湖南省郴州市2021届高三数学上学期第二次质检试题(Word版附答案)

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湖南省郴州市2021届高三数学上学期第二次质检试题(Word版附答案)

郴州市 2021 届高三上学期第二次质检 数学 (试题卷) 注意事项: 1.答题前,考生务必将自己的姓名、准考证号写在答题卡和该试题卷的封面上,并认真核对条 形码上的姓名、准考证号和科目。 2.学生作答时,选择题和非选择题均须作在答题卡上,在本试题卷上作答无效。考生在答题卡 上按答题卡中注意事项的要求答题。 3.考试结束后,将本试题卷和答题卡一并交回。 4.本试题卷共 5 页。如缺页,考生须声明,否则后果自负。 绝密★启用前 郴州市 2021 届高三第二次教学质量监测试卷 数学 一、单选题: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.设集合  2| 2 3 0A x x x    , { | 2 3}B x x    ,则 A B  ( ) A. 1,3 B. 1,3 C. 1,2 D. 2,3 2.若复数 Z 满足 (1 ) 2Z i i  ,则下列说法正确的是( ) A. Z 的虚部为 i B. Z 的共轭复数为 1Z i   C. Z 对应的点在第二象限 D. 2Z  3.已知 a,b 是两条不同的直线,α,β是两个不同的平面,且 a  ,b  ,则“ //a b ”是“ //  ”的 ( ) A 充分不必要条件 B 必要不充分条件 C 充要条件 D 既不充分也不必要条件 4.《易·系辞上》有“河出图,洛出书”之说,河图、洛书是中国古代流传下来的两幅神秘图案,蕴含了深 奥的宇宙星象之理,被誉为“宇宙魔方”,是中华文化阴阳术数之源.河图的排列结构如图 1 所示,一与六共 宗居下,二与七为朋居上,三与八同道居左,四与九为友居右,五与十相守居中,其中白圈数为阳数,黑 点数为阴数,若从阳数和阴数中各取一数,则其差的绝对值大于 5 的概率为( ) A. 4 25 B. 7 25 C. 8 25 D. 2 5 5.已知单位向量 a ,满足等式 2| | | |a b  ,| 2 | 13a b  ,则 a 与b  的夹角为( ) A.120° B.90° C.60° D.30° 6.随着科学技术的发展,放射性同位素技术已经广泛应用于医学、航天等众多领域,并取得了显著经济效益. 假设在放射性同位素钍 234 的衰变过程中,其含量 N(单位:贝克)与时间 t(单位:天)满足函数关系 24 0( ) 2 t N t N   ,其中 0N 为 0t  时钍 234 的含量.已知 24t  时,钍 234 含量的瞬时变化率为 8ln 2 ,则 (96)N  ( ) A.12 贝克 B.12ln 2 贝克 C.24 贝克 D. 24ln2贝克 7.如图 2,设椭圆 1C : 2 2 2 2 1x y a b   ( 0a b  )与双曲线 2C : 2 2 2 2 1x y m n   ( 0m  , 0n  )的公共焦 点为 1F , 2F ,将 1C , 2C 的离心率分别记为 1e , 2e ,点 A 是 1C , 2C 在第一象限的公共点,若点 A 关于 2C 的一条渐近线的对称点为 1F ,则 2 2 1 2 2 2 e e   ( ) A.2 B. 5 2 C. 7 2 D.4 8.已知可导函数 ( )f x 的导函数为 ( )f x ,若对任意的 x  R ,都有 ( ) ( ) 2f x f x   ,且 ( ) 2021f x  为奇 函数,则不等式 ( ) 2019 2xf x e  的解集为( ) A. (0, ) B. ( ,0) C. ( , )e D. 1,e     二、多选题: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。 9.2020 年初,突如其来的疫情改变了人们的消费方式,在目前疫情防控常态化背景下,某大型超市为了解 人们以后消费方式的变化情况,更好的提高服务质量,收集并整理了本超市 2020 年 1 月份到 8 月份的线上 收入和线下收入的数据,并绘制如下的折线图 3.根据折线图,下列结论正确的是( ) A.根据该超市这 8 个月折线图可知,线下收入的平均值在 7,8 内 B.根据该超市这 8 个月折线图可知,线上收入的极差比线下收入的极差大 C.根据该超市这 8 个月折线图可知,每月总收入与时间呈现负相关 D.根据该超市这 8 个月折线图可知,在疫情逐步得到有效控制后,人们比较愿意线下消费 10.已知函数 ( ) sin( )f x x   ( 0  ,| | 2   )的最小正周期为 2 .把函数 ( )f x 的图象向左平移 2 3  个单位长度得到的图象对应的函数为偶函数,则( ) A. 6   B. ,06      是 ( )f x 的图象的对称中心 C. ( )f x 在 0, 2      上单调递增 D. ( )f x 在 0, 上的值域为 1 ,12     11.已知抛物线 C: 2 4y x 的焦点为 F,  1 1,P x y ,  2 2,Q x y 是抛物线上两点,则下列结论正确的是( ) A. C 的准线方程: 1x   B.若直线 PQ过点 F,则 1 2 4x x  C.若 5| | | | 2PF QF  ,则线段 PQ的中点 M 到 y 轴的距离为 1 4 D.若 PF QF ,则 2OPQS △ 12.已知5 3a  ,8 5b  ,则( ) A. a b B. 1 1 2a b   C. 1 1a ba b    D. b aa a b b   三、填空题:本题共 4 小题,每小题 5 分,第 15 题第一问 2 分,第 2 问 3 分,共 20 分。 13. tan 765  __________. 14.已知圆 O 的半径为 5, 4OP  ,过点 P 的 2021 条弦的长度组成一个等差数列 na ,最短弦长为 1a ,最 长弦长为 2021a ,则公差 d  __________. 15.定义:在等式中 2 0 2 1 2 1 2 2 2 2 1 22 n n n n n n n n n n nx x D x D x D x D x D          ( *n N )中,把 0 nD , 1 nD , 2 nD ,… 2n nD 叫做三项式  2 2 n x x  的 n 次系数列(如三项式的 1 次系数列是 1,-1,-2).