数学卷·2018届江苏省苏州市高三期中调研（2017

数学卷·2018届江苏省苏州市高三期中调研（2017

‎2017—2018学年第一学期高三期中调研试卷 ‎ 数 学 2017.11‎ 注意事项：‎ ‎1．本试卷共4页．满分160分，考试时间120分钟．‎ ‎2．请将填空题的答案和解答题的解题过程写在答题卷上，在本试卷上答题无效．‎ ‎3．答题前，务必将自己的姓名、学校、准考证号写在答题纸的密封线内．‎ 一、填空题(本大题共14小题，每小题5分，共70分，请把答案直接填写在答卷纸相应的位置)‎ ‎1．已知集合，则 ▲ ．‎ ‎2．函数的定义域为 ▲ ．‎ ‎3．设命题；命题，那么p是q的 ▲ 条件（选填“充分不必要”、“必要不充分”、“充要”、“既不充分也不必要”）．‎ ‎4．已知幂函数在是增函数，则实数m的值是 ▲ ．‎ ‎5．已知曲线在处的切线的斜率为2，则实数a的值是 ▲ ．‎ ‎6．已知等比数列中，，，则 ▲ ．‎ ‎7．函数图象的一条对称轴是，则的值是 ▲ ．‎ ‎8．已知奇函数在上单调递减，且，则不等式的解集为 ▲ ．‎ ‎9．已知，则的值是 ▲ ．‎ ‎10．若函数的值域为，则实数a的取值范围是 ▲ ．‎ ‎11．已知数列满足，则 ▲ ．‎ ‎12．设的内角的对边分别是，D为的中点，若且，则面积的最大值是 ▲ ．‎ ‎13．已知函数，若对任意的实数，都存在唯一的实数，使，则实数的最小值是 ▲ ．‎ ‎14．已知函数，若直线与交于三个不同的点 （其中），则的取值范围是 ▲ ．‎ 二、解答题(本大题共6个小题，共90分，请在答题卷区域内作答，解答时应写出文字说明、证明过程或演算步骤)‎ ‎15．(本题满分14分)‎ 已知函数的图象与x轴相切，且图象上相邻两个最高点之间的距离为．‎ ‎（1）求的值；‎ ‎（2）求在上的最大值和最小值．‎ ‎16．(本题满分14分)‎ 在中，角A,B,C所对的边分别是a,b,c，已知，且．‎ ‎（1）当时，求的值；‎ ‎（2）若角A为锐角，求m的取值范围．‎ ‎17．(本题满分15分) ‎ 已知数列的前n项和是，且满足，．‎ ‎（1）求数列的通项公式；‎ ‎（2）在数列中，，，若不等式对有解，求实数的取值范围．‎ ‎18．(本题满分15分)‎ 如图所示的自动通风设施．该设施的下部ABCD是等腰梯形，其中为2米，梯形的高为1米，为3米，上部是个半圆，固定点E为CD的中点．MN是由电脑控制可以上下滑动的伸缩横杆（横杆面积可忽略不计），且滑动过程中始终保持和CD平行．当MN位于CD下方和上方时，通风窗的形状均为矩形MNGH（阴影部分均不通风）．‎ ‎（1）设MN与AB之间的距离为且米，试将通风窗的通风面积S（平方米）表示成关于x的函数；‎ ‎（2）当MN与AB之间的距离为多少米时，通风窗的通风面积取得最大值？ ‎ ‎19．(本题满分16分)‎ 已知函数．‎ ‎（1）求过点的的切线方程；‎ ‎（2）当时，求函数在的最大值；‎ ‎（3）证明：当时，不等式对任意均成立（其中为自然对数的底数,）．‎ ‎20．(本题满分16分)‎ 已知数列各项均为正数，,，且对任意恒成立，记的前n项和为．‎ ‎（1）若，求的值；‎ ‎（2）证明：对任意正实数p，成等比数列；‎ ‎（3）是否存在正实数t，使得数列为等比数列．若存在，求出此时和的表达式；若不存在，说明理由．‎ ‎2017—2018学年第一学期高三期中调研试卷 数 学 (附加) 2017.11‎ 注意事项：‎ ‎1．本试卷共2页．满分40分，考试时间30分钟．‎ ‎2．请在答题卡上的指定位置作答，在本试卷上作答无效．‎ ‎3．答题前，请务必将自己的姓名、学校、考试证号填写在答题卡的规定位置．‎ ‎21．【选做题】本题包括A、B、C、D四小题，请选定其中两题，并在相应的答题区域内作答．若多做，则按作答的前两题评分．解答时应写出文字说明、证明过程或演算步骤．‎ A．(几何证明选讲)‎ ‎ （本小题满分10分）‎ 如图，AB为圆O的直径，C在圆O上，于F，点D为线段CF上任意一点，延长AD交圆O于E，．‎ ‎（1）求证：；‎ ‎（2）若，求的值．‎ B．(矩阵与变换)‎ ‎（本小题满分10分）‎ 已知矩阵，，求的值．‎ ‎ C．(极坐标与参数方程)‎ ‎ （本小题满分10分）‎ 在平面直角坐标系中，直线的参数方程为(为参数)，以原点为极点，轴正半轴为极轴建立极坐标系，圆的极坐标方程为．‎ ‎（1）求直线和圆的直角坐标方程；‎ ‎（2）若圆C任意一条直径的两个端点到直线l的距离之和为，求a的值．‎ D．(不等式选讲)‎ ‎（本小题满分10分）‎ 设均为正数，且，求证：．‎ ‎【必做题】第22、23题，每小题10分，共计20分．请在答题卡指定区域内作答，解答时应写出文字说明、证明过程或演算步骤．‎ ‎22．(本小题满分10分)‎ 在小明的婚礼上，为了活跃气氛，主持人邀请10位客人做一个游戏．第一轮游戏中，主持人将标有数字1,2,…,10的十张相同的卡片放入一个不透明箱子中，让客人依次去摸，摸到数字6,7,…,10的客人留下，其余的淘汰，第二轮放入1,2,…,5五张卡片，让留下的客人依次去摸，摸到数字3,4,5的客人留下，第三轮放入1,2,3三张卡片，让留下的客人依次去摸，摸到数字2,3的客人留下，同样第四轮淘汰一位，最后留下的客人获得小明准备的礼物．已知客人甲参加了该游戏．‎ ‎（1）求甲拿到礼物的概率；‎ ‎（2）设表示甲参加游戏的轮数，求的概率分布和数学期望．‎ ‎23．(本小题满分10分)‎ ‎（1）若不等式对任意恒成立，求实数a的取值范围；‎ ‎（2）设，试比较与的大小，并证明你的结论．‎ ‎2017—2018学年第一学期高三期中调研试卷 数 学 参 考 答 案 一、填空题(本大题共14小题，每小题5分，共70分)‎ ‎1． 2． 3．充分不必要 4．1 5． ‎6．4 7． 8． 9． 10． ‎ ‎11． 12． 13． 14． 二、解答题(本大题共6个小题，共90分)‎ ‎15．(本题满分14分)‎ 解：（1）∵图象上相邻两个最高点之间的距离为，‎ ‎∴的周期为，∴，······································································2分 ‎∴，··················································································································4分 此时，‎ 又∵的图象与x轴相切，∴，·······················································6分 ‎∴；··········································································································8分 ‎（2）由（1）可得，‎ ‎∵，∴，‎ ‎∴当，即时，有最大值为；·················································11分 当，即时，有最小值为0．