2009 年高考数学（大纲卷 1理科）

首先通过代数变形隔离出 $\bar{\htmlData{tutor-start=5,tutor-end=6}{z}}$，利用复数乘法法则计算右侧表达式。

详细展开： 原方程为 $\frac{\bar{\htmlData{tutor-start=11,tutor-end=12}{z}}}{\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{+}\mathrm{\htmlData{tutor-start=25,tutor-end=26}{i}}}\htmlData{tutor-start=28,tutor-end=29}{=}\htmlData{tutor-start=29,tutor-end=30}{2}\htmlData{tutor-start=30,tutor-end=31}{+}\mathrm{\htmlData{tutor-start=39,tutor-end=40}{i}}$。 为了求出 $\bar{\htmlData{tutor-start=5,tutor-end=6}{z}}$，我们将方程两边同时乘以分母 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\mathrm{\htmlData{tutor-start=10,tutor-end=11}{i}}$： $\bar{\htmlData{tutor-start=5,tutor-end=6}{z}} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{+}\mathrm{\htmlData{tutor-start=21,tutor-end=22}{i}}\htmlData{tutor-start=23,tutor-end=24}{)}\htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{+}\mathrm{\htmlData{tutor-start=35,tutor-end=36}{i}}\htmlData{tutor-start=37,tutor-end=38}{)}$ 展开右侧乘积： $\bar{\htmlData{tutor-start=5,tutor-end=6}{z}} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{2} \htmlData{tutor-start=12,tutor-end=19}{\times }\htmlData{tutor-start=19,tutor-end=20}{1} \htmlData{tutor-start=21,tutor-end=22}{+} \htmlData{tutor-start=23,tutor-end=24}{2} \htmlData{tutor-start=25,tutor-end=32}{\times }\mathrm{\htmlData{tutor-start=40,tutor-end=41}{i}} \htmlData{tutor-start=43,tutor-end=44}{+} \mathrm{\htmlData{tutor-start=53,tutor-end=54}{i}} \htmlData{tutor-start=56,tutor-end=63}{\times }\htmlData{tutor-start=63,tutor-end=64}{1} \htmlData{tutor-start=65,tutor-end=66}{+} \mathrm{\htmlData{tutor-start=75,tutor-end=76}{i}} \htmlData{tutor-start=78,tutor-end=85}{\times }\mathrm{\htmlData{tutor-start=93,tutor-end=94}{i}}$ $\bar{\htmlData{tutor-start=5,tutor-end=6}{z}} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{2} \htmlData{tutor-start=12,tutor-end=13}{+} \htmlData{tutor-start=14,tutor-end=15}{2}\mathrm{\htmlData{tutor-start=23,tutor-end=24}{i}} \htmlData{tutor-start=26,tutor-end=27}{+} \mathrm{\htmlData{tutor-start=36,tutor-end=37}{i}} \htmlData{tutor-start=39,tutor-end=40}{+} \mathrm{\htmlData{tutor-start=49,tutor-end=50}{i}}^{\htmlData{tutor-start=53,tutor-end=54}{2}}$ 已知 $\mathrm{\htmlData{tutor-start=8,tutor-end=9}{i}}^{\htmlData{tutor-start=12,tutor-end=13}{2}} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{1}$，代入得： $\bar{\htmlData{tutor-start=5,tutor-end=6}{z}} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{2} \htmlData{tutor-start=12,tutor-end=13}{+} \htmlData{tutor-start=14,tutor-end=15}{3}\mathrm{\htmlData{tutor-start=23,tutor-end=24}{i}} \htmlData{tutor-start=26,tutor-end=27}{-} \htmlData{tutor-start=28,tutor-end=29}{1}$ $\bar{\htmlData{tutor-start=5,tutor-end=6}{z}} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{1} \htmlData{tutor-start=12,tutor-end=13}{+} \htmlData{tutor-start=14,tutor-end=15}{3}\mathrm{\htmlData{tutor-start=23,tutor-end=24}{i}}$

根据共轭复数的定义，由 $\bar{\htmlData{tutor-start=5,tutor-end=6}{z}}$ 直接写出 $\htmlData{tutor-start=0,tutor-end=1}{z}$。

详细展开： 若复数 $\htmlData{tutor-start=0,tutor-end=1}{w} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{a} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{b}\mathrm{\htmlData{tutor-start=17,tutor-end=18}{i}}$，则其共轭复数 $\bar{\htmlData{tutor-start=5,tutor-end=6}{w}} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{a} \htmlData{tutor-start=12,tutor-end=13}{-} \htmlData{tutor-start=14,tutor-end=15}{b}\mathrm{\htmlData{tutor-start=23,tutor-end=24}{i}}$。 反之，若已知 $\bar{\htmlData{tutor-start=5,tutor-end=6}{z}} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{a} \htmlData{tutor-start=12,tutor-end=13}{+} \htmlData{tutor-start=14,tutor-end=15}{b}\mathrm{\htmlData{tutor-start=23,tutor-end=24}{i}}$，则 $\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{a} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{b}\mathrm{\htmlData{tutor-start=17,tutor-end=18}{i}}$。 前一步已求得 $\bar{\htmlData{tutor-start=5,tutor-end=6}{z}} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{1} \htmlData{tutor-start=12,tutor-end=13}{+} \htmlData{tutor-start=14,tutor-end=15}{3}\mathrm{\htmlData{tutor-start=23,tutor-end=24}{i}}$。 因此，将虚部符号取反即可得到 $\htmlData{tutor-start=0,tutor-end=1}{z}$： $\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{1} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{3}\mathrm{\htmlData{tutor-start=17,tutor-end=18}{i}}$ 对比选项： (A) $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{3}\mathrm{\htmlData{tutor-start=12,tutor-end=13}{i}}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{3}\mathrm{\htmlData{tutor-start=11,tutor-end=12}{i}}$ (C) $\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{+}\mathrm{\htmlData{tutor-start=10,tutor-end=11}{i}}$ (D) $\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{-}\mathrm{\htmlData{tutor-start=10,tutor-end=11}{i}}$ 结果与选项 (B) 一致。

将绝对值不等式转化为平方不等式或分式不等式组，注意定义域。

详细展开： 原不等式为 $\left|\frac{\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{1}}{\htmlData{tutor-start=17,tutor-end=18}{x}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{1}}\right|\htmlData{tutor-start=28,tutor-end=29}{<}\htmlData{tutor-start=29,tutor-end=30}{1}$。 首先，分母不能为零，即 $\htmlData{tutor-start=0,tutor-end=1}{x} \neq \htmlData{tutor-start=7,tutor-end=8}{1}$。 由于绝对值非负，且右边为 1，两边平方可得等价不等式： $\left(\frac{\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{1}}{\htmlData{tutor-start=17,tutor-end=18}{x}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{1}}\right)^{\htmlData{tutor-start=30,tutor-end=31}{2}} \htmlData{tutor-start=33,tutor-end=34}{<} \htmlData{tutor-start=35,tutor-end=36}{1}$ $\frac{\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{)}^{\htmlData{tutor-start=13,tutor-end=14}{2}}}{\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{x}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{)}^{\htmlData{tutor-start=24,tutor-end=25}{2}}} \htmlData{tutor-start=28,tutor-end=29}{<} \htmlData{tutor-start=30,tutor-end=31}{1}$ 因为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{>} \htmlData{tutor-start=12,tutor-end=13}{0}$（当 $\htmlData{tutor-start=0,tutor-end=1}{x} \neq \htmlData{tutor-start=7,tutor-end=8}{1}$），两边同乘 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}}$ 不改变不等号方向： $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{<} \htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{)}^{\htmlData{tutor-start=19,tutor-end=20}{2}}$ 展开： $\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=12}{+} \htmlData{tutor-start=13,tutor-end=14}{1} \htmlData{tutor-start=15,tutor-end=16}{<} \htmlData{tutor-start=17,tutor-end=18}{x}^{\htmlData{tutor-start=20,tutor-end=21}{2}} \htmlData{tutor-start=23,tutor-end=24}{-} \htmlData{tutor-start=25,tutor-end=26}{2}\htmlData{tutor-start=26,tutor-end=27}{x} \htmlData{tutor-start=28,tutor-end=29}{+} \htmlData{tutor-start=30,tutor-end=31}{1}$ 消去 $\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{1}$： $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{x} \htmlData{tutor-start=3,tutor-end=4}{<} \htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{x}$ $\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{x} \htmlData{tutor-start=3,tutor-end=4}{<} \htmlData{tutor-start=5,tutor-end=6}{0}$ $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{0}$

检验解集是否满足定义域，并写出最终区间表示。

详细展开： 解得 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{0}$。 检查定义域限制 $\htmlData{tutor-start=0,tutor-end=1}{x} \neq \htmlData{tutor-start=7,tutor-end=8}{1}$。 因为 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{0}$ 的所有值都不等于 1，所以定义域限制自然满足。 故解集为 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{x} \htmlData{tutor-start=4,tutor-end=9}{\mid }\htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=12}{<} \htmlData{tutor-start=13,tutor-end=14}{0}\htmlData{tutor-start=14,tutor-end=16}{\}}$。 对比选项： (A) $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{x} \htmlData{tutor-start=4,tutor-end=9}{\mid }\htmlData{tutor-start=9,tutor-end=10}{0}\htmlData{tutor-start=10,tutor-end=11}{<}\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{<}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=16}{\}} \htmlData{tutor-start=17,tutor-end=22}{\cup }\htmlData{tutor-start=22,tutor-end=24}{\{}\htmlData{tutor-start=24,tutor-end=25}{x} \htmlData{tutor-start=26,tutor-end=31}{\mid }\htmlData{tutor-start=31,tutor-end=32}{x}\htmlData{tutor-start=32,tutor-end=33}{>}\htmlData{tutor-start=33,tutor-end=34}{1}\htmlData{tutor-start=34,tutor-end=36}{\}}$ (B) $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{x} \htmlData{tutor-start=4,tutor-end=9}{\mid }\htmlData{tutor-start=9,tutor-end=10}{0}\htmlData{tutor-start=10,tutor-end=11}{<}\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{<}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=16}{\}}$ (C) $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{x} \htmlData{tutor-start=4,tutor-end=9}{\mid }\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{<}\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{<}\htmlData{tutor-start=14,tutor-end=15}{0}\htmlData{tutor-start=15,tutor-end=17}{\}}$ (D) $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{x} \htmlData{tutor-start=4,tutor-end=9}{\mid }\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{<}\htmlData{tutor-start=11,tutor-end=12}{0}\htmlData{tutor-start=12,tutor-end=14}{\}}$ 显然选项 (D) 正确。

