河北省邢台市“五岳联盟”2022届高三上学期10月联考数学试题 含答案

邢台市“五岳联盟”2022届高三上学期10月联考

数学

注意事项：

1.答题前，考生务必将自己的姓名、考生号、考场号、座位号填写在答题卡上。

2.回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑。如需改动，用橡皮擦干净后，再选涂其他答案标号。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。

4.本试卷主要考试内容：人教A版必修1、4、5。

一、选择题：本题共8小题，每小题5分，共40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，，则（    ）

A.  B.

C.  D.

2.若向量，，，则（    ）

A. B. C. D.

3.如图，一个“心形”由两个函数的图象构成，则“心形”上部分的函数解析式可能为（    ）

A. B. C. D.

4.的内角A，B，C的对边分别为a，b，c.已知，则（    ）

A. B. C. D.

5.在等差数列中，，，则的取值范围是（    ）

A. B. C. D.

6.将函数的图象向左平移个单位长度后，得到函数的图象，则（    ）

A.为奇函数  B.的图象关于直线对称

C.的图象关于点对称 D.在上单调递减

7.函数的最小值为（    ）

A. B. C.4 D.

8.根据《民用建筑工程室内环境污染控制标准》，文化娱乐场所室内甲醛浓度为安全范围.已知某新建文化娱乐场所施工中使用了甲醛喷剂，处于良好的通风环境下时，竣工1周后室内甲醛浓度为6.25，3周后室内甲醛浓度为1，且室内甲醛浓度（单位：）与竣工后保持良好通风的时间t（）（单位：周）近似满足函数关系式，则该文化娱乐场所竣工后的甲醛浓度若要达到安全开放标准，至少需要放置的时间约为（    ）

A.5周 B.6周 C.7周 D.8周

二、选择题：本题共4小题，每小题5分，共20分.在每小题给出的选项中，有多项符合题目要求.全部选对的得5分，部分选对的得2分，有选错的得0分.

9.设表示不大于x的最大整数，已知集合，，则（    ）

A.  B.

C. D.

10.下列函数中，定义域与值域相同的是

A.  B.

C.  D.

11.若，则（    ）

A.  B.

C.  D.

12.已知函数的定义域为，，，当时，，则（    ）

A.

B.的图象关于直线对称

C.当时，

D.函数有4个零点

三、填空题：本题共4小题，每小题5分，共20分.把答案填在答题卡的相应位置.

13.设向量，均为单位向量，且，则___________.

14.若，则___________.

15.写出一个同时具有下列四个性质的函数___________.

①定义域为；②单调递增；③；④.

16.一张B4纸的厚度为0.093，将其对折后厚度变为，第2次对折后厚度变为，…，第n次对折后厚度变为，则___________，数列的前n项和为____________.（本题第一空2分，第二空3分）

四、解答题：本题共6小题，共70分.解答应写出必要的文字说明、证明过程或演算步骤.

17.（10分）

如图，在梯形中，.

（1）用，表示，，；

（2）若，且，求的大小.

18.（12分）

已知函数（，）的部分图象如图所示.

（1）求的解析式；

（2）把的图象上所有点的横坐标伸长到原来的m（）倍（纵坐标不变）后，得到函数的图象，若在上有最大值，求m的取值范围.

19.（12分）

已知数列的前n项和为，且.数列的前n项积为，且.

（1）求，的通项公式；

（2）求数列的前n项和.

20.（12分）

已知函数.

（1）若，求；

（2）当时，讨论函数的零点个数.

21.（12分）

如图，点O在点P的正东方向，现有一个圆形音乐喷泉，点O为喷泉中心，用无人机于点P正上空的点处，测得点O的俯角为，点B的俯角为，P，A，O，B四点共线，A，B均在圆O上，且.已知圆O的面积为平方米，且米.

（1）求无人机的飞行高度；

（2）如图，现以A，M，N三点为顶点在音乐喷泉内建造三条排水暗渠，已知暗渠造价为1000元/米，且建造暗渠的预算资金为35000元.若要求，，成等差数列，试问完成三条排水暗渠的建造是否有可能会超预算？说明你的理由.

22.（12分）

已知函数满足.

（1）试问是否存在，使得函数为奇函数？若存在，求a的值；若不存在，请说明理由.

（2）若，，，求m的取值范围.

高三上学期10月联考

数学参考答案

1.B  【解析】本题考查集合的补集，考查数学运算的核心素养.

因为，所以，故.

2.B  【解析】本题考查平面向量的共线问题，考查数学运算的核心素养.

因为，所以，解得.

3.C  【解析】本题考查函数的图象与函数的解析式，考查读图能力与逻辑推理的核心素养.

由图可知，“心形”关于y轴对称，所以上部分的函数为偶函数，排除B，D.又“心形”函数的最大值为1，且，排除A.故选C.

4.A  【解析】本题考查正弦、余弦定理的应用，考查数学运算的核心素养.

由正弦定理可得，整理得，又由余弦定理可知，所以，解得.

5.A  【解析】本题考查等差数列的通项公式，考查逻辑推理与数学运算的核心素养.

设公差为d，因为，，所以，即，从而.

