返回特征解读

2012 年高考数学（大纲卷文科）

exams_raw/普通高考/2012/2012大纲文(贵州,甘肃,青海,西藏,广西).pdf · HS-MATH-1024-v2.1-solution-aware

22 题27 个小问/题组
1

一、选择题 · 集合

已知集合 A={x∣x 是平行四边形}\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=4}{\{}\htmlData{tutor-start=4,tutor-end=5}{x} \htmlData{tutor-start=6,tutor-end=11}{\mid }\htmlData{tutor-start=11,tutor-end=12}{x} \text{ \htmlData{tutor-start=20,tutor-end=21}{是}\htmlData{tutor-start=21,tutor-end=22}{平}\htmlData{tutor-start=22,tutor-end=23}{行}\htmlData{tutor-start=23,tutor-end=24}{四}\htmlData{tutor-start=24,tutor-end=25}{边}\htmlData{tutor-start=25,tutor-end=26}{形}}\htmlData{tutor-start=27,tutor-end=29}{\}}，B={x∣x 是矩形}\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=4}{\{}\htmlData{tutor-start=4,tutor-end=5}{x} \htmlData{tutor-start=6,tutor-end=11}{\mid }\htmlData{tutor-start=11,tutor-end=12}{x} \text{ \htmlData{tutor-start=20,tutor-end=21}{是}\htmlData{tutor-start=21,tutor-end=22}{矩}\htmlData{tutor-start=22,tutor-end=23}{形}}\htmlData{tutor-start=24,tutor-end=26}{\}}，C={x∣x 是正方形}\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=4}{\{}\htmlData{tutor-start=4,tutor-end=5}{x} \htmlData{tutor-start=6,tutor-end=11}{\mid }\htmlData{tutor-start=11,tutor-end=12}{x} \text{ \htmlData{tutor-start=20,tutor-end=21}{是}\htmlData{tutor-start=21,tutor-end=22}{正}\htmlData{tutor-start=22,tutor-end=23}{方}\htmlData{tutor-start=23,tutor-end=24}{形}}\htmlData{tutor-start=25,tutor-end=27}{\}}，D={x∣x 是菱形}\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=4}{\{}\htmlData{tutor-start=4,tutor-end=5}{x} \htmlData{tutor-start=6,tutor-end=11}{\mid }\htmlData{tutor-start=11,tutor-end=12}{x} \text{ \htmlData{tutor-start=20,tutor-end=21}{是}\htmlData{tutor-start=21,tutor-end=22}{菱}\htmlData{tutor-start=22,tutor-end=23}{形}}\htmlData{tutor-start=24,tutor-end=26}{\}}，则 ( )

答案：C⊂B⊂A\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=9}{\subset }\htmlData{tutor-start=9,tutor-end=10}{B}\htmlData{tutor-start=10,tutor-end=18}{\subset }\htmlData{tutor-start=18,tutor-end=19}{A} 且 C⊂D⊂A\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=9}{\subset }\htmlData{tutor-start=9,tutor-end=10}{D}\htmlData{tutor-start=10,tutor-end=18}{\subset }\htmlData{tutor-start=18,tutor-end=19}{A}

题目标签：四边形集合的包含关系

解题过程

按定义判断充分关系

确定四类四边形集合关系

(1)
识别结构并建立关系

正方形同时是特殊矩形和特殊菱形，矩形与菱形又都是特殊平行四边形。

为什么从这里入手：目标是“识别结构并建立关系”，而“正方形同时是特殊矩形和特殊菱形，矩形与菱形又都是特殊平行四边形。”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：因此 C⊂B,C⊂D\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=9}{\subset }\htmlData{tutor-start=9,tutor-end=10}{B}\htmlData{tutor-start=10,tutor-end=11}{,}\htmlData{tutor-start=11,tutor-end=12}{C}\htmlData{tutor-start=12,tutor-end=20}{\subset }\htmlData{tutor-start=20,tutor-end=21}{D}，并且 B⊂A,D⊂A\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=9}{\subset }\htmlData{tutor-start=9,tutor-end=10}{A}\htmlData{tutor-start=10,tutor-end=11}{,}\htmlData{tutor-start=11,tutor-end=12}{D}\htmlData{tutor-start=12,tutor-end=20}{\subset }\htmlData{tutor-start=20,tutor-end=21}{A}；矩形与菱形彼此不包含。

C\subset B\subset A,\quad C\subset D\subset A

(2)
完成计算并核验结论

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：这些都是真包含，因为存在非正方形矩形、非正方形菱形和一般平行四边形。

C⊂B⊂A, C⊂D⊂A\boxed{\htmlData{tutor-start=7,tutor-end=8}{C}\htmlData{tutor-start=8,tutor-end=16}{\subset }\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=25}{\subset }\htmlData{tutor-start=25,tutor-end=26}{A}\htmlData{tutor-start=26,tutor-end=27}{,}\htmlData{tutor-start=27,tutor-end=29}{\ }\htmlData{tutor-start=29,tutor-end=30}{C}\htmlData{tutor-start=30,tutor-end=38}{\subset }\htmlData{tutor-start=38,tutor-end=39}{D}\htmlData{tutor-start=39,tutor-end=47}{\subset }\htmlData{tutor-start=47,tutor-end=48}{A}}
2

一、选择题 · 反函数

函数 y=x+1 (x⩾−1)\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\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=13,tutor-end=15}{\ }\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{x} \htmlData{tutor-start=18,tutor-end=28}{\geqslant }\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{)} 的反函数为 ( )

答案：y=x2−1 (x≥0)\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x}^{\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=11}{\ }\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=16}{\ge}\htmlData{tutor-start=16,tutor-end=17}{0}\htmlData{tutor-start=17,tutor-end=18}{)}

题目标签：根式函数的反函数

解题过程

交换变量并保留值域

求反函数

(1)
识别结构并建立关系

原函数严格递增，值域为 [0,+∞)\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}{)}，它将成为反函数定义域。

为什么从这里入手：含分母、根号或对数的式子必须先合法，后续变形才有意义。“原函数严格递增，值域为 [0,+∞)\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}{)}，它将成为反函数定义域。”正好暴露了限制条件；先识别结构并建立关系，可以防止约分或平方时把禁值偷偷带回答案。

详细展开：由 y=x+1\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{1}} 得 x=y2−1\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=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}，交换 x,y\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{y} 得 y=x2−1\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x}^{\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}。

f−1(x)=x2−1,x≥0\htmlData{tutor-start=0,tutor-end=1}{f}^{\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{x}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{,}\quad \htmlData{tutor-start=24,tutor-end=25}{x}\htmlData{tutor-start=25,tutor-end=28}{\ge}\htmlData{tutor-start=28,tutor-end=29}{0}
(2)
完成计算并核验结论

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：定义域限制不能省略。

y=x2−1 (x≥0)\boxed{\htmlData{tutor-start=7,tutor-end=8}{y}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{x}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=18}{\ }\htmlData{tutor-start=18,tutor-end=19}{(}\htmlData{tutor-start=19,tutor-end=20}{x}\htmlData{tutor-start=20,tutor-end=23}{\ge}\htmlData{tutor-start=23,tutor-end=24}{0}\htmlData{tutor-start=24,tutor-end=25}{)}}
3

一、选择题 · 三角函数

若函数 f(x)=sin⁡x+φ3 (φ∈[0,2π])\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\frac{\htmlData{tutor-start=15,tutor-end=16}{x}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=24}{\varphi}}{\htmlData{tutor-start=26,tutor-end=27}{3}} \htmlData{tutor-start=29,tutor-end=31}{\ }\htmlData{tutor-start=31,tutor-end=32}{(}\htmlData{tutor-start=32,tutor-end=40}{\varphi }\htmlData{tutor-start=40,tutor-end=44}{\in }\htmlData{tutor-start=44,tutor-end=45}{[}\htmlData{tutor-start=45,tutor-end=46}{0}\htmlData{tutor-start=46,tutor-end=47}{,} \htmlData{tutor-start=48,tutor-end=49}{2}\htmlData{tutor-start=49,tutor-end=52}{\pi}\htmlData{tutor-start=52,tutor-end=53}{]}\htmlData{tutor-start=53,tutor-end=54}{)} 是偶函数，则 φ=\htmlData{tutor-start=0,tutor-end=7}{\varphi}\htmlData{tutor-start=7,tutor-end=8}{=} ( )

答案：φ=3π2\htmlData{tutor-start=0,tutor-end=7}{\varphi}\htmlData{tutor-start=7,tutor-end=8}{=}\frac{\htmlData{tutor-start=14,tutor-end=15}{3}\htmlData{tutor-start=15,tutor-end=18}{\pi}}{\htmlData{tutor-start=20,tutor-end=21}{2}}

题目标签：正弦函数偶性确定初相

解题过程

展开相位比较奇偶部分

求 φ\htmlData{tutor-start=0,tutor-end=7}{\varphi}

(1)
识别结构并建立关系

把正弦和角公式展开，偶函数要求奇函数部分系数为零。

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。“把正弦和角公式展开，偶函数要求奇函数部分系数为零。”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：sin⁡(x/3+φ/3)=sin⁡(x/3)cos⁡(φ/3)+cos⁡(x/3)sin⁡(φ/3)\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}{3}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=16}{\varphi}\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}{=}\sin\htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{x}\htmlData{tutor-start=26,tutor-end=27}{/}\htmlData{tutor-start=27,tutor-end=28}{3}\htmlData{tutor-start=28,tutor-end=29}{)}\cos\htmlData{tutor-start=33,tutor-end=34}{(}\htmlData{tutor-start=34,tutor-end=41}{\varphi}\htmlData{tutor-start=41,tutor-end=42}{/}\htmlData{tutor-start=42,tutor-end=43}{3}\htmlData{tutor-start=43,tutor-end=44}{)}\htmlData{tutor-start=44,tutor-end=45}{+}\cos\htmlData{tutor-start=49,tutor-end=50}{(}\htmlData{tutor-start=50,tutor-end=51}{x}\htmlData{tutor-start=51,tutor-end=52}{/}\htmlData{tutor-start=52,tutor-end=53}{3}\htmlData{tutor-start=53,tutor-end=54}{)}\sin\htmlData{tutor-start=58,tutor-end=59}{(}\htmlData{tutor-start=59,tutor-end=66}{\varphi}\htmlData{tutor-start=66,tutor-end=67}{/}\htmlData{tutor-start=67,tutor-end=68}{3}\htmlData{tutor-start=68,tutor-end=69}{)}，故 cos⁡(φ/3)=0\cos\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=12}{\varphi}\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}{=}\htmlData{tutor-start=16,tutor-end=17}{0}。

