmlData{tutor-start=41,tutor-end=42}{.}\htmlData{tutor-start=42,tutor-end=43}{3}$。离差平方和为$\htmlData{tutor-start=0,tutor-end=1}{9}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{4}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{9}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{8}$；离差积和依次相加为$\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{.}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{.}\htmlData{tutor-start=10,tutor-end=11}{7}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{0}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{0}\htmlData{tutor-start=15,tutor-end=16}{.}\htmlData{tutor-start=16,tutor-end=17}{5}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{.}\htmlData{tutor-start=20,tutor-end=21}{8}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{4}\htmlData{tutor-start=23,tutor-end=24}{.}\htmlData{tutor-start=24,tutor-end=25}{8}\htmlData{tutor-start=25,tutor-end=26}{=}\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{4}$。所以$\hat{\htmlData{tutor-start=5,tutor-end=6}{b}}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{/}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{8}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{0}\htmlData{tutor-start=15,tutor-end=16}{.}\htmlData{tutor-start=16,tutor-end=17}{5}$，$\hat{\htmlData{tutor-start=5,tutor-end=6}{a}}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{4}\htmlData{tutor-start=9,tutor-end=10}{.}\htmlData{tutor-start=10,tutor-end=11}{3}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{0}\htmlData{tutor-start=13,tutor-end=14}{.}\htmlData{tutor-start=14,tutor-end=15}{5}\htmlData{tutor-start=15,tutor-end=21}{\times}\htmlData{tutor-start=21,tutor-end=22}{4}\htmlData{tutor-start=22,tutor-end=23}{=}\htmlData{tutor-start=23,tutor-end=24}{2}\htmlData{tutor-start=24,tutor-end=25}{.}\htmlData{tutor-start=25,tutor-end=26}{3}$。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：回归直线在$\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\bar{\htmlData{tutor-start=7,tutor-end=8}{t}}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{4}$处给出$\widehat y=4.3=\bar{y}$，符合最小二乘回归直线必过样本中心点的性质。

怎么想到：年份代号每增加1代表一年，回归斜率的单位就是“千元/年”；2015比2007晚8年，对应$\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{9}$。

为什么从这里入手：统计量都来自明确的样本计数或加权。“怎么想到：年份代号每增加1代表一年，回归斜率的单位就是“千元/年”；2015比2007晚8年，对应$\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{9}$。”给出了数据与权重，先识别结构并建立关系可以避免把频数、频率和概率混为一谈，并能用总数或概率和反查计算。

详细展开：斜率0.5表示在2007至2013年的线性趋势下，人均纯收入平均每年约增加0.5千元。2015年的年份代号为$\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{5}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{7}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{9}$，代入回归方程得到$\widehat y=0.5\times9+2.3=6.8$千元。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：预测值6.8高于2013年实际值5.9，增加约0.9千元，与两年、每年约0.5千元的趋势相符。

怎么想到：$\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{F}_{\htmlData{tutor-start=4,tutor-end=5}{2}}$垂直x轴说明$\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{M}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{c}$；将其代入椭圆可把纵坐标写成$\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{a}$。

为什么从这里入手：解析几何的目的，是把图形关系变成坐标等式。“怎么想到：$\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{F}_{\htmlData{tutor-start=4,tutor-end=5}{2}}$垂直x轴说明$\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{M}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{c}$；将其代入椭圆可把纵坐标写成$\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{a}$。”提供了点、斜率、距离或焦点条件，先识别结构并建立关系后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：取$\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{c}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{b}^{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=11}{/}\htmlData{tutor-start=11,tutor-end=12}{a}\htmlData{tutor-start=12,tutor-end=13}{)}$、$\htmlData{tutor-start=0,tutor-end=1}{F}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{c}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{)}$。直线MN就是MF1，斜率$\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{a}\htmlData{tutor-start=9,tutor-end=10}{c}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{3}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{4}$。令$\htmlData{tutor-start=0,tutor-end=1}{e}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{a}$，用$\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{a}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{e}^{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{)}$得到$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{e}^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{e}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{3}\htmlData{tutor-start=16,tutor-end=17}{/}\htmlData{tutor-start=17,tutor-end=18}{4}$，即$\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{e}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{3}\htmlData{tutor-start=8,tutor-end=9}{e}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{0}$。在$\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{e}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{1}$中只有$\htmlData{tutor-start=0,tutor-end=1}{e}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{2}$。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：舍去不符合离心率范围的根-2；$\htmlData{tutor-start=0,tutor-end=1}{e}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{2}$回代斜率式正好得到3/4。