6.D  【解析】本题考查三角函数的图象及其性质，考查逻辑推理的核心素养.

由题意可知，则不是奇函数，所以A错误；因为，，所以B，C均错误.故选D.

7.C  【解析】本题考查基本不等式的应用，考查逻辑推理与数学运算的核心素养.

因为，当且仅当，即时，等号成立，

所以，当且仅当，即时，等号成立.故的最小值为4.

8.B  【解析】本题考查函数的实际应用，考查数学建模与数学运算的核心素养.

由题意可知，，，，解得.设该文化娱乐场所竣工后放置周后甲醛浓度达到安全开放标准，则，整理得，因为，所以，即.故至少需要放置的时间约为6周.

9.ABD  【解析】本题考查不等式的解集与集合的运算，考查数学抽象与数学运算的核心素养.

因为，所以，所以.

因为，所以.

因为，，

所以，.

10.BCD  【解析】本题考查函数的定义域与值域，考查数学抽象与逻辑推理的核心素养.

的定义域为，值域为.

的定义域和值域均为.

的定义域和值域均为.

的定义域为，因为，且，所以的值域为，则的值域为.

11.BC  【解析】本题考查三角恒等变换，考查数学运算的核心素养.

因为，所以，则，

所以，，

因为，所以，

即，

所以，

则，

故.

12.ACD  【解析】本题考查函数的综合，考查数学抽象与逻辑推理的核心素养.

因为，且，所以，

所以，所以，A正确.

因为，所以的图象关于点对称，B错误.

当时，，.

因为的图象关于点对称，的定义域为，所以，

所以，故当时，，C正确.

由，得，因为的值域为，所以由得，作出，的部分图象，如图所示.由图可知，它们有4个交点，故函数有4个零点，D正确.

13.-7  【解析】本题考查平面向量的数量积，考查数学运算的核心素养.

因为，所以，所以.

14.9  【解析】本题考查三角恒等变换与函数的解析式，考查数学运算与数学抽象的核心素养.

因为，所以，则.

15.（答案不唯一，只要满足（，，）即可）  【解析】本题为开放题，考查基本初等函数的性质，考查数学抽象与逻辑推理的核心素养.

结合前面三个性质及对数函数的性质可以得出的解析式为（，），再根据最后一个性质可得.

16.；  【解析】本题考查数列的综合应用，考查逻辑推理、数学运算的核心素养及化归与转化的数学思想.

因为每对折一次，纸张的厚度增加一倍，所以数列是首项为0.186，公比为2的等比数列，

所以.

因为，所以，

故的前n项和

.

17.解：（1），·····································································1分

，·············································································2分

. ··············································································4分

（2）因为，，所以.································································5分

因为，··········································································7分

且，所以，······································································8分

解得，··········································································9分

故.  ···········································································10分

18.解：（1）由图可知，·······························································1分

则，解得.  ·······································································3分

将点代入，得（），·······························································4分

因为，所以，····································································5分

故的解析式为.  ···································································6分

（2）依题意可得.  ································································8分

因为在上有最大值，且当时，，······················································9分

所以，·········································································11分

又，所以，即m的取值范围是.·······················································12分

19.解：（1）当时，；································································1分

当时，.·········································································2分

经检验，当时，满足，因此.·························································3分

当时，；当时，.··································································5分

当时，满足，因此. ································································6分

（2）由（1）知，

，·············································································7分

，·············································································8分

两式相减得，····································································9分

·············································································10分

，············································································11分

故. ···········································································12分

20.解：（1）∵，，·································································1分

∴，···········································································3分

故. ············································································5分

（2）当时，，···································································6分

，且在上单调递增，·······························································7分

由，得或.·······································································8分

当时，方程只有1个解，的零点个数为1；···············································9分

当时，方程与各有1个解，且这2个解不相等，的零点个数为2；······························10分

当时，方程只有1个解，的零点个数为1.···············································11分

综上，当时，的零点个数为1；当时，的零点个数为2.·····································12分

21.解：（1）设无人机的飞行高度为h米，圆形音乐喷泉的半径为r米，

由题意可知，，则. ································································1分

∵，∴，········································································2分

则，，··········································································3分

则，故无人机的飞行高度为12米. ·····················································4分

（2）∵，，成等差数列，

∴，则.·········································································5分

设，则，.  ·······································································6分

由正弦定理，可得

（米），（米），

（米），········································································8分

∴（米）.·······································································10分

∵，∴，∴，

则，··········································································11分

∵，∴完成三条排水暗渠的建造有可能会超预算.········································12分

22.解：（1）由，得，································································1分

根据这两个等式，消去得.···························································3分

因此（），······································································4分

故存在，使得函数为奇函数.·························································5分

（2）设函数，则.

当或时，；当时，.································································6分

因为，所以.······································································7分

因为，，，

所以对恒成立，···································································8分

所以对恒成立.···································································9分

设函数（），令（），

则， ··········································································10分

当且仅当，即时，等号成立，此时取得最小值，·········································11分

故，m的取值范围为. ······························································12分

[注]另外，本题中的最小值也可以用求导的方法求得，且的极小值点为.