绝密★启用前试卷类型：A2021届高三教学质量检测（一）试卷数学 2020.9注意：1．本试卷共4页，22小题，满分150分，考试用时120分钟．2．答卷前，考生务必将自己的姓名、准考证号填写在答题卡上．正确粘贴条形码． 3．作答选择题时，用2B 铅笔在答题卡上对应答案的选项涂黑．4．非选择题的答案必须写在答题卡各题目的指定区域内相应位置上；不准使用铅笔和涂改液．不按以上要求作答无效．5．考试结束后，考生上交答题卡．一、选择题：本题共8小题，每小题5分，共40分．在每小题给出的四个选项中，只有一项是符合题目要求的． 1．复数21i-的共轭复数是 A ．1i - B ．1i +C ．1i --D ．1i -+2．已知集合20{|}M x x x =-≤，{}sin |,N y y x x ==∈R ，则M N ⋂= A ．[]1,0-B ．(0,1)C ．[0,1]D ．∅3．已知抛物线2:2(0)C x py p =>的准线为l ，圆22:(1)(2)9M x y -+-=与l 相切．则p =A ．1B ．2C ．3D ．44．某学校组织学生参加数学测试，某班成绩的频率分布直方图如图，数据的分组依次为[20,40)，[40,60)，[60,80)，[80,100]．若不低于60分的人数是35人，则该班的学生人数是A ．45B ．50C ．55D ．655．中国古代数学名著《周髀算经》记载的“日月历法”曰：“阴阳之数，日月之法，十九岁为一章，四章为一部，部七十六岁，二十部为一遂，遂千百五二十岁，…．生数皆终，万物复苏，天以更元作纪历”．某老年公寓住有20位老人，他们的年龄（都为正整数）之和恰好为一遂，其中最年长者的年龄在(90,100)之间，其余19人的年龄依次相差一岁，则最年长者的年龄为 A ．94B ．95C ．96D ．986．已知,()0απ∈，()2sin 2cos21παα-=-，则sin α=A ．15B ．5C ．5-D ．57．已知直三棱柱111ABC A B C -的6个顶点都在球O 的球面上．若1AB =，AC =，AB AC ⊥，14AA =，则球O 的表面积为A ．5πB ．10πC ．20πD 8．对于定义在R 上的函数()f x ，且()f x π+为偶函数．当,()0x π∈时，3()cos f x x x =-，设()2a f =，()4b f =，()6c f =，则a ，b ，c 的大小关系为A ．a b c <<B ．b c a <<C ．b a c <<D ．c a b <<二、选择题：本题共4小题，每小题5分，共20分．在每小题给出的选项中，有多项符合题目要求．全部选对的得5分，有选错的得0分，部分选对的得3分． 9．设a ，b ，c 为正实数，且a b >，则 A ．11a b a b->- B ．11a b b a->- C ．()ln 0a b ->D ．()()2211c a b c +>+10．已知曲线12:sin C y x =，2:2sin 23C y x π⎛⎫=+⎪⎝⎭，则 A ．把1C 上各点的横坐标缩短为原来的12倍，纵坐标不变，再把得到的曲线向左平移6π个单位长度，得到曲线2CB ．把1C 上各点的横坐标缩短为原来的12倍，纵坐标不变，再把得到的曲线向右平移56π个单位长度，得到曲线2CC ．把1C 向左平移3π个单位长度，再把得到的曲线上各点的横坐标缩短为原来的12倍．纵坐标不变，得到曲线2CD ．把1C 向左平移6π个单位长度，再把得到的曲线上各点的横坐标缩短为原来的12倍，纵坐标不变，得到曲线2C11．对a ∀，b ∈R ，若函数()f x 同时满足：（1）当0a b +=时，有()()0f a f b +=；（2）当0a b +>时，有()()0f a f b +>，则称()f x 为Ω函数．下列是Ω函数的有 A ．()e e xxf x -=+B ．()e exxf x -=-C ．()sin f x x x =-D ．()0,0,1,0.x f x x x=⎧⎪=⎨-≠⎪⎩12．在长方体1111ABCD A B C D -中，M ，P 是平面11DCC D 内不同的两点，N ，Q 是平面ABCD 内不同的两点，且M ，P ，N ，Q CD ∉，E ，F 分别是线段MN ，PQ 的中点，则下列结论正确的是 A ．若MNPQ ，则EF CD B ．若E ，F 重合，则MPCDC ．若MN 与PQ 相交，且MPCD ，则NQ 可以与CD 相交D ．若MN 与PQ 是异面直线，则EF 不可能与CD 平行 三、填空题：本题共4小题，每小题5分，共20分．13．函数32()2f x x x =-的图象在点()()1,1f 处的切线方程为________．14．1021(2)x x x ⎛⎫-+ ⎪⎝⎭的展开式中8x 的系数为________．（用数字填写答案）15．已知向量()1,m a =，()21,3n b =-（0a >，0b >），若m n ⊥，则12a b+的最小值为________． 16．已知1F ，2F 是双曲线()2222:10,0x y C a b a b-=>>的左，右焦点，以12F F 为直径的圆与C 的左支交于点A ，2AF 与C 的右支交于点B ，123cos 5BF F ∠=-，则C 的离心率为________． 四、解答题：本题共6小题，共70分．解答应写出文字说明、证明过程或演算步骤． 17．（10分）在①sin B C =，②4sin b A =，③2B C A +=这三个条件中任选一个，补充在下面问题中，若问题中的三角形存在，求c 的值；若问题中的三角形不存在，说明理由．问题：是否存在ABC △，它的内角A ，B ，C 的对边分别为a ，b ，c ，且sin cos s n 4i B A A a b =+，2a =，________？注：如果选择多个条件分别解答，按第一个解答计分． 18．（12分）设{}n a 是公比大于1的等比数列，12314a a a ++=，且21a +是1a ，3a 的等差中项． （1）求数列{}n a 的通项公式；（2）若21log 2nn b a ⎛⎫= ⎪⎝⎭，数列{}n b 的前n 项和n T ．19．（12分）如图，在圆柱12O O 中，AB 为圆1O 的直径，C ，D 是弧AB 上的两个三等分点，CF 是圆柱12O O 的母线．