2021仁寿县高二下学期期末模拟考试数学（文）试题含答案

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这是一份2021仁寿县高二下学期期末模拟考试数学（文）试题含答案，共10页。试卷主要包含了06，5亿元等内容，欢迎下载使用。
高2019级仁寿县第四学期期末模拟试卷文科数学 2021.06数学试题卷（文科）共4页．满分150分．考试时间120分钟．注意事项：1．答选择题时，必须使用2B铅笔将答题卡上对应题目的答案标号涂黑，如需改动，用橡皮擦擦干净后，再选涂其他答案标号．2．答非选择题时，必须使用0.5毫米黑色签字笔，将答案书写在答题卡规定的位置上．3．所有题目必须在答题卡上作答，在试题卷上答题无效．4．考试结束后，将答题卡交回．一、选择题：本大题共12小题，每小题5分，共60分．在每小题给出的四个备选项中，只有一项是符合题目要求的．1．已知i为虚数单位，复数，，若z为纯虚数，则（）A． B． C．2 D．2．某学校决定从该校的2000名高一学生中采用系统抽样（等距）的方法抽取50名学生进行体质分析，现将2000名学生从1至2000编号，已知样本中第一个编号为7，则抽取的第26个学生的编号为（）A．997 B．1007 C．1047 D．10873．2020年初，从非洲蔓延到东南亚的蝗虫灾害严重威胁了国际农业生产，影响了人民生活.世界性与区域性温度的异常､早涝频繁发生给蝗灾发生创造了机会.已知蝗虫的产卵量与温度的关系可以用模型拟合，设，其变换后得到一组数据：由上表可得线性回归方程，则（）A． B． C． D．4．甲、乙两名同学在高考前的5次模拟考中的数学成绩如茎叶图所示，记甲、乙两人的平均成绩分别为，下列说法正确的是（）A．，且乙比甲的成绩稳定 B．，且乙比甲的成绩稳定C．，且甲比乙的成绩稳定 D．，且甲比乙的成绩稳定5．2021年电影春节档票房再创新高，其中电影《唐人街探案3》和《你好，李焕英》是今年春节档电影中最火爆的两部电影，这两部电影都是2月12日（大年初一）首映，根据猫眼票房数据得到如下统计图，该图统计了从2月12日到2月18日共计7天的累计票房（单位：亿元），则下列说法中错误的是（）A．这7天电影《你好，李焕英》每天的票房都超过2.5亿元B．这7天两部电影的累计票房的差的绝对值先逐步扩大后逐步缩小C．这7天电影《你好，李焕英》的当日票房占比逐渐增大D．这7天中有4天电影《唐人街探案3》的当日票房占比超过50%6．如图所示的程序框图，若输入x的值为2，输出v的值为16，则判断框内可以填入（）A．k≤3？ B．k≤4？ C．k≥3？ D．k≥4？7．函数的零点个数为（）A． B．或 C．或 D．或或8．苏格兰数学家科林麦克劳林（ColinMaclaurin）研究出了著名的Maclaurin级数展开式，受到了世界上顶尖数学家的广泛认可，下面是麦克劳林建立的其中一个公式：，试根据此公式估计下面代数式的近似值为（）（可能用到数值ln2.414＝0.881，ln3.414＝1.23）A．3.23 B．2.881 C．1.881 D．1.239．函数（其中为自然对数的底数）的图象大致是（）A． B． C． D．10．甲乙两人相约10天内在某地会面，约定先到的人等候另一个人，经过三天后方可离开.若他们在期限内到达目的地是等可能的，则此二人会晤的概率是（）A．0.5 B．0.51 C．0.75 D．0.411．设函数是奇函数的导函数，.当时，，则使得成立的的取值范围是（）A． B． C． D．12．已知函数，，设为实数，若存在实数，使，则实数的取值范围为（）A． B． C． D．二、填空题:本大题共4小题，每小题5分，共20分. 请将答案填在答题卷中的相应位置.13．的边，为边上一动点，则的概率为__▲___14．复数（是虚数单位）是方程的一个根，则实数___▲___15．已知函数的定义域为，且的图像如右图所示，记的导函数为，则不等式的解集是____▲___16．若函数图象在点处的切线方程为，则的最小值为___▲___三、解答题：本大题共6小题，共70分，解答应写出文字说明、证明过程或推演步骤．17．（本小题10分）（1）用反证法证明：在一个三角形中，至少有一个内角大于或等于．（2）用分析法证明：当时，．18．（本小题12分）某企业有A，B两个分厂生产某种产品，规定该产品的某项质量指标值不低于120的为优质品．分别从A，B两厂中各随机抽取100件产品统计其质量指标值，得到如下频率分布直方图：（1）根据频率分布直方图，分别求出B分厂的质量指标值的中位数和平均数的估计值；（2）填写下面列联表，并根据列联表判断是否有99%的把握认为这两个分厂的产品质量有差异？ 19．（本小题12分）已知函数．（1）求曲线在处的切线方程；（2）求曲线过点的切线方程． 20．（本小题12分）某工厂为了对新研发的产品进行合理定价，将该产品按事先拟定的价格进行试销，得到一组检测数据如表所示：已知变量具有线性负相关关系，且，现有甲､乙､丙三位同学通过计算求得其回归直线方程分别为：甲；乙；丙，其中有且仅有一位同学的计算结果是正确的.（1）试判断谁的计算结果正确？并求出的值；（2）若由线性回归方程得到的估计数据与检测数据的误差不超过1，则该检测数据是“理想数据”．现从检测数据中随机抽取2个，求“理想数据”至少有一个的概率． 21．（本小题12分）已知函数．（1）求函数的最值；（2）求证：． 22．（本小题12分）已知函数为自然对数的底数）．（1）若是的极值点，求的取值；（2）若只有一个零点，求的取值范围．高2019级仁寿县第四学期期末模拟试卷文科数学参考答案一、选择题1.C 2.B 3.B 4.A 5.D 6.A 7.A 8.B 9.C 10.B 11.C 12.B二、填空题 13． 14． 15． 16．三、解答题17．解：（1）证明：假设三角形三个角都小于·············································1分那么····························································································2分而三角形内角和与假设矛盾·····························································3分所以三角形至少有一个内角大于或等于·····························································4分（2）证明：要证（），只需证·····················································5分则需证，即··········································································7分令，则，当时，；当时，，所以，即，则原不等式得证··················································10分 18．