山东省胶州市2020-2021学年高一上学期期中学业水平检测——数学试题含答案

这是一份山东省胶州市2020-2021学年高一上学期期中学业水平检测——数学试题含答案，共8页。试卷主要包含了作答选择题时，已知，则的大小关系为，下列说法正确的是等内容，欢迎下载使用。
2020-2021学年度第一学期期中学业水平检测高一数学本试卷4页，22小题，满分150分．考试用时120分钟．   注意事项：1．答题前，考生务必将自己的姓名、考生号、考场号和座号填写在试题卷和答题卡上，并将准考证号条形码粘贴在答题卡上的指定位置；2．作答选择题时：选出每小题答案后，用2B铅笔在答题卡上对应题目选项的答案信息点涂黑；如需要改动，用橡皮擦干净后，再选涂其它答案，答案不能答在试卷上；非选择题必须用黑色字迹的专用签字笔作答，答案必须写在答题卡各题目指定区域内相应位置上；如需改动，先划掉原来的答案，然后再写上新答案；不准使用铅笔和涂改液．不按以上要求作答无效；3．考生必须保证答题卡的整洁，考试结束后，请将答题卡上交．  一、单项选择题：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．若函数的定义域为集合，则（    ）A． B． C． D．2．下列函数中与函数是同一函数的是（    ） A．          B． C． D．3．十六世纪中叶，英国数学家雷科德在《砺智石》一书中首先把“”作为等号使用，后来英国数学家哈利奥特首次使用“”和“”符号，并逐渐被数学界接受，不等号的引入对不等式的发展影响深远．若，则下列结论正确的是（    ） A． B． C． D．4．专家对某地区新冠肺炎爆发趋势进行研究发现，从确诊第一名患者开始累计时间（单位：天）与病情爆发系数之间，满足函数模型：，当时，标志着疫情将要大面积爆发，则此时约为（    ）(参考数据：) A． B． C． D．5．若关于的方程有两个正根，则的最小值为（    ） A． B． C． D．6．若函数是上的单调递增函数，则实数的取值范围是（    ） A． B． C． D．7．已知，则的大小关系为（    ） A． B． C． D．8．已知奇函数在上单调递减，若，则满足的的取值区间是（    ）A． B． C． D． 二、多项选择题：本题共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3分，有选错的得0分。 9．下列说法正确的是（    ） A．“对任意一个无理数，也是无理数”是真命题 B．“”是“”的充要条件 C．命题“”的否定是“” D．若“”的必要不充分条件是“”，则实数的取值范围是 10．“双”购物节中，某电商对顾客实行购物优惠活动，规定一次购物付款总额满一定额度，可以给与优惠： （1）如果购物总额不超过元，则不给予优惠；（2）如果购物总额超过元但不超过元，可以使用一张元优惠劵；（3）如果购物总额超过元但不超过元，则按标价给予折优惠；（4）如果购物总额超过元，其中元内的按第（3）条给予优惠，超过元的部分给予折优惠．某人购买了部分商品，则下列说法正确的是（    ） A．如果购物总额为元，则应付款为元 B．如果购物总额为元，则应付款为元 C．如果购物总额为元，则应付款为元 D．如果购物时一次性全部付款元，则购物总额为元11．下列函数是偶函数且在上具有单调性的函数是（    ） A． B． C．  D． 12．若，则下列选项成立的是（    ） A．  B．若，则C．的最小值为 D．若，则三、填空题：本题共4个小题，每小题5分，共20分。 13．已知集合，则集合的个数为            个．14．已知关于的不等式的解集为，则              ．15．            ．16．一位少年能将圆周率准确记忆到小数点后面位，更神奇的是提问小数点后面的位数时，这位少年都能准确地说出该数位上的数字．记圆周率小数点后第位上的数字为，则是的函数，设，．则（1）的值域为                            ；（2）函数与函数的交点有             个．（第一空2分，第二空3分）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．（10分）已知全集，集合，集合．（1）求；（2）求；（3）设集合，若，求实数的取值范围． 18．（12分）已知函数的定义域为，当时，函数．（1）若，利用定义研究在区间上的单调性；（2）若是偶函数，求的解析式． 19．（12分）某地区上年度电价为元，年用电量为，本年度计划将电价下降到元至元之间，而用户期望的电价为元．经测算，下调电价后新增用电量和实际电价与用户的期望电价的差成反比（比例系数为）．该地区的电力成本价为元．（1）写出本年度电价下调后电力部门的收益（单位：元）关于实际电价（单位：元）的函数解析式；（收益实际电量（实际电价成本价））（2）设，当电价最低定为多少时，仍可保证电力部门的收益比上年至少增长？20．（12分）已知函数，．（1）若在上单调递增，求实数的取值区间；（2）求关于的不等式的解集．   21．（12分）已知函数是奇函数，．（1）求的值，并求关于的不等式的解集；（2）求函数图象的对称中心．   22．（12分）已知函数．（1）直接写出在上的单调区间（无需证明）；（2）求在上的最大值；（3）设函数的定义域为，若存在区间，满足：，，使得，则称区间为的“区间”．已知（），若是函数的“区间”，求实数的最大值．
2020-2021学年度第一学期期中学业水平检测高一数学参考答案一、单项选择题：本题共8小题，每小题5分，共40分。  1—8: D A C B   B C C A  二、多项选择题：本题共4小题，每小题5分，共20分。9．CD;        10．ABD;        11． BC;         12．ABD;            三、填空题：本题共4个小题，每小题5分，共20分。13. ；     14. ；     15. ；     16. （1）；（2）；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.（10分）解：（1）由题解得，，················································2分所以·······························································3分（2）所以或·························································5分所以或·····························································6分（3）因为，所以且····················································8分所以的取值范围为：··················································10分 18.