（1）求证：1CO 平面AFD ；（2）设AC =45FBC ∠=︒，求二面角B AF C --的余弦值．20．（12分）某市广电运营商为了解该市广电网络从业人员的业务水平与服务水平．随机调查了140名客户，结果显示：业务水平的满意率为67，服务水平的满意率为57，两者都满意的有90名客户．（1）根据上述结果完成下面22⨯列联表．并判断是否有97.5%的把握认为该市广电网络从业人员的业务水平与服务水平有关；（2）若从对服务水平不满意的客户中，随机抽取2名征求改进意见，用X 表示对业务水平不满意的人数，求X 的分布列与期望；（3）若用频率代替概率，假定在业务服务协议终止时，对业务水平和服务水平两项都满意的客户流失率为5%，只对其中一项不满意的客户流失率为40%，对两项都不满意的客户流失率为75%，从该社区中任选4名客户，估计在业务服务协议终止时至少有2名客户流失的概率为多少？ 附：22()()()()()n ad bc K a b c d a c b d ⨯-=++++，n a b c d =+++21．（12分）已知椭圆C 的两个焦点分别是()1,0-，()1,0，并且经过点1,2⎛⎫ ⎪ ⎪⎝⎭． （1）求椭圆C 的标准方程；（2）已知点()0,2Q ，若C 上总存在两个点A 、B 关于直线y x m =+对称，且4QA QB ⋅<，求实数m 的取值范围． 22．（12分）已知函数()()ln 1f x x a x =-+，a ∈R ． （1）讨论()f x 的单调性；（2）设()()1g x f x x =++，若函数()g x 有两个不同的零点1x ，2x ，求a 的取值范围．2020学年第一学期高三调研考试数学试题参考答案与评分标准评分说明：1．参考答案与评分标准给出了一种或几种解法供参考，如果考生的解法与参考答案不同，可根据试题主要考查的知识点和能力对照评分标准给以相应的分数．2．对解答题中的计算题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度，可视影响的程度决定后继部分的得分，但所给分数不得超过该部分正确解答应得分数的一半．如果后继部分的解答有较严重的错误，就不再给分．3．解答右端所注分数，表示考生正确做到这一步应得的累加分数． 4．只给整数分数，选择题和填空题不给中间分．一、选择题：本题共12小题，每小题5分，共60分．1～8小题为单项选择题，在每小题给出的四个选项中，只有一项是符合题目要求的；9～12小题为多项选择题，在每小题给出的选项中，有多项符合题目要求，全部选对的得5分，有选错的得0分，部分选对的得3分．8．解，因为函数()f x π+为偶函数，所以()()f x f x ππ-+=+， 即函数()f x 的图象关于直线xπ=对称，即()2()f x f x π=-．又因为当,()0x π∈时，()3cos f x x x =-，所以函数()f x 在(0,)π上单调递减，因而在(),2ππ上单调递增，因为4226π<-<，所以()()(426)2f f f π<-<，即()()426()f f f <<，即b a c <<．故选C ． 12．解：若MNPQ ，则M 、N 、P 、Q 四点共面γ，当MN PQ <时，平面11DCC D 、ABCD 、γ两两相交有三条交线，分别为MP 、NQ 、CD ，则三条交线交于一点O ，则CD 与平面γ交于点O ，则EF 与CD 不平行．故A 错误： 若E ，F 两点重合，则MPNQ ，M 、N 、P ﹑Q 四点共面γ，平面11DCC D 、ABCD 、γ两两相交有三条交线，分别为MP 、NQ 、CD ，由MPNQ ，得MPNQCD ，故B 正确；若MN 与PQ 相交，确定平面γ，平面11DCC D 、ABCD 、γ两两相交有三条交线，分别为MP 、NQ 、CD ，因为MP CD ，所以MP NQ CD ，所以NQ 与CD 不可能相交．故C 错误；当MN 与PQ 是异面直线时，如图，连接NP ，取NP 中点G ，连接EG ，FG ．则EG MP ，因为MP ⊂平面11DCC D ，EG ⊄平面11DCC D ，则EG 平面11DCC D ．假设EFCD ，因为CD ⊂平面11DCC D ，EF ⊄平面11DCC D ，所以EF 平面11DCC D ．又EF EG E ⋂=，∴平面EFG平面11DCC D ，同理可得，平面EFG平面ABCD ，则平面11DCCD 平面ABCD ，与平面11DCC D ⋂平面ABCD CD =矛盾．所以假设错误，EF 不可能与CD 平行，故D 正确，故选BD ．二、填空题：本题共4小题，每小题5分，共20分． 13．0x y +=14．2515．7+1616．解：由题意知1290F AF ∠=︒，123cos 5BF F ∠=-， 所以13cos 5ABF ∠=，即1||3||5AB BF =，易得11::3:4:5AB AF BF =． 设||3AB =，1145B AF F ==，2BF x =．由双曲线的定义得：345x x +-=-，解得：3x =，所以12F F c ==⇒=，因为2521a x a =-=⇒=，所以离心率e =．三、解答题；本大题共6小题，满分70分．解答应写出文字说明，证明过程或演算步骤． 17．（10分）解：已知sin cos 4sin B A a b A =+，由正弦定理，得4sin sin sin sin cos B A A A B B =+， ······································································ 2分因为B 为三角形内角，sin 0B ≠， ······················································································· 3分所以sin s i 4s n A A A =+，即3sin os A A = ···························································· 4分所以tan 3A =， ········································································································· 