*补充验证*： 取 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{2}$，左边 $\htmlData{tutor-start=0,tutor-end=1}{=} \htmlData{tutor-start=2,tutor-end=3}{|}\frac{\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}}{\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{3}}\htmlData{tutor-start=16,tutor-end=17}{|} \htmlData{tutor-start=18,tutor-end=19}{=} \frac{\htmlData{tutor-start=26,tutor-end=27}{1}}{\htmlData{tutor-start=29,tutor-end=30}{3}} \htmlData{tutor-start=32,tutor-end=33}{<} \htmlData{tutor-start=34,tutor-end=35}{1}$，成立。 取 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{.}\htmlData{tutor-start=6,tutor-end=7}{5}$，左边 $\htmlData{tutor-start=0,tutor-end=1}{=} \htmlData{tutor-start=2,tutor-end=3}{|}\frac{\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{.}\htmlData{tutor-start=11,tutor-end=12}{5}}{\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{0}\htmlData{tutor-start=16,tutor-end=17}{.}\htmlData{tutor-start=17,tutor-end=18}{5}}\htmlData{tutor-start=19,tutor-end=20}{|} \htmlData{tutor-start=21,tutor-end=22}{=} \htmlData{tutor-start=23,tutor-end=24}{|}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{3}\htmlData{tutor-start=26,tutor-end=27}{|} \htmlData{tutor-start=28,tutor-end=29}{=} \htmlData{tutor-start=30,tutor-end=31}{3} \htmlData{tutor-start=32,tutor-end=33}{>} \htmlData{tutor-start=34,tutor-end=35}{1}$，不成立。（排除 A, B） 取 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{.}\htmlData{tutor-start=7,tutor-end=8}{5}$，左边 $\htmlData{tutor-start=0,tutor-end=1}{=} \htmlData{tutor-start=2,tutor-end=3}{|}\frac{\htmlData{tutor-start=9,tutor-end=10}{0}\htmlData{tutor-start=10,tutor-end=11}{.}\htmlData{tutor-start=11,tutor-end=12}{5}}{\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{.}\htmlData{tutor-start=17,tutor-end=18}{5}}\htmlData{tutor-start=19,tutor-end=20}{|} \htmlData{tutor-start=21,tutor-end=22}{=} \frac{\htmlData{tutor-start=29,tutor-end=30}{1}}{\htmlData{tutor-start=32,tutor-end=33}{3}} \htmlData{tutor-start=35,tutor-end=36}{<} \htmlData{tutor-start=37,tutor-end=38}{1}$，成立。（C 是 D 的子集，D 更完整） 取 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{2}$ 在 D 中但不在 C 中，故 C 不全。

写出双曲线渐近线方程，联立抛物线方程，利用判别式为零求解。

详细展开： 双曲线 $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{a}^{\htmlData{tutor-start=16,tutor-end=17}{2}}}\htmlData{tutor-start=19,tutor-end=20}{-}\frac{\htmlData{tutor-start=26,tutor-end=27}{y}^{\htmlData{tutor-start=29,tutor-end=30}{2}}}{\htmlData{tutor-start=33,tutor-end=34}{b}^{\htmlData{tutor-start=36,tutor-end=37}{2}}}\htmlData{tutor-start=39,tutor-end=40}{=}\htmlData{tutor-start=40,tutor-end=41}{1}$ 的渐近线方程为 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=8}{\pm }\frac{\htmlData{tutor-start=14,tutor-end=15}{b}}{\htmlData{tutor-start=17,tutor-end=18}{a}}\htmlData{tutor-start=19,tutor-end=20}{x}$。 由于抛物线 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x}^{\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{1}$ 关于 $\htmlData{tutor-start=0,tutor-end=1}{y}$ 轴对称，且开口向上，两条渐近线与抛物线的相切情况是对称的，只需考虑其中一条，例如 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{b}}{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{x}$。 联立方程组： $\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{y} \htmlData{tutor-start=16,tutor-end=17}{=} \frac{\htmlData{tutor-start=24,tutor-end=25}{b}}{\htmlData{tutor-start=27,tutor-end=28}{a}}\htmlData{tutor-start=29,tutor-end=30}{x} \\ \htmlData{tutor-start=34,tutor-end=35}{y} \htmlData{tutor-start=36,tutor-end=37}{=} \htmlData{tutor-start=38,tutor-end=39}{x}^{\htmlData{tutor-start=41,tutor-end=42}{2}} \htmlData{tutor-start=44,tutor-end=45}{+} \htmlData{tutor-start=46,tutor-end=47}{1} \end{cases}$ 消去 $\htmlData{tutor-start=0,tutor-end=1}{y}$ 得： $\frac{\htmlData{tutor-start=6,tutor-end=7}{b}}{\htmlData{tutor-start=9,tutor-end=10}{a}}\htmlData{tutor-start=11,tutor-end=12}{x} \htmlData{tutor-start=13,tutor-end=14}{=} \htmlData{tutor-start=15,tutor-end=16}{x}^{\htmlData{tutor-start=18,tutor-end=19}{2}} \htmlData{tutor-start=21,tutor-end=22}{+} \htmlData{tutor-start=23,tutor-end=24}{1}$ 整理为一元二次方程： $\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \frac{\htmlData{tutor-start=14,tutor-end=15}{b}}{\htmlData{tutor-start=17,tutor-end=18}{a}}\htmlData{tutor-start=19,tutor-end=20}{x} \htmlData{tutor-start=21,tutor-end=22}{+} \htmlData{tutor-start=23,tutor-end=24}{1} \htmlData{tutor-start=25,tutor-end=26}{=} \htmlData{tutor-start=27,tutor-end=28}{0}$ 因为直线与抛物线相切，所以该方程有且仅有一个实根，即判别式 $\htmlData{tutor-start=0,tutor-end=7}{\Delta }\htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{0}$。 $\htmlData{tutor-start=0,tutor-end=7}{\Delta }\htmlData{tutor-start=7,tutor-end=8}{=} \left(\htmlData{tutor-start=15,tutor-end=16}{-}\frac{\htmlData{tutor-start=22,tutor-end=23}{b}}{\htmlData{tutor-start=25,tutor-end=26}{a}}\right)^{\htmlData{tutor-start=36,tutor-end=37}{2}} \htmlData{tutor-start=39,tutor-end=40}{-} \htmlData{tutor-start=41,tutor-end=42}{4} \htmlData{tutor-start=43,tutor-end=50}{\times }\htmlData{tutor-start=50,tutor-end=51}{1} \htmlData{tutor-start=52,tutor-end=59}{\times }\htmlData{tutor-start=59,tutor-end=60}{1} \htmlData{tutor-start=61,tutor-end=62}{=} \htmlData{tutor-start=63,tutor-end=64}{0}$ $\frac{\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{a}^{\htmlData{tutor-start=16,tutor-end=17}{2}}} \htmlData{tutor-start=20,tutor-end=21}{-} \htmlData{tutor-start=22,tutor-end=23}{4} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{0}$ $\frac{\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{a}^{\htmlData{tutor-start=16,tutor-end=17}{2}}} \htmlData{tutor-start=20,tutor-end=21}{=} \htmlData{tutor-start=22,tutor-end=23}{4}$ 即 $\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{4}\htmlData{tutor-start=9,tutor-end=10}{a}^{\htmlData{tutor-start=12,tutor-end=13}{2}}$。

利用双曲线离心率公式及 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{b}$ 关系求解。

详细展开： 双曲线离心率公式为 $\htmlData{tutor-start=0,tutor-end=1}{e} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{c}}{\htmlData{tutor-start=13,tutor-end=14}{a}} \htmlData{tutor-start=16,tutor-end=17}{=} \sqrt{\htmlData{tutor-start=24,tutor-end=25}{1} \htmlData{tutor-start=26,tutor-end=27}{+} \frac{\htmlData{tutor-start=34,tutor-end=35}{b}^{\htmlData{tutor-start=37,tutor-end=38}{2}}}{\htmlData{tutor-start=41,tutor-end=42}{a}^{\htmlData{tutor-start=44,tutor-end=45}{2}}}}$。 由上一步得 $\frac{\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{a}^{\htmlData{tutor-start=16,tutor-end=17}{2}}} \htmlData{tutor-start=20,tutor-end=21}{=} \htmlData{tutor-start=22,tutor-end=23}{4}$。 代入公式： $\htmlData{tutor-start=0,tutor-end=1}{e} \htmlData{tutor-start=2,tutor-end=3}{=} \sqrt{\htmlData{tutor-start=10,tutor-end=11}{1} \htmlData{tutor-start=12,tutor-end=13}{+} \htmlData{tutor-start=14,tutor-end=15}{4}} \htmlData{tutor-start=17,tutor-end=18}{=} \sqrt{\htmlData{tutor-start=25,tutor-end=26}{5}}$。 故选项 (C) 正确。

分析第一种互斥情况：选出的唯一一名女同学来自甲组，意味着乙组选出的全是男同学。

详细展开： 题目要求从甲、乙两组各选出 2 名同学，共 4 人。 条件限制：4 人中恰有 1 名女同学。 这 1 名女同学可能来自甲组，也可能来自乙组。这两类情况互斥。

情形 1：女同学在甲组，男同学在乙组（且甲组另 1 人为男，乙组 2 人均为男）。 1. 从甲组选 1 名女同学：甲组有 3 名女同学，选法为 $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{3}}^{\htmlData{tutor-start=7,tutor-end=8}{1}}$。 2. 从甲组选 1 名男同学：甲组有 5 名男同学，选法为 $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{5}}^{\htmlData{tutor-start=7,tutor-end=8}{1}}$。 此时甲组选毕，共选出 1 女 1 男。 3. 从乙组选 2 名男同学：乙组有 6 名男同学，选法为 $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{6}}^{\htmlData{tutor-start=7,tutor-end=8}{2}}$。 注意：乙组不能选女同学，否则女同学总数会超过 1 名。

