最新苏州市届高三上学期期中考试数学试题优秀名师资料.doc

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20172018学年第一学期高三期中调研试卷数学(附加题部分)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）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20172018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 41 564 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：12一年级数学下册教材共六个单元和一个总复习，分别从数与代数、空间图形、实践活动等方面对学生进行教育。31、会数、会读、会写100以内的数;在具体情境中把握数的相对大小关系;能够运用数进行表达和交流，体会数与日常生活的密切联系。·············································8分4弓形：弦及所对的弧组成的图形叫做弓形。P2. 俯角：当从高处观测低处的目标时，视线与水平线所成的锐角称为俯角8.解直角三角形：在直角三角形中，除直角外，一共有五个元素，即三条边和二个锐角。由直角三角形中除直角外的已知元素，求出所有未知元素的过程，叫做解直角三角形（须知一条边）。所以····································································10分23(本题满分10分)九年级数学下册知识点归纳2. 图像性质：解：（1）原问题等价于对任意恒成立，第一章 直角三角形边的关系令，则，当时，恒成立，即在上单调递增，经过不在同一直线上的三点,能且仅能作一个圆.(1)三角形的外接圆: 经过一个三角形三个顶点的圆叫做这个三角形的外接圆.（2）抛物线的描述：开口方向、对称性、y随x的变化情况、抛物线的最高（或最低）点、抛物线与x轴的交点。恒成立；(一)教学重点当时，令，则，第三章 圆在上单调递减，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.正弦：（2）法一：在（1）中取，得，令，上式即为，33.123.18加与减（一）3 P13-17(二)教学难点（4）二次函数的图象：是以直线为对称轴，顶点坐标为（，）的抛物线。（开口方向和大小由a来决定）即，·····························································································7分3、学习并掌握100以内加减法(包括不进位、不退位与进位、退位)计算方法，并能正确计算;能根据具体问题，估计运算的结果;初步学会应用加减法解决生活中简单问题，感受加减法与日常生活的密切联系。（6）二次函数的图象：是以直线x=h为对称轴，顶点坐标为（h，k）的抛物线。（开口方向和大小由a来决定）94.234.29加与减（二）4 P49-56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：当时，成立；·············································································6分假设当时结论成立，即，在（1）中取，得，令，有，·······································································8分那么，当时，也成立；由可知，·····································································10分