φ3=π2\frac\varphi3=\frac\pi2
(2)
完成计算并核验结论

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：结合给定范围，得到 φ=3π/2\htmlData{tutor-start=0,tutor-end=7}{\varphi}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{3}\htmlData{tutor-start=9,tutor-end=12}{\pi}\htmlData{tutor-start=12,tutor-end=13}{/}\htmlData{tutor-start=13,tutor-end=14}{2}。

3π2\boxed{\frac{\htmlData{tutor-start=13,tutor-end=14}{3}\htmlData{tutor-start=14,tutor-end=17}{\pi}}{\htmlData{tutor-start=19,tutor-end=20}{2}}}
4

一、选择题 · 三角函数

已知 α\htmlData{tutor-start=0,tutor-end=6}{\alpha} 为第二象限角，sin⁡α=35\sin\htmlData{tutor-start=4,tutor-end=10}{\alpha}\htmlData{tutor-start=10,tutor-end=11}{=}\frac{\htmlData{tutor-start=17,tutor-end=18}{3}}{\htmlData{tutor-start=20,tutor-end=21}{5}}，则 sin⁡2α=\sin \htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=12}{\alpha}\htmlData{tutor-start=12,tutor-end=13}{=} ( )

答案：−2425\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{4}}{\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{5}}

题目标签：第二象限角的二倍角

解题过程

先定余弦符号

求 sin⁡2α\sin\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=11}{\alpha}

(1)
识别结构并建立关系

第二象限余弦为负。

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。“第二象限余弦为负。”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：由 sin⁡α=3/5\sin\htmlData{tutor-start=4,tutor-end=10}{\alpha}\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}{5} 得 cos⁡α=−4/5\cos\htmlData{tutor-start=4,tutor-end=10}{\alpha}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{4}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{5}，所以 sin⁡2α=2sin⁡αcos⁡α=−24/25\sin\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=11}{\alpha}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{2}\sin\htmlData{tutor-start=17,tutor-end=23}{\alpha}\cos\htmlData{tutor-start=27,tutor-end=33}{\alpha}\htmlData{tutor-start=33,tutor-end=34}{=}\htmlData{tutor-start=34,tutor-end=35}{-}\htmlData{tutor-start=35,tutor-end=36}{2}\htmlData{tutor-start=36,tutor-end=37}{4}\htmlData{tutor-start=37,tutor-end=38}{/}\htmlData{tutor-start=38,tutor-end=39}{2}\htmlData{tutor-start=39,tutor-end=40}{5}。

sin⁡2α=−2425\sin\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=11}{\alpha}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{-}\frac{\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{4}}{\htmlData{tutor-start=23,tutor-end=24}{2}\htmlData{tutor-start=24,tutor-end=25}{5}}
(2)
完成计算并核验结论

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：负号来自象限条件。

−2425\boxed{\htmlData{tutor-start=7,tutor-end=8}{-}\frac{\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{4}}{\htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{5}}}
5

一、选择题 · 椭圆

椭圆的中心在原点，焦距为 4，一条准线为 x=−4\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}{4}，则该椭圆的方程为 ( )

答案：x28+y24=1\frac{\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{8}}\htmlData{tutor-start=15,tutor-end=16}{+}\frac{\htmlData{tutor-start=22,tutor-end=23}{y}^{\htmlData{tutor-start=25,tutor-end=26}{2}}}{\htmlData{tutor-start=29,tutor-end=30}{4}}\htmlData{tutor-start=31,tutor-end=32}{=}\htmlData{tutor-start=32,tutor-end=33}{1}

题目标签：由焦距和准线确定椭圆

解题过程

联立焦距与准线公式

求椭圆方程

(1)
识别结构并建立关系

焦距为 4 给出 c=2\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}，准线位置使用 x=±a2/c\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=6}{\pm }\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}{c}。

为什么从这里入手：目标是“识别结构并建立关系”，而“焦距为 4 给出 c=2\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}，准线位置使用 x=±a2/c\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=6}{\pm }\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}{c}。”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：a2/c=4\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{c}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{4}，故 a2=8\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{8}；再由 b2=a2−c2\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}{c}^{\htmlData{tutor-start=15,tutor-end=16}{2}} 得 b2=4\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}。

a2=8,b2=4\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{8}\htmlData{tutor-start=7,tutor-end=8}{,}\quad \htmlData{tutor-start=14,tutor-end=15}{b}^{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{4}
(2)
完成计算并核验结论

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：长轴在 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴，方程为 x2/8+y2/4=1\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{8}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{y}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\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}。

x28+y24=1\boxed{\frac{\htmlData{tutor-start=13,tutor-end=14}{x}^{\htmlData{tutor-start=16,tutor-end=17}{2}}}{\htmlData{tutor-start=20,tutor-end=21}{8}}\htmlData{tutor-start=22,tutor-end=23}{+}\frac{\htmlData{tutor-start=29,tutor-end=30}{y}^{\htmlData{tutor-start=32,tutor-end=33}{2}}}{\htmlData{tutor-start=36,tutor-end=37}{4}}\htmlData{tutor-start=38,tutor-end=39}{=}\htmlData{tutor-start=39,tutor-end=40}{1}}
6

一、选择题 · 数列

已知数列 {an}\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=9}{\}} 的前 n\htmlData{tutor-start=0,tutor-end=1}{n} 项和为 Sn\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}，a1=1\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}，Sn=2an+1\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{n}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{1}}，则 Sn=\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{=} ( )

答案：Sn=(32)n−1\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{(}\frac{\htmlData{tutor-start=13,tutor-end=14}{3}}{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{)}^{\htmlData{tutor-start=21,tutor-end=22}{n}\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{1}}

题目标签：由前项和关系求数列

解题过程

相邻前项和作差

求 Sn\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}

(1)
识别结构并建立关系

先由 n=1\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 求第二项，再对相邻关系作差。

为什么从这里入手：目标是“识别结构并建立关系”，而“先由 n=1\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 求第二项，再对相邻关系作差。”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：a2=1/2\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\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}；当 n≥2\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{2} 时，an=Sn−Sn−1=2an+1−2an\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}{S}_{\htmlData{tutor-start=9,tutor-end=10}{n}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{S}_{\htmlData{tutor-start=15,tutor-end=16}{n}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{1}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{2}\htmlData{tutor-start=21,tutor-end=22}{a}_{\htmlData{tutor-start=24,tutor-end=25}{n}\htmlData{tutor-start=25,tutor-end=26}{+}\htmlData{tutor-start=26,tutor-end=27}{1}}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{2}\htmlData{tutor-start=30,tutor-end=31}{a}_{\htmlData{tutor-start=33,tutor-end=34}{n}}，故 an+1=3an/2\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}{2}。

Sn=2an+1=(32)n−1\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{n}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{1}}\htmlData{tutor-start=14,tutor-end=15}{=}\left(\frac{\htmlData{tutor-start=27,tutor-end=28}{3}}{\htmlData{tutor-start=30,tutor-end=31}{2}}\right)^{\htmlData{tutor-start=41,tutor-end=42}{n}\htmlData{tutor-start=42,tutor-end=43}{-}\htmlData{tutor-start=43,tutor-end=44}{1}}
(2)
完成计算并核验结论

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：n=1\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 时同样成立。

Sn=(32)n−1\boxed{\htmlData{tutor-start=7,tutor-end=8}{S}_{\htmlData{tutor-start=10,tutor-end=11}{n}}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{(}\frac{\htmlData{tutor-start=20,tutor-end=21}{3}}{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{)}^{\htmlData{tutor-start=28,tutor-end=29}{n}\htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=31}{1}}}
7

一、选择题 · 排列组合

6 位选手依次演讲，其中选手甲不在第一个也不在最后一个演讲，则不同的演讲次序共有 ( )

答案：480\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{8}\htmlData{tutor-start=2,tutor-end=3}{0}

题目标签：有限制的全排列

解题过程

安排甲的位置再排其余人

求演讲次序数

(1)
识别结构并建立关系

甲只能选中间四个位置。

为什么从这里入手：目标是“识别结构并建立关系”，而“甲只能选中间四个位置。”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：甲的位置有 4 种，其余 5 人任意排列有 5!\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=2}{!} 种。

N=4⋅5!=480\htmlData{tutor-start=0,tutor-end=1}{N}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=8}{\cdot}\htmlData{tutor-start=8,tutor-end=9}{5}\htmlData{tutor-start=9,tutor-end=10}{!}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{4}\htmlData{tutor-start=12,tutor-end=13}{8}\htmlData{tutor-start=13,tutor-end=14}{0}
(2)
完成计算并核验结论

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：所以共有 480 种。

480\boxed{\htmlData{tutor-start=7,tutor-end=8}{4}\htmlData{tutor-start=8,tutor-end=9}{8}\htmlData{tutor-start=9,tutor-end=10}{0}}
8

一、选择题 · 空间向量

已知正四棱柱 ABCD−A1B1C1D1\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{D}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{A}_{\htmlData{tutor-start=8,tutor-end=9}{1}}\htmlData{tutor-start=10,tutor-end=11}{B}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{C}_{\htmlData{tutor-start=18,tutor-end=19}{1}}\htmlData{tutor-start=20,tutor-end=21}{D}_{\htmlData{tutor-start=23,tutor-end=24}{1}} 中，AB=2\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{2}，CC1=22\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{C}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{2}\sqrt{\htmlData{tutor-start=14,tutor-end=15}{2}}，E\htmlData{tutor-start=0,tutor-end=1}{E} 为 CC1\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{C}_{\htmlData{tutor-start=4,tutor-end=5}{1}} 的中点，则直线 AC1\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}_{\htmlData{tutor-start=4,tutor-end=5}{1}} 到平面 BED\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{E}\htmlData{tutor-start=2,tutor-end=3}{D} 的距离为 ( )