则(1)三项 式 2 2 n x x  的 2 次系数列各项之和等于__________;(2) 9 5D  __________. 16.如图 4,已知球 O 是直三棱柱 1 1 1ABC A B C 的外接球, 1 6AC BC CC   , AC BC ,E,F 分别 为 1BB , 1 1AC 的中点,过点 A,E,F 作三棱柱的截面α,若α交 1 1B C 于 M,过点 M 作球 O 的截面,则所得 截面圆面积的最小值是__________. 四、解答题:本大题共 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.(本小题满分 10 分) 设 na 为等差数列, nb 是正项等比数列,且 1 1 2a b  , 3 22a b .在① 5 3 112b b b  ,② 5 42a b  , ③ 2 2 1log log 1n nb b   , 2n  , *n N 这三个条件中任选一个,求解下列问题: (Ⅰ)写出你选择的条件并求数列 na 和 nb 的通项公式; (Ⅱ)在(Ⅰ)的条件下,若 n n nc a b  ( *n N ),求数列 nc 的前 n 项和 nS . 18.(本小题满分 12 分) 如图 5,在平面四边形 ABCD 中, AB AD , 1AB  , 3AD  , 2BC  . (I)若 2 2CD  ,求四边形 ABCD 的面积; (Ⅱ)若 7cos 5BCD  , 0, 2ADC      ,求 sin ADC . 19.(本小题满分 12 分) 如图 6,在直四棱柱 1 1 1 1ABCD A B C D 中,四边形 ABCD 为平行四边形,M 为 1AA 的中点, 1BC BD  , 1 2AB AA  . (Ⅰ)求证:平面 BMD  平面 BDC ; (Ⅱ)求二面角 1 1D MC C  的正弦值. 20.(本小题满分 12 分) 已知椭圆 C: 2 2 2 2 1x y a b   ( 0a b  )的离心率为 3 3 ,直线 5 0x y   与椭圆 C 有且只有一个公共 点. (Ⅰ)求椭圆 C 的标准方程 (Ⅱ)设点 ( 3,0)A  , ( 3,0)B ,P 为椭圆 C 上一点,且直线 PA 与 PB 的斜率乘积为 2 3  ,点 M,N 是椭圆 C 上不同于 A,B 的两点,且满足 //AP OM , //BP ON ,求证: OMN△ 的面积为定值. 21.(本小题满分 12 分) 已知函数 ( ) 1 lnxf x ae x  (Ⅰ)当 1a  ,讨论函数 ( )f x 的单调性; (Ⅱ)若不等式  ( ) x af x e x x  ( 0a  ),对 (1, )x  恒成立,求实数 a 的取值范围. 22.(本小题满分 12 分) 垃圾分类,是指按一定规定或标准将垃圾分类储存、分类投放和分类搬运,从而转变成公共资源的一系列 活动的总称.分类的目的是提高垃圾的资源价值和经济价值,力争物尽其用.垃圾分类后,大部分运往垃圾处 理厂进行处理.为了监测垃圾处理过程中对环境造成的影响,某大型垃圾处理厂为此建立了 5 套环境监测系 统,并制定如下方案:每年工厂的环境监测费用预算定为 80 万元,日常全天候开启 3 套环境监测系统,若 至少有 2 套系统监测出排放超标,则立即检查污染处理系统;若有且只有 1 套系统监测出排放超标,则立 即同时启动另外 2 套系统进行 1 小时的监测,且后启动的这 2 套监测系统中只要有 1 套系统监测出排放超 标,也立即检查污染处理系统.设每个时间段(以 1 小时为计量单位)被每套系统监测出排放超标的概率均 为 p( 0 1p  ),且各个时间段每套系统监测出排放超标情况相互独立. (Ⅰ)当 1 2p  时,求某个时间段需要检查污染处理系统的概率; (Ⅱ)若每套环境监测系统运行成本为 20 元/小时(不启动则不产生运行费用),除运行费用外,所有的环 境监测系统每年的维修和保养费用需要 6 万元.现以此方案实施,问该工厂的环境监测费用是否会超过预算 (全年按 9000 小时计算)?并说明理由. 郴州市 2021 届高三第二次教学质量监测试卷 数学参考答案及评分细则 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。 1-5 BCDAC 6-8 CDA 二、多选题:本题共 4 小题,每小题 5 分,共 20 分。在每小题给出的四个选项中,多项是符合题目要求的。 全部选对得 5 分,有选错得 0 分,部分选对得 3 分。 9.AD 10.BCD 11.ACD 12.ABD 三、填空题:本题共 4 小题,每小题 5 分,第 16 题第一问 2 分,第 2 问 3 分,共 20 分。 13.1 14. 1 505 15.(1)4 (2)-80 16.8 四、解答题:本大题共 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.(Ⅰ)选择①:设 na 的公差为 d, nb 的公比为 q( 0q  ). 则根据题意有 4 2 2 2 4 2 2 24 d q q q      ,·················································································(2 分) 解得 2 3 q d    ············································································································· (4 分) 所以, 2 3( 1) 3 1na n n     12 2 2n n nb    ·························································(5 分) 选择②:设 na 的公差为 d, nb 的公比为 q( 0q  ). 则根据题意有 3 2 2 4 2 4 2 2 d q d q       ,··············································································· (2 分) 解得 2 3 q d    ············································································································· (4 分) 所以 2 3( 1) 3 1na n n     , 12 2 2n n nb    .