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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.·································
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