(完整word版)高中数学必修一至五模块综合测试(word文档良心出品).pdf

资源ID：83124980 资源大小：122.21KB 全文页数：6页
资源格式： PDF 下载积分：4.3金币

快捷下载

会员登录下载

微信登录下载

三方登录下载：

微信扫一扫登录

下载资源需要4.3金币

邮箱/手机：
温馨提示：	快捷下载时，用户名和密码都是您填写的邮箱或者手机号，方便查询和重复下载（系统自动生成）。如填写123，账号就是123，密码也是123。
支付方式：
验证码：	换一换

账号：
密码：
验证码：	换一换
当日自动登录忘记密码？

友情提示

1、下载资料失败解决办法

2、PDF文件下载后，可能会被浏览器默认打开，此种情况可以点击浏览器菜单，保存网页到桌面，就可以正常下载了。

3、本站不支持迅雷下载，请使用电脑自带的IE浏览器，或者360浏览器、谷歌浏览器下载即可。

4、本站资源下载后的文档和图纸-无水印,预览文档经过压缩，下载后原文更清晰。

5、试题试卷类文档，如果标题没有明确说明有答案则都视为没有答案，请知晓。

网站客服

侵权投诉

(完整word版)高中数学必修一至五模块综合测试(word文档良心出品).pdf

第 1 页共 6 页6 5 主视图6 5 侧视图俯视图高中数学必修模块综合测试卷一、选择题：本大题共10 小题，每小题5 分，共 50 分1.已知集合11 2,1,0,1,2|28R2xMNxx，则MNIA0,1B1 0，C 1,0,1D2,1,0,1,22.某社区现有480个住户，其中中等收入家庭200 户、低收入家庭160 户，其他为高收入家庭。在建设幸福广东的某次分层抽样调查中，高收入家庭被抽取了6 户，则该社区本次被抽取的总户数为A20B24C30D363.已知实数列1,2a b c成等比数列，则abc等于（）A4 B4 C22D224.过点(1,1),(1,1)AB-且圆心在直线20 xy+-=上的圆的方程是A22(3)(1)4xy-+=B.22(3)(1)4xy+-=C22(1)(1)4xy-+-=D.22(1)(1)4xy+=5.已知向量a与b的夹角为120o，且|1abrr|，则|abrr等于A1 B3C2 D3 6.已知1,4,20,xyxyy则24xy的最小值是A8 B9 C10 D13 7.有一个几何体的三视图及其尺寸如图所示（单位：cm），则该几何体的表面积为A212 cmB.215 cmC.224 cmD.236 cm8.设,x yR?则“2x 3且2y 3”是“224xy+?”的A充分而不必要条件B必要而不充分条件C充分必要条件D即不充分也不必要条件9.若23x，12xP，2logQx，Rx,则P，Q，R的大小关系是AQPRBQRPCPRQDPQR10.一个三角形同时满足：三边是连续的三个自然数；最大角是最小角的2 倍，则这个三角形第 2 页共 6 页最小角的余弦值为A3 78B34C74D18二、填空题：本大题共4 小题，每小题5 分，共 20 分11.sin(30)sin(30)cosaaa+-oo的值为.12.如右图所示，函数2xfx，2g xx，若输入的x值为 3，则输出的h x的值为.13.若函数2213fxaxax是偶函数，则函数fx的单调递减区间为14.已知数列na满足12a,*121()nnaanN,则4a，该数列的通项公式na三、解答题：本大题共6 小题，共 80 分15.（本题满分12 分）有四个数，已知前三个成等比数列，且和为19,后三个成等差数列，且和为12,求此四数。16.（本题满分13 分）设(,)a b是有序数对，其中a是从区间(3,1)A=-中任取的一个整数，b是从区间(2,3)B=-中任取的一个整数。（1）请列举出(,)a b的各种情况；（2）任取（1）中的一组(,)a b，求使得ba-为正整数的概率。否是开始()()h xf x()()f xg x输出()h x输入x结束()()h xg x文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9第 3 页共 6 页17.（本题满分13 分）如图，在RtAOB中，6OAB，斜边4ABRtAOC可以通过RtAOB以直线AO为轴旋转得到，且二面角BAOC是直二面角动点D的斜边AB上（I）求证：平面COD平面AOB；（II）当D为AB的中点时，求异面直线AO与CD所成角的正切值；（III）求CD与平面AOB所成角的最大值的正切值18.（本题满分14 分）已知向量2(2cos,3)ax=r，(1,sin 2)bx=r，函数()f xa br r，2()g xb=r（）求函数)(xg的最小正周期；（）在ABC中，cba,分别是角CBA,的对边，且3)(Cf，1c，32ab，且ba，求ba,的值OCADBE文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9第 4 页共 6 页19.（本题满分14 分）直线ykxb与圆224xy交于A、B两点，记AOB的面积为S（其中O为坐标原点）（）当0k，02b时，求S的最大值；（）当2b，1S时，求实数k的值20.