答案：D；1\htmlData{tutor-start=0,tutor-end=1}{1}

题目标签：平行直线到平面的距离

解题过程

建立坐标求平面方程

求直线到平面的距离

(1)
识别结构并建立关系

正四棱柱适合按底面边和侧棱建立直角坐标。

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。“正四棱柱适合按底面边和侧棱建立直角坐标。”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 A(0,0,0),B(2,0,0),D(0,2,0),C1(2,2,22),E(2,2,2)\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\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}{,}\htmlData{tutor-start=9,tutor-end=10}{B}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{,}\htmlData{tutor-start=13,tutor-end=14}{0}\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}{,}\htmlData{tutor-start=18,tutor-end=19}{D}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{0}\htmlData{tutor-start=21,tutor-end=22}{,}\htmlData{tutor-start=22,tutor-end=23}{2}\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}{C}_{\htmlData{tutor-start=30,tutor-end=31}{1}}\htmlData{tutor-start=32,tutor-end=33}{(}\htmlData{tutor-start=33,tutor-end=34}{2}\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}{2}\sqrt{\htmlData{tutor-start=44,tutor-end=45}{2}}\htmlData{tutor-start=46,tutor-end=47}{)}\htmlData{tutor-start=47,tutor-end=48}{,}\htmlData{tutor-start=48,tutor-end=49}{E}\htmlData{tutor-start=49,tutor-end=50}{(}\htmlData{tutor-start=50,tutor-end=51}{2}\htmlData{tutor-start=51,tutor-end=52}{,}\htmlData{tutor-start=52,tutor-end=53}{2}\htmlData{tutor-start=53,tutor-end=54}{,}\sqrt{\htmlData{tutor-start=60,tutor-end=61}{2}}\htmlData{tutor-start=62,tutor-end=63}{)}。平面 BED\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{E}\htmlData{tutor-start=2,tutor-end=3}{D} 方程为 x+y−2z=2\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}{-}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{z}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{2}；AC1\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}_{\htmlData{tutor-start=4,tutor-end=5}{1}} 方向与法向量数量积为零，故直线平行该平面。

d=∣0−2∣1+1+2=1\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\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}{2}\htmlData{tutor-start=12,tutor-end=13}{|}}{\sqrt{\htmlData{tutor-start=21,tutor-end=22}{1}\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}{2}}}\htmlData{tutor-start=28,tutor-end=29}{=}\htmlData{tutor-start=29,tutor-end=30}{1}
(2)
完成计算并核验结论

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：用直线上的点 A\htmlData{tutor-start=0,tutor-end=1}{A} 计算即得距离 1。

1\boxed{\htmlData{tutor-start=7,tutor-end=8}{1}}
9

一、选择题 · 平面向量

△ABC\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C} 中，AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 边的高为 CD\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D}，若 CB→=a\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{C}\htmlData{tutor-start=17,tutor-end=18}{B}}\htmlData{tutor-start=19,tutor-end=20}{=}\boldsymbol{\htmlData{tutor-start=32,tutor-end=33}{a}}，CA→=b\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{C}\htmlData{tutor-start=17,tutor-end=18}{A}}\htmlData{tutor-start=19,tutor-end=20}{=}\boldsymbol{\htmlData{tutor-start=32,tutor-end=33}{b}}，a⋅b=0\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}} \htmlData{tutor-start=15,tutor-end=21}{\cdot }\boldsymbol{\htmlData{tutor-start=33,tutor-end=34}{b}}\htmlData{tutor-start=35,tutor-end=36}{=}\htmlData{tutor-start=36,tutor-end=37}{0}，∣a∣=1\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{|}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{1}，∣b∣=2\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{b}}\htmlData{tutor-start=15,tutor-end=16}{|}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{2}，则 AD→=\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{D}}\htmlData{tutor-start=19,tutor-end=20}{=} ( )

答案：D；45(a−b)\frac{4}{5}(\boldsymbol a-\boldsymbol b)

题目标签：直角三角形高的向量表示

解题过程

设垂足在线段上的参数

求 AD→\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{D}}

(1)
识别结构并建立关系

以 C\htmlData{tutor-start=0,tutor-end=1}{C} 为原点，则 A\htmlData{tutor-start=0,tutor-end=1}{A}、B\htmlData{tutor-start=0,tutor-end=1}{B} 的位置向量分别是 b,a\boldsymbol b,\boldsymbol a。

为什么从这里入手：空间关系只靠观察容易漏条件，而“以 C\htmlData{tutor-start=0,tutor-end=1}{C} 为原点，则 A\htmlData{tutor-start=0,tutor-end=1}{A}、B\htmlData{tutor-start=0,tutor-end=1}{B} 的位置向量分别是 b,a\boldsymbol b,\boldsymbol a。”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：设 AD→=t(a−b)\overrightarrow{AD}=t(\boldsymbol a-\boldsymbol b)，则 D\htmlData{tutor-start=0,tutor-end=1}{D} 的位置向量为 b+t(a−b)\boldsymbol b+t(\boldsymbol a-\boldsymbol b)。由 CD⊥AB\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=8}{\perp }\htmlData{tutor-start=8,tutor-end=9}{A}\htmlData{tutor-start=9,tutor-end=10}{B} 得 −4+5t=0\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{4}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{5}\htmlData{tutor-start=4,tutor-end=5}{t}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{0}。

t=45\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{4}}{\htmlData{tutor-start=11,tutor-end=12}{5}}
(2)
完成计算并核验结论

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：所以 AD→=45(a−b)\overrightarrow{AD}=\frac{4}{5}(\boldsymbol a-\boldsymbol b)，选择 D。

45(a−b)\boxed{\frac{4}{5}(\boldsymbol a-\boldsymbol b)}
10

一、选择题 · 双曲线

已知 F1\htmlData{tutor-start=0,tutor-end=1}{F}_{\htmlData{tutor-start=3,tutor-end=4}{1}}、F2\htmlData{tutor-start=0,tutor-end=1}{F}_{\htmlData{tutor-start=3,tutor-end=4}{2}} 为双曲线 C:x2−y2=2\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=4}{x}^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{y}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{2} 的左、右焦点，点 P\htmlData{tutor-start=0,tutor-end=1}{P} 在 C\htmlData{tutor-start=0,tutor-end=1}{C} 上，∣PF1∣=2∣PF2∣\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{F}_{\htmlData{tutor-start=5,tutor-end=6}{1}}\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}{P}\htmlData{tutor-start=12,tutor-end=13}{F}_{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{|}，则 cos⁡∠F1PF2=\cos \htmlData{tutor-start=5,tutor-end=12}{\angle }\htmlData{tutor-start=12,tutor-end=13}{F}_{\htmlData{tutor-start=15,tutor-end=16}{1}}\htmlData{tutor-start=17,tutor-end=18}{P}\htmlData{tutor-start=18,tutor-end=19}{F}_{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{=} ( )

答案：C；34\frac{\htmlData{tutor-start=6,tutor-end=7}{3}}{\htmlData{tutor-start=9,tutor-end=10}{4}}

题目标签：双曲线焦点三角形

解题过程

用定义求两焦半径

求焦点夹角余弦

(1)
识别结构并建立关系

双曲线 a=2,c=2\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=11}{,}\htmlData{tutor-start=11,tutor-end=12}{c}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{2}，右支上两焦半径之差为 2a\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{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。“双曲线 a=2,c=2\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=11}{,}\htmlData{tutor-start=11,tutor-end=12}{c}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{2}，右支上两焦半径之差为 2a\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{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：设 PF2=d\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{F}_{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{d}，则 PF1=2d\htmlData{tutor-start=0,tutor-end=1}{P}\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}{2}\htmlData{tutor-start=8,tutor-end=9}{d} 且 2d−d=22\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{d}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{d}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{2}\sqrt{\htmlData{tutor-start=12,tutor-end=13}{2}}，所以两边为 22,42\htmlData{tutor-start=0,tutor-end=1}{2}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{4}\sqrt{\htmlData{tutor-start=17,tutor-end=18}{2}}；焦距 F1F2=4\htmlData{tutor-start=0,tutor-end=1}{F}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{F}_{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{4}。

cos⁡∠F1PF2=32+8−162⋅42⋅22=34\cos\htmlData{tutor-start=4,tutor-end=11}{\angle }\htmlData{tutor-start=11,tutor-end=12}{F}_{\htmlData{tutor-start=14,tutor-end=15}{1}}\htmlData{tutor-start=16,tutor-end=17}{P}\htmlData{tutor-start=17,tutor-end=18}{F}_{\htmlData{tutor-start=20,tutor-end=21}{2}}\htmlData{tutor-start=22,tutor-end=23}{=}\frac{\htmlData{tutor-start=29,tutor-end=30}{3}\htmlData{tutor-start=30,tutor-end=31}{2}\htmlData{tutor-start=31,tutor-end=32}{+}\htmlData{tutor-start=32,tutor-end=33}{8}\htmlData{tutor-start=33,tutor-end=34}{-}\htmlData{tutor-start=34,tutor-end=35}{1}\htmlData{tutor-start=35,tutor-end=36}{6}}{\htmlData{tutor-start=38,tutor-end=39}{2}\htmlData{tutor-start=39,tutor-end=44}{\cdot}\htmlData{tutor-start=44,tutor-end=45}{4}\sqrt{\htmlData{tutor-start=51,tutor-end=52}{2}}\htmlData{tutor-start=53,tutor-end=58}{\cdot}\htmlData{tutor-start=58,tutor-end=59}{2}\sqrt{\htmlData{tutor-start=65,tutor-end=66}{2}}}\htmlData{tutor-start=68,tutor-end=69}{=}\frac{\htmlData{tutor-start=75,tutor-end=76}{3}}{\htmlData{tutor-start=78,tutor-end=79}{4}}
(2)
完成计算并核验结论

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：故选择 C。

34\boxed{\frac{\htmlData{tutor-start=13,tutor-end=14}{3}}{\htmlData{tutor-start=16,tutor-end=17}{4}}}
11

一、选择题 · 函数估值

已知 x=ln⁡π\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\ln \htmlData{tutor-start=6,tutor-end=9}{\pi}，y=log⁡52\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\log_{\htmlData{tutor-start=8,tutor-end=9}{5}} \htmlData{tutor-start=11,tutor-end=12}{2}，z=e−12\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\text{\htmlData{tutor-start=8,tutor-end=9}{e}}^{\htmlData{tutor-start=12,tutor-end=13}{-}\frac{\htmlData{tutor-start=19,tutor-end=20}{1}}{\htmlData{tutor-start=22,tutor-end=23}{2}}}，则 ( )