························································14分 ‎16．(本题满分14分)‎ 解：由题意得，．···············································································2分 ‎（1）当时，，‎ 解得或；································································································6分 ‎（2），····························8分 ‎∵A为锐角，∴，∴，····················································11分 又由可得，·························································································13分 ‎∴．·····································································································14分 ‎17．(本题满分15分)‎ 解：（1）∵，∴，‎ ‎∴，························································‎ ‎·································2分 又当时，由得符合，∴，······························3分 ‎∴数列是以1为首项，3为公比的等比数列，通项公式为；·····················5分 ‎（2）∵，∴是以3为首项，3为公差的等差数列，····················7分 ‎∴，·····················································································9分 ‎∴，即，即对有解，··································10分 设，‎ ‎∵，‎ ‎∴当时，，当时，，‎ ‎∴，‎ ‎∴，···························································································14分 ‎∴．·············································································································15分 ‎18．(本题满分15分)‎ 解：（1）当时，过作于（如上图），‎ 则，，，‎ 由，得，‎ ‎∴，‎ ‎∴；·······························································4分 当时，过作于，连结（如下图），‎ 则，，‎ ‎∴，‎ ‎∴，······································································8分 综上：；·································································9分 ‎（2）当时，在上递减，‎ ‎∴；································································································11分 当时，， ‎ 当且仅当，即时取“”，‎ ‎∴，此时，∴的最大值为，············································14分 答：当MN与AB之间的距离为米时，通风窗的通风面积 取得最大值．····················15分 ‎19．(本题满分16分)‎ 解：（1）设切点坐标为，则切线方程为，‎ 将代入上式，得，，‎ ‎∴切线方程为；·······························································································2分 ‎（2）当时，，‎ ‎∴，············································································3分 当时，，当时，，‎ ‎∴在递增，在递减，·············································································5分 ‎∴当时，的最大值为；‎ 当时，的最大值为；········································································7分 ‎（3）可化为，‎ 设，要证时对任意均成立，‎ 只要证，下证此结论成立．‎ ‎∵，∴当时，，·······················································8分 设，则，∴在递增，‎ 又∵在区间上的图象是一条不间断的曲线，且，，‎ ‎∴使得，即，，····················································11分 当时，，；当时，，；‎ ‎∴函数在递增，在递减，‎ ‎∴，····························14分 ‎∵在递增，∴，即，‎ ‎∴当时，不等式对任意均成立．··························16分 ‎20．(本题满分16分)‎ 解：（1）∵，∴，又∵，∴；·······································2分 ‎（2）由，两式相乘得，‎ ‎∵，∴，‎ 从而的奇数项和偶数项均构成等比数列，···································································4分 设公比分别为，则，，······································5分 又∵，∴，即，···························································6分 设，则，且恒成立，‎ 数列是首项为，公比为的等比数列，问题得证；····································8分 ‎（3）法一：在（2）中令，则数列是首项为，公比为的等比数列，‎ ‎∴，‎ ，·····································································10分 且，‎ ‎∵数列为等比数列，∴ 即，即 解得（舍去），·························································································13分 ‎∴，，‎ 从而对任意有，‎ 此时，为常数，满足成等比数列，‎ 当时，，又，∴，‎ 综上，存在使数列为等比数列，此时．