计算该情形下的选法数 $\htmlData{tutor-start=0,tutor-end=1}{N}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$： $\htmlData{tutor-start=0,tutor-end=1}{N}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{C}_{\htmlData{tutor-start=11,tutor-end=12}{3}}^{\htmlData{tutor-start=15,tutor-end=16}{1}} \htmlData{tutor-start=18,tutor-end=25}{\times }\htmlData{tutor-start=25,tutor-end=26}{C}_{\htmlData{tutor-start=28,tutor-end=29}{5}}^{\htmlData{tutor-start=32,tutor-end=33}{1}} \htmlData{tutor-start=35,tutor-end=42}{\times }\htmlData{tutor-start=42,tutor-end=43}{C}_{\htmlData{tutor-start=45,tutor-end=46}{6}}^{\htmlData{tutor-start=49,tutor-end=50}{2}}$ $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{3}}^{\htmlData{tutor-start=7,tutor-end=8}{1}} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{3}$ $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{5}}^{\htmlData{tutor-start=7,tutor-end=8}{1}} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{5}$ $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{6}}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{=} \frac{\htmlData{tutor-start=18,tutor-end=19}{6} \htmlData{tutor-start=20,tutor-end=27}{\times }\htmlData{tutor-start=27,tutor-end=28}{5}}{\htmlData{tutor-start=30,tutor-end=31}{2} \htmlData{tutor-start=32,tutor-end=39}{\times }\htmlData{tutor-start=39,tutor-end=40}{1}} \htmlData{tutor-start=42,tutor-end=43}{=} \htmlData{tutor-start=44,tutor-end=45}{1}\htmlData{tutor-start=45,tutor-end=46}{5}$ $\htmlData{tutor-start=0,tutor-end=1}{N}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{3} \htmlData{tutor-start=10,tutor-end=17}{\times }\htmlData{tutor-start=17,tutor-end=18}{5} \htmlData{tutor-start=19,tutor-end=26}{\times }\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{5} \htmlData{tutor-start=29,tutor-end=30}{=} \htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{5} \htmlData{tutor-start=34,tutor-end=41}{\times }\htmlData{tutor-start=41,tutor-end=42}{1}\htmlData{tutor-start=42,tutor-end=43}{5} \htmlData{tutor-start=44,tutor-end=45}{=} \htmlData{tutor-start=46,tutor-end=47}{2}\htmlData{tutor-start=47,tutor-end=48}{2}\htmlData{tutor-start=48,tutor-end=49}{5}$ 种。

分析第二种互斥情况：选出的唯一一名女同学来自乙组，意味着甲组选出的全是男同学。最后将两类情况相加。

详细展开： 情形 2：女同学在乙组，男同学在甲组（且乙组另 1 人为男，甲组 2 人均为男）。 1. 从乙组选 1 名女同学：乙组有 2 名女同学，选法为 $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{2}}^{\htmlData{tutor-start=7,tutor-end=8}{1}}$。 2. 从乙组选 1 名男同学：乙组有 6 名男同学，选法为 $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{6}}^{\htmlData{tutor-start=7,tutor-end=8}{1}}$。 此时乙组选毕，共选出 1 女 1 男。 3. 从甲组选 2 名男同学：甲组有 5 名男同学，选法为 $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{5}}^{\htmlData{tutor-start=7,tutor-end=8}{2}}$。 注意：甲组不能选女同学。

计算该情形下的选法数 $\htmlData{tutor-start=0,tutor-end=1}{N}_{\htmlData{tutor-start=3,tutor-end=4}{2}}$： $\htmlData{tutor-start=0,tutor-end=1}{N}_{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{C}_{\htmlData{tutor-start=11,tutor-end=12}{2}}^{\htmlData{tutor-start=15,tutor-end=16}{1}} \htmlData{tutor-start=18,tutor-end=25}{\times }\htmlData{tutor-start=25,tutor-end=26}{C}_{\htmlData{tutor-start=28,tutor-end=29}{6}}^{\htmlData{tutor-start=32,tutor-end=33}{1}} \htmlData{tutor-start=35,tutor-end=42}{\times }\htmlData{tutor-start=42,tutor-end=43}{C}_{\htmlData{tutor-start=45,tutor-end=46}{5}}^{\htmlData{tutor-start=49,tutor-end=50}{2}}$ $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{2}}^{\htmlData{tutor-start=7,tutor-end=8}{1}} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{2}$ $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{6}}^{\htmlData{tutor-start=7,tutor-end=8}{1}} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{6}$ $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{5}}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{=} \frac{\htmlData{tutor-start=18,tutor-end=19}{5} \htmlData{tutor-start=20,tutor-end=27}{\times }\htmlData{tutor-start=27,tutor-end=28}{4}}{\htmlData{tutor-start=30,tutor-end=31}{2} \htmlData{tutor-start=32,tutor-end=39}{\times }\htmlData{tutor-start=39,tutor-end=40}{1}} \htmlData{tutor-start=42,tutor-end=43}{=} \htmlData{tutor-start=44,tutor-end=45}{1}\htmlData{tutor-start=45,tutor-end=46}{0}$ $\htmlData{tutor-start=0,tutor-end=1}{N}_{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{2} \htmlData{tutor-start=10,tutor-end=17}{\times }\htmlData{tutor-start=17,tutor-end=18}{6} \htmlData{tutor-start=19,tutor-end=26}{\times }\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{0} \htmlData{tutor-start=29,tutor-end=30}{=} \htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{2}\htmlData{tutor-start=33,tutor-end=34}{0}$ 种。

总选法数 $\htmlData{tutor-start=0,tutor-end=1}{N}$ 为两种情形之和： $\htmlData{tutor-start=0,tutor-end=1}{N} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{N}_{\htmlData{tutor-start=7,tutor-end=8}{1}} \htmlData{tutor-start=10,tutor-end=11}{+} \htmlData{tutor-start=12,tutor-end=13}{N}_{\htmlData{tutor-start=15,tutor-end=16}{2}} \htmlData{tutor-start=18,tutor-end=19}{=} \htmlData{tutor-start=20,tutor-end=21}{2}\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{5} \htmlData{tutor-start=24,tutor-end=25}{+} \htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{2}\htmlData{tutor-start=28,tutor-end=29}{0} \htmlData{tutor-start=30,tutor-end=31}{=} \htmlData{tutor-start=32,tutor-end=33}{3}\htmlData{tutor-start=33,tutor-end=34}{4}\htmlData{tutor-start=34,tutor-end=35}{5}$ 种。

对比选项： (A) 150 (B) 180 (C) 300 (D) 345 结果与选项 (D) 一致。

展开数量积表达式，利用已知条件简化。

详细展开： 已知 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{|}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{|}\boldsymbol{\htmlData{tutor-start=30,tutor-end=31}{b}}\htmlData{tutor-start=32,tutor-end=33}{|}\htmlData{tutor-start=33,tutor-end=34}{=}\htmlData{tutor-start=34,tutor-end=35}{|}\boldsymbol{\htmlData{tutor-start=47,tutor-end=48}{c}}\htmlData{tutor-start=49,tutor-end=50}{|}\htmlData{tutor-start=50,tutor-end=51}{=}\htmlData{tutor-start=51,tutor-end=52}{1}$，且 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}} \htmlData{tutor-start=15,tutor-end=21}{\cdot }\boldsymbol{\htmlData{tutor-start=33,tutor-end=34}{b}}\htmlData{tutor-start=35,tutor-end=36}{=}\htmlData{tutor-start=36,tutor-end=37}{0}$。 目标式：$\htmlData{tutor-start=0,tutor-end=1}{E} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\boldsymbol{\htmlData{tutor-start=17,tutor-end=18}{a}}\htmlData{tutor-start=19,tutor-end=20}{-}\boldsymbol{\htmlData{tutor-start=32,tutor-end=33}{c}}\htmlData{tutor-start=34,tutor-end=35}{)} \htmlData{tutor-start=36,tutor-end=42}{\cdot }\htmlData{tutor-start=42,tutor-end=43}{(}\boldsymbol{\htmlData{tutor-start=55,tutor-end=56}{b}}\htmlData{tutor-start=57,tutor-end=58}{-}\boldsymbol{\htmlData{tutor-start=70,tutor-end=71}{c}}\htmlData{tutor-start=72,tutor-end=73}{)}$ 展开： $\htmlData{tutor-start=0,tutor-end=1}{E} \htmlData{tutor-start=2,tutor-end=3}{=} \boldsymbol{\htmlData{tutor-start=16,tutor-end=17}{a}} \htmlData{tutor-start=19,tutor-end=25}{\cdot }\boldsymbol{\htmlData{tutor-start=37,tutor-end=38}{b}} \htmlData{tutor-start=40,tutor-end=41}{-} \boldsymbol{\htmlData{tutor-start=54,tutor-end=55}{a}} \htmlData{tutor-start=57,tutor-end=63}{\cdot }\boldsymbol{\htmlData{tutor-start=75,tutor-end=76}{c}} \htmlData{tutor-start=78,tutor-end=79}{-} \boldsymbol{\htmlData{tutor-start=92,tutor-end=93}{c}} \htmlData{tutor-start=95,tutor-end=101}{\cdot }\boldsymbol{\htmlData{tutor-start=113,tutor-end=114}{b}} \htmlData{tutor-start=116,tutor-end=117}{+} \boldsymbol{\htmlData{tutor-start=130,tutor-end=131}{c}} \htmlData{tutor-start=133,tutor-end=139}{\cdot }\boldsymbol{\htmlData{tutor-start=151,tutor-end=152}{c}}$ 代入已知条件： $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}} \htmlData{tutor-start=15,tutor-end=21}{\cdot }\boldsymbol{\htmlData{tutor-start=33,tutor-end=34}{b}} \htmlData{tutor-start=36,tutor-end=37}{=} \htmlData{tutor-start=38,tutor-end=39}{0}$ $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{c}} \htmlData{tutor-start=15,tutor-end=21}{\cdot }\boldsymbol{\htmlData{tutor-start=33,tutor-end=34}{c}} \htmlData{tutor-start=36,tutor-end=37}{=} \htmlData{tutor-start=38,tutor-end=39}{|}\boldsymbol{\htmlData{tutor-start=51,tutor-end=52}{c}}\htmlData{tutor-start=53,tutor-end=54}{|}^{\htmlData{tutor-start=56,tutor-end=57}{2}} \htmlData{tutor-start=59,tutor-end=60}{=} \htmlData{tutor-start=61,tutor-end=62}{1}$ 所以： $\htmlData{tutor-start=0,tutor-end=1}{E} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{0} \htmlData{tutor-start=6,tutor-end=7}{-} \boldsymbol{\htmlData{tutor-start=20,tutor-end=21}{a}} \htmlData{tutor-start=23,tutor-end=29}{\cdot }\boldsymbol{\htmlData{tutor-start=41,tutor-end=42}{c}} \htmlData{tutor-start=44,tutor-end=45}{-} \boldsymbol{\htmlData{tutor-start=58,tutor-end=59}{b}} \htmlData{tutor-start=61,tutor-end=67}{\cdot }\boldsymbol{\htmlData{tutor-start=79,tutor-end=80}{c}} \htmlData{tutor-start=82,tutor-end=83}{+} \htmlData{tutor-start=84,tutor-end=85}{1}$ $\htmlData{tutor-start=0,tutor-end=1}{E} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{1} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{(}\boldsymbol{\htmlData{tutor-start=21,tutor-end=22}{a}} \htmlData{tutor-start=24,tutor-end=25}{+} \boldsymbol{\htmlData{tutor-start=38,tutor-end=39}{b}}\htmlData{tutor-start=40,tutor-end=41}{)} \htmlData{tutor-start=42,tutor-end=48}{\cdot }\boldsymbol{\htmlData{tutor-start=60,tutor-end=61}{c}}$ 要求 $\htmlData{tutor-start=0,tutor-end=1}{E}$ 的最小值，即求 $\htmlData{tutor-start=0,tutor-end=1}{(}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}} \htmlData{tutor-start=16,tutor-end=17}{+} \boldsymbol{\htmlData{tutor-start=30,tutor-end=31}{b}}\htmlData{tutor-start=32,tutor-end=33}{)} \htmlData{tutor-start=34,tutor-end=40}{\cdot }\boldsymbol{\htmlData{tutor-start=52,tutor-end=53}{c}}$ 的最大值。