怎么想到：F1在弦MN内部，$\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{N}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{5}\htmlData{tutor-start=4,tutor-end=5}{F}_{\htmlData{tutor-start=7,tutor-end=8}{1}}\htmlData{tutor-start=9,tutor-end=10}{N}$等价于$\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{F}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{:}\htmlData{tutor-start=7,tutor-end=8}{F}_{\htmlData{tutor-start=10,tutor-end=11}{1}}\htmlData{tutor-start=12,tutor-end=13}{N}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{4}\htmlData{tutor-start=15,tutor-end=16}{:}\htmlData{tutor-start=16,tutor-end=17}{1}$，可直接用向量分点写N。

为什么从这里入手：解析几何的目的，是把图形关系变成坐标等式。“怎么想到：F1在弦MN内部，$\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{N}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{5}\htmlData{tutor-start=4,tutor-end=5}{F}_{\htmlData{tutor-start=7,tutor-end=8}{1}}\htmlData{tutor-start=9,tutor-end=10}{N}$等价于$\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{F}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{:}\htmlData{tutor-start=7,tutor-end=8}{F}_{\htmlData{tutor-start=10,tutor-end=11}{1}}\htmlData{tutor-start=12,tutor-end=13}{N}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{4}\htmlData{tutor-start=15,tutor-end=16}{:}\htmlData{tutor-start=16,tutor-end=17}{1}$，可直接用向量分点写N。”提供了点、斜率、距离或焦点条件，先识别结构并建立关系后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：$\htmlData{tutor-start=0,tutor-end=1}{N}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{F}_{\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{-}\frac{\htmlData{tutor-start=14,tutor-end=15}{1}}{\htmlData{tutor-start=17,tutor-end=18}{4}}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{M}\htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{F}_{\htmlData{tutor-start=25,tutor-end=26}{1}}\htmlData{tutor-start=27,tutor-end=28}{)}\htmlData{tutor-start=28,tutor-end=29}{=}\htmlData{tutor-start=29,tutor-end=30}{(}\htmlData{tutor-start=30,tutor-end=31}{-}\htmlData{tutor-start=31,tutor-end=32}{3}\htmlData{tutor-start=32,tutor-end=33}{c}\htmlData{tutor-start=33,tutor-end=34}{/}\htmlData{tutor-start=34,tutor-end=35}{2}\htmlData{tutor-start=35,tutor-end=36}{,}\htmlData{tutor-start=36,tutor-end=37}{-}\htmlData{tutor-start=37,tutor-end=38}{b}^{\htmlData{tutor-start=40,tutor-end=41}{2}}\htmlData{tutor-start=42,tutor-end=43}{/}\htmlData{tutor-start=43,tutor-end=44}{(}\htmlData{tutor-start=44,tutor-end=45}{4}\htmlData{tutor-start=45,tutor-end=46}{a}\htmlData{tutor-start=46,tutor-end=47}{)}\htmlData{tutor-start=47,tutor-end=48}{)}$。直线MF1的y截距为$\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{a}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{2}$，故$\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{a}$。把N代入椭圆并令$\htmlData{tutor-start=0,tutor-end=1}{e}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{a}$，得$\htmlData{tutor-start=0,tutor-end=1}{9}\htmlData{tutor-start=1,tutor-end=2}{e}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{4}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{e}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{)}\htmlData{tutor-start=18,tutor-end=19}{/}\htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{6}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{1}$，所以$\htmlData{tutor-start=0,tutor-end=1}{e}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{7}$。再由$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{e}^{\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{b}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{a}^{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{4}\htmlData{tutor-start=21,tutor-end=22}{/}\htmlData{tutor-start=22,tutor-end=23}{a}$得到$\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{7}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{b}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{8}$。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：回代后截距为2，N满足椭圆方程，且向量长度比例给出$\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{N}\htmlData{tutor-start=2,tutor-end=3}{:}\htmlData{tutor-start=3,tutor-end=4}{F}_{\htmlData{tutor-start=6,tutor-end=7}{1}}\htmlData{tutor-start=8,tutor-end=9}{N}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{5}\htmlData{tutor-start=11,tutor-end=12}{:}\htmlData{tutor-start=12,tutor-end=13}{1}$，所有题设同时成立。