江苏盐城2021-2022学年第二学期期终高二数学试题与答案

2021/2022学年度第二学期高二年级期终考试

数学参考答案

1.C  2.C 3.A 4.D

5.B  6.D 7.B 8.C

9.BC  10.ABD 11.AD 12.ACD

13.  14. 15. 16.

17．解：（1）提出假设:是否有兴趣收看天宫课堂与性别无关．···································1分

根据列联表中的数据，可以求得，···············································3分

因为，而，

所以没有95%的把握认为“是否有兴趣收看天宫课堂与性别有关”．·····················5分

（2）依题意，随机变量X的可能取值为0，1，2，····································6分

，，，

··········································································9分

随机变量X的概率分布表如下

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

18.解：（1）根据题设条件，设数列的公差为,数列的公比为，

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

所以所以,=2，所以.····························································6分

（2）根据题设条件，当，

由即，解得且，所以

=+········································································8分

=+·······································································10分

=374.·····································································12分

19.（1）证明：由底面ABCD为矩形可知，

又因为，平面，，

所以平面，·································································2分

又因为平面，故，

满足，∴，

又因为，平面，，故平面，····················································4分

又因为平面，故 PD  AC.······················································6分

（2）解：由（1）可知平面，又底面 ABCD 为矩形，故以为基底建立如图所示空间直角坐标系，则，

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

设平面ADE的一个法向量为，

由，可取，································································10分

，

则直线PB与平面ADE所成角的正弦值为.··········································12分

20.解：（1）计算可得，··························································2分

则，······································································4分

，

则y关于x的线性回归方程为.····················································6分

(2)

由上表可知只有最后一位同学体重超标了，因为用频率估计概率，故可认为从高二男生中任选一人，体重超标的概率为，则，·······7分

，，

，，

故随机变量X的概率分布表为

·········································································11分

其数学期望为（人）.························································12分

21.解：（1）设点，则，

当时，取得最小值为，························································1分

，

则当时，取得最大值，························································2分

解得，则椭圆方程为.·························································4分

注：由条件直接写出，也可，扣1分.

（2）设点，当或时，易得过点Q作椭圆的两条切线并不垂直，

故可设过点Q的椭圆的切线方程为，

联立方程组，消元可得，

由可得，··································································6分

又直线过点，则，于是，

化简可得，

由两条切线互相垂直可知，该方程的两根之积，·····································8分

则，即点Q在圆上，·························································10分

由解得，故存在点满足题意.···················································12分

22.解：（1）当时，，，·························································1分

，，则函数在处的切线方程为，·················································3分

切线与坐标轴的交点为，与坐标轴围成的三角形的面积为.······························4分

（2），因为函数有两个极值点，

所以方程有两个不相等实数根，

故且，即，·································································6分

且，

则，不妨设，·······························································8分

据上表可知，在处取得极大值，在处取得极小值，

，

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

设，由于在上恒成立，

故在上递增，故，

则的取值范围为.····························································12分