答案：D；y<z<x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{z}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{x}

题目标签：对数指数大小比较

解题过程

用熟悉常数作界

比较三个数

(1)
识别结构并建立关系

分别估计三个正数所在的简单区间。

为什么从这里入手：目标是“识别结构并建立关系”，而“分别估计三个正数所在的简单区间。”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：x=ln⁡π>1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\ln\htmlData{tutor-start=5,tutor-end=8}{\pi}\htmlData{tutor-start=8,tutor-end=9}{>}\htmlData{tutor-start=9,tutor-end=10}{1}；y=log⁡52<1/2\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\log_{\htmlData{tutor-start=8,tutor-end=9}{5}}\htmlData{tutor-start=10,tutor-end=11}{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}{2}，因为 2<5\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{<}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{5}}；z=e−1/2=1/e\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{e}^{\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=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{/}\sqrt{\htmlData{tutor-start=19,tutor-end=20}{e}} 介于 1/2\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{2} 与 1 之间。

y<12<z<1<x\htmlData{tutor-start=0,tutor-end=1}{y}\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}{<}\htmlData{tutor-start=14,tutor-end=15}{z}\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}{x}
(2)
完成计算并核验结论

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：所以 y<z<x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{z}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{x}，选择 D。

y<z<x\boxed{\htmlData{tutor-start=7,tutor-end=8}{y}\htmlData{tutor-start=8,tutor-end=9}{<}\htmlData{tutor-start=9,tutor-end=10}{z}\htmlData{tutor-start=10,tutor-end=11}{<}\htmlData{tutor-start=11,tutor-end=12}{x}}
12

一、选择题 · 反射与周期

正方形 ABCD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{D} 的边长为 1，点 E\htmlData{tutor-start=0,tutor-end=1}{E} 在边 AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 上，点 F\htmlData{tutor-start=0,tutor-end=1}{F} 在边 BC\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C} 上，AE=BF=13\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}{B}\htmlData{tutor-start=4,tutor-end=5}{F}\htmlData{tutor-start=5,tutor-end=6}{=}\frac{\htmlData{tutor-start=12,tutor-end=13}{1}}{\htmlData{tutor-start=15,tutor-end=16}{3}}。动点 P\htmlData{tutor-start=0,tutor-end=1}{P} 从 E\htmlData{tutor-start=0,tutor-end=1}{E} 出发沿直线向 F\htmlData{tutor-start=0,tutor-end=1}{F} 运动，每当碰到正方形的边时反弹，反弹时反射角等于入射角，当点 P\htmlData{tutor-start=0,tutor-end=1}{P} 第一次碰到 E\htmlData{tutor-start=0,tutor-end=1}{E} 时，P\htmlData{tutor-start=0,tutor-end=1}{P} 与正方形的边碰撞的次数为 ( )

答案：6\htmlData{tutor-start=0,tutor-end=1}{6} 次

题目标签：正方形反射路径

解题过程

展开镜面为平铺直线

求首次返回的碰撞次数

(1)
识别结构并建立关系

把每次反射改为把正方形镜像展开，折线路径变为斜率 1/2\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{2} 的直线。

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。“把每次反射改为把正方形镜像展开，折线路径变为斜率 1/2\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{2} 的直线。”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：从 E=(1/3,0)\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}{1}\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}{0}\htmlData{tutor-start=8,tutor-end=9}{)} 出发，返回同名边界点的首个镜像为 (13/3,2)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{3}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{)}。途中穿过 4 条竖直网格线和 2 条水平网格线，且没有同时穿过格点。

N=4+2=6\htmlData{tutor-start=0,tutor-end=1}{N}\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}{2}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{6}
(2)
完成计算并核验结论

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：最后一次碰撞正是回到 E\htmlData{tutor-start=0,tutor-end=1}{E}，计入总数，故为 6 次。

6\boxed{\htmlData{tutor-start=7,tutor-end=8}{6}}
13

二、填空题 · 二项式定理

(x+12x)8\left(\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{+}\frac{\htmlData{tutor-start=14,tutor-end=15}{1}}{\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{x}}\right)^{\htmlData{tutor-start=29,tutor-end=30}{8}} 的展开式中 x2\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} 的系数为______。

答案：7\htmlData{tutor-start=0,tutor-end=1}{7}

题目标签：二项展开指定幂系数

解题过程

由指数方程确定项号

求 x2\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} 系数

(1)
识别结构并建立关系

通项选取第二项 k\htmlData{tutor-start=0,tutor-end=1}{k} 次时，x\htmlData{tutor-start=0,tutor-end=1}{x} 的指数为 8−2k\htmlData{tutor-start=0,tutor-end=1}{8}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{k}。

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。“通项选取第二项 k\htmlData{tutor-start=0,tutor-end=1}{k} 次时，x\htmlData{tutor-start=0,tutor-end=1}{x} 的指数为 8−2k\htmlData{tutor-start=0,tutor-end=1}{8}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{k}。”提示了差、比或可累加关系，先识别结构并建立关系能把第 n\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：令 8−2k=2\htmlData{tutor-start=0,tutor-end=1}{8}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{k}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{2} 得 k=3\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}，系数为 (83)(1/2)3=56/8\binom83(1/2)^{3}=56/8。

(83)23=7\frac{\binom83}{2^{3}}=7
(2)
完成计算并核验结论

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：所以系数为 7。

7\boxed{\htmlData{tutor-start=7,tutor-end=8}{7}}
14

二、填空题 · 线性规划

若 x\htmlData{tutor-start=0,tutor-end=1}{x}、y\htmlData{tutor-start=0,tutor-end=1}{y} 满足约束条件 {x−y+1⩾0,x+y−3⩽0,x+3y−3⩾0,\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{y}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{1} \htmlData{tutor-start=20,tutor-end=30}{\geqslant }\htmlData{tutor-start=30,tutor-end=31}{0}\htmlData{tutor-start=31,tutor-end=32}{,} \\ \htmlData{tutor-start=36,tutor-end=37}{x}\htmlData{tutor-start=37,tutor-end=38}{+}\htmlData{tutor-start=38,tutor-end=39}{y}\htmlData{tutor-start=39,tutor-end=40}{-}\htmlData{tutor-start=40,tutor-end=41}{3} \htmlData{tutor-start=42,tutor-end=52}{\leqslant }\htmlData{tutor-start=52,tutor-end=53}{0}\htmlData{tutor-start=53,tutor-end=54}{,} \\ \htmlData{tutor-start=58,tutor-end=59}{x}\htmlData{tutor-start=59,tutor-end=60}{+}\htmlData{tutor-start=60,tutor-end=61}{3}\htmlData{tutor-start=61,tutor-end=62}{y}\htmlData{tutor-start=62,tutor-end=63}{-}\htmlData{tutor-start=63,tutor-end=64}{3} \htmlData{tutor-start=65,tutor-end=75}{\geqslant }\htmlData{tutor-start=75,tutor-end=76}{0}\htmlData{tutor-start=76,tutor-end=77}{,} \end{cases} 则 z=3x−y\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{y} 的最小值为______。

答案：−1\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}

题目标签：三约束线性规划

解题过程

求可行域顶点并代入

求 z=3x−y\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{y} 最小值

(1)
识别结构并建立关系

三条边界直线两两相交构成三角形，线性函数的极值在顶点取得。

为什么从这里入手：函数的局部变化由导数控制。“三条边界直线两两相交构成三角形，线性函数的极值在顶点取得。”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：顶点为 (0,1),(1,2),(3,0)\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}{1}\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}{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}{,}\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}{0}\htmlData{tutor-start=16,tutor-end=17}{)}，目标值分别是 −1,1,9\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}{1}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{9}。

zmin⁡=−1\htmlData{tutor-start=0,tutor-end=1}{z}_{\min}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}
(2)
完成计算并核验结论

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：所以最小值在 (0,1)\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}{1}\htmlData{tutor-start=4,tutor-end=5}{)} 取得。

−1\boxed{\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}}
15

二、填空题 · 三角函数

当函数 y=sin⁡x−3cos⁡x (0⩽x<2π)\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\sin \htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{-}\sqrt{\htmlData{tutor-start=15,tutor-end=16}{3}}\cos \htmlData{tutor-start=22,tutor-end=23}{x} \htmlData{tutor-start=24,tutor-end=26}{\ }\htmlData{tutor-start=26,tutor-end=27}{(}\htmlData{tutor-start=27,tutor-end=28}{0} \htmlData{tutor-start=29,tutor-end=39}{\leqslant }\htmlData{tutor-start=39,tutor-end=40}{x} \htmlData{tutor-start=41,tutor-end=42}{<} \htmlData{tutor-start=43,tutor-end=44}{2}\htmlData{tutor-start=44,tutor-end=47}{\pi}\htmlData{tutor-start=47,tutor-end=48}{)} 取得最大值时，x=\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}______。

答案：x=5π6\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{5}\htmlData{tutor-start=9,tutor-end=12}{\pi}}{\htmlData{tutor-start=14,tutor-end=15}{6}}

题目标签：辅助角公式求最大点

解题过程

化成单一正弦

求最大值点

(1)
识别结构并建立关系

把两个同频三角函数合成为一个振幅为 2 的正弦。

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。“把两个同频三角函数合成为一个振幅为 2 的正弦。”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：sin⁡x−3cos⁡x=2sin⁡(x−π/3)\sin \htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{-}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}\cos \htmlData{tutor-start=20,tutor-end=21}{x}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{2}\sin\htmlData{tutor-start=27,tutor-end=28}{(}\htmlData{tutor-start=28,tutor-end=29}{x}\htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=33}{\pi}\htmlData{tutor-start=33,tutor-end=34}{/}\htmlData{tutor-start=34,tutor-end=35}{3}\htmlData{tutor-start=35,tutor-end=36}{)}，最大值要求 x−π/3=π/2+2kπ\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}{3}\htmlData{tutor-start=7,tutor-end=8}{=}\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}{+}\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{k}\htmlData{tutor-start=16,tutor-end=19}{\pi}。

x=5π6+2kπ\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{5}\htmlData{tutor-start=9,tutor-end=12}{\pi}}{\htmlData{tutor-start=14,tutor-end=15}{6}}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{k}\htmlData{tutor-start=19,tutor-end=22}{\pi}
(2)
完成计算并核验结论