计算向量 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}\htmlData{tutor-start=14,tutor-end=15}{+}\boldsymbol{\htmlData{tutor-start=27,tutor-end=28}{b}}$ 的模，利用点积定义求最大值。

详细展开： 令 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{u}} \htmlData{tutor-start=15,tutor-end=16}{=} \boldsymbol{\htmlData{tutor-start=29,tutor-end=30}{a}} \htmlData{tutor-start=32,tutor-end=33}{+} \boldsymbol{\htmlData{tutor-start=46,tutor-end=47}{b}}$。 因为 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}} \htmlData{tutor-start=15,tutor-end=21}{\perp }\boldsymbol{\htmlData{tutor-start=33,tutor-end=34}{b}}$ 且均为单位向量，所以 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{u}}$ 的模为： $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{u}}\htmlData{tutor-start=15,tutor-end=16}{|} \htmlData{tutor-start=17,tutor-end=18}{=} \sqrt{\htmlData{tutor-start=25,tutor-end=26}{|}\boldsymbol{\htmlData{tutor-start=38,tutor-end=39}{a}}\htmlData{tutor-start=40,tutor-end=41}{|}^{\htmlData{tutor-start=43,tutor-end=44}{2}} \htmlData{tutor-start=46,tutor-end=47}{+} \htmlData{tutor-start=48,tutor-end=49}{|}\boldsymbol{\htmlData{tutor-start=61,tutor-end=62}{b}}\htmlData{tutor-start=63,tutor-end=64}{|}^{\htmlData{tutor-start=66,tutor-end=67}{2}} \htmlData{tutor-start=69,tutor-end=70}{+} \htmlData{tutor-start=71,tutor-end=72}{2}\boldsymbol{\htmlData{tutor-start=84,tutor-end=85}{a}}\htmlData{tutor-start=86,tutor-end=91}{\cdot}\boldsymbol{\htmlData{tutor-start=103,tutor-end=104}{b}}} \htmlData{tutor-start=107,tutor-end=108}{=} \sqrt{\htmlData{tutor-start=115,tutor-end=116}{1} \htmlData{tutor-start=117,tutor-end=118}{+} \htmlData{tutor-start=119,tutor-end=120}{1} \htmlData{tutor-start=121,tutor-end=122}{+} \htmlData{tutor-start=123,tutor-end=124}{0}} \htmlData{tutor-start=126,tutor-end=127}{=} \sqrt{\htmlData{tutor-start=134,tutor-end=135}{2}}$。 根据点积定义： $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{u}} \htmlData{tutor-start=15,tutor-end=21}{\cdot }\boldsymbol{\htmlData{tutor-start=33,tutor-end=34}{c}} \htmlData{tutor-start=36,tutor-end=37}{=} \htmlData{tutor-start=38,tutor-end=39}{|}\boldsymbol{\htmlData{tutor-start=51,tutor-end=52}{u}}\htmlData{tutor-start=53,tutor-end=54}{|} \htmlData{tutor-start=55,tutor-end=56}{|}\boldsymbol{\htmlData{tutor-start=68,tutor-end=69}{c}}\htmlData{tutor-start=70,tutor-end=71}{|} \cos \htmlData{tutor-start=77,tutor-end=84}{\theta }\htmlData{tutor-start=84,tutor-end=85}{=} \sqrt{\htmlData{tutor-start=92,tutor-end=93}{2}} \htmlData{tutor-start=95,tutor-end=102}{\times }\htmlData{tutor-start=102,tutor-end=103}{1} \htmlData{tutor-start=104,tutor-end=111}{\times }\cos \htmlData{tutor-start=116,tutor-end=123}{\theta }\htmlData{tutor-start=123,tutor-end=124}{=} \sqrt{\htmlData{tutor-start=131,tutor-end=132}{2}} \cos \htmlData{tutor-start=139,tutor-end=145}{\theta}$ 其中 $\htmlData{tutor-start=0,tutor-end=6}{\theta}$ 为向量 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{u}}$ 与 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{c}}$ 的夹角。 要使 $\htmlData{tutor-start=0,tutor-end=1}{E} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{1} \htmlData{tutor-start=6,tutor-end=7}{-} \boldsymbol{\htmlData{tutor-start=20,tutor-end=21}{u}} \htmlData{tutor-start=23,tutor-end=29}{\cdot }\boldsymbol{\htmlData{tutor-start=41,tutor-end=42}{c}}$ 最小，需使 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{u}} \htmlData{tutor-start=15,tutor-end=21}{\cdot }\boldsymbol{\htmlData{tutor-start=33,tutor-end=34}{c}}$ 最大。 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{u}} \htmlData{tutor-start=15,tutor-end=21}{\cdot }\boldsymbol{\htmlData{tutor-start=33,tutor-end=34}{c}}$ 的最大值为 $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{2}}$（当 $\cos \htmlData{tutor-start=5,tutor-end=12}{\theta }\htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{1}$，即 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{c}}$ 与 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}\htmlData{tutor-start=14,tutor-end=15}{+}\boldsymbol{\htmlData{tutor-start=27,tutor-end=28}{b}}$ 同向时）。 所以 $\htmlData{tutor-start=0,tutor-end=1}{E}_{\min} \htmlData{tutor-start=9,tutor-end=10}{=} \htmlData{tutor-start=11,tutor-end=12}{1} \htmlData{tutor-start=13,tutor-end=14}{-} \sqrt{\htmlData{tutor-start=21,tutor-end=22}{2}}$。 故选项 (D) 正确。

根据棱柱性质，侧棱平行且相等，将异面直线 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{C}_{\htmlData{tutor-start=4,tutor-end=5}{1}}$ 的夹角转化为 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}_{\htmlData{tutor-start=4,tutor-end=5}{1}}$ 的夹角。

详细展开： 在三棱柱 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{A}_{\htmlData{tutor-start=7,tutor-end=8}{1}}\htmlData{tutor-start=9,tutor-end=10}{B}_{\htmlData{tutor-start=12,tutor-end=13}{1}}\htmlData{tutor-start=14,tutor-end=15}{C}_{\htmlData{tutor-start=17,tutor-end=18}{1}}$ 中，侧棱 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{C}_{\htmlData{tutor-start=4,tutor-end=5}{1}} \htmlData{tutor-start=7,tutor-end=17}{\parallel }\htmlData{tutor-start=17,tutor-end=18}{A}\htmlData{tutor-start=18,tutor-end=19}{A}_{\htmlData{tutor-start=21,tutor-end=22}{1}}$。 因此，异面直线 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{C}_{\htmlData{tutor-start=4,tutor-end=5}{1}}$ 所成的角等于直线 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}_{\htmlData{tutor-start=4,tutor-end=5}{1}}$ 所成的角（或其补角）。 设该角为 $\htmlData{tutor-start=0,tutor-end=6}{\theta}$，则 $\cos \htmlData{tutor-start=5,tutor-end=12}{\theta }\htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{|}\cos \htmlData{tutor-start=20,tutor-end=28}{\langle }\vec{\htmlData{tutor-start=33,tutor-end=34}{A}\htmlData{tutor-start=34,tutor-end=35}{B}}\htmlData{tutor-start=36,tutor-end=37}{,} \vec{\htmlData{tutor-start=43,tutor-end=44}{A}\htmlData{tutor-start=44,tutor-end=45}{A}_{\htmlData{tutor-start=47,tutor-end=48}{1}}} \htmlData{tutor-start=51,tutor-end=58}{\rangle}\htmlData{tutor-start=58,tutor-end=59}{|}$。 我们需要在 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{A}_{\htmlData{tutor-start=15,tutor-end=16}{1}}$ 或相关几何结构中求出这个角的余弦值。

利用射影条件确定顶点位置，计算 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{A}_{\htmlData{tutor-start=15,tutor-end=16}{1}}$ 的三边长，使用余弦定理求解。

