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2019届山东省青岛市高三3月教学质量检测(一模)数学(文)试题(解析版).pdf

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1、第 1 页共 21 页2019届山东省青岛市高三3 月教学质量检测一模数学文试题一、单项选择题1已知集合，集合，则ABCD【答案】B【解析】现根据题干得到集合B 的元素，再由集合交集的概念得到结果.【详解】集合，集合，则.故答案为：B.【点睛】这个题目考查了集合的交集的运算，属于简单题目.2已知为虚数单位，复数满足，则在复平面内对应的点位于A第一象限B第二象限C第三象限D第四象限【答案】A【解析】根据复数的四则运算得到复数的化简结果，进而得到在复平面内所对应的点.【详解】复数满足,在复平面内对应的点位：，在第一象限.故答案为：A.【点睛】如果是复平面内表示复数的点，则当，时，点位于第一象限

2、；当，时，点位于第二象限；当，时，点位于第三象限；当，时，点位于第四象限当时，点位于实轴上方的半平面内；当时，点位于实轴下方的半平面内3“结绳计数”是远古时期人类智慧的结晶，即人们通过在绳子上打结来记录数量.如下图的是一位农民记录自己采摘果实的个数.在从右向左依次排列的不同绳子上打结，满四进一.根据图示可知，农民采摘的果实的个数是第 2 页共 21 页A493 B 383 C183 D123【答案】C【解析】根据题意将四进制数转化为十进制数即可.【详解】根据题干知满四进一,则表示四进制数,将四进制数转化为十进制数,得到故答案为:C.【点睛】此题以数学文化为载体，考查了进位制等基础知识，注

3、意运用四进制转化为十进制数，考查运算能力，属于基础题4调查机构对某高科技行业进行调查统计，得到该行业从业者学历分布饼状图，从事该行业岗位分布条形图，如下图.给出以下三种说法：该高科技行业从业人员中学历为博士的占一半以上；该高科技行业中从事技术岗位的人数超过总人数的；该高科技行业中从事运营岗位的人员主要是本科生，其中正确的个数为A0 个B 1 个C2 个D3 个【答案】C【解析】利用饼状图、行业岗位分布条形图得到相应命题的真假.【详解】根据饼状图得到从事该行行业的人群中有百分之五十五的人是博士，故正确；从条形图中可得到从事技术岗位的占总的百分之三十九点六，故正确；而从条形图中看不出来从事各个岗位

4、的人的学历，故得到错误.文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文

5、档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1

6、F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q

7、6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N1

8、0文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2

9、A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F

10、5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10文档编码:CD1T3F2A1F4 HS6Y6M1F5Q6 ZQ2H7I9X1N10第 3 页共 21 页故答案为：C.【点睛】此题考查命题真假的判断，考查饼状图、条形图的性质等基础知识，考查运算求解能力，是基础题5执行如下图的程序框图，则输出的值为A7 B 6 C5 D4【答案】C【解析】根据框图，依次进入循环，直到不

11、满足判断框内的条件为止.【详解】K=9,s=1,进入循环得,k=8,再进入循环,k=7,进入循环得到，不满足判断框的条件，故此时输出k 值，得到k=5.故答案为：C.【点睛】对于程序框图的读图问题，一般按照从左到右、从上到下的顺序，理清算法的输入、输出、条件结构、循环结构等基本单元，并注意各要素之间的流向是如何建立的特别地，当程序框图中含有循环结构时，需首先明确循环的判断条件是什么，以决定循环的次数6在中，则ABCD【答案】A 文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档

12、编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y

13、6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7

14、 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4

15、文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P

16、4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5

17、S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6

18、Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4第 4 页共 21 页【解析】根据向量减法的三角形法则得到，再由向量的减法法则，以和为基底表示向量.【详解】根据向量的减法法则得到,又因为，故得到，代入上式得到.故答案为：A.【点睛】这个题目考查的是向量基本定理的应用；解决向量的小题常用方法有：数形结合，向量的三角形法则，平行四边形法则等；建系将向量坐标化；向量基底化，选基底时一般选择已知大小和方向的向量为基底。7已知数列为等比数列，满足；数列为等差数列，其前项和为，且，则A

19、13 B 48 C78 D156【答案】C【解析】由等比数列的性质可得a76，再由等差数列的求和公式和中项性质，可得所求和【详解】等比数列 an 中，a3a11a72，可得 a726a7，解得 a76，数列 bn是等差数列中b7a76，根据等差数列的前n 项和与等差中项的性质得到：S1313 b1+b13 13b713b7代入求得结果为：78.故选：C【点睛】此题考查等差数列和等比数列的通项公式、求和公式的运用，考查方程思想和运算能力，属于基础题8已知双曲线：，为坐标原点，过的右顶点且垂直于轴的文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D

