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1、1/8绝密 启用前2018 年普通高等学校招生全国统一考试文科数学注意事项:1答卷前,考生务必将自己的姓名和准考证号填写在答题卡上。2回答选择题时,选出每小题答案后,用铅笔把答题卡对应题目的答案标号涂黑。如需改动,用橡皮擦干净后,再选涂其它答案标号。回答非选择题时,将答案写在答题卡上。写在本试卷上无效。3考试结束后,将本试卷和答题卡一并交回。一、选择题:本题共12 小题,每小题5 分,共 60 分。在每小题给出的四个选项中,只有一项是符合题目要求的。1已知集合02A,21012B,则ABIA02,B12,C0D21012,2设1i2i1iz,则 zA0 B12C1 D23某地区经过一年的新农村
2、建设,农村的经济收入增加了一倍实现翻番为更好地了解该地区农村的经济收入变化情况,统计了该地区新农村建设前后农村的经济收入构成比例得到如下饼图:则下面结论中不正确的是A新农村建设后,种植收入减少B新农村建设后,其他收入增加了一倍以上C新农村建设后,养殖收入增加了一倍D新农村建设后,养殖收入与第三产业收入的总和超过了经济收入的一半4已知椭圆C:22214xya的一个焦点为(2 0),则C的离心率为A13B12C22D2 235已知圆柱的上、下底面的中心分别为1O,2O,过直线12O O的平面截该圆柱所得的截面是面积为8的正方形,则该圆柱的表面积为A12 2B12C8 2D10精品资料-欢迎下载-欢
3、迎下载 名师归纳-第 1 页,共 8 页 -2/86设函数321fxxaxax若fx 为奇函数,则曲线yfx在点00,处的切线方程为A2yxByxC2yxD yx7在 ABC中,AD为 BC 边上的中线,E为AD的中点,则EBu uu rA3144ABACuu u ru uu rB1344ABACuuu ru uu rC3144ABACu u u ruu u rD1344ABACuuu ruuu r8已知函数222cossin2fxxx,则A()f x的最小正周期为,最大值为3 B()f x的最小正周期为,最大值为4 C()f x的最小正周期为2,最大值为3 D()f x的最小正周期为2,最大
4、值为4 9某圆柱的高为2,底面周长为16,其三视图如右图圆柱表面上的点M在正视图上的对应点为A,圆柱表面上的点N 在左视图上的对应点为B,则在此圆柱侧面上,从M到 N 的路径中,最短路径的长度为A2 17B2 5C3 D2 10在长方体1111ABCDA B C D中,2ABBC,1AC与平面11BBC C所成的角为30,则该长方体的体积为A8 B6 2C8 2D8 311已知角的顶点为坐标原点,始边与x轴的非负半轴重合,终边上有两点1Aa,2Bb,且2cos23,则abA15B55C2 55D1 12设函数2010 xxfxx,则满足12fxfx的 x 的取值范围是A1,B0,C10,D0,
5、二、填空题(本题共4 小题,每小题5 分,共 20 分)13已知函数22logfxxa,若31f,则a_14若xy,满足约束条件220100 xyxyy,则32zxy 的最大值为 _15直线1yx与圆22230 xyy交于AB,两点,则AB_16 ABC的 内 角ABC,的 对 边 分 别 为 abc,已 知sinsin4 sinsinbCcBaBC,2228bca,则 ABC的面积为 _精品资料-欢迎下载-欢迎下载 名师归纳-第 2 页,共 8 页 -文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J
6、8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B
7、2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X
8、1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:
9、CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3
10、T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q
11、2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X
12、2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X13/8三、解答题:共70 分。解答应写出文字说明、证明过程或演算步骤。第1721 题为必考题,每个试题考生都必须作答。第22、23 题为选考题,考生根据要求作答。(一)必考题:共60 分。17(12 分)已知数列na满足11a,121nnnana,设nnabn(1)求123bbb,;(2)判断数列nb是否为等
13、比数列,并说明理由;(3)求na的通项公式18(12 分)如 图,在 平 行 四 边 形 ABCM 中,3ABAC,90ACM,以AC为折痕将 ACM折起,使点M到达点D的位置,且ABDA(1)证明:平面ACD 平面 ABC;(2)Q 为线段AD上一点,P为线段BC 上一点,且23BPDQDA,求三棱锥QABP 的体积19(12 分)某家庭记录了未使用节水龙头50 天的日用水量数据(单位:m3)和使用了节水龙头50 天的日用水量数据,得到频数分布表如下:未使用节水龙头50 天的日用水量频数分布表日用水量00.1,0.10.2,0.20.3,0.30.4,0.40.5,0.50.6,0.60.7
14、,频数1 3 2 4 9 26 5 使用了节水龙头50 天的日用水量频数分布表日用水量00.1,0.10.2,0.20.3,0.30.4,0.40.5,0.50.6,频数1 5 13 10 16 5(1)在答题卡上作出使用了节水龙头50 天的日用水量数据的频率分布直方图:精品资料-欢迎下载-欢迎下载 名师归纳-第 3 页,共 8 页 -文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:
15、CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3
16、T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q
17、2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X
18、2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J
19、8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B
20、2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X
21、1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X14/8(2)估计该家庭使用节水龙头后,日用水量小于0.35 m3的概率;(3)估计该家庭使用节水龙头后,一年能节省多少水?(一年按365 天计算,同一组中的数据以这组数据所在区间中点的值作代表)20(12 分)设抛物线22Cyx:,点20A,20B,过点A的直线 l 与 C 交于M,N 两点(1)当 l 与x轴垂直时,求直线BM的方程;(2)证明:ABMABN21(12 分)已知函数eln1xfxax(1)设2x是fx的极值点