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：结合 0≤x<2π\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=5}{\le }\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=11}{\pi}，唯一解为 5π/6\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=4}{\pi}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{6}。

5π6\boxed{\frac{\htmlData{tutor-start=13,tutor-end=14}{5}\htmlData{tutor-start=14,tutor-end=17}{\pi}}{\htmlData{tutor-start=19,tutor-end=20}{6}}}
16

二、填空题 · 空间向量

已知正方体 ABCD−A1B1C1D1\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{D}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{A}_{\htmlData{tutor-start=8,tutor-end=9}{1}}\htmlData{tutor-start=10,tutor-end=11}{B}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{C}_{\htmlData{tutor-start=18,tutor-end=19}{1}}\htmlData{tutor-start=20,tutor-end=21}{D}_{\htmlData{tutor-start=23,tutor-end=24}{1}} 中，E\htmlData{tutor-start=0,tutor-end=1}{E}、F\htmlData{tutor-start=0,tutor-end=1}{F} 分别为 BB1\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{B}_{\htmlData{tutor-start=4,tutor-end=5}{1}}、CC1\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{C}_{\htmlData{tutor-start=4,tutor-end=5}{1}} 的中点，那么异面直线 AE\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{E} 与 D1F\htmlData{tutor-start=0,tutor-end=1}{D}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{F} 所成角的余弦值为______。

答案：35\frac{\htmlData{tutor-start=6,tutor-end=7}{3}}{\htmlData{tutor-start=9,tutor-end=10}{5}}

题目标签：正方体异面直线夹角

解题过程

用方向向量数量积

求异面直线夹角余弦

(1)
识别结构并建立关系

令正方体棱长为 1，建立标准坐标。

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。“令正方体棱长为 1，建立标准坐标。”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：可取 AE→=(1,0,1/2)\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{E}}\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}{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}{2}\htmlData{tutor-start=28,tutor-end=29}{)}，D1F→=(1,0,−1/2)\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{D}_{\htmlData{tutor-start=19,tutor-end=20}{1}}\htmlData{tutor-start=21,tutor-end=22}{F}}\htmlData{tutor-start=23,tutor-end=24}{=}\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}{,}\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}\htmlData{tutor-start=33,tutor-end=34}{)}。数量积为 3/4\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{4}，两模均为 5/2\sqrt{\htmlData{tutor-start=6,tutor-end=7}{5}}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{2}。

cos⁡theta=3/45/4=35\cos\\\htmlData{tutor-start=6,tutor-end=7}{t}\htmlData{tutor-start=7,tutor-end=8}{h}\htmlData{tutor-start=8,tutor-end=9}{e}\htmlData{tutor-start=9,tutor-end=10}{t}\htmlData{tutor-start=10,tutor-end=11}{a}\htmlData{tutor-start=11,tutor-end=12}{=}\frac{\htmlData{tutor-start=18,tutor-end=19}{3}\htmlData{tutor-start=19,tutor-end=20}{/}\htmlData{tutor-start=20,tutor-end=21}{4}}{\htmlData{tutor-start=23,tutor-end=24}{5}\htmlData{tutor-start=24,tutor-end=25}{/}\htmlData{tutor-start=25,tutor-end=26}{4}}\htmlData{tutor-start=27,tutor-end=28}{=}\frac{\htmlData{tutor-start=34,tutor-end=35}{3}}{\htmlData{tutor-start=37,tutor-end=38}{5}}
(2)
完成计算并核验结论

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：异面直线角取锐角，所以答案为 3/5\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{5}。

35\boxed{\frac{\htmlData{tutor-start=13,tutor-end=14}{3}}{\htmlData{tutor-start=16,tutor-end=17}{5}}}
17

三、解答题 · 解三角形

△ABC\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C} 的内角 A\htmlData{tutor-start=0,tutor-end=1}{A}、B\htmlData{tutor-start=0,tutor-end=1}{B}、C\htmlData{tutor-start=0,tutor-end=1}{C} 成等差数列，其对边 a\htmlData{tutor-start=0,tutor-end=1}{a}、b\htmlData{tutor-start=0,tutor-end=1}{b}、c\htmlData{tutor-start=0,tutor-end=1}{c} 满足 2b2=3ac\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{b}^{\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}{a}\htmlData{tutor-start=9,tutor-end=10}{c}，求 A\htmlData{tutor-start=0,tutor-end=1}{A}。

答案：A=π6A=\frac\pi6 或 π2\frac\pi2

题目标签：角成等差的三角形

解题过程

先确定中间角再用正弦定理

求角 A\htmlData{tutor-start=0,tutor-end=1}{A}

(1)
识别结构并建立关系

三个角成等差且和为 π\htmlData{tutor-start=0,tutor-end=3}{\pi}，故中间角 B=π/3\htmlData{tutor-start=0,tutor-end=1}{B}\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}{3}。

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。“三个角成等差且和为 π\htmlData{tutor-start=0,tutor-end=3}{\pi}，故中间角 B=π/3\htmlData{tutor-start=0,tutor-end=1}{B}\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}{3}。”提示了差、比或可累加关系，先识别结构并建立关系能把第 n\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：由正弦定理把 2b2=3ac\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{b}^{\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}{a}\htmlData{tutor-start=9,tutor-end=10}{c} 化为 2sin⁡2B=3sin⁡Asin⁡C\htmlData{tutor-start=0,tutor-end=1}{2}\sin^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{B}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{3}\sin \htmlData{tutor-start=17,tutor-end=18}{A}\sin \htmlData{tutor-start=23,tutor-end=24}{C}，得到 sin⁡Asin⁡C=1/2\sin \htmlData{tutor-start=5,tutor-end=6}{A}\sin \htmlData{tutor-start=11,tutor-end=12}{C}\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}{2}；又 A+C=2π/3\htmlData{tutor-start=0,tutor-end=1}{A}\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}{2}\htmlData{tutor-start=5,tutor-end=8}{\pi}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{3}，从而 cos⁡(A−C)=1/2\cos\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{C}\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}{2}。

∣A−C∣=π3|A-C|=\frac\pi3
(2)
完成计算并核验结论

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：联立得到 (A,C)=(π/6,π/2)\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}{C}\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=10}{\pi}\htmlData{tutor-start=10,tutor-end=11}{/}\htmlData{tutor-start=11,tutor-end=12}{6}\htmlData{tutor-start=12,tutor-end=13}{,}\htmlData{tutor-start=13,tutor-end=16}{\pi}\htmlData{tutor-start=16,tutor-end=17}{/}\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{)} 或互换，所以 A\htmlData{tutor-start=0,tutor-end=1}{A} 有两个可能值。

A=π6 或 π2\boxed{A=\frac\pi6\text{ 或 }\frac\pi2}
18

三、解答题 · 数列

已知数列 {an}\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=9}{\}} 中，a1=1\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}，前 n\htmlData{tutor-start=0,tutor-end=1}{n} 项和 Sn=n+23an\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{=}\frac{\htmlData{tutor-start=12,tutor-end=13}{n}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{2}}{\htmlData{tutor-start=17,tutor-end=18}{3}}\htmlData{tutor-start=19,tutor-end=20}{a}_{\htmlData{tutor-start=22,tutor-end=23}{n}}。 (1) 求 a2,a3\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{3}}； (2) 求 {an}\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=9}{\}} 的通项公式。

答案：a2=3,a3=6\htmlData{tutor-start=0,tutor-end=1}{a}_{\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}{a}_{\htmlData{tutor-start=11,tutor-end=12}{3}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{6}；an=n(n+1)2a_{n}=\frac{n(n+1)}2

题目标签：前项和与通项的比例递推

解题过程

（1）直接代入小下标

求 a2,a3\htmlData{tutor-start=0,tutor-end=1}{a}_{\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}{3}}

(1)
识别结构并建立关系

用 Sn=a1+⋯+an\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{1}}\htmlData{tutor-start=11,tutor-end=12}{+}\cdots\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{a}_{\htmlData{tutor-start=22,tutor-end=23}{n}} 分别代入 n=2,3\htmlData{tutor-start=0,tutor-end=1}{n}\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}{3}。

为什么从这里入手：代数题要寻找能减少未知量或降低次数的关系。“用 Sn=a1+⋯+an\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{1}}\htmlData{tutor-start=11,tutor-end=12}{+}\cdots\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{a}_{\htmlData{tutor-start=22,tutor-end=23}{n}} 分别代入 n=2,3\htmlData{tutor-start=0,tutor-end=1}{n}\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}{3}。”正好提供了因式、根或参数之间的等式，所以先识别结构并建立关系，再代入、消元或比较系数，计算链条会更短也更容易复核。

详细展开：1+a2=4a2/3\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}{2}}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{4}\htmlData{tutor-start=9,tutor-end=10}{a}_{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{/}\htmlData{tutor-start=15,tutor-end=16}{3} 得 a2=3\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{3}；1+3+a3=5a3/3\htmlData{tutor-start=0,tutor-end=1}{1}\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}{a}_{\htmlData{tutor-start=7,tutor-end=8}{3}}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{5}\htmlData{tutor-start=11,tutor-end=12}{a}_{\htmlData{tutor-start=14,tutor-end=15}{3}}\htmlData{tutor-start=16,tutor-end=17}{/}\htmlData{tutor-start=17,tutor-end=18}{3} 得 a3=6\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}{6}。

a2=3,a3=6\htmlData{tutor-start=0,tutor-end=1}{a}_{\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}{,}\quad \htmlData{tutor-start=14,tutor-end=15}{a}_{\htmlData{tutor-start=17,tutor-end=18}{3}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{6}
(2)
完成计算并核验结论

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：两项均满足原比例。

a2=3,a3=6\boxed{\htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{2}}\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}{a}_{\htmlData{tutor-start=18,tutor-end=19}{3}}\htmlData{tutor-start=20,tutor-end=21}{=}\htmlData{tutor-start=21,tutor-end=22}{6}}

（2）相邻前项和相减

求通项公式

(1)
识别结构并建立关系

把 an=Sn−Sn−1\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}{S}_{\htmlData{tutor-start=9,tutor-end=10}{n}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{S}_{\htmlData{tutor-start=15,tutor-end=16}{n}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{1}} 代入题设关系。