（本题满分14 分）设a为实数，函数2()2()|f xxxaxa.(1)若(0)1f，求a的取值范围；(2)求()f x的最小值；(3)设函数()(),(,)h xf x xa，直接写出(不需给出演算步骤)不等式()1h x的解集.文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9第 5 页共 6 页参考答案1-10：CBCCB CCACB 11.1 12.9 13.),0(14.23，1231n15.解:设此四数分别为2(),adad a ada-+，则()()312adaada-+=故4a=，四数为2(4),4,4,44ddd-+，所以2(4)(4)4194dd-+-+=，解得142dd=-或，故当14d=时此四数为25,10-,4,18;当2d=-时，此四数为9，6，4，2.16.解:依题意知a可取集合 A 的2,1,0-三数之一，b可取集合 B 的1,0,1,2-四数之一，（1）(,)a b的各种情况有：(2,1),(2,0),(2,1),(2,2),(1,1),(1,0),(1,1),(1,2),-(0,1),(0,0),(0,1),(0,2)-共 12 种（2）使得ba-为整数的情况有(2,1),(2,0),(2,1),(2,2),(1,0),(1,1),(1,2),(0,1),(0,2)-共 9种，故使得ba-为整数的概率为93124P=。17.解：（I）由题意，COAO，BOAO，BOC是二面角BAOC是直二面角，又Q二面角BAOC是直二面角，COBO，又AOBOOQI，CO平面AOB，又CO平面COD平面COD平面AOB（II）作DEOB，垂足为E，连结CE，则DEAO，CDE是异面直线AO与CD所成的角在RtCOE中，2COBO，112OEBO，225CECOOE又132DEAO.在RtCDE中，515tan33CECDEDE，异面直线AO与CD所成角的正切值为153（III）由（I）知，CO平面AOB，CDO是CD与平面AOB所成的角，且2tanOCCDOODOD 当OD最小时，CDO最大，这时，ODAB，垂足为D，3OA OBODABg，2 3tan3CDO，CD与平面AOB所成角的最大值的正切值为2 3318.解：（）221cos413()1sin 21cos4222xg xbxxr函数)(xg的最小周期242T（）()f xa br r2(2cos,3)(1,sin 2)xx22cos3 sin2xxcos213sin 2xx2sin(2)16x31)62sin(2)(CCf1)62sin(CC是三角形内角，)613,6(62C，262C即：6C232cos222abcabC即：722ba，又32ab可得：71222aa解之得：432或a，23或a所以当3a时，2b；当2a，3b，ba2a，3b文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9第 6 页共 6 页19,解：（1）当0k时，直线方程为yb，设点A的坐标为1()xb，点B的坐标为2()xb，由224xb，解得21 24xb，所以2212 4ABxxb所以12SAB bgg24bb22422bb当且仅当24bb，即2b时，S取得最大值2（2）设圆心O到直线2ykx的距离为d，则221dk 因为圆的半径为2R，所以2222244211ABkRdkk于是222241212111kkSABdkkk，即2410kk，解得23k故实数k的值为23，23，23，2320.解:（1）若(0)1f，则20|111aa aaa（2）当xa时，22()32,f xxaxa22min(),02,0()2(),0,033f a aaaf xaafaa当xa时，22()2,f xxaxa2min2(),02,0()(),02,0faaaaf xf aaaa综上22min2,0()2,03aaf xaa（3）(,)xa时，()1h x得223210 xaxa，222412(1)128aaa当6622aa或时，0,(,)xa；当6622a时，0,得：223232()()033aaaaxxxa讨论得：当26(,)22a时，解集为(,)a;当62(,)22a时，解集为223232(,)33aaaaa;当22,22a时，解集为232,)3aa.文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9文档编码:CT8A2T4R8T6 HP5M5E10H2R5 ZM10E10A4I3J9

注意事项

本文（(完整word版)高中数学必修一至五模块综合测试(word文档良心出品).pdf）为本站会员（H****o）主动上传，淘文阁 - 分享文档赚钱的网站仅提供信息存储空间，仅对用户上传内容的表现方式做保护处理，对上载内容本身不做任何修改或编辑。若此文所含内容侵犯了您的版权或隐私，请立即通知淘文阁 - 分享文档赚钱的网站（点击联系客服），我们立即给予删除！

温馨提示：如果因为网速或其他原因下载失败请重新下载，重复下载不扣分。