2021年高三2月阶段性检测数学文试题 含答案

秘密★启用前
2021年高三2月阶段性检测数学文试题含答案
注意事项
1、本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分,考试结束后，请将本试卷和答题卡一并交回．满分150分，考试时间120分钟．
2、每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑，如需改动，用橡皮擦擦干净后，再选涂其它答案标号，不能答在试题卷上．
3、答题前认真阅读答题卡上的“注意事项”．
参考公式：
样本数据的标准差
其中为样本平均数
柱体体积公式
其中为底面面积，为高
锥体体积公式
其中为底面面积，为高
球的表面积，体积公式
，
其中为球的半径
第Ⅰ卷(选择题，共60分)
一、选择题：本大题共12题，每小题5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、设全集,集合,，则
A．B．C．D．
2、若复数满足，则对应的点位于
A．第一象限B．第二象限C．第三象限D．第四象限3、已知,，则的值等于
A．B．C．D．
4、设是等差数列的前项和，，则
5、“”是“直线:与:平行”的
A ．充分不必要条件
B ．必要不充分条件
C ．充分必要条件
D ．既不充分也不必要条件
6、已知等比数列的首项公比，则 A ．
B ．
C ．
D ．
7、执行如图所示的程序框图．若输出， 则框图中①处可以填入 A ．
B ．
C ．
D ．
8、某四面体的三视图如图所示，该四面体的六条棱的长度中最大的是 A ．
B ．
C ．
D ．
9、设，， , 则
A ．
B ．
C ．
D ．
10、已知函数的最大值为，最小值为，最小正周期为，直线是其图像的一条对称轴，则下面各式中符合条件的函数解析式是 A ． B ． C ． D ．
11、已知双曲线（，），过其右焦点且垂直于实轴的直线与双曲线交于两点，为坐标原点，若，则双曲线的离心率为
正（主）视图
2
2
12、已知是圆：上的两个点，是线段上的动点，当的面积最大时，则的最大值是 A ． B ． C ． D ．
第Ⅱ卷 (非选择题，共90分)
本卷包括必考题和选考题两部分．第13题第21题为必考题，每个试题考生都必须作答．第22题第24题为选考题，考生根据要求作答．
二、填空题：本大题共4个小题，每小题5分．
13、已知实数满足，则目标函数的最大值为 ．
14、下面茎叶图表示的是甲、乙两人在次综合测评中的成绩，其中一个数字被污损．则甲的平均成绩超过乙的平均成绩的概率为 ．
15、一个所有棱长均为的正三棱锥（底面是正三角形，顶点在底面的射影是底面的中心）的顶点与底面的三个顶点均在某个球的球面上，则此球的体积为 ．
16、若函数对任意的()0,,(cos2)(sin 2)0f f m θπθθ∈+-≤恒成立，则的最大值是___________．
三、解答题：本大题共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、（本小题满分12分）已知某校在一次考试中，名学生的历史和语文成绩如下表：
（Ⅰ）若在本次考试中，规定历史成绩在以上（包括分）且语文成绩在分以上
（包括分）的为优秀，计算这五名同学的优秀率；
（Ⅱ）根据上表利用最小二乘法，求出关于的线性回归方程，其中
；
（III）利用（Ⅱ）中的线性回归方程，试估计历史分的同学的语文成绩．（四舍五入到整数）
18、（本小题满分12分）在△中,三个内角,,的对边分别为,,，且
．
（Ⅰ）求角；
（Ⅱ）若，求△面积的最大值．
19、（本小题满分12分）在直角梯形中，，，，如图，把沿翻折，使得平面平面．
（Ⅰ）求证：；
（Ⅱ）若点为线段中点，求点到平面的距离．
B B
20、（本小题满分12分）已知函数
（Ⅰ）求曲线在点处的切线方程；
（Ⅱ）对恒成立，求实数的取值范围．
21、（本小题满分12分）椭圆短轴的一个端点与两个焦点的连线构成面积为的等腰直角三角形．
（Ⅰ）求椭圆的方程；
（Ⅱ）过点的直线与椭圆相交于，两点．点，记直线的斜率分别为，当最大时，求直线的方程．
请考生在第22、23、24三题中任选一题作答，如果多做，则按所做的第一题计分．作答时用2B铅笔在答题卡上把所选题目的题号涂黑．
22、（本小题满分10分）【选修4—1：几何证明选讲】
如图，是⊙的一条切线，切点为，都是⊙的割线，已知．
（Ⅰ）证明：；
23、（本小题满分10分）【选修4—4：坐标系与参数
方程】
已知在直角坐标系中，直线的参数方程是（为参数），在极
坐标系（与直角坐标系取相同的长度单位，且以原点为极点，以轴正半轴为极轴）中，圆的极坐标方程为．
（Ⅰ）将圆的极坐标方程化为直角坐标方程；
（Ⅱ）若直线与圆相交于两点，点的坐标为，试求的值．
24、（本小题满分10分）【选修4—5：不等式选讲】
设不等式的解集与关于的不等式的解集相同．
（Ⅰ）求，的值；
（Ⅱ）求函数的最大值，以及取得最大值时的值．
凯里一中xx届高三年级2月阶段性检测
数学参考答案（文）
一、选择题：本大题共12题，每小题5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．
题号 1 2 3 4 5 6 7 8 9 10 11 12 选项 C A A B A D B C C D D C
二、填空题：本大题共4个小题，每小题5分．
13、14、15、16、
三、解答题：本大题共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、解：（Ⅰ）这名学生中有名历史成绩在以上且语文成绩在分以上，
所以这五名同学的优秀率为．······························································································（Ⅱ）······································································································································ ·················································································································································
（III ）
所以历史分的同学的语文成绩是 ·························································································· 18、解：（Ⅰ）由得：, ···················································································································
结合正弦定理有：, ··············································································································· 因为在△中，,
所以． ······································································································································ 又,
所以． ······································································································································ （Ⅱ）由余弦定理 , 因为,,
所以． ······································································································································ 因为,
所以． （当且仅当时取等号） ······························································································ 因此，． ·································································································································· 19、解：（Ⅰ）证明：因为，，
