2021年2021年2019届天津市七校(静海一中、宝坻一中、杨村一中等)高三上学期期末考试数学(理)试题(解析版).pdf

《2021年2021年2019届天津市七校(静海一中、宝坻一中、杨村一中等)高三上学期期末考试数学(理)试题(解析版).pdf》由会员分享，可在线阅读，更多相关《2021年2021年2019届天津市七校(静海一中、宝坻一中、杨村一中等)高三上学期期末考试数学(理)试题(解析版).pdf（21页珍藏版）》请在淘文阁 - 分享文档赚钱的网站上搜索。

1、第 1 页 共 21 页2019 届天津市七校（静海一中、宝坻一中、杨村一中等）高三上学期期末考试数学（理）试题一、单选题1已知集合，则（）ABCD【答案】B【解析】求解集合A，然后根据补集的运算求解，再根据集合的交集的运算，即可求解.【详解】由题意或，所以，所以，故选 B.【点睛】本题主要考查了绝对值不等式的求解，以及集合的混合运算问题，其中解答总正确求解集合 A，准确利用集合的运算是解答的关键，着重考查了推理与计算能力，属于基础题.2设，直线：，直线：，则“”是“”的（）A充分而不必要条件B必要而不充分条件C充要条件D既不充分也不必要条件【答案】C【解析】根据直线平行的等价条件求出得取值范

2、围，结合充分条件和必要条件的定义，进行判定，即可得到答案.【详解】由题意，当时，两直线，此时两直线不平行，当时，若，则满足，由得，解得或，第 2 页 共 21 页当时，成立，当时，成立，即两直线是重合的（舍去），故所以是的充要条件，故选C.【点睛】本题主要考查了充分条件和必要条件的判定，以及两直线位置关系的应用，其中解答中根据直线平行的等价条件求出得值是解答的本题的关键，着重考查了推理与运算能力，属于基础题.3设变量满足约束条件，则目标函数的最小值是（）A-5 B1 C2 D 7【答案】B【解析】由约束条件作出可行域，化目标函数为直线方程的斜截式，结合图象确定目标函数的最优解，代入目标函数，即

3、可求解.【详解】由题意，画出约束条件所表示的平面区域，如图所示，由目标函数，可得，由图象可知，当直线过点时，直线在y 轴上的截距最小，此时目标函数取得最小值，最小值为，故选 B.【点睛】本题主要考查简单线性规划求解目标函数的最值问题其中解答中正确画出不等式组表示的可行域，利用“一画、二移、三求”，确定目标函数的最优解是解答的关键，着重考查了数形结合思想，及推理与计算能力，属于基础题文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N

4、4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1

5、O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O

6、2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S

7、2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4

8、C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C

9、2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y

10、9S2N4B1 ZI4K5I4C1O3文档编码:CD3C2O2B6Z10 HS9Y9S2N4B1 ZI4K5I4C1O3第 3 页 共 21 页4执行如图所示的程序框图，输出的值为（）A7 B14 C30 D41【答案】C【解析】由已知中的程序语句可知，该程序的功能是利用循环结构计算并输出变量S的值，模拟程序运行的过程，分析循环中各变量的变化情况，即可求解.【详解】由题意，模拟程序的运行，可得，不满足条件，执行循环体，满足条件能被整除，；不满足条件，执行循环体，满足条件能被整除，；不满足条件，执行循环体，满足条件能被整除，；不满足条件，执行循环体，满足条件能被整除，；此时，满足，推出循环，输出

11、S的值为 30，故选 C.【点睛】本题主要考查了循环结构的程序框图的计算与输出问题，其中利用循环结构表示算法，一定要先确定是用当型循环结构，还是用直到型循环结构；当型循环结构的特点是先判断再循环，直到型循环结构的特点是先执行一次循环体，再判断；注意输入框、处理框、判断框的功能，不可混用，着重考查了分析问题和解答问题的能力，属于基础题.5已知，则的大小关系为（）ABCD【答案】D【解析】现判断函数是奇函数，同时又是增函数，结合指数幂和对数的性质判断，文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 Z

12、X5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7

13、Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档

14、编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1

15、H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V

16、4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 H

17、F2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8

18、W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3第 4 页 共 21 页三个变量的大小，结合单调性进行判定，即可得到答案.【详解】函数是奇函数，当时，为增函数，又由，则，所以，故选 D.【点睛】本题主要考查了函数值的比较大小问题，其中解答中熟练应用函数的奇偶性和函数的单调性，合理得到的取值范围是解答的关键，着重考查了分析问题和解答问题的能力，属于基础题.6己知函