5分因为0A π<<，所以6A π= ······························································································ 6分 选择条件①的解析：解法一：由sin B C =及正弦定理，可得b =， ························································ 7分由余弦定理2222cos a b c bc A =+-，则)22422c c ⎛⎫=+-⋅ ⎝⋅⎪⎪⎭， ············································································· 9分解得2c =． ·················································································································· 10分解法二：由sin B C =，又因为6A π=，所以56B C π=-， ············································ 7分则5sin 6C C π⎛⎫-=⎪⎝⎭，展开得，cos C C =， ······················································ 8分所以tan C =，6C π= 所以A C =， ·················································································································· 9分 所以2c =． ·················································································································· 10分 选择条件②的解析：解法一：由4sin b A =，可得4sin26b π==， ······································································· 7分由余弦定理2222cos a b c bc A =+-得，·············································································· 8分2222222cos6c c π=+-⨯⨯⨯，······················································································ 9分解得c = ·············································································································· 0分 解法二：由4sin b A =得4sin26b π==． ············································································· 7分因为2a =，所以，ABC △是以C 为顶角的等边三角形． 所以6A B π==，所以23C π=． ····················································································· 8分由正弦定理sin sin a cA C=得，22sinsin63cππ=， ··································································· 9分解得c = ·············································································································· 10分 选择条件③的解析：解法一：由2B C A +=，又因为A B C π++=，则3A π=， ·················································· 8分与6A π=矛盾，则问题中的三角形不存在． ········································································· 10分解法二：由2B C A +=，则263B C ππ+=⨯=，则632A B C ππππ++=+=<，························································································ 8分与三角形内角和等于π矛盾，因而三角形不存在． ································································ 10分 18．（12分）解：（1）设等比数列{}n a 的公比为q ．