书签分享收藏举报版权申诉 / 9

立即下载

当前位置：首页 > 教育专区 > 高考资料 > 2022年东城高三一模数学文含答案 .pdf

2022年东城高三一模数学文含答案 .pdf

上传人：Q****o

文档编号：56653045

上传时间：2022-11-02

格式：PDF

页数：9

大小：167.02KB

( 4.5 )

《2022年东城高三一模数学文含答案 .pdf》由会员分享，可在线阅读，更多相关《2022年东城高三一模数学文含答案 .pdf（9页珍藏版）》请在淘文阁 - 分享文档赚钱的网站上搜索。

1、北京市东城区 2011-2012 学年第二学期综合练习（一）高三数学（文科）学校 _班级 _姓名 _考号 _ 第卷（选择题共 40 分）一、本大题共8 小题，每小题5 分，共 40 分。在每小题列出的四个选项中，选出符合题目要求的一项。（1）若a，bR，i是虚数单位，且(2)i1iba，则ab的值为（A）1（B）2（C）3（D）4（2）若集合,02mA，2,1B，则“1m”是“2,1,0BA”的（A）充分不必要条件（B）必要不充分条件（C）充分必要条件（D）既不充分也不必要条件（3）若点(,)P x y在不等式组,2yxyxx表示的平面区域内，则2zxy的最大值为（A）0（B）2（C）4（D）

2、6（4）已知x，y，zR，若1，x，y，z，3成等差数列，则xyz的值为（A）2（B）4（C）6（D）8（5）右图给出的是计算1001.81614121的值的一个程序框图，其中判断框内应填入的条件是（A）50i（B）25i（C）50i（D）25i（6）已知2sin(45)10o，且090oo，则cos的值为（A）513（B）1213（C）35（D）45（7）已知函数()()()f xxaxb(其中)ab的图象如右图所示，则函数()xg xab的图象大致为（A）（B）（C）（D）俯视图左视图主视图21122（8）设集合10,)2A，1,12B，函数1,()22(1),.xxAfxxxB若0

3、xA，且0()ff xA，则0 x的取值范围是（A）(41,0 （B）(21,41 （C）(21,41)（D）0，83 第卷（共 110 分）二、填空题：本大题共6 小题，每小题5 分，共 30 分。（9）已知一个四棱锥的三视图如图所示，则该四棱锥的体积是.(10)命题“000(0,),tansin2xxx”的否定是 .(11)在如图所示的茎叶图中，乙组数据的中位数是；若从甲、乙两组数据中分别去掉一个最大数和一个最小数后，两组数据的平均数中较大的一组是组(12)双曲线222xy的离心率为；若抛物线2yax的焦点恰好为该双曲线的右焦点，则a的值为.(13)已知ABC中，ADBC于D，2ADBD，

4、1CD，则AB ACuuu r uu u r_(14)已知数列na，1am，mN，1,21,2nnnnnaaaaa为偶数,为奇数.若na中有且只有5个不同的数字，则m的不同取值共有个甲乙0 7 9 5 4 5 5 1 8 4 4 6 4 7 1 9 3 文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文

5、档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2

6、文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C

7、2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10

8、C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F1

9、0C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F

10、10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2

11、F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2三、解答题：本大题共6 小题，共80 分。解答应写出文字说明，演算步骤或证明过程。（15）（本小题共13 分）已知函数22()(sin2cos2)2sin 2f xxxx.（）求()f x的最小正周期；（）若函数()yg x的图象是由()yf x的图象向右平移8个单位长度得到的，当x0,4时，求()yg x的最大值和最小值.（16）（本小题共13 分）某班同学利用寒假在5 个居民小区内选择两个小区逐户进行一次“低碳

12、生活习惯”的调查，以计算每户的碳月排放量若月排放量符合低碳标准的称为“低碳族”，否则称为“非低碳族”若小区内有至少%75的住户属于“低碳族”，则称这个小区为“低碳小区”，否则称为“非低碳小区”已知备选的5 个居民小区中有三个非低碳小区，两个低碳小区.（）求所选的两个小区恰有一个为“非低碳小区”的概率；（）假定选择的“非低碳小区”为小区A，调查显示其“低碳族”的比例为21，数据如图 1 所示，经过同学们的大力宣传，三个月后，又进行了一次调查，数据如图2 所示，问这时小区A是否达到“低碳小区”的标准？O月排放量（百千克/户频率组距0.46 0.23 0.10 0.07 1 2 3 4 5 图 2

13、O月排放量（百千克/户频率组距0.30 0.25 0.20 0.15 0.05 1 2 3 4 5 图 1 6 0.14（17）（本小题共14 分）如图1，在边长为3的正三角形ABC中，E，F，P分别为AB，AC，BC上的点，且满足1AEFCCP.将AEF沿EF折起到1A EF的位置，使平面1A EF平面文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4

14、HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4

15、 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V

16、4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3

17、V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T

18、3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8

19、T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O

20、8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2PFEABCQFA1CPBEEFB，连结1A B，1A P.（如图2）（）若Q为1A B中点，求证：PQ平面1A EF；（）求证：1A EEP.图 1 图 2（18）（本小题共13 分）已知1x是函数()(2)exf xax的一个极值点(aR)（）求a的值；（）当1x，20,2x时，证明：12()()ef xf x（19）（本小题共13 分）已知椭圆2222:1(

