2019届高三数学期中调研试题.doc

- 1 -20192019 学年第一学期高三期中调研试卷学年第一学期高三期中调研试卷数数 学学 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1,2,3,4,5,1,3,2,3UAB，则()UAB ðI 2函数1 ln(1)yx的定义域为 3设命题:4p x ；命题2:540q xx ，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22*()m myxmN在(0,)是增函数，则实数m的值是 5已知曲线3( )lnf xaxx在(1,(1)f处的切线的斜率为 2，则实数a的值是 6已知等比数列na中，32a ，4616a a ，则7935aa aa 7函数sin(2)(0)2yx图象的一条对称轴是12x，则的值是 8已知奇函数( )f x在(,0)上单调递减，且(2)0f，则不等式( )01f x x的解集为 9已知tan()24，则cos2的值是 10若函数8,2( )log5,2axxf xxx (01)aa且的值域为6,)，则实数a的取值范围是 11已知数列, nnab满足1111,1,(*)21nnn naabbnaN，则122017b bbL 12设ABC的内角, ,A B C的对边分别是, ,a b c，D为AB的中点，若cossinbaCcA且- 2 -2CD ，则ABC面积的最大值是 13已知函数( )sin()6f xx，若对任意的实数5,62 ，都存在唯一的实数0,m，使( )( )0ff，则实数m的最小值是 14已知函数ln ,0( )21,0x xf xxx，若直线yax与( )yf x交于三个不同的点( ,( ), ( ,( ),A m f mB n f n( ,( )C t f t（其中mnt），则12nm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21( )sin(2)(0,0)242f xaxb ab 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2（1）求, a b的值；（2）求( )f x在0,4上的最大值和最小值16(本题满分 14 分)在ABC中，角A,B,C所对的边分别是a,b,c，已知sinsinsin()BCmA mR，且240abc（1）当52,4am时，求, b c的值；（2）若角A为锐角，求m的取值范围- 3 -17(本题满分 15 分) 已知数列na的前n项和是nS，且满足11a ，* 131()nnSSnN（1）求数列na的通项公式；（2）在数列 nb中，13b ，*1 1()n nn nabbna N，若不等式2 nnabn对*nN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AB为 2 米，梯形的高为 1 米，CD为 3 米，上部ACmD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5(02xx 且1)x 米，试将通风窗的通风面积S（平方米）表示成关于x的函数( )yS x；（2）当MN与AB之间的距离为多少米时，通风窗的通风面积S取得最大值？ - 4 -19(本题满分 16 分)已知函数2( )ln , ( )f xx g xxxm（1）求过点(0, 1)P的( )f x的切线方程；（2）当0m时，求函数( )( )( )F xf xg x在, 0(a的最大值；（3）证明：当3m-时，不等式2( )( )(2)exf xg xxx对任意1 ,12x均成立（其中e为自然对数的底数,e2.718.）20(本题满分 16 分)已知数列na各项均为正数，11a ,22a ，且312nnnna aaa对任意*nN恒成立，记na的前n项和为nS（1）若33a ，求5a的值；- 5 -（2）证明：对任意正实数p，221nnapa成等比数列；（3）是否存在正实数t，使得数列nSt为等比数列若存在，求出此时na和nS的表达式；若不存在，说明理由- 6 -20172018 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，CFAB于F，点D为线段CF上任意一点，延长AD交圆O于E，030AEC（1）求证：AFFO；（2）若3CF ，求AD AE的值B(矩阵与变换)（本小题满分 10 分）已知矩阵12 21A，4 2 ，求49 A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l的参数方程为425 2 5xtyt (t为参数)，以原点O为极点，x轴正半轴为极轴建立极坐标系，圆C的极坐标方程为2 cos()(0)4aa（1）求直线l和圆C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5，求a的值DEFAOBC- 7 -D(不等式选讲)（本小题满分 10 分）设, x y均为正数，且xy，求证：2212232xyxxyy【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( )E23(本小题满分 10 分)（1）若不等式(1)ln(1)xxax对任意0,)x恒成立，求实数a的取值范围；（2）设*nN，试比较111 231nL与ln(1)n的大小，并证明你的结论- 8 -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)11 2(1,2)(2,)U 3充分不必要 41 51 364 738( 2,0)(1,2)U 94 5 10(1,2 111 20181221 132141(1,e)e二、解答题(本大题共 6 个小题，共 90 分)15(本题满分 14 分)解：（1）( )f x图象上相邻两个最高点之间的距离为2，( )f x的周期为2，202|2aa且，······································································2分2a ，··················································································································4 分此时21( )sin(4)242f xxb ，又( )f x的图象与x轴相切，12|022bb且，·······················································6 分21 22b ；··········································································································8 分（2）由（1）可得22( )sin(4)242f xx ，- 9 -0,4x，4,444x ，当444x，即4x时，( )f x有最大值为21 2；·················································11 分当442x，即16x时，( )f x有最小值为0························································14 分16(本题满分 14 分)解：由题意得bcma，240abc···············································································2 分（1）当52,4am时，5,12bcbc，解得21 2bc或1 2 2bc ；································································································6 分（2）2 2222222 2 2()()22cos2322 2amaabcabcbcaAmabcbc，····························8 分A为锐角，2cos23(0,1)Am ，2322m，····················································11 分又由bcma可得0m ，·························································································13 分- 10 -622m·····································································································14 分17(本题满分 15 分)解：（1）* 131()nnSSnN，* 