为什么从这里入手：代数题要寻找能减少未知量或降低次数的关系。“把 an=Sn−Sn−1\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}{S}_{\htmlData{tutor-start=9,tutor-end=10}{n}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{S}_{\htmlData{tutor-start=15,tutor-end=16}{n}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{1}} 代入题设关系。”正好提供了因式、根或参数之间的等式，所以先识别结构并建立关系，再代入、消元或比较系数，计算链条会更短也更容易复核。

详细展开：整理得 (n−1)an=(n+1)an−1\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{n}\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}{a}_{\htmlData{tutor-start=8,tutor-end=9}{n}}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{n}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{)}\htmlData{tutor-start=16,tutor-end=17}{a}_{\htmlData{tutor-start=19,tutor-end=20}{n}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{1}}，故 an/an−1=(n+1)/(n−1)\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}{a}_{\htmlData{tutor-start=9,tutor-end=10}{n}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{n}\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}{(}\htmlData{tutor-start=21,tutor-end=22}{n}\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{)}。连乘并使用 a1=1\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}。

an=∏k=2nk+1k−1=n(n+1)2a_{n}=\prod_{k=2}^{n}\frac{k+1}{k-1}=\frac{n(n+1)}2
(2)
完成计算并核验结论

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：公式对 n=1\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 也成立。

an=n(n+1)2\boxed{a_{n}=\frac{n(n+1)}2}
19

三、解答题 · 立体几何

如图，四棱锥 P−ABCD\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}{B}\htmlData{tutor-start=4,tutor-end=5}{C}\htmlData{tutor-start=5,tutor-end=6}{D} 中，底面 ABCD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{D} 为菱形，PA⊥\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{A} \htmlData{tutor-start=3,tutor-end=8}{\perp} 底面 ABCD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{D}，AC=22\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}{2}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{2}}，PA=2\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{2}，E\htmlData{tutor-start=0,tutor-end=1}{E} 是 PC\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{C} 上的一点，PE=2EC\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{E}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{E}\htmlData{tutor-start=5,tutor-end=6}{C}。 (1) 证明：PC⊥\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{C} \htmlData{tutor-start=3,tutor-end=8}{\perp} 平面 BED\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{E}\htmlData{tutor-start=2,tutor-end=3}{D}； (2) 设二面角 A−PB−C\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{P}\htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{C} 为 90∘90^\circ，求 PD\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{D} 与平面 PBC\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C} 所成角的大小。

原卷图示 1
原卷图示 1原卷第 1 页 · qwen3.7-plus_layout_detection · 需复核

答案：PC⊥\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=7}{\perp} 平面 BED\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{E}\htmlData{tutor-start=2,tutor-end=3}{D}；线面角 30∘30^\circ

题目标签：菱形四棱锥的垂直与线面角

解题过程

（1）利用菱形对角线和分点

证明 PC⊥\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=7}{\perp} 平面 BED\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{E}\htmlData{tutor-start=2,tutor-end=3}{D}

(1)
识别结构并建立关系

菱形对角线 AC⊥BD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=8}{\perp }\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{D}，且 PA⊥\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=7}{\perp} 底面，所以 PC\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{C} 与 BD\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{D} 垂直。

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。“菱形对角线 AC⊥BD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=8}{\perp }\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{D}，且 PA⊥\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=7}{\perp} 底面，所以 PC\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{C} 与 BD\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{D} 垂直。”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：取 A(−2,0,0),C(2,0,0),B(0,b,0),D(0,−b,0),P(−2,0,2)\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\sqrt{\htmlData{tutor-start=9,tutor-end=10}{2}}\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}{,}\htmlData{tutor-start=17,tutor-end=18}{C}\htmlData{tutor-start=18,tutor-end=19}{(}\sqrt{\htmlData{tutor-start=25,tutor-end=26}{2}}\htmlData{tutor-start=27,tutor-end=28}{,}\htmlData{tutor-start=28,tutor-end=29}{0}\htmlData{tutor-start=29,tutor-end=30}{,}\htmlData{tutor-start=30,tutor-end=31}{0}\htmlData{tutor-start=31,tutor-end=32}{)}\htmlData{tutor-start=32,tutor-end=33}{,}\htmlData{tutor-start=33,tutor-end=34}{B}\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}{b}\htmlData{tutor-start=38,tutor-end=39}{,}\htmlData{tutor-start=39,tutor-end=40}{0}\htmlData{tutor-start=40,tutor-end=41}{)}\htmlData{tutor-start=41,tutor-end=42}{,}\htmlData{tutor-start=42,tutor-end=43}{D}\htmlData{tutor-start=43,tutor-end=44}{(}\htmlData{tutor-start=44,tutor-end=45}{0}\htmlData{tutor-start=45,tutor-end=46}{,}\htmlData{tutor-start=46,tutor-end=47}{-}\htmlData{tutor-start=47,tutor-end=48}{b}\htmlData{tutor-start=48,tutor-end=49}{,}\htmlData{tutor-start=49,tutor-end=50}{0}\htmlData{tutor-start=50,tutor-end=51}{)}\htmlData{tutor-start=51,tutor-end=52}{,}\htmlData{tutor-start=52,tutor-end=53}{P}\htmlData{tutor-start=53,tutor-end=54}{(}\htmlData{tutor-start=54,tutor-end=55}{-}\sqrt{\htmlData{tutor-start=61,tutor-end=62}{2}}\htmlData{tutor-start=63,tutor-end=64}{,}\htmlData{tutor-start=64,tutor-end=65}{0}\htmlData{tutor-start=65,tutor-end=66}{,}\htmlData{tutor-start=66,tutor-end=67}{2}\htmlData{tutor-start=67,tutor-end=68}{)}。由 PE:EC=2:1\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{E}\htmlData{tutor-start=2,tutor-end=3}{:}\htmlData{tutor-start=3,tutor-end=4}{E}\htmlData{tutor-start=4,tutor-end=5}{C}\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}{1} 得 E(2/3,0,2/3)\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{(}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{2}}\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}{0}\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}{3}\htmlData{tutor-start=18,tutor-end=19}{)}；计算可得 PC→⋅BE→=0\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{P}\htmlData{tutor-start=17,tutor-end=18}{C}}\htmlData{tutor-start=19,tutor-end=24}{\cdot}\overrightarrow{\htmlData{tutor-start=40,tutor-end=41}{B}\htmlData{tutor-start=41,tutor-end=42}{E}}\htmlData{tutor-start=43,tutor-end=44}{=}\htmlData{tutor-start=44,tutor-end=45}{0}，且 PC⊥BD\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=8}{\perp }\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{D}。

PC\perp BE,\quad PC\perp BD

(2)
完成计算并核验结论

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：BE,BD\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{E}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{D} 在平面内相交，故 PC⊥\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=7}{\perp} 平面 BED\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{E}\htmlData{tutor-start=2,tutor-end=3}{D}。

PC⊥平面 BED\boxed{\htmlData{tutor-start=7,tutor-end=8}{P}\htmlData{tutor-start=8,tutor-end=9}{C}\htmlData{tutor-start=9,tutor-end=14}{\perp}\text{\htmlData{tutor-start=20,tutor-end=21}{平}\htmlData{tutor-start=21,tutor-end=22}{面} }\htmlData{tutor-start=24,tutor-end=25}{B}\htmlData{tutor-start=25,tutor-end=26}{E}\htmlData{tutor-start=26,tutor-end=27}{D}}

（2）用二面角条件定短对角线

求 PD\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{D} 与平面 PBC\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C} 所成角

(1)
识别结构并建立关系

把二面角垂直转成两个平面法向量垂直，可先求出参数 b\htmlData{tutor-start=0,tutor-end=1}{b}。

为什么从这里入手：空间关系只靠观察容易漏条件，而“把二面角垂直转成两个平面法向量垂直，可先求出参数 b\htmlData{tutor-start=0,tutor-end=1}{b}。”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：平面 APB,CPB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{C}\htmlData{tutor-start=5,tutor-end=6}{P}\htmlData{tutor-start=6,tutor-end=7}{B} 法向量数量积为零给出 b=2\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{2}}。此时 PD→=(2,−2,−2)\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{P}\htmlData{tutor-start=17,tutor-end=18}{D}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{(}\sqrt{\htmlData{tutor-start=27,tutor-end=28}{2}}\htmlData{tutor-start=29,tutor-end=30}{,}\htmlData{tutor-start=30,tutor-end=31}{-}\sqrt{\htmlData{tutor-start=37,tutor-end=38}{2}}\htmlData{tutor-start=39,tutor-end=40}{,}\htmlData{tutor-start=40,tutor-end=41}{-}\htmlData{tutor-start=41,tutor-end=42}{2}\htmlData{tutor-start=42,tutor-end=43}{)}，平面 PBC\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C} 法向量可取 (2,2,2)\htmlData{tutor-start=0,tutor-end=1}{(}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{,}\sqrt{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{,}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{)}。

sin⁡theta=∣PD→⋅n∣∣PD∣∣n∣=12\sin\\theta=\frac{|\overrightarrow{PD}\cdot\boldsymbol n|}{|PD||\boldsymbol n|}=\frac{1}{2}
(2)
完成计算并核验结论

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：线面角取锐角，故 θ=30∘\theta=30^\circ。

30∘\boxed{30^\circ}
20

解答题 · 相互独立事件

乒乓球比赛规则规定：一局比赛，双方比分在10平前，一方连续发球2次后，对方再连续发球2次，依次轮换. 每次发球，胜方得1分，负方得0分. 设在甲、乙的比赛中，每次发球，发球方得1分的概率为0.6，各次发球的胜负结果相互独立. 甲、乙的一局比赛中，甲先发球. (1) 求开始第4次发球时，甲、乙的比分为1比2的概率； (2) 求开始第5次发球时，甲得分领先的概率.