解：（1）B分厂的质量指标值；由，则的中位数为······································2分的平均数为·········6分（2）列联表：···············································································7分由列联表可知的观测值为：··································11分所以有99%的把握认为两个分厂的产品质量有差异．················································12分 19．解：(1)由已知得,则，所以切线斜率，···························2分因为，所以切点坐标为，·····················································4分所以所求直线方程为，故曲线在处的切线方程为.·······················································5分(2)由已知得，设切点为,···················································6分则，即，得或，所以切点为或，切线的斜率为或，·················································10分所以切线方程为或即切线方程为或·······························································12分 20．解：（1）∵变量具有线性负相关关系，∴甲是错误的.······································1分又∵∴，满足方程，故乙是正确的.··········3分由得··································································5分（2）由计算可得“理想数据”有个，即，则用字母代表“理想数据”，其余检测数据用代表，·················································································7分所以从检测数中随机抽取个有，，，，，，，，，，，，，，共种，······························································································10分满足题意的有，，，，，，，，，，，共种，·········11分所以“理想数据”至少有一个的概率为····························································12分 21．解：（1）由题可知··································································1分所以，当时，，单调递减；当时，，单调递增，所以··································································4分（2）方法一：，令，，故在上单调递增.············································5分又，又在上连续，使得，即，.（*）·····························7分随的变化情况如下：↘极小值↗. 由（*）式得，代入上式得. 令，，故在上单调递减.，又，.即.·················································································12分方法二：，································································8分易得，，所以，得证······················································12分 22．解：（1），······································································1分当时，，；，，此时恒成立，则不是函数的极值点·······················································································3分所以···············································································4分（2）只有一个零点，显然是，所以分为两种情况第1种情况：满足，此时·····················································6分第2种情况：无解，令，；①当时，，单调递增，故在上存在使得；·························································································8分②当时，方程显然无解；·······························································9分③当时，解得，当时，，单调递减；当时，，单调递增，则，即，所以·································································································11分综上所述：················································································12分