苏州市2018届高三上学期期中考试数学试题

高三数学期中试卷第4页共14页苏州市2018届高三第一学期期中调研试卷数学一、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QUOTE中，角A,B,C所对的边分别是a,b,c，已知，且QUOTEQUOTE．（1）当时，求的值；（2）若角AQUOTE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？【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4．15．6．47．8．9．10．11．12．13．14．二、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QUOTE时，QUOTE，解得或；································································································6分（2），····························8分∵AQUOTE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：123················································································8分P所以．·································