21、0)xyCabab过点0,1，且离心率为32.（）求椭圆C的方程；（）12,AA为椭圆C的左、右顶点，直线:2 2lx与x轴交于点D，点P是椭圆C上异于12,A A的动点，直线12,A P A P分别交直线l于,E F两点.证明：DEDF恒为定值.（20）(本小题共14 分)对于函数()f x，若00()fxx，则称0 x为()f x的“不动点”；若00()ff xx，则称0 x为()f x的“稳定点”.函数()f x的“不动点”和“稳定点”的集合分别记为A和B,即()Ax f xx,()Bx ff xx.文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文

22、档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2

23、文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C

24、2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10

25、C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F1

26、0C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F

27、10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2

28、F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2（）设函数()34f xx，求集合A和B；（）求证：AB；（）设函数2()(0)f xaxbxc a，且A，求证：B.北京市东城区 2011-2012 学年第二

29、学期综合练习（一）高三数学参考答案及评分标准（文科）一、选择题（本大题共8 小题，每小题5 分，共 40 分）（1）D（2）A（3）D（4）C（5）B（6）D（7）A（8）C二、填空题（本大题共6 小题，每小题5 分，共 30 分）（9）43（10）(0,),tansin2xxx（11）84 乙（12）28（13）2（14）8注：两个空的填空题第一个空填对得3 分，第二个空填对得2 分三、解答题（本大题共6 小题，共80 分）（15）（共 13 分）解：（）因为22()(sin 2cos2)2sin2f xxxxsin 4cos4xx2sin(4)4x,6 分所以函数()f x的最小正周期为2

30、.8 分（）依题意,()yg x2 sin4()8x4 2 sin(4)4x.10 分因为04x，所以34444x.11 分当442x，即316x时，()g x取最大值2；当444x，即0 x时，()g x取最小值1.13 分文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3

31、V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T

32、3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8

33、T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O

34、8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5

35、O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P

36、5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10

37、P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2MQFA1CPBEDPFEACB（16）（共 13 分）解：（）设三个“非低碳小区”为CBA,，两个“低碳小区”为,m n 2 分用),(yx表示选定的两个小区，,x yA B C m n，则从 5 个小区中任选两个小区,所有可能的结果有10 个，它们是(,)A B，(,)A C，(,)A m，(,)A n,(,)B C,(,)B m,(,)B n,(,)C m,(,)C n，(,)m n.5 分用D表示：“选出的两个小区恰有一个为非低碳小区

38、”这一事件，则D中的结果有6 个，它们是：(,)A m，(,)A n,(,)B m,(,)B n，(,)C m,(,)C n.7 分故所求概率为63()105P D.8 分（II）由图 1 可知月碳排放量不超过300千克的成为“低碳族”.10 分由图 2 可知，三个月后的低碳族的比例为0.070.230.460.760.75，12 分所以三个月后小区A达到了“低碳小区”标准.13 分（17）（共 14 分）证明：（）取1A E中点M，连结,QMMF在1A BE中，,Q M分别为11,A B A E的中点，所以QMBE，且12QMBE因为12CFCPFAPB，所以PFBE,且12PFBE，所以Q

39、MPF，且QMPF所以四边形PQMF为平行四边形所以PQFM 5 分又因为FM平面1A EF，且PQ平面1A EF，所以PQ平面1A EF 7 分（）取BE中点D，连结DF.因为1AECF，1DE，文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O

40、6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4

41、O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F

42、4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3

43、F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R

44、3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10

45、R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M1

46、0R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2所以2AFAD，而60Ao，即ADF是正三角形.又因为1AEED,所以EFAD.所以在图2 中有1A EEF.9 分因为平面1A EF平面EFB，平面1A EF I平面EFBEF，所以1A E平面BEF.12 分又EP平面BEF，所以1A EEP.14 分（18）（共 13 分）（）解：()(2)exfxaxa，2 分由已知得0)1(f，解得1a 4 分当1a时，()(2)exf xx，在1x处取得极小值所以1a.5 分（）证明：由（）知，()(2)exf xx，()(

47、1)exfxx.当1,0 x时，0)1()(xexxf，)(xf在区间0,1单调递减；当1,2x时，()(1)0 xfxxe，)(xf在区间1,2单调递增.8分所以在区间0,2上，()f x的最小值为(1)ef，又(0)2f，(2)0f，所以在区间0,2上，()f x的最大值为(2)0f.12 分对于12,0,2x x，有12maxmin()()()()f xf xfxfx所以12()()0(e)ef xf x.13 分（19）（共 13 分）（）解：由题意可知，1b，32ca，解得2a.4分文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD1

48、0P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD

49、10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:C

50、D10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:CD10P5O8T3V4 HT7M10R3F4O6 ZY1M6J2F10C2文档编码:

文档加载中……请稍候！
如果长时间未打开，您也可以点击刷新试试。

下载文档到电脑，查找使用更方便

4.3 金币

版权申诉 word格式文档无特别注明外均可编辑修改；预览文档经过压缩，下载后原文更清晰！ 立即下载

配套讲稿：: 如PPT文件的首页显示word图标，表示该PPT已包含配套word讲稿。双击word图标可打开word文档。
特殊限制：: 部分文档作品中含有的国旗、国徽等图片，仅作为作品整体效果示例展示，禁止商用。设计者仅对作品中独创性部分享有著作权。
关键词：: 2022年东城高三一模数学文含答案 2022 东城高三一模数学答案

淘文阁 - 分享文档赚钱的网站所有资源均是用户自行上传分享，仅供网友学习交流，未经上传用户书面授权，请勿作他用。

限制150内

关于本文

本文标题：2022年东城高三一模数学文含答案 .pdf
链接地址：https://www.taowenge.com/p-56653045.html