详细展开： 设底面边长和侧棱长均为 $\htmlData{tutor-start=0,tutor-end=1}{a}$。 取 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}$ 中点 $\htmlData{tutor-start=0,tutor-end=1}{D}$，连接 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}$。因为 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C}$ 是等边三角形，所以 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{B}\htmlData{tutor-start=10,tutor-end=11}{C}$，且 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D} \htmlData{tutor-start=3,tutor-end=4}{=} \frac{\sqrt{\htmlData{tutor-start=17,tutor-end=18}{3}}}{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{a}$。 题设 $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 在底面 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}$ 上的射影为 $\htmlData{tutor-start=0,tutor-end=1}{D}$，即 $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{D} \htmlData{tutor-start=7,tutor-end=12}{\perp}$ 平面 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}$。 在 Rt$\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{D}\htmlData{tutor-start=16,tutor-end=17}{A}$ 中，$\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{D} \htmlData{tutor-start=7,tutor-end=8}{=} \sqrt{\htmlData{tutor-start=15,tutor-end=16}{A}\htmlData{tutor-start=16,tutor-end=17}{A}_{\htmlData{tutor-start=19,tutor-end=20}{1}}^{\htmlData{tutor-start=23,tutor-end=24}{2}} \htmlData{tutor-start=26,tutor-end=27}{-} \htmlData{tutor-start=28,tutor-end=29}{A}\htmlData{tutor-start=29,tutor-end=30}{D}^{\htmlData{tutor-start=32,tutor-end=33}{2}}} \htmlData{tutor-start=36,tutor-end=37}{=} \sqrt{\htmlData{tutor-start=44,tutor-end=45}{a}^{\htmlData{tutor-start=47,tutor-end=48}{2}} \htmlData{tutor-start=50,tutor-end=51}{-} \htmlData{tutor-start=52,tutor-end=53}{(}\frac{\sqrt{\htmlData{tutor-start=65,tutor-end=66}{3}}}{\htmlData{tutor-start=69,tutor-end=70}{2}}\htmlData{tutor-start=71,tutor-end=72}{a}\htmlData{tutor-start=72,tutor-end=73}{)}^{\htmlData{tutor-start=75,tutor-end=76}{2}}} \htmlData{tutor-start=79,tutor-end=80}{=} \sqrt{\htmlData{tutor-start=87,tutor-end=88}{a}^{\htmlData{tutor-start=90,tutor-end=91}{2}} \htmlData{tutor-start=93,tutor-end=94}{-} \frac{\htmlData{tutor-start=101,tutor-end=102}{3}}{\htmlData{tutor-start=104,tutor-end=105}{4}}\htmlData{tutor-start=106,tutor-end=107}{a}^{\htmlData{tutor-start=109,tutor-end=110}{2}}} \htmlData{tutor-start=113,tutor-end=114}{=} \frac{\htmlData{tutor-start=121,tutor-end=122}{1}}{\htmlData{tutor-start=124,tutor-end=125}{2}}\htmlData{tutor-start=126,tutor-end=127}{a}$。 接下来计算 $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{B}$ 的长度。连接 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{D}$。在 Rt$\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{D}\htmlData{tutor-start=16,tutor-end=17}{B}$ 中，$\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{D} \htmlData{tutor-start=3,tutor-end=4}{=} \frac{\htmlData{tutor-start=11,tutor-end=12}{1}}{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{a}$，$\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{D} \htmlData{tutor-start=7,tutor-end=8}{=} \frac{\htmlData{tutor-start=15,tutor-end=16}{1}}{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{a}$。 所以 $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{B} \htmlData{tutor-start=7,tutor-end=8}{=} \sqrt{\htmlData{tutor-start=15,tutor-end=16}{A}_{\htmlData{tutor-start=18,tutor-end=19}{1}}\htmlData{tutor-start=20,tutor-end=21}{D}^{\htmlData{tutor-start=23,tutor-end=24}{2}} \htmlData{tutor-start=26,tutor-end=27}{+} \htmlData{tutor-start=28,tutor-end=29}{B}\htmlData{tutor-start=29,tutor-end=30}{D}^{\htmlData{tutor-start=32,tutor-end=33}{2}}} \htmlData{tutor-start=36,tutor-end=37}{=} \sqrt{\frac{\htmlData{tutor-start=50,tutor-end=51}{1}}{\htmlData{tutor-start=53,tutor-end=54}{4}}\htmlData{tutor-start=55,tutor-end=56}{a}^{\htmlData{tutor-start=58,tutor-end=59}{2}} \htmlData{tutor-start=61,tutor-end=62}{+} \frac{\htmlData{tutor-start=69,tutor-end=70}{1}}{\htmlData{tutor-start=72,tutor-end=73}{4}}\htmlData{tutor-start=74,tutor-end=75}{a}^{\htmlData{tutor-start=77,tutor-end=78}{2}}} \htmlData{tutor-start=81,tutor-end=82}{=} \frac{\sqrt{\htmlData{tutor-start=95,tutor-end=96}{2}}}{\htmlData{tutor-start=99,tutor-end=100}{2}}\htmlData{tutor-start=101,tutor-end=102}{a}$。 现在考察 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{A}_{\htmlData{tutor-start=15,tutor-end=16}{1}}$，其三边长分别为： $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{a}$ $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}_{\htmlData{tutor-start=4,tutor-end=5}{1}} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{a}$ $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{B} \htmlData{tutor-start=7,tutor-end=8}{=} \frac{\sqrt{\htmlData{tutor-start=21,tutor-end=22}{2}}}{\htmlData{tutor-start=25,tutor-end=26}{2}}\htmlData{tutor-start=27,tutor-end=28}{a}$ 由余弦定理： $\cos \htmlData{tutor-start=5,tutor-end=12}{\angle }\htmlData{tutor-start=12,tutor-end=13}{A}_{\htmlData{tutor-start=15,tutor-end=16}{1}}\htmlData{tutor-start=17,tutor-end=18}{A}\htmlData{tutor-start=18,tutor-end=19}{B} \htmlData{tutor-start=20,tutor-end=21}{=} \frac{\htmlData{tutor-start=28,tutor-end=29}{A}\htmlData{tutor-start=29,tutor-end=30}{B}^{\htmlData{tutor-start=32,tutor-end=33}{2}} \htmlData{tutor-start=35,tutor-end=36}{+} \htmlData{tutor-start=37,tutor-end=38}{A}\htmlData{tutor-start=38,tutor-end=39}{A}_{\htmlData{tutor-start=41,tutor-end=42}{1}}^{\htmlData{tutor-start=45,tutor-end=46}{2}} \htmlData{tutor-start=48,tutor-end=49}{-} \htmlData{tutor-start=50,tutor-end=51}{A}_{\htmlData{tutor-start=53,tutor-end=54}{1}}\htmlData{tutor-start=55,tutor-end=56}{B}^{\htmlData{tutor-start=58,tutor-end=59}{2}}}{\htmlData{tutor-start=62,tutor-end=63}{2} \htmlData{tutor-start=64,tutor-end=70}{\cdot }\htmlData{tutor-start=70,tutor-end=71}{A}\htmlData{tutor-start=71,tutor-end=72}{B} \htmlData{tutor-start=73,tutor-end=79}{\cdot }\htmlData{tutor-start=79,tutor-end=80}{A}\htmlData{tutor-start=80,tutor-end=81}{A}_{\htmlData{tutor-start=83,tutor-end=84}{1}}} \htmlData{tutor-start=87,tutor-end=88}{=} \frac{\htmlData{tutor-start=95,tutor-end=96}{a}^{\htmlData{tutor-start=98,tutor-end=99}{2}} \htmlData{tutor-start=101,tutor-end=102}{+} \htmlData{tutor-start=103,tutor-end=104}{a}^{\htmlData{tutor-start=106,tutor-end=107}{2}} \htmlData{tutor-start=109,tutor-end=110}{-} \frac{\htmlData{tutor-start=117,tutor-end=118}{1}}{\htmlData{tutor-start=120,tutor-end=121}{2}}\htmlData{tutor-start=122,tutor-end=123}{a}^{\htmlData{tutor-start=125,tutor-end=126}{2}}}{\htmlData{tutor-start=129,tutor-end=130}{2} \htmlData{tutor-start=131,tutor-end=137}{\cdot }\htmlData{tutor-start=137,tutor-end=138}{a} \htmlData{tutor-start=139,tutor-end=145}{\cdot }\htmlData{tutor-start=145,tutor-end=146}{a}} \htmlData{tutor-start=148,tutor-end=149}{=} \frac{\frac{\htmlData{tutor-start=162,tutor-end=163}{3}}{\htmlData{tutor-start=165,tutor-end=166}{2}}\htmlData{tutor-start=167,tutor-end=168}{a}^{\htmlData{tutor-start=170,tutor-end=171}{2}}}{\htmlData{tutor-start=174,tutor-end=175}{2}\htmlData{tutor-start=175,tutor-end=176}{a}^{\htmlData{tutor-start=178,tutor-end=179}{2}}} \htmlData{tutor-start=182,tutor-end=183}{=} \frac{\htmlData{tutor-start=190,tutor-end=191}{3}}{\htmlData{tutor-start=193,tutor-end=194}{4}}$。 因为 $\frac{\htmlData{tutor-start=6,tutor-end=7}{3}}{\htmlData{tutor-start=9,tutor-end=10}{4}} \htmlData{tutor-start=12,tutor-end=13}{>} \htmlData{tutor-start=14,tutor-end=15}{0}$，所以该角为锐角，即为异面直线所成角。

余弦函数 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\cos \htmlData{tutor-start=7,tutor-end=8}{z}$ 的图象关于点 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{k}\htmlData{tutor-start=2,tutor-end=6}{\pi }\htmlData{tutor-start=6,tutor-end=7}{+} \frac{\htmlData{tutor-start=14,tutor-end=17}{\pi}}{\htmlData{tutor-start=19,tutor-end=20}{2}}\htmlData{tutor-start=21,tutor-end=22}{,} \htmlData{tutor-start=23,tutor-end=24}{0}\htmlData{tutor-start=24,tutor-end=25}{)}$ 中心对称，代入题设点坐标求解 $\htmlData{tutor-start=0,tutor-end=7}{\varphi}$。