20、6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3

21、G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2

22、W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW

23、2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3

24、J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10

25、J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:

26、CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4第 5 页共 21 页直线交的渐近线于，过的右焦点且垂直于轴的直线交的渐近线于，假设与的面积比为，则双曲线的渐近线方程为ABCD【答案】B【解析】由三角形的面积比等于相似比的平方，可得，即可求出渐近线方程【详解】由三角形的面积比等于相似比的平方，则，C 的渐近线方程为yx，故选：B【点睛】这个题目考查了双曲线的几何意义的应用，考查了三角形面积之比等于相似比这一

27、转化，题目比较基础.9某几何体的三视图如下图其中正视图中的曲线为两个四分之一圆弧，则该几何体的体积为文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码

28、:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6

29、HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 Z

30、J10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档

31、编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y

32、6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7

33、 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4第 6 页共 21 页ABCD【答案】B【解析】根据三视图得到原图是一个棱长为4 的正方体，挖去了两个圆柱，圆柱的底面圆的半径为2，让正方体的体积减去半个圆柱的体

34、积即可.【详解】根据三视图得到原图是一个棱长为4 的正方体，挖去了两个圆柱，圆柱的底面圆的半径为 2，故得到的体积为正方体的体积减去半个圆柱的体积，故答案为：B.【点睛】思考三视图复原空间几何体首先应深刻理解三视图之间的关系，遵循“长对正，高平齐，宽相等”的基本原则，其内涵为正视图的高是几何体的高，长是几何体的长；俯视图的长是几何体的长，宽是几何体的宽；侧视图的高是几何体的高，宽是几何体的宽.由三视图画出直观图的步骤和思考方法：1、首先看俯视图，根据俯视图画出几何体地面的直观图；2、观察正视图和侧视图找到几何体前、后、左、右的高度；3、画出整体，然后再根据三视图进行调整.10已知函数在一个周期

35、内的图象如下图，则的解析式是AB文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B

36、6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6

37、T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G

38、6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W

39、5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2

40、D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J

41、3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4第 7 页共 21 页CD【答案】B【解析】由函数的图像得到函数的周期排除AC，再由图像的到在处取得最值，从而得到答案.【详解】根据图像得到三角函数的周期为，由周期的公式知.此时排除 AC.又因为图像中函数在处取得最大值，代入BD

42、发现 D 不合题意故舍去.故答案为：B。【点睛】这个题目考查了三角函数的图像的性质的应用，知图求式，比较好的方法有：根据图像中的特殊点或者图像中表达出来的函数的定义域，进行选项排除.11已知函数，假设，则，的大小关系是ABCD【答案】D【解析】可以得出，从而得出ca，同样的方法得出ab，从而得出 a，b，c的大小关系【详解】,根据对数函数的单调性得到ac,又因为，再由对数函数的单调性得到 ab,ca，且 ab；cab故选：D【点睛】考查对数的运算性质，对数函数的单调性比较两数的大小常见方法有：做差和 0 比较，做商和 1 比较，或者构造函数利用函数的单调性得到结果.12已知函数，假设方程为常

43、数有两个不相等的根，则实数的取值范围是文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2

44、W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW

45、2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3

46、J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10

47、J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:

48、CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 H

49、T3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4文档编码:CW2D6T9P4Y6 HT3J3G6M5S7 ZJ10J2W5B6Q4第 8 页共 21 页ABCD【答案】D【解析】求出当 x0 时，函数的导数，研究函数的极值和图象，作出函数fx的图象，由数形结合进行求解即可【详解】当 x0 时，函数f x 2 lnx+1 1lnx，由 f x 0 得 1

50、lnx0 得 lnx 1，得 0 xe，由 f x 0 得 1lnx 0得 lnx1，得 xe，当 x 值趋向于正无穷大时，y 值也趋向于负无穷大，即当xe时，函数fx取得极大值，极大值为fe 2eelne2eee，当 x0 时，fx x2 x x+2+，是二次函数，在轴处取得最大值，作出函数fx的图象如图：要使 fx aa 为常数有两个不相等的实根，则 a 0或ae，即实数 a 的取值范围是，0，故选：D【点睛】此题主要考查函数与方程的应用，利用分段函数的表达式作出函数的图象，利用数形结合是解决此题的关键已知函数零点(方程根)的个数，求参数取值范围的三种常用的方法：(1)直接法，直接根据题设

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关键词：: 2019 山东省青岛市教学质量检测数学试题解析

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本文标题：2019届山东省青岛市高三3月教学质量检测(一模)数学(文)试题(解析版).pdf
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