2022湖北省云学新高考联盟学校高一下学期5月联考数学试题（含答案）

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2022年湖北云学新高考联盟学校高一年级5月联考数学试卷命题学校：洪湖一中  命题人：路明琴  审题人：洪湖一中  杨前军  嘉鱼一中  甘业明、郑世波考试时间：2022年5月30日09：50—11：50一、选择题（本大题共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.下列命题正确的是（    ）A.有两个面平行，其余各面都是平行四边形的几何体是棱柱B.有一个面是平面多边形，其余各面都是三角形的几何体是棱锥C.等腰梯形绕它上、下底边中点的连线旋转180°可以得到一个圆台D.在圆柱的上、下底面的圆周上各取一个点，则这两点的连线是圆柱的母线2.已知向量，，且，则（    ）A.6 B. C. D.3.设，为两个不同的平面，则的一个充分条件是（    ）A.内有无数条直线与平行 B.，平行于同一个平面C.，平行于同一条直线 D.，垂直于同一个平面4.顺次连接点，，，所构成的图形是（    ）A.等腰梯形 B.平行四边形 C.菱形 D.矩形5.设，，，则a，b，c大小关系正确的是（    ）A. B. C. D.6.设函数，若，，则函数的零点个数为（    ）A.1 B.2 C.3 D.47.在中，设，则动点M的轨迹必通过的（    ）A.外心 B.内心 C.重心 D.垂心8.已知四面体的所有棱长均为2，M，N分别为棱，的中点，F为棱上异于A，B的动点.有下列结论：①线段的长度为；②若点G为线段上的动点，则无论点F与G如何运动，直线与直线都是异面直线；③异面直线和所成的角为；④的最小值为2.其中正确的结论为（    ）A.①③④ B.②③ C.②③④ D.①④二、选择题（本大题共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2分，有选错的得0分。）9.已知i为虚数单位，以下四个说法中正确的是（    ）A.B.C.若，则复平面内对应的点位于第二象限D.已知复数z满足，则z在复平面内对应的点的轨迹为直线10.已知向量，（），则下列命题正确的是（    ）A.若，则B.存在，使得C.若向量在方向上的投影向量为，则向量与的夹角为D.与向量共线的单位向量是11.已知正方体的棱长为1，下面选项正确的是（    ）A.直线与平面不垂直 B.四面体的体积为C.异面直线与直线所成角的为 D.直线与平面所成的角为12.已知正四棱台，下底面边长为4，上底面边长为2，侧棱长为2，则（    ）A.它的表面积是 B.它的外接球球心在该四棱台的内部C.侧棱与下底面所成的角为 D.它的体积比半径为的球的体积小三、填空题（本大题共4小题，每小题5分，共20分）13.已知虚数z的实部不为0，模为2，则符合要求的一个虚数_______________.14.已知四棱锥中，侧棱平面，底面是矩形，则该四棱锥的4个侧面中直角三角形的个数是______________.15.如图，矩形是水平放置的一个平面图形由斜二测画法得到的直观图，其中，，则原四边形的周长是16.已知一圆锥底面直径是，圆锥的高是，在该圆锥内放置一个棱长为a的正四面体，且正四面体可以在该圆锥内任意转动，则a的最大值为_____________.四、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10分）平行四边形中，点M在上，且，点N在上，且，记，（1）以，为基底表示；（2）求证：M、N、C三点共线.18.（本小题满分12分）已知关于x的不等式的解集为（）.（1）求a，b的值；（2）当，，且满足时，有恒成立，求k的取值范围.19.（本小题满分12分）某科技企业生产一种电子设备的年固定成本为600万元，除此之外每台机器的额外生产成本与产量满足一定的关系式.设年产量为x（，）台，若年产量不足70台，则每台设备的额外成本为万元；若年产量大于等于70台不超过200台，则每台设备的额外成本为万元.每台设备售价为100万元，通过市场分析，该企业生产的电子设备能全部售完.（1）写出年利润W（万元）关于年产量x（台）的关系式；（2）当年产量为多少台时，年利润最大，最大值为多少？20.（本小题满分12分）设函数.（Ⅰ）求函数的周期及图象的对称轴；（Ⅱ）在锐角中，若，且能盖住的最小圆的面积为，求的取值范围.21.（本小题满分12分）已知四棱锥的底面是菱形，平面，，，F，G分别为，中点，.（Ⅰ）求证：平面；（Ⅱ）求三棱锥的体积；（Ⅲ）求证：与不垂直.22.（本小题满分12分）对于函数，若在定义域内存在实数x，满足，则称为“局部奇函数”.（1）已知二次函数（），试判断是否为“局部奇函数”，并说明理由；（2）若是定义在区间上的“局部奇函数”，求实数m的取值范围；（3）若为定义在R上的“局部奇函数”，求实数m的取值范围.              2022年湖北云学新高考联盟学校高一年级5月联考数学评分细则一、单选题1.C  2.D  3.B  4.B  5.C  6.B  7.A  8.A二、多选题9.BD  10.AC  11.BCD  12.AD三、填空题13.（答案不唯一）  14.4  15.  16.4四、解答题17.（1）解：；······································································5分（2）证明：∵Mb，，∴，∴且与有公共点M，所以M、N、C三点共线.····················································10分18.解：（1）因为不等式的解集为（），所以1和a是方程的两个实数根且，··············································2分所以，解得.