详细展开： 函数 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}\cos\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=18}{\varphi}\htmlData{tutor-start=18,tutor-end=19}{)}$ 的对称中心满足相位 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=11}{\varphi }\htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{k}\htmlData{tutor-start=14,tutor-end=18}{\pi }\htmlData{tutor-start=18,tutor-end=19}{+} \frac{\htmlData{tutor-start=26,tutor-end=29}{\pi}}{\htmlData{tutor-start=31,tutor-end=32}{2}}\htmlData{tutor-start=33,tutor-end=34}{,} \htmlData{tutor-start=35,tutor-end=36}{k} \htmlData{tutor-start=37,tutor-end=41}{\in }\mathbf{\htmlData{tutor-start=49,tutor-end=50}{Z}}$。 此时函数值为 0，符合中心对称点纵坐标为 0 的特征。 将点 $\htmlData{tutor-start=0,tutor-end=1}{(}\frac{\htmlData{tutor-start=7,tutor-end=8}{4}\htmlData{tutor-start=8,tutor-end=11}{\pi}}{\htmlData{tutor-start=13,tutor-end=14}{3}}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{0}\htmlData{tutor-start=18,tutor-end=19}{)}$ 的横坐标 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{4}\htmlData{tutor-start=11,tutor-end=14}{\pi}}{\htmlData{tutor-start=16,tutor-end=17}{3}}$ 代入上述条件： $\htmlData{tutor-start=0,tutor-end=1}{2} \htmlData{tutor-start=2,tutor-end=9}{\times }\frac{\htmlData{tutor-start=15,tutor-end=16}{4}\htmlData{tutor-start=16,tutor-end=19}{\pi}}{\htmlData{tutor-start=21,tutor-end=22}{3}} \htmlData{tutor-start=24,tutor-end=25}{+} \htmlData{tutor-start=26,tutor-end=34}{\varphi }\htmlData{tutor-start=34,tutor-end=35}{=} \htmlData{tutor-start=36,tutor-end=37}{k}\htmlData{tutor-start=37,tutor-end=41}{\pi }\htmlData{tutor-start=41,tutor-end=42}{+} \frac{\htmlData{tutor-start=49,tutor-end=52}{\pi}}{\htmlData{tutor-start=54,tutor-end=55}{2}}$ $\frac{\htmlData{tutor-start=6,tutor-end=7}{8}\htmlData{tutor-start=7,tutor-end=10}{\pi}}{\htmlData{tutor-start=12,tutor-end=13}{3}} \htmlData{tutor-start=15,tutor-end=16}{+} \htmlData{tutor-start=17,tutor-end=25}{\varphi }\htmlData{tutor-start=25,tutor-end=26}{=} \htmlData{tutor-start=27,tutor-end=28}{k}\htmlData{tutor-start=28,tutor-end=32}{\pi }\htmlData{tutor-start=32,tutor-end=33}{+} \frac{\htmlData{tutor-start=40,tutor-end=43}{\pi}}{\htmlData{tutor-start=45,tutor-end=46}{2}}$ 解得： $\htmlData{tutor-start=0,tutor-end=8}{\varphi }\htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{k}\htmlData{tutor-start=11,tutor-end=15}{\pi }\htmlData{tutor-start=15,tutor-end=16}{+} \frac{\htmlData{tutor-start=23,tutor-end=26}{\pi}}{\htmlData{tutor-start=28,tutor-end=29}{2}} \htmlData{tutor-start=31,tutor-end=32}{-} \frac{\htmlData{tutor-start=39,tutor-end=40}{8}\htmlData{tutor-start=40,tutor-end=43}{\pi}}{\htmlData{tutor-start=45,tutor-end=46}{3}} \htmlData{tutor-start=48,tutor-end=49}{=} \htmlData{tutor-start=50,tutor-end=51}{k}\htmlData{tutor-start=51,tutor-end=55}{\pi }\htmlData{tutor-start=55,tutor-end=56}{+} \frac{\htmlData{tutor-start=63,tutor-end=64}{3}\htmlData{tutor-start=64,tutor-end=68}{\pi }\htmlData{tutor-start=68,tutor-end=69}{-} \htmlData{tutor-start=70,tutor-end=71}{1}\htmlData{tutor-start=71,tutor-end=72}{6}\htmlData{tutor-start=72,tutor-end=75}{\pi}}{\htmlData{tutor-start=77,tutor-end=78}{6}} \htmlData{tutor-start=80,tutor-end=81}{=} \htmlData{tutor-start=82,tutor-end=83}{k}\htmlData{tutor-start=83,tutor-end=87}{\pi }\htmlData{tutor-start=87,tutor-end=88}{-} \frac{\htmlData{tutor-start=95,tutor-end=96}{1}\htmlData{tutor-start=96,tutor-end=97}{3}\htmlData{tutor-start=97,tutor-end=100}{\pi}}{\htmlData{tutor-start=102,tutor-end=103}{6}}$。

遍历整数 $\htmlData{tutor-start=0,tutor-end=1}{k}$，找到使 $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=8}{\varphi}\htmlData{tutor-start=8,tutor-end=9}{|}$ 最小的值。

详细展开： $\htmlData{tutor-start=0,tutor-end=7}{\varphi}_{\htmlData{tutor-start=9,tutor-end=10}{k}} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{k}\htmlData{tutor-start=15,tutor-end=19}{\pi }\htmlData{tutor-start=19,tutor-end=20}{-} \frac{\htmlData{tutor-start=27,tutor-end=28}{1}\htmlData{tutor-start=28,tutor-end=29}{3}\htmlData{tutor-start=29,tutor-end=32}{\pi}}{\htmlData{tutor-start=34,tutor-end=35}{6}}$。 当 $\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}$ 时，$\htmlData{tutor-start=0,tutor-end=7}{\varphi}_{\htmlData{tutor-start=9,tutor-end=10}{1}} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=18}{\pi }\htmlData{tutor-start=18,tutor-end=19}{-} \frac{\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{3}\htmlData{tutor-start=28,tutor-end=31}{\pi}}{\htmlData{tutor-start=33,tutor-end=34}{6}} \htmlData{tutor-start=36,tutor-end=37}{=} \frac{\htmlData{tutor-start=44,tutor-end=45}{6}\htmlData{tutor-start=45,tutor-end=49}{\pi }\htmlData{tutor-start=49,tutor-end=50}{-} \htmlData{tutor-start=51,tutor-end=52}{1}\htmlData{tutor-start=52,tutor-end=53}{3}\htmlData{tutor-start=53,tutor-end=56}{\pi}}{\htmlData{tutor-start=58,tutor-end=59}{6}} \htmlData{tutor-start=61,tutor-end=62}{=} \htmlData{tutor-start=63,tutor-end=64}{-}\frac{\htmlData{tutor-start=70,tutor-end=71}{7}\htmlData{tutor-start=71,tutor-end=74}{\pi}}{\htmlData{tutor-start=76,tutor-end=77}{6}}$，$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=8}{\varphi}_{\htmlData{tutor-start=10,tutor-end=11}{1}}\htmlData{tutor-start=12,tutor-end=13}{|} \htmlData{tutor-start=14,tutor-end=15}{=} \frac{\htmlData{tutor-start=22,tutor-end=23}{7}\htmlData{tutor-start=23,tutor-end=26}{\pi}}{\htmlData{tutor-start=28,tutor-end=29}{6}}$。 当 $\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}$ 时，$\htmlData{tutor-start=0,tutor-end=7}{\varphi}_{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=19}{\pi }\htmlData{tutor-start=19,tutor-end=20}{-} \frac{\htmlData{tutor-start=27,tutor-end=28}{1}\htmlData{tutor-start=28,tutor-end=29}{3}\htmlData{tutor-start=29,tutor-end=32}{\pi}}{\htmlData{tutor-start=34,tutor-end=35}{6}} \htmlData{tutor-start=37,tutor-end=38}{=} \frac{\htmlData{tutor-start=45,tutor-end=46}{1}\htmlData{tutor-start=46,tutor-end=47}{2}\htmlData{tutor-start=47,tutor-end=51}{\pi }\htmlData{tutor-start=51,tutor-end=52}{-} \htmlData{tutor-start=53,tutor-end=54}{1}\htmlData{tutor-start=54,tutor-end=55}{3}\htmlData{tutor-start=55,tutor-end=58}{\pi}}{\htmlData{tutor-start=60,tutor-end=61}{6}} \htmlData{tutor-start=63,tutor-end=64}{=} \htmlData{tutor-start=65,tutor-end=66}{-}\frac{\htmlData{tutor-start=72,tutor-end=75}{\pi}}{\htmlData{tutor-start=77,tutor-end=78}{6}}$，$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=8}{\varphi}_{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{|} \htmlData{tutor-start=14,tutor-end=15}{=} \frac{\htmlData{tutor-start=22,tutor-end=25}{\pi}}{\htmlData{tutor-start=27,tutor-end=28}{6}}$。 当 $\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}$ 时，$\htmlData{tutor-start=0,tutor-end=7}{\varphi}_{\htmlData{tutor-start=9,tutor-end=10}{3}} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{3}\htmlData{tutor-start=15,tutor-end=19}{\pi }\htmlData{tutor-start=19,tutor-end=20}{-} \frac{\htmlData{tutor-start=27,tutor-end=28}{1}\htmlData{tutor-start=28,tutor-end=29}{3}\htmlData{tutor-start=29,tutor-end=32}{\pi}}{\htmlData{tutor-start=34,tutor-end=35}{6}} \htmlData{tutor-start=37,tutor-end=38}{=} \frac{\htmlData{tutor-start=45,tutor-end=46}{1}\htmlData{tutor-start=46,tutor-end=47}{8}\htmlData{tutor-start=47,tutor-end=51}{\pi }\htmlData{tutor-start=51,tutor-end=52}{-} \htmlData{tutor-start=53,tutor-end=54}{1}\htmlData{tutor-start=54,tutor-end=55}{3}\htmlData{tutor-start=55,tutor-end=58}{\pi}}{\htmlData{tutor-start=60,tutor-end=61}{6}} \htmlData{tutor-start=63,tutor-end=64}{=} \frac{\htmlData{tutor-start=71,tutor-end=72}{5}\htmlData{tutor-start=72,tutor-end=75}{\pi}}{\htmlData{tutor-start=77,tutor-end=78}{6}}$，$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=8}{\varphi}_{\htmlData{tutor-start=10,tutor-end=11}{3}}\htmlData{tutor-start=12,tutor-end=13}{|} \htmlData{tutor-start=14,tutor-end=15}{=} \frac{\htmlData{tutor-start=22,tutor-end=23}{5}\htmlData{tutor-start=23,tutor-end=26}{\pi}}{\htmlData{tutor-start=28,tutor-end=29}{6}}$。 比较可知，当 $\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}$ 时，$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=8}{\varphi}\htmlData{tutor-start=8,tutor-end=9}{|}$ 取得最小值 $\frac{\htmlData{tutor-start=6,tutor-end=9}{\pi}}{\htmlData{tutor-start=11,tutor-end=12}{6}}$。 对比选项，(A) 符合。