答案：44125\frac{\htmlData{tutor-start=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{4}}{\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{5}}；192625\frac{\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{9}\htmlData{tutor-start=8,tutor-end=9}{2}}{\htmlData{tutor-start=11,tutor-end=12}{6}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{5}}

题目标签：轮换发球的领先概率

解题过程

（1）按前三球胜者枚举

求开始第 4 次发球时比分为 1:2\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{:}\htmlData{tutor-start=2,tutor-end=3}{2} 的概率

(1)
识别结构并建立关系

前三次中前两次由甲发球，甲得分概率均为 0.6；第三次由乙发球，甲得分概率为 0.4。

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。“前三次中前两次由甲发球，甲得分概率均为 0.6；第三次由乙发球，甲得分概率为 0.4。”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：甲恰得 1 分的三种互斥情形概率相加为 0.6⋅0.4⋅0.6+0.4⋅0.6⋅0.6+0.4⋅0.4⋅0.4=0.352\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{6}\htmlData{tutor-start=3,tutor-end=8}{\cdot}\htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{.}\htmlData{tutor-start=10,tutor-end=11}{4}\htmlData{tutor-start=11,tutor-end=16}{\cdot}\htmlData{tutor-start=16,tutor-end=17}{0}\htmlData{tutor-start=17,tutor-end=18}{.}\htmlData{tutor-start=18,tutor-end=19}{6}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{0}\htmlData{tutor-start=21,tutor-end=22}{.}\htmlData{tutor-start=22,tutor-end=23}{4}\htmlData{tutor-start=23,tutor-end=28}{\cdot}\htmlData{tutor-start=28,tutor-end=29}{0}\htmlData{tutor-start=29,tutor-end=30}{.}\htmlData{tutor-start=30,tutor-end=31}{6}\htmlData{tutor-start=31,tutor-end=36}{\cdot}\htmlData{tutor-start=36,tutor-end=37}{0}\htmlData{tutor-start=37,tutor-end=38}{.}\htmlData{tutor-start=38,tutor-end=39}{6}\htmlData{tutor-start=39,tutor-end=40}{+}\htmlData{tutor-start=40,tutor-end=41}{0}\htmlData{tutor-start=41,tutor-end=42}{.}\htmlData{tutor-start=42,tutor-end=43}{4}\htmlData{tutor-start=43,tutor-end=48}{\cdot}\htmlData{tutor-start=48,tutor-end=49}{0}\htmlData{tutor-start=49,tutor-end=50}{.}\htmlData{tutor-start=50,tutor-end=51}{4}\htmlData{tutor-start=51,tutor-end=56}{\cdot}\htmlData{tutor-start=56,tutor-end=57}{0}\htmlData{tutor-start=57,tutor-end=58}{.}\htmlData{tutor-start=58,tutor-end=59}{4}\htmlData{tutor-start=59,tutor-end=60}{=}\htmlData{tutor-start=60,tutor-end=61}{0}\htmlData{tutor-start=61,tutor-end=62}{.}\htmlData{tutor-start=62,tutor-end=63}{3}\htmlData{tutor-start=63,tutor-end=64}{5}\htmlData{tutor-start=64,tutor-end=65}{2}。

P=0.352=44125\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{.}\htmlData{tutor-start=4,tutor-end=5}{3}\htmlData{tutor-start=5,tutor-end=6}{5}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{=}\frac{\htmlData{tutor-start=14,tutor-end=15}{4}\htmlData{tutor-start=15,tutor-end=16}{4}}{\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{5}}
(2)
完成计算并核验结论

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：所以第一问概率为 44/125\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{4}\htmlData{tutor-start=2,tutor-end=3}{/}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{5}。

44125\boxed{\frac{\htmlData{tutor-start=13,tutor-end=14}{4}\htmlData{tutor-start=14,tutor-end=15}{4}}{\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{5}}}

（2）按四球中甲胜三次或四次

求开始第 5 次发球时甲领先概率

(1)
识别结构并建立关系

前两球甲发球，甲胜率 0.6；后两球乙发球，甲胜率 0.4。

为什么从这里入手：目标是“识别结构并建立关系”，而“前两球甲发球，甲胜率 0.6；后两球乙发球，甲胜率 0.4。”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：甲四战全胜概率为 0.620.42\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{6}^{\htmlData{tutor-start=5,tutor-end=6}{2}} \htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{.}\htmlData{tutor-start=10,tutor-end=11}{4}^{\htmlData{tutor-start=13,tutor-end=14}{2}}。恰胜三球时，分别漏掉一个甲发球局或一个乙发球局，概率和为 2(0.4⋅0.6⋅0.42)+2(0.62⋅0.6⋅0.4)\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{.}\htmlData{tutor-start=4,tutor-end=5}{4}\htmlData{tutor-start=5,tutor-end=10}{\cdot}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{.}\htmlData{tutor-start=12,tutor-end=13}{6}\htmlData{tutor-start=13,tutor-end=18}{\cdot}\htmlData{tutor-start=18,tutor-end=19}{0}\htmlData{tutor-start=19,tutor-end=20}{.}\htmlData{tutor-start=20,tutor-end=21}{4}^{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{)}\htmlData{tutor-start=26,tutor-end=27}{+}\htmlData{tutor-start=27,tutor-end=28}{2}\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}^{\htmlData{tutor-start=34,tutor-end=35}{2}}\htmlData{tutor-start=36,tutor-end=41}{\cdot}\htmlData{tutor-start=41,tutor-end=42}{0}\htmlData{tutor-start=42,tutor-end=43}{.}\htmlData{tutor-start=43,tutor-end=44}{6}\htmlData{tutor-start=44,tutor-end=49}{\cdot}\htmlData{tutor-start=49,tutor-end=50}{0}\htmlData{tutor-start=50,tutor-end=51}{.}\htmlData{tutor-start=51,tutor-end=52}{4}\htmlData{tutor-start=52,tutor-end=53}{)}。

P=0.3072=192625\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{.}\htmlData{tutor-start=4,tutor-end=5}{3}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{7}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{=}\frac{\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{9}\htmlData{tutor-start=17,tutor-end=18}{2}}{\htmlData{tutor-start=20,tutor-end=21}{6}\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{5}}
(2)
完成计算并核验结论

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：四球后领先等价于至少赢三球，故结果如上。

192625\boxed{\frac{\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{9}\htmlData{tutor-start=15,tutor-end=16}{2}}{\htmlData{tutor-start=18,tutor-end=19}{6}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{5}}}
21

解答题 · 导数

已知函数 f(x)=13x3+x2+ax\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=5,tutor-end=6}{=} \frac{\htmlData{tutor-start=13,tutor-end=14}{1}}{\htmlData{tutor-start=16,tutor-end=17}{3}}\htmlData{tutor-start=18,tutor-end=19}{x}^{\htmlData{tutor-start=21,tutor-end=22}{3}} \htmlData{tutor-start=24,tutor-end=25}{+} \htmlData{tutor-start=26,tutor-end=27}{x}^{\htmlData{tutor-start=29,tutor-end=30}{2}} \htmlData{tutor-start=32,tutor-end=33}{+} \htmlData{tutor-start=34,tutor-end=35}{a}\htmlData{tutor-start=35,tutor-end=36}{x}. (1) 讨论 f(x)\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}{)} 的单调性； (2) 设 f(x)\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}{)} 有两个极值点 x1,x2\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{x}_{\htmlData{tutor-start=10,tutor-end=11}{2}}，若过两点 (x1,f(x1)),(x2,f(x2))\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{f}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{x}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{)}\htmlData{tutor-start=16,tutor-end=17}{)}\htmlData{tutor-start=17,tutor-end=18}{,} \htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{x}_{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{,} \htmlData{tutor-start=27,tutor-end=28}{f}\htmlData{tutor-start=28,tutor-end=29}{(}\htmlData{tutor-start=29,tutor-end=30}{x}_{\htmlData{tutor-start=32,tutor-end=33}{2}}\htmlData{tutor-start=34,tutor-end=35}{)}\htmlData{tutor-start=35,tutor-end=36}{)} 的直线 l\htmlData{tutor-start=0,tutor-end=1}{l} 与 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴的交点在曲线 f(x)\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}{)} 上，求 a\htmlData{tutor-start=0,tutor-end=1}{a} 的值.

答案：单调性见过程；a=0,23,34\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\frac{\htmlData{tutor-start=10,tutor-end=11}{2}}{\htmlData{tutor-start=13,tutor-end=14}{3}}\htmlData{tutor-start=15,tutor-end=16}{,}\frac{\htmlData{tutor-start=22,tutor-end=23}{3}}{\htmlData{tutor-start=25,tutor-end=26}{4}}

题目标签：含参三次函数极值弦

解题过程

（1）按导数判别式分类

讨论单调性

(1)
识别结构并建立关系

f′(x)=x2+2x+a=(x+1)2+a−1\htmlData{tutor-start=0,tutor-end=1}{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=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{a}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{x}\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=24,tutor-end=25}{2}}\htmlData{tutor-start=26,tutor-end=27}{+}\htmlData{tutor-start=27,tutor-end=28}{a}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{1}。

为什么从这里入手：目标是“识别结构并建立关系”，而“f′(x)=x2+2x+a=(x+1)2+a−1\htmlData{tutor-start=0,tutor-end=1}{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=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{a}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{x}\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=24,tutor-end=25}{2}}\htmlData{tutor-start=26,tutor-end=27}{+}\htmlData{tutor-start=27,tutor-end=28}{a}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{1}。”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：若 a≥1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{1}，导数非负，函数在实数上递增；若 a<1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{1}，令 x1,2=−1∓1−a\htmlData{tutor-start=0,tutor-end=1}{x}_{\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=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=13}{\mp}\sqrt{\htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{a}}，函数在两根外递增、两根之间递减。

a<1:↑(−∞,x1),↓(x1,x2),↑(x2,+∞)\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=12}{\uparrow}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=20}{\infty}\htmlData{tutor-start=20,tutor-end=21}{,}\htmlData{tutor-start=21,tutor-end=22}{x}_{\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=38}{\downarrow}\htmlData{tutor-start=38,tutor-end=39}{(}\htmlData{tutor-start=39,tutor-end=40}{x}_{\htmlData{tutor-start=42,tutor-end=43}{1}}\htmlData{tutor-start=44,tutor-end=45}{,}\htmlData{tutor-start=45,tutor-end=46}{x}_{\htmlData{tutor-start=48,tutor-end=49}{2}}\htmlData{tutor-start=50,tutor-end=51}{)}\htmlData{tutor-start=51,tutor-end=52}{,}\htmlData{tutor-start=52,tutor-end=60}{\uparrow}\htmlData{tutor-start=60,tutor-end=61}{(}\htmlData{tutor-start=61,tutor-end=62}{x}_{\htmlData{tutor-start=64,tutor-end=65}{2}}\htmlData{tutor-start=66,tutor-end=67}{,}\htmlData{tutor-start=67,tutor-end=68}{+}\htmlData{tutor-start=68,tutor-end=74}{\infty}\htmlData{tutor-start=74,tutor-end=75}{)}
(2)
完成计算并核验结论

