山东省临沂第十八中学2022届高三上学期期末学业水平检测数学试卷
展开第一学期期末学业水平检测
高三数学
本试卷6页,22小题,满分150分.考试用时120分钟.
注意事项:
1.答题前,考生务必将自己的姓名、考生号、考场号和座号填写在试题卷和答题卡上,并将准考证号条形码粘贴在答题卡上的指定位置;
2.作答选择题时:选出每小题答案后,用2B铅笔在答题卡上对应题目选项的答案信息点涂黑;如需要改动,用橡皮擦干净后,再选涂其它答案,答案不能答在试卷上;非选择题必须用黑色字迹的专用签字笔作答,答案必须写在答题卡各题目指定区域内相应位置上;如需改动,先划掉原来的答案,然后再写上新答案;不准使用铅笔和涂改液.不按以上要求作答无效;
3.考生必须保证答题卡的整洁,考试结束后,请将答题卡上交.
一、单项选择题: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.集合,集合,则( )
A. B. C. D.
2.已知是虚数单位,复数为纯虚数的充要条件是( )
A. B. C. D.
3.某校高三年级的学生参加了一次数学测试,学生的成绩全部介于分到分之间(满分分),为统计学生的这次考试情况,从中随机抽取名学生的考试成绩作为样本进行统计.将这名学生的测试成绩的统计结果按如下方式分成八组:第一组,第二组,第三组,…….如图是按上述分组方法得到的频率分布直方图的一部分.则第七组的频数为( )
A. B. C. D.
4.设函数的定义域为,满足,且.
则( )
A. B. C. D.
5.在直角梯形中,,,,,为的中点,则( )
A. B. C. D.
6.已知函数,则不等式的解集为( )
A. B. C. D.
7.三棱锥的底面是边长为的等边三角形,该三棱锥的所有顶点均在半径为 的球上,则三棱锥的体积最大值为( )
A. B. C. D.
8.已知定义在上函数的图象是连续不断的,满足,,且在上单调递增,若,,,则( )
A. B. C. D.
二、多项选择题:本题共4小题,每小题5分,共20分。在每小题给出的四个选项中,有多项符合题目要求。全部选对的得5分,部分选对的得3分,有选错的得0分。
9.已知点为曲线的焦点,则曲线的方程可能为( )
A. B.
C. D.
10.在棱长为的正方体中,点在棱上,则下列结论正确的是( )
A.直线与平面平行
B.平面截正方体所得的截面为三角形
C.异面直线与所成的角为
D.的最小值为
11.对于函数(其中),下列结论正确的是( )
A.若,,则的最小值为;
B.若,则函数的图象向右平移个单位可以得到函数的图象;
C.若,则函数在区间上单调递增;
D.若函数的一个对称中心到与它最近一条对称轴的距离为,则.
12.如图,是以为直径的圆上一段圆弧,是以 为直径的圆上一段圆弧,是以为直径的圆上一段圆弧,三段弧构成曲线.则下述正确的是( )
A.曲线与轴围成的面积等于;
B.曲线上有个整点(横纵坐标均为整数的点);
C.所在圆的方程为:;
D.与的公切线方程为:.
三、填空题:本题共4个小题,每小题5分,共20分。
13.若命题“,”为假命题,则实数的取值范围是 .
14.已知等比数列的前项和为,.若,则 .
15.(第一空2分,第二空3分)若二项式的展开式中所有项的系数和为,则:(1) ;(2)该二项式展开式中含有项的系数为 .
16.黄金分割比被誉为“人间最巧的比例”.离心率的椭圆被称为“优美椭圆”,在平面直角坐标系中的“优美椭圆”的左右顶点分别为,“优美椭圆”上动点(异于椭圆的左右顶点),设直线的斜率分别为,则 .
四、解答题:本题共6小题,共70分。解答应写出文字说明,证明过程或演算步骤。
17.(10分)
已知数列,满足:,,.
(1)证明:数列为等差数列,数列为等比数列;
(2)记数列的前项和为,求及使得的的取值范围.
18.(12分)
在中,内角的对边分别为,已知.
(1)求;
(2)若,求取最小值时的面积.
19.(12分)
如图,在三棱台中,,分别为上的点,
平面平面,.
(1)证明:平面平面;
(2)若,求二面角的大小.
20.(12分)
有甲、乙两家公司都需要招聘求职者,这两家公司的聘用信息如下:
甲公司 | 乙公司 | ||||||||
职位 | 职位 | ||||||||
月薪(元) | 月薪(元) | ||||||||
获得相应职位概率 | 获得相应职位概率 |
(1)根据以上信息,如果你是该求职者,你会选择哪一家公司?并说明理由;
(2)某课外实习作业小组调查了名职场人士,就选择这两家公司的意愿做了统计,得到以下数据分布:
选择意愿 人员结构 | 岁以上(含岁)男性 | 岁以上(含岁)女性 | 岁以下男性 | 岁以下女性 |
选择甲公司 | ||||
选择乙公司 |
若分析选择意愿与年龄这两个分类变量,计算得到的的观测值为,测得出“选择意愿与年龄有关系”的结论犯错误的概率的上限是多少?并用统计学知识分析,选择 意愿与年龄变量和性别变量哪一个关联性更大?