19、数图象的两条相邻的对称轴之间的距离为2，将的图象向右平移个单位长度，得到函数的图象，则下列是函数的单调递增区间的为（）ABCD【答案】B【解析】由函数图象的两条相邻的对称轴之间的距离为2，求得，所以，将函数的图象向右平移个单位长度，得到函数，利用三角函数的性质，即可求解.【详解】函数图象的两条相邻的对称轴之间的距离为2，所以函数的最小正周期为4，则，解得，所以，将函数的图象向右平移个单位长度，得到函数，令，解得，文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1

20、H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V

21、4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 H

22、F2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8

23、W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 Z

24、X5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7

25、Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档

26、编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3第 5 页 共 21 页当时，函数的单调单调递增区间为，故选 B.【点睛】本题主要考查了函数的图象变换的应用，以及正弦型函数的图象与性质的应用，其中解答中根据三角函数的图象变换得到函数的解析式，再根据三角函数的图象与性质求解是解答的关键，着重考查了推理与运算能力，属于基础题.7已知双曲线的左、右焦点分别为，过作圆的切线，交双曲线右支于点，若，则双曲线

27、的离心率为（）AB2 CD【答案】A【解析】设切点为N，连接 ON，作作，垂足为 A，由，得到，在直角三角形中，可得，得到，再由双曲线的定义，解得，利用双曲线的离心率的定义，即可求解.【详解】设切点为N，连接 ON，作作，垂足为 A，由，且为的中位线，可得，即有，在直角三角形中，可得，即有，由双曲线的定义可得，可得，所以，所以，故选 A.【点睛】本题考查了双曲线的几何性质离心率的求解，其中求双曲线的离心率(或离心率的文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:C

28、L1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C1

29、0V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10

30、 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5

31、K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2

32、 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6

33、E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3

34、文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3第 6 页 共 21 页取值范围)，常见有两种方法：求出，代入公式；只需要根据一个条件得到关于的齐次式，转化为的齐次式，然后转化为关于的方程(不等式)，解方程(不等式)，即可得(的取值范围)8定义域为的函数满足，当时，.若时，恒成立，则实数的取值范围是（）A，BC，D【答案】C【解析】根据函数的性质和函数的解析式，求得，则若时，恒成立转化为且，即

35、可求解.【详解】当时，则，当时，则，则当时，当时，当时，当时，所以时，文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10

36、V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10

37、HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K

38、8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2

39、ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E

40、7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文

41、档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3第 7 页 共 21 页所以，“若时，恒成立”等价于，解得，故选 C.【点睛】本题主要考查了函数的

42、最值的求解，以及函数的恒成立问题的求解，其中解答中根据题意根据函数的性质和函数的解析式，求得函数的最值，再把恒成立问题转化为不等式组求解是解答的关键，着重考查了转化思想，以及分析问题和解答问题的能力，属于中档试题.二、填空题9已知复数(是虚数单位)，则复数的虚部为 _.【答案】2【解析】根据复数的代数形式的四则运算，化简复数，即可得到答案.【详解】由题意，复数，所以复数的虚部为.【点睛】本题主要考查了复数的四则运算，以及复数的基本概念的应用，其中解答中熟练应用复数的运算法则化简是解答的关键，着重考查了计算能力，属于基础题.10若二项式的展开式中的常数项为，则_.【答案】124【解析】根据题意，

43、利用二项式求得的值，再求出被积分函数的原函数，即可求解.【详解】由题意，二项展开式的通项为，文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档

44、编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1

45、H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V

46、4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 H

47、F2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8

48、W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 Z

49、X5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3文档编码:CL1H3C10V4S10 HF2A5K8W5F2 ZX5I6E7Y9O3第 8 页 共 21 页由，得，所以，则.【点睛】本题主要考查了二项式定理的

50、应用，以及定积分的计算问题，其中解答中根据二项展开式的通项，求得的值，再根据定积分的计算求解是解答的关键，着重考查了推理与运算能力.11已知正方体中，四面体的表面积为，则该正方体的体积是 _.【答案】8【解析】由已知画出图形，设正方体的棱长为，由四面体的表面积为求的 的值，则正方体的体积，即可求解.【详解】如图所示，设正方体的棱长为，则四面体的棱长为，其表面积，得，所以正方体的体积是.【点睛】本题主要考查了多面体的体积的计算问题，其中解答中熟记正四面体的表面积的计算公式，准确计算是解答的关键，着重考查了空间想象能力，以及推理与运算能力，属于基础题.12已知抛物线的参数方程为（为参数，），其焦点