2021湖州德清县三中高二下学期返校考试数学试题含答案

这是一份2021湖州德清县三中高二下学期返校考试数学试题含答案，共12页。
德清三中2020学年第二学期返校考试卷高二  数学本试卷满分150分，考试时间120分钟选择题部分（共40分）一、选择题：本大题共10小题，每小题4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．直线的倾斜角是（    ）A.45° B.60° C.120° D.135°2．在空间直角坐标系中，，，则A，B两点的距离是（    ）A.6 B.4 C.6 D.23．准线为的抛物线标准方程是（    ）A． B． C． D．4．圆：，圆：，则圆与圆的位置关系为（    ）A．相交 B．相离 C．内切 D．外切5.已知空间中不过同一点的三条直线m，n，l，则“m，n，l在同一平面”是“m，n，l两两相交”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.已知双曲线G：的左､右焦点分别为、，若点P在G的右支上，且，则（    ）A.3 B.5 C. D.7．已知过点的直线l被圆截得的弦长为，则直线l的方程是（    ）A.  B.C.或 D.或8．已知m，n是两条直线，，是两个平面，则下列命题中错误的是（    ）A．若，，，则B．若，，则C．若，，，则D．若，，，则9.如图，在棱长为1的正方体中，点M是底面正方形的中心，点P是底面所在平面内的一个动点，且满足，则动点P的轨迹为（    ）A.圆 B.抛物线 C.双曲线 D.椭圆9．已知直线：与直线：分别过定点A，B，且交于点P，则的最大值是（    ）A.5 B．5 C．8 D．1010.如图，已知正方体的棱长为4，E为棱的中点，点P在侧面上运动.当平面与平面、平面所成的角相等时，则的最小值为（    ）A. B. C. D.非选择题部分（共110分）二、填空题：本大题共7小题，多空题每题6分，单空题每题4分，共36分.12.双曲线的焦距是_________，渐近线方程是_________．13.直线：，直线：，若，则________；若，则________．14．若某几何体的三视图（单位：）如图所示，则该几何体的体积是_________，最长的棱长是_________．15.已知过点，且斜率为k的动直线l与抛物线C：相交于B，C两点，则k的取值范围为_________；若N为抛物线C上一动点，M为线段中点，则点M的轨迹方程为____________．16.长、宽、高分别为2，1，2的长方体的每个顶点都在同一个球面上，则该球的表面积为__________．17．如图，在侧棱垂直于底面的三棱柱中，，，E，F分别是，的中点，则异面直线与所成角的余弦值是________．18．已知，是椭圆C：的焦点，若椭圆C上存在点P，使，则椭圆C的离心率的取值范围是________．三、解答题：本大题共5小题，共74分.解答应写出文字说明、证明过程或演算步骤.19．（本题满分14分）设圆C的半径为r，圆心C是直线与直线的交点．（Ⅰ）若圆C过原点O，求圆C的方程；（Ⅱ）已知点，若圆C上存在点M，使，求r的取值范围．20．（本题满分15分）如图，已知三棱锥﹐，是边长为的正三角形，，，点F为线段的中点．（Ⅰ）证明：平面；（Ⅱ）求直线与平面所成角的大小．21.（本题满分15分）已知抛物线，与圆F：，直线：与抛物线相交于M，N两点.（1）求证：.（2）若直线与圆F相切，求的面积S.22．（本题满分15分）如图，在三棱锥中，，，，E，F分别是，的中点，M是上一点．（Ⅰ）求证：平面；（Ⅱ）求直线与平面所成角的正弦值的最大值．23.（本题满分15分）已知椭圆E：（）的左右焦点分别为，，其离心率为，点在椭圆E上.（1）求椭圆E的标准方程；（2）经过椭圆E的左焦点作斜率之积为的两条直线，，直线交椭圆E于A，B，直线交椭圆E于C，D，G，H分别是线段，的中点，求面积的最大值.答案一、选择题1.【答案】A 2.【答案】C 3.【答案】A 4.【答案】D5.【答案】B 6.【答案】B 7.【答案】D 8.【答案】C9.【答案】D 10.【答案】D11.【答案】A解：如图，设点F为的中点，点P在平面内的射影为，则在平面内的射影为，在平面内的射影为.由于平面与平面、平面所成的角相等，则，故.由于，从而，即点P到直线的距离.设点由此点M为的中点，则点P在线段上运动.故的最小值即为点到直线的距离.二、填空题12.【答案】；13.【答案】-1；414.【答案】20；15.【答案】（1）或  （2）16.【答案】17.【答案】18.【答案】三、解答题19.解：（1）由，得，所以圆心.又∵圆C过原点O，∴∴圆C的方程为：································································7分.（2）设，由，得：，化简，得：.∴点M在以为圆心，半径为2的圆上.又∵点M在圆C：上，∴，即，∴.·········································································14分.20.解（1）∵，，∴，∴∵∴平面.······································································7分（2）∵平面，∴平面平面作，交于H，连接，∵平面，平面平面∴平面∴就是直线与平面所成角．·······················································11分∵为正三角形，∴且H是中点，∴，∴．·····································································15分21．解（1）设，联立，∴  ∵，∴，即.··········································································7分（2）∵直线与圆相切，  ∴，∴原点到直线的距离，···························································10分，.············································································15分22.解法一：（1）∵，F是中点；∴∵，E，F分别是，的中点；∴又∵，∴面····································································7分（2）过A作于H∵面，面，∴∵，E为中点，∴又∵  ∴面∴，又，∴面∴在平面上的射影是∴即为与平面所成的角中，，中，，∴.··········································································14分解法二：解：（Ⅰ）建系如图，则，，∵，设则，解得，∴，∴，，∴，∴平面.·······································································7分（Ⅱ），，设平面的一个法向量设则设直线与平面所成角为············································································14分23.解（1）因为，得，则，又椭圆经过点，则，即，故椭圆E的标准方程为.····························································6分.（2）设直线的斜率为，则：，设，，联立得，，，，··········································································8分.的中点，同理可得的中点，，所以，，···································································10分.则：.令得，所以在x轴上的交点为，·····················································12分.所以，令，，因为，，即面积的最大值.·························································15分.