所以,,
,2CD =
=
所以 ，
所以 ． ·································································································································· 又平面平面，平面平面
所以平面 ·································································································································· 又平面，
所以·········································································································································· （Ⅱ）由已知条件可得，，所以． 由（Ⅰ）知平面，即为三棱锥的高，
又，所以 ························································································································ 由平面，知
所以2
2
2
2
2
6,2,8CA DA DC AC BC =+===
于是·········································································································································· 设点到平面的距离为，
因为点为线段中点，求点到平面的距离为 ·········································································· 20、解：（Ⅰ） ································································································································
················································································································································· 所以曲线在点处的切线方程为
即 ············································································································································· （Ⅱ）依题意对恒成立等价于在上恒成立
可得在上恒成立， ······································································································ 令
····································································································································
，得 列表：
函数的最小值为, ··············································································· 根据题意，． ················································································ 21、解：（Ⅰ）由题：
()22222221
22
a a c a a a
b
c ⎧+=⎪⎪⨯⨯=⎨⎪⎪=+⎩
·
···············································································································
解得
所以椭圆的方程为 ·················································································································· （Ⅱ）①当直线的斜率为时，容易求得 ············································································· ②当直线的斜率不为时，设，直线的方程为 将代入，整理得 ．
则 ·················································································································································
又， 所以
()()121212
2
121212
93334493y y y y y y x x m y y m y y -++--⋅=---++ ··············································································································································· 令，则 当时，
当时，（仅当，即时，取等号） 此时
由①②得，当最大时，直线的方程为． ··········································································· 22、证明：（Ⅰ）∵是⊙O 的一条切线，为割线，
∴， ········································································································································ 又∵，
∴；·········································································································································· （Ⅱ）由（Ⅰ）有，
∵∠EAC=∠DAC ，∴△ADC ∽△ACE ，
∴∠ADC =∠ACE ， ··········································································································· ∵∠ADC =∠EGF ， ∴∠EGF =∠ACE ，
∴GF ∥AC ． ······························································································································· 23、解：（Ⅰ）由得，，
所以，······································································································································ ∴，
即圆的直角坐标系方程为：
① ··················································································
（Ⅱ）设两点对应的参数为，与①式联立得
所以·········································································································································· 根据参数的意义可知：
121212
12
2
t t t t t t t t +-=
=
=
=
································································· 24、解：（Ⅰ）不等式的解集为或。