辽宁省沈阳市郊联体2020-2021学年高二下学期开学初数学试题（图片版）

沈阳市郊联体2020-2021学年度第二学期开学初高二年级 数学答案 一、单选题： 1. D 2.B 3.C 4.C 5.B 6. B 7.A 8.D 二、多选题： 9. ABC 10. BD 11. BCD 12. BD 三、填空题： 13. 14.60 15． 16. 四、解答题： 17. 由命题直线经过第二､三､四象限，可得， 解得；··························································· 4分 由命题：方程表示双曲线，可得，解得，··········· 8分 因为都为真命题，可得， 即实数的取值范围.··················································10分 18． （1）证明：以为 原点，以所在的直线分别为轴，如图建立空间直角坐标系， ， ，所以， 所以.·························································6分 （2）， 设平面的法向量为， 则，，，令，则.·······9分 设与平面所成角为， ，······················11分 所以与平面所成角的正弦值为.·······························12分 19. （1）因为，且， 所以，解得或（舍），·········3分 故的展开式中二项式系数最大的项为第5项， 即；················································6分 （2）令，可知，················································7分 令，得，····················8分 所以，·······························10分 故.································12分 20. （1）由题意得，，····················1分 ， ，，········································2分 因而相关系数.·····················3分 由于很接近1，说明x，y线性相关性很强， 因而可以用线性回归方程模型拟合y与x的关系.由于，故其关系为负相关. ····4分 （2）由（1）知，，∴， 则所求的回归方程是.·······································6分 当特征量x为12时，可预测特征量.··················8分 （3）由（1）知，，又由， 得，·····························································10分 从而.··12分 21. （1）根据题意列出列联表如下： 位置 类型 糟糕 良好 合计 电信 3 2 5 网通 2 3 5 合计 5 5 10 ····································································2分 ，故在犯错误的概率不超过的前提下，不能说明游戏的网络状况与网络的类型有关.·································4分 （2）依题意，所求概率.··································6分 （3）随机变量的所有可能取值为1，2，3， ；；.···8分 故的分布列为 1 2 3 ······································································10分 ·∴.··································12分 22. （1）∵椭圆的离心率为，∴， ∵四边形的面积为，∴， 又，解得：，，， ∴椭圆的方程为.····································4分 （2）设，，则的周长为， ，即， ·····6分 当时，的方程为，，. 当与轴不垂直时，设， 由，得，·························8分 ∴，， ，··········································10分 令，， ， ∵，∴，∴. 综上可知：.···················································12分 $$ 1 2 3 $$