（12分）解：（1）当时，······················································1分设且·······························································2分则·································································3分························································5分因为所以，·····························································6分所以， 即所以在区间为单调递增函数··············································7分（2）令，则，·······················································8分所以······························································10分因为是偶函数所以······························································11分所以函数在上的解析式为：··································································12分  19．（12分）解：（1），·························································4分（2）由（1）知：时，·················································6分依题意得：且························································9分化简得····························································10分解得······························································11分所以当电价最低定为元时，仍可保证电力部门的收益比上年至少增长··································12分 20.（12分）解：（1）因为函数的图象为开口向上的抛物线，其对称轴为直线·················1分由二次函数图象可知，的单调增区间为·····································2分因为在上单调递增，所以···············································4分所以，所以实数的取值区间是············································5分（2）由得：·························································7分或·································································8分①当时，，不等式的解集是··············································9分②当时，，不等式的解集是··············································10分③当时，，不等式的解集是··············································11分综上，①当时，不等式的解集是②当时，不等式的解集是③当时，不等式的解集是··········································12分 21.（12分）解：（1）由题意得，函数的定义域为······································1分因为函数是奇函数，所以，··············································2分所以，解得 检验可知，当时，函数为奇函数，满足题意·································3分由得，所以， 即 ···························································5分所以，解得，所以该不等式的解集为·······································6分（2）由题知：·······················································9分所以函数的图象是由的图象向上平移一个单位得到的··························10分因为为奇函数，所以其图象的对称中心为··································11分所以图象的对称中心是················································12分22.（12分）解：（1）在区间上单调递减·············································2分在区间上单调递增··············································4分（2）由题意知，······················································5分①若，则在上单调递减，所以的最大值为····································6分②若，则在上单调递减，在上单调递增因为此时，所以的最大值为··············································8分③若，则在上单调递减，在上单调递增因为此时，所以的最大值为·············································10分综上知：若，则的最大值为；若，则的最大值为（3）由（1）（2）知：①当时，在上的值域为，在上的值域为，因为，所以满足，，使得所以此时是的“区间”··················································11分②当时，在上的值域为，在上的值域为，因为当时，，所以，使得，即，，所以此时不是的“区间”所以实数的最大值为··················································12分