2022湖北省九师联盟高三上学期10月质量检测数学试题含答案

这是一份2022湖北省九师联盟高三上学期10月质量检测数学试题含答案，共15页。试卷主要包含了本试卷分选择题和非选择题两部分，答题前，考生务必用直径0，本卷命题范围，已知向量，满足，，，则，函数的单调递增区间是，设，是复数，则等内容，欢迎下载使用。
九师联盟2022届高三上学期10月质量检测数学考生注意：1.本试卷分选择题和非选择题两部分。满分150分，考试时间120分钟。2.答题前，考生务必用直径0.5毫米黑色墨水签字笔将密封线内项目填写清楚。3.考生作答时，请将答案答在答题卡上。选择题每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑；非选择题请用直径0.5毫米黑色墨水签字笔在答题卡上各题的答题区域内作答，超出答题区域书写的答案无效，在试题卷、草稿纸上作答无效。4.本卷命题范围：函数与导数、复数、三角函数、解三角形、平面向量、不等式。一、选择题：本题共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.不等式的解集为（    ）A.  B.C.  D.2.已知复数，则下列说法正确的是（    ）A. z的模为  B. z的虚部为C. z的共轭复数为 D. z的共轭复数在复平面内对应的点在第四象限3.若，则（    ）A.-4 B. C.-3 D.4.一种药在病人血液中的量保持在1500以上时才有疗效，而低于600时病人就有危险.现给某病人的静脉注射了这种药2400，如果药在血液中以每小时20%的比例衰减，要使病人没有危险，再次注射该药的时间不能超过（，结果精确到1h）（    ）A.5h B.6h C.7h D.8h5.已知向量，满足，，，则（    ）A. B. C. D.6.函数的单调递增区间是（    ）A.， B.，C.， D.，7.把一条线段分为两部分，使其中一部分与全长之比等于另一部分与这部分之比，其比值是一个无理数，由于按此比例设计的造型十分美丽柔和，因此称为黄金分割，黄金分割不仅仅体现在诸如绘画、雕塑、音乐、建筑等艺术领域，而且在管理、工程设计等方面也有着不可忽视的作用.在中，点D为线段的黄金分割点（），，，，则（    ）A. B. C. D.8.已知函数的定义城为，则满足的实数a的取值范围是（    ）A. B. C. D.二、选择题：本题共4小题，每小题5分，共20分。在每小题给出的选项中，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.为了得到函数的图象，只需将函数图象上（    ）A.所有点的横坐标伸长到原来的2倍，纵坐标不变B.所有点的横坐标缩短到原来的倍，纵坐标不变C.所有点沿y轴向下平移1个单位长度D.所有点沿x轴向右平移个单位长度10.设，是复数，则（    ）A.  B.若，则C.若，则 D.若，则11.下列命题成立的是（    ）A.若，，则B.若不等式的解集是，则C.若，，则D.若a，b满足，则的取值范围是12.已知函数，则（    ）A.当时，B.当时，C.当时，D.方程有两个不同的解三、填空题：本题共4小题，每小题5分，共20分。13.若是奇函数，则__________.14.已知O是的外心，且，则__________.15.已知函数（，）满足，其图象与x轴在原点右侧的第一个交点的坐标为，则函数的一个解析式为__________.16.拿破仑·波拿巴，十九世纪法国伟大的军事家、政治家，对数学很有兴趣，他发现并证明了著名的拿破仑定理：“以任意三角形的三条边为边向外构造三个等边三角形，则这三个三角形的中心恰为另一个等边三角形的顶点”.如图，在中，，以，，为边向外作三个等边三角形，其中心依次为D，E，F，若，则__________，的最大值为__________.四、解答题：本题共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）在①，②，③这三个条件中任选一个，补充在下面的问题中，并进行解答；的内角A，B，C的对边分别为a，b，c，若的面积，，_________，求b.注：如果选择多个条件分别解答，按第一个解答计分.18.（本小题满分12分）已知，，方程的一个根为，复数，满足.（1）求复数；（2）若，求复数.19.（本小题满分12分）已知函数.（1）若，解关于x的不等式；（2）若存在，使得成立，求整数a的最大值.20.（本小题满分12分）已知向量，，（1）若，且，求x的值；（2）若函数，且，求的值.21.（本小题满分12分）已知函数，.（1）求函数的极值；（2）证明：有且只有两条直线与函数，的图象都相切.22.（本小题满分12分）（1）已知函数（），求证：；（2）若函数在上为减函数，求实数k的取值范围.高三数学参考答案、提示及评分细则1.B  由，得，所以.故选B.2.A  ，z的模为，故Ａ正确；z的虚部为，故B错误；z的共轭复数为，故C错误；z的共轭复数在复平面内对应的点为，在第一象限，故D错误.故选A.3.D  ，.故选D.4.B  血液中含药量y（单位：）与注射后的时间t（单位：h）的关系式为，由题意，得，即，两边取对数，得.故选B.5.B  由已知，得，结合，，解得，所以，即.故选B.6.A  ，设，则，，则当]时，y是关于t的减函数；当，时，t是关于x的减函数，根据复合函数的单调性法则，函数的单调递增区间是，.故选A.7.A  点D为线段的黄金分割点，则，所以，则.故选A.8.C  函数和的图象在上都关于直线对称，且它们都在上递增，在上递减，则函数的图象在上关于直线对称，且在上递增，在上递减.由，得，即，从而，解得.故选C.9.AC  函数图象上所有点的横坐标伸长到原来的2倍，纵坐标不变，可得函数的图象，则A正确；函数图象上所有点的横坐标缩短到原来的倍，纵坐标不变，可得函数的图象，则B错误；，将图象上的所有点沿y轴向下平移1个单位长度，就得到函数的图象，则C正确；函数图象上所有点沿x轴向右平移个单位长度，可得函数的图象，则D错误.故选AC.10.AC  设，，a，b，x，，则，A成立；，则，所以，，从而，所以，C成立；对于B，取，，满足，但结论不成立；对于D，取，，满足，但结论不成立.故选AC.11.BC  对于A，取，，，，则，则A错误；对于B，方程的两根分别为1和2，则，，解得，，所以，则B正确；因为，，所以，则C正确；由，，得，又，所以，即的取值范围是，则D错误.故选BC.12.BC  在定义域内单调递增，则，A错误；，令，，当时，，则在单调递增，所以B正确；由，可得，令，在上小于0，所以在单调递减，则当时，，即，所以C正确；，当时，，而函数在上单调递增，函数的图象与直线仅有一个公共点，如图所示：则方程仅有一个解，故D错误.另解：方程的解的个数，即为的解的个数，即为函数与图象交点的个数，作出函数与图象如图所示：由图象可知方程只有一个解，故D错误.故选BC13.-1  .由题意，得对任意的恒成立对任意的恒成立对任意的恒成立，考虑到，于是.14.-1  取的中点D，则.15.（或）  设的最小正周期为T，由题意，得，即，则，将代入，得，解得，，考虑到，得或，所以或.16.（2分）；（3分）  设，，.如图，连接，.由拿破仑定理知，为等边三角形.在中，，，设，由余弦定理，得，解得，即，，同理；又，，所以，在中，由余弦定理，得，即，化简得，由基本不等式得，解得（当且仅当时取等号），所以.17.解：选择条件①：由，根据正弦定理，有，···············································2分即，  ··········································································4分由，，所以，故.  ············································································6分由的面积，解得.  ·································································8分根据余弦定理，得，故.  ···········································································10分选择条件②：由，根据正弦定理，有，·················································2分由，得，所以，  ··································································4分则，又，所以. ···································································6分由的面积，解得. ··································································8分根据余弦定理，得，故. ···········································································10分选择条件③：，即，·······························································2分整理得，解得或-1.  ································································4分由，得，所以. ···································································6分由的面积，解得.  ·································································8分根据余弦定理，得，故.  ···········································································10分18.解；（1）依题意，得，即，···········································································2分由复数相等的定义及a，，得，解得.  ··········································································3分故复数.  ········································································4分（2）设（，），由，得， ··························································6分，·············································································8分又，得，即，所以，·········································································10分解得，所以. ··········································································12分19.解；（1）由，得，，······························································1分当，即，或时，的根，原不等式的解集为；·······························································3分当，即，或时，的根，原不等式的解集为；·······························································5分当，即时，原不等式的解集为. ·······················································6分（2）由，得，再由，得， ·····································································8分所以存在，使得成立就等价于.而（当且仅当时等号成立），·······················································10分所以，解得，故整数a的最大值为-1.  ····························································12分20.解：（1）由，得，································································1分即，所以或.·········································································2分当时，，则；····································································3分当时，得，，则.综上，x的值为或.·································································4分（2） ··········································································6分. ··············································································7分由，得，········································································8分所以 ··········································································10分. ·············································································12分21.（1）解：的定义域为，····························································1分且. ············································································3分当时，；当时，，所以在上单调递增，在上单调递减，···················································4分所以是的极大值点，故的极大值为，没有极小值. ·························································5分（2）证明：设直线l分别切，的图象于点，，由，得l的方程为，即l：；由，得l的方程为，即l：.比较l的方程，得，消去，得. ·······································································7分令（），则.当时，；当时，，所以在上单调递减，在上单调递增，所以.···········································································9分因为，所以在上有一个零点；·······················································10分由，得，所以在上有一个零点.所以在上有两个零点，····························································11分故有且只有两条直线与函数，的图象都相切. ············································12分22.（1）证明：.  ····································································1分因为，所以，，所以，所以在上为减函数，··························································3分于是，，故.·············································································4分（2）解；设，则，从而在上为增函数，由，得，即. ·····································································5分（i）当时，，则，从而，因为函数在上为减函数，所以，即对恒成立，即对恒成立，根据（1），，所以.再结合，此时，. ··································································7分（ii）当时，，则，从而，因为函数在上为减函数，所以，即对恒成立，即对恒成立，根据（1），，所以.再结合，此时. ···································································9分（iii）当时，则存在唯一的，使得，从而.当时，，即存在，且，使得，这与“在上为减函数”矛盾，此时不合题意. ·····················11分综上，实数k的取值范围是. ·························································12分