131(,2)nnSSnnN，* 13(,2)nnaa nnN，·························································································2 分又当1n 时，由2131SS得23a 符合213aa，* 13()nnaa nN，······························3分数列na是以 1 为首项，3 为公比的等比数列，通项公式为1*3()n nanN；·····················5 分（2）*1 13()n nn nabbna N， nb是以 3 为首项，3 为公差的等差数列，····················7 分*33(1)3 ()nbnn nN，·····················································································9 分2 nnabn，即1233nnn，即213 3nnn对*nN有解，··································10 分设2 * 13( )()3nnnf nnN，2221(1)3(1)32(41)(1)( )333nnnnnnnnnf nf n，当4n时，(1)( )f nf n，当4n 时，(1)( )f nf n，(1)(2)(3)(4)(5)(6)ffffff，max4 ( )(4)27f nf，···································································- 11 -························14 分4 27·············································································································15 分18(本题满分 15 分)解：（1）当01x 时，过A作AKCD于K（如上图），则1AK ，1 22CDABDK，1HMx ，由2AKMH DKDH，得1 22HMxDH，322HGDHx，2( )(1)(2)2S xHM HGxxxx ；·······························································4 分当512x时，过E作ETMN于T，连结EN（如下图），则1ETx，2 2239(1)(1)224MNTNxx，292(1)4MNx，29( )2(1)(1)4S xMN ETxx，······································································8 分综上：222 , 01 ( )952(1)(1) ,142xxx S x xxx ；·································································9 分（2）当01x 时，2219( )2()24S xxxx 在0,1)上递减，max( )(0)2S xS；····································································- 12 -····························11 分2当512x时，2229(1)(1)994( )2(1)(1)2424xx S xxx ， 当且仅当29(1)(1)4xx，即3 251(1, )42x 时取“”，max9( )4S x，此时max9( )24S x，( )S x的最大值为9 4，············································14 分答：当MN与AB之间的距离为3 214米时，通风窗的通风面积S取得最大值····················15 分19(本题满分 16 分)解：（1）设切点坐标为00(,ln)xx，则切线方程为00 01ln()yxxxx，将(0, 1)P代入上式，得0ln0x ，01x ，切线方程为1yx；·······························································································2 分（2）当0m 时，2( )ln,(0,)F xxxx x，(21)(1)( ),(0,)xxF xxx ，············································································3 分当01x时，( )0F x，当1x 时，( )0F x，( )F x在(0,1)递增，在(1,)递减，·············································································5 分当01a时，( )F x的最大值为2( )lnF aaaa；当1a 时，( )F x的最大值为(1)0F；- 13 -········································································7 分（3）2( )( )(2)exf xg xxx可化为(2)elnxmxxx，设1( )(2)eln, ,12xh xxxx x，要证3m-时( )mh x对任意1 ,12x均成立，只要证max( )3h x ，下证此结论成立1( )(1)(e)xh xxx，当112x时，10x ，·······················································8 分设1( )exu xx，则21( )e0xu xx，( )u x在1( ,1)2递增，又( )u x在区间1 ,12上的图象是一条不间断的曲线，且1( )e202u，(1)e10u ，01( ,1)2x使得0()0u x，即001exx，00ln xx ，····················································11 分当01( ,)2xx时，( )0u x ，( )0h x；当0(,1)xx时，( )0u x ，( )0h x；函数( )h x在01 ,2x递增，在0,1x递减，0 max0000000 0012( )()(2)eln(2)212xh xh xxxxxxxxx ，····························14 分212yxx 在1( ,1)2x递增，00 02()121223h xxx ，即max( )3h x ，当3m-时，不等式2( )( )(2)exf xg xxx对任意1 ,12x均成立··························16 分20(本题满分 16 分)解：（1）1423a aa a，46a ，又- 14 -2534a aa a，54392aa；·······································2 分（2）由3121423nnnnnnnna aaaaaaa ，两式相乘得2 134123nnnnnnna aaaaaa，0na ，2* 42()nnna aanN，从而na的奇数项和偶数项均构成等比数列，···································································4 分设公比分别为12,q q，则11 22222nn naa qq，11 211 11nn naa qq ，······································5 分又312=nnnnaa aa，42231122aaq aaq，即12qq，···························································6 分设12qqq，则2212223()nnnnapaq apa，且2210nnapa恒成立，数列221nnapa是首项为2p，公比为q的等比数列，问题得证；····································8 分（3）法一：在（2）中令1p ，则数列221nnaa是首项为3，公比为q的等比数列，22212223213 ,1 ()()()3(1),11k kkkkkkq Saaaaaaqqq ，12122132 ,13(1)2,11kkkkkkkqq SSaqqqq ，·····································································10 分且12341,3,3,33SSSq Sq，- 15 -数列nSt为等比数列，2 2132 324()()(),()()(),StSt StStSt St即22(3)(1)(3),(3)(3)(33),ttqtqttqt，即26(1),3,tqttq 