依题意，有()21321a a a +=+ ········································································································ 1分 将()13221a a a +=+代入12314a a a ++=得()222114a a ++=，得24a =． (2)分联立1232144a a a a ++==⎧⎨⎩得21111144a a q a q a q ++==⎧⎨⎩两式两边相除消去1a 得22520q q -+=， 解得2q =或12q =（舍去）， ····························································································· 3分 所以1422a ==． ··········································································································· 4分 所以111222n n nn a a q --==⨯=． ························································································ 5分（2）解法一：因为21log 22nn n n b a n ⎛⎫==-⋅ ⎪⎝⎭······································································ 6分所以，231222322n n T n -=⨯+⨯+⨯+⨯……① ······························································· 7分()32142122232122n n n T n n +-=⨯+⨯+⨯+-⨯+⨯……② ················································· 8分 ①-②，得21322222n n n T n +=++++-⨯ ······································································ 9分1112(12)222212n n n n n n +++-=-⨯=-⋅--． ········································································· 11分所以，数列{}n b 的前n 项和11222n n n T n ++=-⋅-． ····························································· 12分解法二：因为21log 2(2)22nn n n n b a n n n ⎛⎫==-⋅=-+⋅ ⎪⎝⎭[]{}2(1)4(2)2n n n =-++--+⋅所以()()112222n nn b n n +⋅=-++--⎦+⋅⎡⎤⎣ ········································································· 8分 进而得2132(22)2(12)2(32)2(22)2n T ⎡⎤⎡⎤=-+⋅--+⋅+-+⋅--+⋅+⋯+⎣⎦⎣⎦[]{}1(1)22(2)2n n n n +-++⋅--+⋅[]11(1)22222(1)n n n n ++=-++⋅--⋅-= ········································································· 11分 所以数列{}n b 的前n 项和为()1122n n T n +⋅=-- ································································· 12分 19．（12分）解：（1）连接1O C ，1O D ， ····························································································· 1分因为C ，D 是半圆AB 上的两个三等分点， 所以11160AO D DO C CO B ∠=∠=∠=︒， 又1111O A O B O C O D ===，所以1AO D △，1CO D △，1BO C △均为等边三角形．所以11O A AD DC CO ===， ··························································································· 2分 所以四边形1ADCO 是平行四边形． ···················································································· 3分 所以1CO AD ， ·········································································································· 4分因为1CO ⊄平面AFD ，AD ⊂平面AFD ， 所以1CO 平面AFD ． ···································································································· 5分（2）因为FC 是圆柱12O O 的母线，所以FC ⊥平面ABC ，BC ⊂平面ABC ，所以FC BC ⊥ ······················································ 6分 因为AB 为圆1O 的直径，所以90ACB ∠=︒，在Rt ABC △中，60ABC ∠=︒，AC =，所以1tan 60ACBC ==︒，所以在Rt FBC △中，tan451FC BC =︒= ··········································································· 7分 （方法一）因为BC AC ⊥，BC FC ⊥，AC FC C ⋂=， 所以BC ⊥平面FAC ， 又FA ⊂平面FAC ， 所以BC FA ⊥．在FAC △内，作CH FA ⊥于点H ，连接BH ．因为BC CH C ⋂=，BC ，CH ⊂平面BCH ，所以FA ⊥平面BCH ， ····································································································· 8分 又BH⊂平面BCH ，所以FA BH ⊥，。