····················································· (5 分 ) 选择③:由 2 2 1log log 1n nb b   , 2n  , *n N 得 2 2 1log log 1n nb b   ∴ 1 2n n b b   ·············································································· (2 分) ∴ nb 的公比为 2q  又 2 2 4d q  ∴ 3d  ·····························································(4 分) 所以 2 3( 1) 3 1na n n     , 12 2 2n n nb    .··························································· (5 分) (Ⅱ)由(Ⅰ)可知 (3 1) 2n nc n   ··········································································· (6 分) 2 32 2 5 2 8 2 (3 1) 2n nS n          ① 2 3 12 2 2 5 2 (3 4)2 (3 1)2n n nS n n          ②···················································· (7 分) ①-②得 2 3 12 2 3 2 3 2 3 2 (3 1) 2n n nS n              ············································(8 分)  2 3 14 3 2 2 2 (3 1)2n nn        ········································································ (9 分) ∴ 2 12 2 24 3 (3 1)21 2 n n nS n        1(3 4)2 8nn    ··································································································· (10 分) 18.(Ⅰ)连接 BD,在 Rt ABD△ 中 由勾股定理得: 2 2 2 4BD AB AD   ,所以 2BD  ················································· (1 分) 在 BCD△ 中,由余弦定理知: 2 2 2 3cos 2 4 BC CD BDC BC CD    ·····································(3 分) ∴ 2 7sin 1 cos 4C C   ··················································································· (4 分) 所以 1 7sin2 2BCDS BC CD C  △ , 1 3 2 2ABDS AB AD  △ ···································· (5 分) 所以 ABCD 的面积 3 7 2ABD BCDS S S   △ △ ························································· (6 分) (Ⅱ)由 7cos 5BCD  得 3 2sin 5BCD  ·····························································(7 分) 在 BCD△ 中,由正弦定理知: sin sin BC BD BDC BCD   所以 sin 3sin 5 BC BCDBDC BD     ········································································· (8 分) 因为 0, 2ADC      ,所以 0, 2BDC      , 4cos 5BDC  ····································(9 分) 在 Rt ABD△ 中, 3tan 3 ABADB AD    ,所以 6ADB   ······································(10 分) 所以 3 3 4 1 4 3 3sin sin 6 5 2 5 2 10ADC BDC              ·································· (12 分) 19.(Ⅰ)因为 1BC BD  , 2CD AB  .可得 2 2 2BC BD CD  , ∴ BD BC ,又∵ //AD BC ,∴ BD AD . 又∵ 1 1 1 1ABCD A B C D 是直四棱柱, ∴ 1DD  平面 ABCD .∴ 1DD BD . 1DD AD D ,∴ BD  平面 1 1ADD A , ∴ BD MD .·········································································································· (2 分) 法一:取 1BB 中点 N,连接 NC , MN , ∵ //MN DC ,∴四边形 MNCD 为平行四边形,∴ //MD NC , ∵ 1 2 2 NB BC BC CC   ,∴ 1NBC BCC∽△ △ ,∴ 1 90C BC BCN     ,∴ 1BC CN , 又∵ //MD NC ,∴ 1MD BC .··················································································· (4 分) 又 1BC BD B ,∴ MD  平面 1BDC .