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：这也说明有两个极值点当且仅当 a<1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{1}。

单调性如上\boxed{\text{\htmlData{tutor-start=13,tutor-end=14}{单}\htmlData{tutor-start=14,tutor-end=15}{调}\htmlData{tutor-start=15,tutor-end=16}{性}\htmlData{tutor-start=16,tutor-end=17}{如}\htmlData{tutor-start=17,tutor-end=18}{上}}}

（2）平移变量求极值点弦方程

求参数 a\htmlData{tutor-start=0,tutor-end=1}{a}

(1)
识别结构并建立关系

令 u=x+1,s2=1−a\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}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{s}^{\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}{a}，三次式写成 f=13u3−s2u+23−a\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{3}}\htmlData{tutor-start=13,tutor-end=14}{u}^{\htmlData{tutor-start=16,tutor-end=17}{3}}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{s}^{\htmlData{tutor-start=22,tutor-end=23}{2}}\htmlData{tutor-start=24,tutor-end=25}{u}\htmlData{tutor-start=25,tutor-end=26}{+}\frac{\htmlData{tutor-start=32,tutor-end=33}{2}}{\htmlData{tutor-start=35,tutor-end=36}{3}}\htmlData{tutor-start=37,tutor-end=38}{-}\htmlData{tutor-start=38,tutor-end=39}{a}。

为什么从这里入手：目标是“识别结构并建立关系”，而“令 u=x+1,s2=1−a\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}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{s}^{\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}{a}，三次式写成 f=13u3−s2u+23−a\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{3}}\htmlData{tutor-start=13,tutor-end=14}{u}^{\htmlData{tutor-start=16,tutor-end=17}{3}}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{s}^{\htmlData{tutor-start=22,tutor-end=23}{2}}\htmlData{tutor-start=24,tutor-end=25}{u}\htmlData{tutor-start=25,tutor-end=26}{+}\frac{\htmlData{tutor-start=32,tutor-end=33}{2}}{\htmlData{tutor-start=35,tutor-end=36}{3}}\htmlData{tutor-start=37,tutor-end=38}{-}\htmlData{tutor-start=38,tutor-end=39}{a}。”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：两个极值点为 u=±s\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=6}{\pm }\htmlData{tutor-start=6,tutor-end=7}{s}，连接它们的直线与 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴交点横坐标为 x0=−a/[2(1−a)]\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=9}{/}\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}{1}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{a}\htmlData{tutor-start=15,tutor-end=16}{)}\htmlData{tutor-start=16,tutor-end=17}{]}。要求该交点也在曲线上，即 f(x0)=0\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=5,tutor-end=6}{0}}\htmlData{tutor-start=7,tutor-end=8}{)}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{0}。

a(12a2−17a+6)=0\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}{2}\htmlData{tutor-start=4,tutor-end=5}{a}^{\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=12}{7}\htmlData{tutor-start=12,tutor-end=13}{a}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{6}\htmlData{tutor-start=15,tutor-end=16}{)}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{0}
(2)
完成计算并核验结论

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：解得 a=0,2/3,3/4\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0}\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}{3}\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}{4}，都满足 a<1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{1}。

a=0,23,34\boxed{\htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{0}\htmlData{tutor-start=10,tutor-end=11}{,}\frac{\htmlData{tutor-start=17,tutor-end=18}{2}}{\htmlData{tutor-start=20,tutor-end=21}{3}}\htmlData{tutor-start=22,tutor-end=23}{,}\frac{\htmlData{tutor-start=29,tutor-end=30}{3}}{\htmlData{tutor-start=32,tutor-end=33}{4}}}
22

解答题 · 解析几何

已知抛物线 C:y=(x+1)2\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=4}{y} \htmlData{tutor-start=5,tutor-end=6}{=} \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}{)}^{\htmlData{tutor-start=14,tutor-end=15}{2}} 与圆 M:(x−1)2+(y−12)2=r2\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{x}\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=10,tutor-end=11}{2}} \htmlData{tutor-start=13,tutor-end=14}{+} \left(\htmlData{tutor-start=21,tutor-end=22}{y}\htmlData{tutor-start=22,tutor-end=23}{-}\frac{\htmlData{tutor-start=29,tutor-end=30}{1}}{\htmlData{tutor-start=32,tutor-end=33}{2}}\right)^{\htmlData{tutor-start=43,tutor-end=44}{2}} \htmlData{tutor-start=46,tutor-end=47}{=} \htmlData{tutor-start=48,tutor-end=49}{r}^{\htmlData{tutor-start=51,tutor-end=52}{2}} (r>0\htmlData{tutor-start=0,tutor-end=1}{r}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0}) 有一个公共点 A\htmlData{tutor-start=0,tutor-end=1}{A}，且在 A\htmlData{tutor-start=0,tutor-end=1}{A} 处两曲线的切线为同一直线 l\htmlData{tutor-start=0,tutor-end=1}{l}. (1) 求 r\htmlData{tutor-start=0,tutor-end=1}{r}； (2) 设 m,n\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{n} 是异于 l\htmlData{tutor-start=0,tutor-end=1}{l} 且与 C\htmlData{tutor-start=0,tutor-end=1}{C} 及 M\htmlData{tutor-start=0,tutor-end=1}{M} 都相切的两条直线，m,n\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{n} 的交点为 D\htmlData{tutor-start=0,tutor-end=1}{D}，求 D\htmlData{tutor-start=0,tutor-end=1}{D} 到 l\htmlData{tutor-start=0,tutor-end=1}{l} 的距离.

答案：r=52\htmlData{tutor-start=0,tutor-end=1}{r}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\sqrt{\htmlData{tutor-start=14,tutor-end=15}{5}}}{\htmlData{tutor-start=18,tutor-end=19}{2}}；距离 65\frac6{\sqrt{5}}

题目标签：抛物线与圆的公切线

解题过程

（1）用切线斜率和半径垂直

求圆半径

(1)
识别结构并建立关系

设公共点横坐标满足 t=x+1\htmlData{tutor-start=0,tutor-end=1}{t}\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}，抛物线切线方向为 (1,2t)\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}{2}\htmlData{tutor-start=4,tutor-end=5}{t}\htmlData{tutor-start=5,tutor-end=6}{)}。

为什么从这里入手：函数的局部变化由导数控制。“设公共点横坐标满足 t=x+1\htmlData{tutor-start=0,tutor-end=1}{t}\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}，抛物线切线方向为 (1,2t)\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}{2}\htmlData{tutor-start=4,tutor-end=5}{t}\htmlData{tutor-start=5,tutor-end=6}{)}。”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：圆心到切点的半径与切线垂直，得到 2t3−2=0\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{t}^{\htmlData{tutor-start=4,tutor-end=5}{3}}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{0}，故 t=1\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}，公共点 A=(0,1)\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}{1}\htmlData{tutor-start=6,tutor-end=7}{)}。

r=∣MA∣=1+14=52\htmlData{tutor-start=0,tutor-end=1}{r}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{|}\htmlData{tutor-start=3,tutor-end=4}{M}\htmlData{tutor-start=4,tutor-end=5}{A}\htmlData{tutor-start=5,tutor-end=6}{|}\htmlData{tutor-start=6,tutor-end=7}{=}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{+}\frac{\htmlData{tutor-start=21,tutor-end=22}{1}}{\htmlData{tutor-start=24,tutor-end=25}{4}}}\htmlData{tutor-start=27,tutor-end=28}{=}\frac{\sqrt{\htmlData{tutor-start=40,tutor-end=41}{5}}}{\htmlData{tutor-start=44,tutor-end=45}{2}}
(2)
完成计算并核验结论

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：公共切线为 l:y=2x+1\htmlData{tutor-start=0,tutor-end=1}{l}\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}{2}\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{1}。

r=52\boxed{\htmlData{tutor-start=7,tutor-end=8}{r}\htmlData{tutor-start=8,tutor-end=9}{=}\frac{\sqrt{\htmlData{tutor-start=21,tutor-end=22}{5}}}{\htmlData{tutor-start=25,tutor-end=26}{2}}}

（2）参数化抛物线切线

求另两公切线交点到 l\htmlData{tutor-start=0,tutor-end=1}{l} 的距离

(1)
识别结构并建立关系

抛物线在参数 t\htmlData{tutor-start=0,tutor-end=1}{t} 处切线为 y=2tx−t2+2t\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}{t}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{t}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{t}。

为什么从这里入手：函数的局部变化由导数控制。“抛物线在参数 t\htmlData{tutor-start=0,tutor-end=1}{t} 处切线为 y=2tx−t2+2t\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}{t}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{t}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{t}。”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：它与圆相切的条件化为 (t−1)2(t2−6t−1)=0\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{t}\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=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{t}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{6}\htmlData{tutor-start=17,tutor-end=18}{t}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{)}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{0}。除去 l\htmlData{tutor-start=0,tutor-end=1}{l} 对应的 t=1\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}，另两根为 3±10\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=4}{\pm}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{0}}；两条切线交于 D=(2,−1)\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{=}\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=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{)}。

d(D,l)=∣2⋅2−(−1)+1∣5\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{D}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{l}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{=}\frac{\htmlData{tutor-start=13,tutor-end=14}{|}\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=20}{\cdot}\htmlData{tutor-start=20,tutor-end=21}{2}\htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{-}\htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{)}\htmlData{tutor-start=26,tutor-end=27}{+}\htmlData{tutor-start=27,tutor-end=28}{1}\htmlData{tutor-start=28,tutor-end=29}{|}}{\sqrt{\htmlData{tutor-start=37,tutor-end=38}{5}}}
(2)
完成计算并核验结论

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：所以距离为 6/5\htmlData{tutor-start=0,tutor-end=1}{6}\htmlData{tutor-start=1,tutor-end=2}{/}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{5}}。

65\boxed{\frac6{\sqrt{5}}}