设切点为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}_{\htmlData{tutor-start=4,tutor-end=5}{0}}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{y}_{\htmlData{tutor-start=11,tutor-end=12}{0}}\htmlData{tutor-start=13,tutor-end=14}{)}$，利用导数几何意义和点在直线上、曲线上三个条件列式。

详细展开： 曲线方程：$\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \ln\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{a}\htmlData{tutor-start=11,tutor-end=12}{)}$，定义域 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{a}$。 直线方程：$\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{1}$，斜率 $\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}$。 设切点横坐标为 $\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}$。 1. 导数条件：$\htmlData{tutor-start=0,tutor-end=1}{y}' \htmlData{tutor-start=3,tutor-end=4}{=} \frac{\htmlData{tutor-start=11,tutor-end=12}{1}}{\htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{a}}$。在切点处，$\htmlData{tutor-start=0,tutor-end=1}{y}'\htmlData{tutor-start=2,tutor-end=3}{|}_{\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{x}_{\htmlData{tutor-start=10,tutor-end=11}{0}}} \htmlData{tutor-start=14,tutor-end=15}{=} \frac{\htmlData{tutor-start=22,tutor-end=23}{1}}{\htmlData{tutor-start=25,tutor-end=26}{x}_{\htmlData{tutor-start=28,tutor-end=29}{0}}\htmlData{tutor-start=30,tutor-end=31}{+}\htmlData{tutor-start=31,tutor-end=32}{a}} \htmlData{tutor-start=34,tutor-end=35}{=} \htmlData{tutor-start=36,tutor-end=37}{1}$。 由此得：$\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{a} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{1} \implies \htmlData{tutor-start=23,tutor-end=24}{x}_{\htmlData{tutor-start=26,tutor-end=27}{0}} \htmlData{tutor-start=29,tutor-end=30}{=} \htmlData{tutor-start=31,tutor-end=32}{1} \htmlData{tutor-start=33,tutor-end=34}{-} \htmlData{tutor-start=35,tutor-end=36}{a}$。 2. 点在直线上：$\htmlData{tutor-start=0,tutor-end=1}{y}_{\htmlData{tutor-start=3,tutor-end=4}{0}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{x}_{\htmlData{tutor-start=11,tutor-end=12}{0}} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{1}$。 3. 点在曲线上：$\htmlData{tutor-start=0,tutor-end=1}{y}_{\htmlData{tutor-start=3,tutor-end=4}{0}} \htmlData{tutor-start=6,tutor-end=7}{=} \ln\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{x}_{\htmlData{tutor-start=15,tutor-end=16}{0}} \htmlData{tutor-start=18,tutor-end=19}{+} \htmlData{tutor-start=20,tutor-end=21}{a}\htmlData{tutor-start=21,tutor-end=22}{)}$。 将 $\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}$ 代入曲线方程，得 $\htmlData{tutor-start=0,tutor-end=1}{y}_{\htmlData{tutor-start=3,tutor-end=4}{0}} \htmlData{tutor-start=6,tutor-end=7}{=} \ln\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{)} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{0}$。 所以切点纵坐标 $\htmlData{tutor-start=0,tutor-end=1}{y}_{\htmlData{tutor-start=3,tutor-end=4}{0}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{0}$。

利用切点坐标满足直线方程求解 $\htmlData{tutor-start=0,tutor-end=1}{a}$。

详细展开： 已知 $\htmlData{tutor-start=0,tutor-end=1}{y}_{\htmlData{tutor-start=3,tutor-end=4}{0}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{0}$，代入直线方程 $\htmlData{tutor-start=0,tutor-end=1}{y}_{\htmlData{tutor-start=3,tutor-end=4}{0}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{x}_{\htmlData{tutor-start=11,tutor-end=12}{0}} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{1}$： $\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{x}_{\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=23}{} + 1 \implies }\htmlData{tutor-start=23,tutor-end=24}{x}_{\htmlData{tutor-start=26,tutor-end=27}{0}} \htmlData{tutor-start=29,tutor-end=30}{=} \htmlData{tutor-start=31,tutor-end=32}{-}\htmlData{tutor-start=32,tutor-end=33}{1}$。 此前已得关系式 $\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{1} \htmlData{tutor-start=10,tutor-end=11}{-} \htmlData{tutor-start=12,tutor-end=13}{a}$。 将 $\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{1}$ 代入： $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{1} \htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=23}{ a \implies a =} \htmlData{tutor-start=24,tutor-end=25}{2}$。 检验：当 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}$ 时，曲线 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\ln\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{)}$，切点 $\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{1}$，$\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{>}\htmlData{tutor-start=10,tutor-end=11}{0}$，定义域满足。 故 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}$。

确定 P, Q 在各自平面内相对于棱 l 的位置约束。

详细展开： 设二面角 $\htmlData{tutor-start=0,tutor-end=6}{\alpha}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{l}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=14}{\beta}$ 的棱为 $\htmlData{tutor-start=0,tutor-end=1}{l}$。 $\htmlData{tutor-start=0,tutor-end=1}{P} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=12}{\alpha}$，$\htmlData{tutor-start=0,tutor-end=1}{P}$ 到 $\htmlData{tutor-start=0,tutor-end=5}{\beta}$ 的距离为 $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{3}}$。 作 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{M} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=14}{\beta}$ 于 $\htmlData{tutor-start=0,tutor-end=1}{M}$，则 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{M} \htmlData{tutor-start=3,tutor-end=4}{=} \sqrt{\htmlData{tutor-start=11,tutor-end=12}{3}}$。 过 $\htmlData{tutor-start=0,tutor-end=1}{P}$ 作 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{H} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{l}$ 于 $\htmlData{tutor-start=0,tutor-end=1}{H}$，连接 $\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{H}$。由三垂线定理逆定理，$\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{H} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{l}$。 $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{P}\htmlData{tutor-start=8,tutor-end=9}{H}\htmlData{tutor-start=9,tutor-end=10}{M}$ 即为二面角的平面角，大小为 $60^\circ$。 在 Rt$\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{P}\htmlData{tutor-start=11,tutor-end=12}{M}\htmlData{tutor-start=12,tutor-end=13}{H}$ 中，$PM = PH \sin 60^\circ$。 即 $\sqrt{3} = PH \cdot \frac{\sqrt{3}}{2} \implies PH = 2$。 这说明点 $\htmlData{tutor-start=0,tutor-end=1}{P}$ 在平面 $\htmlData{tutor-start=0,tutor-end=6}{\alpha}$ 内，且到棱 $\htmlData{tutor-start=0,tutor-end=1}{l}$ 的距离恒为 2。即 $\htmlData{tutor-start=0,tutor-end=1}{P}$ 在 $\htmlData{tutor-start=0,tutor-end=6}{\alpha}$ 内平行于 $\htmlData{tutor-start=0,tutor-end=1}{l}$ 且距离为 2 的直线上运动。 同理，$\htmlData{tutor-start=0,tutor-end=1}{Q} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=11}{\beta}$，$\htmlData{tutor-start=0,tutor-end=1}{Q}$ 到 $\htmlData{tutor-start=0,tutor-end=6}{\alpha}$ 的距离为 $\htmlData{tutor-start=0,tutor-end=1}{2}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{3}}$。 设 $\htmlData{tutor-start=0,tutor-end=1}{Q}$ 到 $\htmlData{tutor-start=0,tutor-end=1}{l}$ 的距离为 $\htmlData{tutor-start=0,tutor-end=1}{Q}\htmlData{tutor-start=1,tutor-end=2}{K}$（$\htmlData{tutor-start=0,tutor-end=1}{K}$ 在 $\htmlData{tutor-start=0,tutor-end=1}{l}$ 上）。 $\htmlData{tutor-start=0,tutor-end=1}{Q}$ 到 $\htmlData{tutor-start=0,tutor-end=6}{\alpha}$ 的距离 $d_{Q} = QK \sin 60^\circ = 2\sqrt{3}$。 $\htmlData{tutor-start=0,tutor-end=1}{Q}\htmlData{tutor-start=1,tutor-end=2}{K} \htmlData{tutor-start=3,tutor-end=9}{\cdot }\frac{\sqrt{\htmlData{tutor-start=21,tutor-end=22}{3}}}{\htmlData{tutor-start=25,tutor-end=26}{2}} \htmlData{tutor-start=28,tutor-end=29}{=} \htmlData{tutor-start=30,tutor-end=31}{2}\sqrt{\htmlData{tutor-start=37,tutor-end=38}{3}} \implies \htmlData{tutor-start=49,tutor-end=50}{Q}\htmlData{tutor-start=50,tutor-end=51}{K} \htmlData{tutor-start=52,tutor-end=53}{=} \htmlData{tutor-start=54,tutor-end=55}{4}$。 这说明点 $\htmlData{tutor-start=0,tutor-end=1}{Q}$ 在平面 $\htmlData{tutor-start=0,tutor-end=5}{\beta}$ 内，且到棱 $\htmlData{tutor-start=0,tutor-end=1}{l}$ 的距离恒为 4。

