2017-2018学年第一学期江苏省苏州市高三期中调研试卷数学（理）

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 QUOTE 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 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
·············································8分