······················16分 法二：由（2）知，则，，且，‎ ‎∵数列为等比数列，∴ 即，即 解得（舍去），·······················································································11分 ‎∴，，从而对任意有，····································13分 ‎∴，‎ 此时，为常数，满足成等比数列，‎ 综上，存在使数列为等比数列，此时．······················16分 ‎21．【选做题】本题包括A、B、C、D四小题，请选定其中两题，并在相应的答题区域内作答．若多做，则按作答的前两题评分．解答时应写出文字说明、证明过程或演算步骤．‎ A．(几何证明选讲，本小题满分10分)‎ 解：（1）证明 ：连接，∵，∴，‎ 又，∴为等边三角形，‎ ‎∵，∴为中边上的中线，‎ ‎∴；······································································5分 ‎（2）解：连接BE，‎ ‎∵，是等边三角形，‎ ‎∴可求得，，‎ ‎∵为圆O的直径，∴，∴，‎ 又∵，∴∽，∴，‎ 即 ．··················································································10分 B．(矩阵与变换，本小题满分10分)‎ 解：矩阵A的特征多项式为，‎ 令，解得矩阵A的特征值，····························································2分 当时特征向量为，当时特征向量为，·····································6分 又∵，······························································································8分 ‎∴．···········································································10分 C．(极坐标与参数方程，本小题满分10分)‎ 解：（1）直线的普通方程为；··········································································3分 圆C的直角坐标方程为；·······························································6分 ‎（2）∵圆C任意一条直径的两个端点到直线l的距离之和为，‎ ‎∴圆心C到直线l的距离为，即，·······················································8分 解得或 ．·······························································································10分 D．(不等式选讲，本小题满分10分)‎ 证：∵，‎ ‎∴ ，‎ ‎∴．····················································································10分 ‎22．(本题满分10分)‎ 解：（1）甲拿到礼物的事件为，‎ 在每一轮游戏中，甲留下的概率和他摸卡片的顺序无关，‎ 则，‎ 答：甲拿到礼物的概率为；·······················································································3分 ‎（2）随机变量的所有可能取值是1,2,3,4．·····································································4分 ，‎ ，‎ ，‎ ，‎ 随机变量的概率分布列为：‎ ‎1‎ ‎2‎ ‎3‎ ‎·············································8分 ‎4‎ P 所以．····································································10分 ‎23．(本题满分10分)‎ 解：（1）原问题等价于对任意恒成立，‎ 令，则，‎ 当时，恒成立，即在上单调递增，‎ ‎∴恒成立；‎ 当时，令，则，‎ ‎∴在上单调递减，在上单调递增，‎ ‎∴，即存在使得，不合题意；‎ 综上所述，a的取值范围是．················································································4分 ‎（2）法一：在（1）中取，得，‎ 令，上式即为，‎ 即，·····························································································7分 ‎∴ 上述各式相加可得．····················································10分 法二：注意到，，……，‎ 故猜想，····································································5分 下面用数学归纳法证明该猜想成立．‎ 证明：①当时，，成立；·············································································6分 ‎②假设当时结论成立，即，‎ 在（1）中取，得，‎ 令，有，·······································································8分 那么，当时，‎ ，也成立；‎ 由①②可知，．·····································································10分