将二面角沿棱展开为平面图形，利用两点之间线段最短求解。

详细展开： 将平面 $\htmlData{tutor-start=0,tutor-end=6}{\alpha}$ 和 $\htmlData{tutor-start=0,tutor-end=5}{\beta}$ 沿棱 $\htmlData{tutor-start=0,tutor-end=1}{l}$ 展开铺平。由于二面角为 $60^\circ$，展开后两平面构成一个 $60^\circ$ 的角？不，是将两个半平面展成一个平面，此时原来的二面角角度体现在坐标系的旋转或向量夹角上。 更直观的方法：建立局部坐标系。 设棱 $\htmlData{tutor-start=0,tutor-end=1}{l}$ 为 $\htmlData{tutor-start=0,tutor-end=1}{z}$ 轴。 在垂直于 $\htmlData{tutor-start=0,tutor-end=1}{l}$ 的截面中，$\htmlData{tutor-start=0,tutor-end=1}{P}$ 的投影 $\htmlData{tutor-start=0,tutor-end=1}{P}'$ 到原点（棱）距离为 2，$\htmlData{tutor-start=0,tutor-end=1}{Q}$ 的投影 $\htmlData{tutor-start=0,tutor-end=1}{Q}'$ 到原点距离为 4。 $\htmlData{tutor-start=0,tutor-end=1}{P}'$ 与 $\htmlData{tutor-start=0,tutor-end=1}{Q}'$ 的夹角为 $60^\circ$。 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{Q}$ 在 $\htmlData{tutor-start=0,tutor-end=1}{z}$ 方向（平行于棱方向）可以任意移动。要使 $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{Q}\htmlData{tutor-start=3,tutor-end=4}{|}$ 最小，应使 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{Q}$ 在平行于棱的方向上坐标相同（即投影重合）。 此时 $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{Q}\htmlData{tutor-start=3,tutor-end=4}{|}$ 的最小值即为截面中 $\htmlData{tutor-start=0,tutor-end=1}{P}'$ 与 $\htmlData{tutor-start=0,tutor-end=1}{Q}'$ 之间的距离。 在 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{P}'\htmlData{tutor-start=12,tutor-end=13}{O}\htmlData{tutor-start=13,tutor-end=14}{Q}'$ 中，$OP'=2, OQ'=4, \angle P'OQ' = 60^\circ$。 由余弦定理： $|P'Q'|^{2} = OP'^{2} + OQ'^{2} - 2 OP' \cdot OQ' \cos 60^\circ$ $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{P}'\htmlData{tutor-start=3,tutor-end=4}{Q}'\htmlData{tutor-start=5,tutor-end=6}{|}^{\htmlData{tutor-start=8,tutor-end=9}{2}} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{2}^{\htmlData{tutor-start=16,tutor-end=17}{2}} \htmlData{tutor-start=19,tutor-end=20}{+} \htmlData{tutor-start=21,tutor-end=22}{4}^{\htmlData{tutor-start=24,tutor-end=25}{2}} \htmlData{tutor-start=27,tutor-end=28}{-} \htmlData{tutor-start=29,tutor-end=30}{2} \htmlData{tutor-start=31,tutor-end=37}{\cdot }\htmlData{tutor-start=37,tutor-end=38}{2} \htmlData{tutor-start=39,tutor-end=45}{\cdot }\htmlData{tutor-start=45,tutor-end=46}{4} \htmlData{tutor-start=47,tutor-end=53}{\cdot }\frac{\htmlData{tutor-start=59,tutor-end=60}{1}}{\htmlData{tutor-start=62,tutor-end=63}{2}} \htmlData{tutor-start=65,tutor-end=66}{=} \htmlData{tutor-start=67,tutor-end=68}{4} \htmlData{tutor-start=69,tutor-end=70}{+} \htmlData{tutor-start=71,tutor-end=72}{1}\htmlData{tutor-start=72,tutor-end=73}{6} \htmlData{tutor-start=74,tutor-end=75}{-} \htmlData{tutor-start=76,tutor-end=77}{8} \htmlData{tutor-start=78,tutor-end=79}{=} \htmlData{tutor-start=80,tutor-end=81}{1}\htmlData{tutor-start=81,tutor-end=82}{2}$。 $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{Q}\htmlData{tutor-start=3,tutor-end=4}{|}_{\min} \htmlData{tutor-start=12,tutor-end=13}{=} \sqrt{\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{2}} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{2}\sqrt{\htmlData{tutor-start=33,tutor-end=34}{3}}$。

将 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}$ 为奇函数转化为 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 的对称性。

详细展开： 1. $\htmlData{tutor-start=0,tutor-end=1}{g}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{f}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{)}$ 是奇函数 $\implies \htmlData{tutor-start=9,tutor-end=10}{g}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{)} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{g}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{x}\htmlData{tutor-start=21,tutor-end=22}{)}$。 即 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{)} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{f}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{)}$。 令 $\htmlData{tutor-start=0,tutor-end=1}{t} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{1} \htmlData{tutor-start=8,tutor-end=23}{\implies x = t-}\htmlData{tutor-start=23,tutor-end=24}{1}$，则 $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{1} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{t}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{)}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{1} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{t}$。 所以 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{t}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{f}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{t}\htmlData{tutor-start=13,tutor-end=14}{)}$。 这表明 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 关于点 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}$ 中心对称。 即 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{f}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{)}$ 或 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{f}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{)}$。

2. $\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{f}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{)}$ 是奇函数 $\implies \htmlData{tutor-start=9,tutor-end=10}{h}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{)} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{h}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{x}\htmlData{tutor-start=21,tutor-end=22}{)}$。 即 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{)} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{f}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{)}$。 令 $\htmlData{tutor-start=0,tutor-end=1}{t} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1} \htmlData{tutor-start=8,tutor-end=23}{\implies x = t+}\htmlData{tutor-start=23,tutor-end=24}{1}$，则 $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{t}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{)}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{1} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{t}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{2}$。 所以 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{t}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{)} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{f}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{t}\htmlData{tutor-start=14,tutor-end=15}{)}$。 这表明 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 关于点 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)}$ 中心对称。 即 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{)} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{f}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{x}\htmlData{tutor-start=17,tutor-end=18}{)}$ 或 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{f}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{)}$。

利用两个对称中心推导周期性，并逐项验证。

详细展开： 由 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{f}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{)}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{)} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{f}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{)}$，可得： $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{f}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{)}$。 令 $\htmlData{tutor-start=0,tutor-end=1}{t} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{x}$，则 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{t}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{f}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{t}\htmlData{tutor-start=15,tutor-end=16}{)}$。 即 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{t}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{f}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{t}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{)}$。 令 $\htmlData{tutor-start=0,tutor-end=1}{u} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{t}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{2}$，则 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{u}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{4}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{f}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{u}\htmlData{tutor-start=12,tutor-end=13}{)}$。 所以 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 是周期为 $\htmlData{tutor-start=0,tutor-end=1}{T}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{4}$ 的周期函数。

现在验证选项： (A) $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 是偶函数？不一定。例如 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\sin\htmlData{tutor-start=9,tutor-end=10}{(}\frac{\htmlData{tutor-start=16,tutor-end=19}{\pi}}{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{x}\htmlData{tutor-start=24,tutor-end=25}{)}$ 满足题意（对称中心 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}$ 等），它是奇函数。排除。 (B) $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 是奇函数？不一定。虽然上面的例子是，但可以构造非奇非偶的周期函数满足对称性吗？ 由 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 关于 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{)}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}$ 对称，且周期为4。 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{f}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{)} \htmlData{tutor-start=13,tutor-end=14}{=} \htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{f}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{0}\htmlData{tutor-start=21,tutor-end=22}{)}\htmlData{tutor-start=22,tutor-end=23}{)} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{f}\htmlData{tutor-start=27,tutor-end=28}{(}\htmlData{tutor-start=28,tutor-end=29}{0}\htmlData{tutor-start=29,tutor-end=30}{)}$，无矛盾。 若 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 关于原点对称，则需 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{f}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{)}$。 已知 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{f}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{4}\htmlData{tutor-start=14,tutor-end=15}{)}$? 不直接。 让我们看 (D)。

(D) $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{3}\htmlData{tutor-start=5,tutor-end=6}{)}$ 是奇函数？ 令 $\htmlData{tutor-start=0,tutor-end=1}{F}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{f}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{3}\htmlData{tutor-start=12,tutor-end=13}{)}$。需验证 $\htmlData{tutor-start=0,tutor-end=1}{F}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{F}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{)}$，即 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{)} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{f}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{3}\htmlData{tutor-start=16,tutor-end=17}{)}$。 这等价于 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 关于点 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{3}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}$ 中心对称。 已知 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 关于 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}$ 对称，且 $\htmlData{tutor-start=0,tutor-end=1}{T}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{4}$。 对称中心每隔 $\htmlData{tutor-start=0,tutor-end=1}{T}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{2} \htmlData{tutor-start=4,tutor-end=5}{=} \htmlData{tutor-start=6,tutor-end=7}{2}$ 重复出现。 对称中心序列：$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{3}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{3}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{5}\htmlData{tutor-start=15,tutor-end=16}{,} \dots$ 以及 $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{,} \dots$。 因为 $\htmlData{tutor-start=0,tutor-end=1}{1}$ 是对称中心，所以 $\htmlData{tutor-start=0,tutor-end=1}{1} \htmlData{tutor-start=2,tutor-end=3}{+} \frac{\htmlData{tutor-start=10,tutor-end=11}{T}}{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{2}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{3}$ 也是对称中心。 所以 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 关于 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{3}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{)}$ 对称。 故 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{f}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{3}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{)}$ 成立。 即 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{)}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{3}\htmlData{tutor-start=10,tutor-end=11}{)} \htmlData{tutor-start=12,tutor-end=13}{.}\htmlData{tutor-start=13,tutor-end=14}{.}\htmlData{tutor-start=14,tutor-end=15}{.}$ 简单说，$\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{f}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{)}$ 的图象是将 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 左移3单位。 原对称中心 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{3}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{)}$ 移至 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{)}$。 所以新函数关于原点对称，是奇函数。 故 (D) 正确。

(C) $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{f}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{)}$？ 我们推出的是 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{4}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{f}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{)}$。且 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{f}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{)}$（由半周期对称性）。除非 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=11}{\equiv }\htmlData{tutor-start=11,tutor-end=12}{0}$，否则 (C) 错。