参考公式:
21.(12分)
已知函数.
(1)证明:;
(2)数列满足:,.
(ⅰ)证明:;
(ⅱ)证明:,.
22.(12分)
已知椭圆的短轴长和焦距相等,左、右焦点分别为、,点满足:.已知直线与椭圆相交于两点.
(1)求椭圆的标准方程;
(2)若直线过点,且,求直线的方程;
(3)若直线与曲线相切于点,且中点的横坐标等于.
证明:符合题意的点有两个,并任求出其中一个的坐标.
第一学期期末学业水平检测高三数学参考答案
一、单项选择题:本题共8小题,每小题5分,共40分。
18:D C A B A D C D
二、多项选择题:本题共4小题,每小题5分,共20分。
9. AD; 10. ACD; 11. AD; 12. BCD;
三、填空题:本题共4个小题,每小题5分,共20分。
13. ; 14. ; 15. (1);(2); 16. ;
四、解答题:本题共6小题,共70分。解答应写出文字说明,证明过程或演算步骤。
17.(10分)
解:(1)由和相加得:
所以,因此数列是以为公差的等差数列·····································2分
又由和相减得:,所以,
又,因此数列是以为公比的等比数列·······································4分
(2)由(1)知:
两式相加得:························································6分
所以·······························································8分
因为,所以··························································9分
又因为,
所以使得的的取值范围为···············································10分
18.(12分)
解:(1)因为,
所以,即, ·························································1分
由正弦定理得,······················································2分
由于为的内角,所以,·················································3分
所以,即····························································4分
由于为的内角,∴,所以···············································5分
又因为,所以,;·····················································6分
(2)在中由余弦定理知:
···································································9分
所以,等号当仅当时等号成立···········································11分
此时······························································12分
19.(12分)
解: (1)因为平面平面,
平面平面
平面平面
所以·································2分
因为,所以四边形为平行四边形
所以,因为
所以,为的中点·························3分
同理为的中点,所以
因为,所以····························4分
又且,所以四边形是平行四边形,所以,
又,所以.···························································5分
又平面,,所以平面,
又平面,所以平面平面·················································6分
(2)由(1)知,,因为,,,所以
分别以所在的直线为轴,轴,轴,建立如图所示的空间直角坐标系,
则 7分
设平面的一个法向量为,因为
则,取,得 ·························································9分
设平面的一个法向量为,因为
则,取,得 ·························································11分
所以,则二面角的大小为···············································12分
20.(12分)
解:(1)设甲公司与乙公司的月薪分别为随机变量,,
则,
,
,
,
则,,··························································4分
我希望不同职位的月薪差距小一些,故选择甲公司;或我希望不同职位的月薪差距大一些,故选择乙公司;
(只要言之有理即给分)············································6分
(2)因为,根据表中对应值,
得出“选择意愿与年龄有关系”的结论犯错的概率的上限是,··················7分
由数据分布可得选择意愿与性别两个分类变量的列联表如下:
| 选择甲公司 | 选择乙公司 | 总计 |
男 | |||
女 | |||
总计 |
计算,且,
对照临界值表得出结论“选择意愿与性别有关”的犯错误的概率上限为,
由,所以与年龄相比,选择意愿与性别关联性更大.··························12分
21.(12分)
解:(1)由题意知,,·················································1分
当时,,
所以在区间上单调递减·················································2分
当时,令,因为
所以在区间上单调递增,因此············································4分
故当时,,
所以在区间上单调递增·················································5分
因此当时,··························································6分
所以
(2)(ⅰ)由(1)知,在区间上单调递增,
因为
故·································································7分
所以·······························································8分
因此当时,,又因为,
所以·······························································9分
(ⅱ)函数,则
令,则····························································10分
所以在区间上单调递增;
因此 ······························································11分
所以在区间上单调递减,所以
因此
所以,····························································12分
22.(12分)
解:(1)设椭圆焦距为,因为椭圆的短轴长和焦距相等,
所以,①····························································1分
因为,所以点在椭圆上
将代入得:②························································2分
由①②解得:························································3分
所以椭圆的方程为·····················································4分
(2)设,,由题意,则可设直线的方程为:,
由得:,
所以,·····························································5分
又因为,所以,
所以,解得:,······················································6分
所以·······························································7分
所以,解得:
所以直线的方程为:或·················································8分
(3)设,,由题意直线的斜率存在,设直线的方程为:,
由得:,则··························································9分
因为直线与曲线相切于点,所以,
所以,整理得·······················································10分
令,所以
因为在上单调递增;且
所以,存在使得······················································11分
因此在上单调递减,在上单调递增;所以
又因为,所以,
又因为,
因此除零点外,在上还有一个零点
所以,符合题意的点有两个,其中一个的坐标为·····························12分
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