解得1 4t q（3t 舍去），·························································································13 分2 24121kk kS ，21 2121k kS ，从而对任意*nN有21n nS ，此时2nnSt ，12nnSt St为常数，满足nSt成等比数列，当2n时，11 1222nnn nnnaSS ，又11a ，1*2()n nanN，综上，存在1t 使数列nSt为等比数列，此时1*2,21()nn nnaSnN······················16 分法二：由（2）知，则1 22n naq，1 21n naq ，且12341,3,3,33SSSq Sq，数列nSt为等比数列，2 2132 324()()(),()()(),StSt StStSt St即22(3)(1)(3),(3)(3)(33),ttqtqttqt，即26(1),3,tqttq 解得1 4t q（3t 舍去），·······················································································11 分121 222nn naq，22 212n na ，从而对任意*nN有12n na，····································13 分- 16 -01211222222112n nn nS，此时2nnSt ，12nnSt St为常数，满足nSt成等比数列，综上，存在1t 使数列nSt为等比数列，此时1*2,21()nn nnaSnN······················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,OC AC，030AEC，0260AOCAEC ，又OAOC，AOC为等边三角形，CFAB，CF为AOC中AO边上的中线，AFFO；······································································5 分（2）解：连接BE，3CF ，AOC是等边三角形，可求得1AF ，4AB ，AB为圆O的直径，90AEBo，AEBAFD ，又BAEDFA ，AEBAFD，ADAF ABAE，即4 14AD AEAB AF ··················································································10 分B(矩阵与变换，本小题满分 10 分)解：矩阵A A的特征多项式为212( )2321f，令( )0f，解得矩阵A A的特征值121,3 ，····························································2 分DFEAOBC- 17 -当11 时特征向量为11 1 ，当23时特征向量为21 1 ，·····································6 分又12432 ，······························································································8 分50 494949 11225031331 A···········································································10 分C(极坐标与参数方程，本小题满分 10 分)解：（1）直线l的普通方程为220xy；··········································································3 分圆C的直角坐标方程为2 22()()222aaaxy；·······························································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5，圆心C到直线l的距离为5 2，即|2|52 25aa ，········································ ···············8 分解得3a 或1 3a ········································································ ·······················10 分D(不等式选讲，本小题满分 10 分)证：0,0,0xyxy，22211222()2()xyxyxxyyxy- 18 -232211()()3 ()3()()xyxyxyxyxy，2212232xyxxyy····················································································10 分22(本题满分 10 分)解：（1）甲拿到礼物的事件为A，在每一轮游戏中，甲留下的概率和他摸卡片的顺序无关，则13211( )253210P A ，答：甲拿到礼物的概率为1 10；·······················································································3 分（2）随机变量的所有可能取值是1,2,3,4·····································································4 分112P ， 1212255P ， 1311325310P ， 132142535P ，随机变量的概率分布列为：1234P1 21 51 101 5所以1111( )1234225105E ····································································10 分23(本题满分 10 分)·············································8 分- 19 -解：（1）原问题等价于ln(1)01axxx对任意0,)x恒成立，令( )ln(1)1axg xxx，则21'( )(1)xag xx ，当1a时，21'( )0(1)xag xx 恒成立，即( )g x在0,)上单调递增，( )(0)0g xg恒成立；当1a 时，令'( )0g x ，则10xa ，( )g x在(0,1)a 上单调递减，在(1,)a 上单调递增，(1)(0)0g ag，即存在0x 使得( )0g x ，不合题意；综上所述，a的取值范围是(,1··········································································· ·····4 分（2）法一：在（1）中取1a ，得ln(1)(0,)1xxxx，令*1()xnnN，上式即为11ln()1n nn，即1ln(1)ln1nnn，·····························································································7 分1ln2ln1,2 1ln3ln2,31ln(1)ln,1nnn 上述各式相加可得111ln(1)231nnL*()nN····················································10 分法二：注意到1ln22，11ln323，故猜想111ln(1)231nnL*()nN，····································································5 分- 20 -下面用数学归纳法证明该猜想成立证明：当1n 时，1ln22，成立；·············································································6 分假设当nk时结论成立，即111ln(1)231kkL，在（1）中取1a ，得ln(1)(0,)1xxxx，令*1()1xkkN，有12ln()21k kk，·······································································8 分那么，当1nk时，11111ln(1)ln(1)23122ln()ln(2)21kkkkkkkkL，也成立；由可知， 111ln(1)231nnL··································································· ··10 分