解题方法
1 . 如图,已知正方体
的棱长为1,E为
的中点,P为对角线上
的一个动点,过P作与平面ACE平行的平面,则此平面截正方体所得的截面( )
![](https://img.xkw.com/dksih/QBM/editorImg/2022/11/27/1205f59e-9d55-4904-923a-80e402aec6b5.png?resizew=185)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/4a46a98f42df8dd1d7ba89fe7e6961a5.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/394c5d2f55221975503be8aa18022480.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/69a7bcc1efb8a2ff57d64b6d057da463.png)
![](https://img.xkw.com/dksih/QBM/editorImg/2022/11/27/1205f59e-9d55-4904-923a-80e402aec6b5.png?resizew=185)
A.截面不可能是五边形 |
B.截面可以是正六边形 |
C.P从D点向![]() |
D.截面面积的最大值为![]() |
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2 . 群的概念由法国天才数学家伽罗瓦(1811-1832)在19世纪30年代开创,群论虽起源于对代数多项式方程的研究,但在量子力学、晶体结构学等其他学科中也有十分广泛的应用.设
是一个非空集合,“
”是一个适用于
中元素的运算,若同时满足以下四个条件,则称
对“
”构成一个群:(1)封闭性,即若
,则存在唯一确定的
,使得
;(2)结合律成立,即对
中任意元素
都有
;(3)单位元存在,即存在
,对任意
,满足
,则
称为单位元;(4)逆元存在,即任意
,存在
,使得
,则称
与
互为逆元,
记作
.一般地,
可简记作
可简记作
可简记作
,以此类推.正八边形
的中心为
.以
表示恒等变换,即不对正八边形作任何变换;以
表示以点
为中心,将正八边形逆时针旋转
的旋转变换;以
表示以
所在直线为轴,将正八边形进行轴对称变换.定义运算“
”表示复合变换,即
表示将正八边形先进行
变换再进行
变换的变换.以形如
,并规定
的变换为元素,可组成集合
,则
对运算“
”可构成群,称之为“正八边形的对称变换群”,记作
.则以下关于
及其元素的说法中,正确的有( )
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A.![]() ![]() |
B.![]() ![]() |
C.![]() |
D.![]() |
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