22、求a,并求fx的单调区间;(2)证明:当1ea时,0fx(二)选考题:共10 分。请考生在第22、23 题中任选一题作答。如果多做,则按所做的第一题计分。22选修 44:坐标系与参数方程(10 分)在直角坐标系xOy 中,曲线1C的方程为2yk x以坐标原点为极点,x轴正半轴为极轴建立极坐标系,曲线2C的极坐标方程为22cos30(1)求2C的直角坐标方程;精品资料-欢迎下载-欢迎下载 名师归纳-第 4 页,共 8 页 -文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编
23、码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10
24、S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F
25、5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX
26、1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V
27、5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV
28、1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J
29、4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X15/8(2)若1C与2C有且仅有三个公共点,求1C的方程23选修 45:不等式选讲(10 分)已知11fxxax(1)当1a时,求不等式1fx的解集;(2)若01x,时不等式fxx成立,求a的取值范围绝密 启用前2018 年普通高等学校招生全国统一考试文科数学试题参考答案一、选择题1A 2C 3A 4C 5B 6D 7A 8B 9B 10C
30、 11B 12 D 二、填空题13-7 146 15 2216233三、解答题17解:(1)由条件可得an+1=2(1)nnan将 n=1 代入得,a2=4a1,而 a1=1,所以,a2=4将 n=2 代入得,a3=3a2,所以,a3=12从而 b1=1,b2=2,b3=4(2)bn是首项为 1,公比为 2 的等比数列由条件可得121nnaann,即 bn+1=2bn,又 b1=1,所以 bn 是首项为1,公比为2 的等比数列(3)由(2)可得12nnan,所以 an=n 2n-118解:(1)由已知可得,BAC=90,BAAC又 BAAD,所以 AB平面 ACD 又 AB平面 ABC,所以平
31、面 ACD平面 ABC精品资料-欢迎下载-欢迎下载 名师归纳-第 5 页,共 8 页 -文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码
32、:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S
33、3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5
34、Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1
35、X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5
36、J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1
37、B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X16/8(2)由已知可得,DC=CM=AB=3,DA=3 2 又23BPDQDA,
38、所以2 2BP作 QEAC,垂足为 E,则 QE P13DC 由已知及(1)可得 DC平面 ABC,所以 QE平面 ABC,QE=1因此,三棱锥QABP 的体积为1111322sin 451332QABPABPVQES19解:(1)(2)根据以上数据,该家庭使用节水龙头后50 天日用水量小于0.35m3的频率为0.2 0.1+1 0.1+2.60.1+2 0.05=0.48,因此该家庭使用节水龙头后日用水量小于0.35m3的概率的估计值为0.48(3)该家庭未使用节水龙头50 天日用水量的平均数为11(0.05 10.1530.25 20.35 40.45 90.55 260.65 5)0.4
39、850 x精品资料-欢迎下载-欢迎下载 名师归纳-第 6 页,共 8 页 -文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S
40、3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5
41、Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1
42、X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5
43、J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1
44、B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4
45、X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X17/8该家庭使用了节水龙头后50 天日用水量的平均数为21(0.05 10.1550.25
46、130.35 100.45 160.55 5)0.3550 x估计使用节水龙头后,一年可节省水3(0.480.35)36547.45(m)20解:(1)当 l 与 x 轴垂直时,l 的方程为 x=2,可得 M 的坐标为(2,2)或(2,2)所以直线 BM 的方程为 y=112x或112yx(2)当 l 与 x 轴垂直时,AB 为 MN 的垂直平分线,所以ABM=ABN当 l 与 x 轴不垂直时,设l 的方程为(2)(0)yk xk,M(x1,y1),N(x2,y2),则 x10,x20由2(2)2yk xyx,得 ky2 2y 4k=0,可知 y1+y2=2k,y1y2=4直线 BM,BN 的
47、斜率之和为1221121212122()22(2)(2)BMBNyyx yx yyykkxxxx将112yxk,222yxk及 y1+y2,y1y2的表达式代入式分子,可得121221121224()882()0y yk yyx yx yyykk所以 kBM+kBN=0,可知 BM,BN 的倾斜角互补,所以ABM+ABN综上,ABM=ABN21解:(1)f(x)的定义域为(0),f(x)=aex1x由题设知,f(2)=0,所以 a=212e从而 f(x)=21eln12exx,f(x)=211e2exx当 0 x2 时,f(x)2 时,f(x)0所以 f(x)在(0,2)单调递减,在(2,+)
48、单调递增(2)当 a1e时,f(x)eln1exx设 g(x)=eln1exx,则e1()exg xx当 0 x1 时,g(x)1 时,g(x)0所以 x=1 是 g(x)的最小值点故当 x0 时,g(x)g(1)=0因此,当1ea时,()0f x22选修 4-4:坐标系与参数方程(10 分)精品资料-欢迎下载-欢迎下载 名师归纳-第 7 页,共 8 页 -文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2
49、G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1
50、文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:CP10S3T10F5Q2 HX1X2A2V5J8 ZV1B2G4J4X1文档编码:C