·······························································4分（2）由（1）知，且，故，····································································7分当且仅当，即时，等号成立.··················································8分依题意有，即，得，···································································11分所以k的取值范围为.························································12分19.解：（1）当，时，；······································································3分当，时，.∴.······································································6分（2）①当，时，，∴当时，y取得最大值，最大值为1200万元.·······································8分②当，时，，·····································································10分当且仅当，即时，y取得最大值1320，··········································11分∵，∴当年产量为80台时，年利润最大，最大值为1320万元.····························12分20.解：（Ⅰ）因为，······································································2分所以函数的周期····························································3分令（），解得（）所以函数的周期是，对称轴方程是（）.·········································5分（Ⅱ）因为，所以.又因为为锐角三角形，所以，.所以，故有.·······························································6分已知能盖住的最小圆为的外接圆，而其面积为，设半径为。所以，得，·······························································7分的角A，B，C所对的边分别为a，b，c.由正弦定理.所以，，，······································································9分因为为锐角三角形，所以.···················································10所以，则，故，···································································11分所以的取值范围是.·························································12分21.（Ⅰ）证明：如图，连接，，∵O是中点，F是中点，∴，平面，平面，则平面.∵O是中点，G是中点，∴，平面，平面，则平面.又，，平面，∴平面平面，又平面，则平面.··································································4分（注：也可构造线线平行证明.）（Ⅱ）解：∵底面，底面，∴，又四边形为菱形，∴，又，、平面，∴平面，且，·····························································6分而F为的中点，∴；····································································8分（Ⅲ）证明：假设，又，且，，平面，∴平面，而平面，则，与矛盾.∴假设错误，故与不垂直.···················································12分22.解：为“局部奇函数”等价于关于x的方程有解.（1）当（）时，方程，即有解，而，所以，从而为“局部奇函数”.······································2分（2）当时，可化为.·························································3分因为的定义域为，所以方程在上有解.令，则，上式化为.·························································4分设，则在上为单调减函数；在上为单调增函数.因为，，，所以时，.所以，即.································································7分（3）当时，······························································8分可化为.设，则，则，从而只需要关于t的方程在上有解即可.···········································9分令.①       当时，在上有解，由，即，解得；②       当时，在上有解等价于，解得.·································································11分（注：也可转化为大根不小于2求解）综上，所求实数m的取值范围为.···············································12分