······································································(5 分) 又 MD  平面 MBD ,所以平面 BMD  平面 1BDC .······················································ (6 分) 法二:连接 1 1AC ,可计算得 1 22 2C M  , 6 2MD  , 1 2C D  , 所以 2 2 2 1 1MD C D MC  ,由勾股定理的逆定理得: 1MD DC ,余同法一 (Ⅱ)以 DA为 x 轴, DB为 y 轴, 1DD 为 z 轴,建立如图所示的坐标系, 则 (0,1,0)B , 1( 1,1, 2)C  , 1(0,0, 2)D , 21,0, 2M       ∴ 1 22,1, 2MC        , 1 (0,0, 2)CC  , 1 1 ( 1,1,0)D C   ············································ (7 分) 设 平 面 1C CM 的 法 向 量 为 ( , , )n x y z , 由 1 1 0 0 CC n MC n          得 2 0 2 3 0 z x y z       可 求 得 一 个 法 向 量 (1,2,0)n  ············································································································· (8 分) 同理可得平面 1 1D C M 的一个法向量 (1,1, 2)m  ························································· (10 分) 设二面角 1 1D MC C  的大小为θ 所以 | | 3 3 5| cos | | cos , | | || | 102 5 m nm n m n              ·····················································(11 分) 则 55sin 10   ,即二面角 1M BC D  的正弦值为 55 10 .·············································(12 分) 20.解:(Ⅰ)∵直线 5 0x y   与椭圆有且只有一个公共点, ∴直线 5 0x y   与椭圆 C: 2 2 2 2 1x y a b   ( 0a b  )相切, ∴ 2 2 2 2 5 0 1 x y x y a b        2 2 2 2 2 2 22 5 5 0b a x a x a a b      ·········································· (3 分) ∴ 2 20 5a b     又∵ 3 3 c a  ,∴ 3a  ,∴ 2 2 2 2b a c   , 椭圆 C 的方程为 2 2 13 2 x y  .······················································································(5 分) (Ⅱ)证明:由题意 M、N 是椭圆 C 上不同于 A,B 的两点, 由题意知,直线 AP , BP 斜率存在且不为 0,又由已知 2 3AP BPk k   . 由 //AP OM , //BP ON ,所以 2 3OM ONk k   ······························································ (6 分) 设直线 MN 的方程为 x my t  , 代入椭圆方程得 2 2 22 3 4 2 6 0m y mty t     ①······················································(7 分) 设  1 1,M x y ,  2 2,N x y , 则 1 2 2 4 2 3 mty y m     , 2 1 2 2 2 6 2 3 ty y m   ······································································ (8 分) 又   2 1 2 1 2 2 2 2 2 1 2 1 2 1 2 2 6 2 3 6 3OM ON y y y y tk k x x m y y mt y y t t m          ··································· (9 分) 得 2 22 2 3t m  ·····································································································(10 分) 所以 2 2 1 2 2 2 1 1 48 24 72 1 2 6 | | 6| | | | | |2 2 2 3 2 2 2MON m t tS t y y t tm t      △ 即 MON△ 的面积为定值 6 2 ·················································································(12 分) 21.解:(Ⅰ) ( )f x 的定义域为 (0, ) , 1( ) lnxf x e x x       ,······································ (1 分) 令 1( ) lng x x x   ,则 2 2 1 1 1( ) xg x x x x     ,·····························································(2 分) 当 (0,1)x 时, ( ) 0g x  , ( )g x 单调递减, 当 (1, )x  时, ( ) 0g x  , ( )g x 单调递增,······························································(3 分) ∴ 1x  时, ( )g x 取得极小值即最小值 (1) 1g  , ∴ ( ) 0f x  在 (0, ) 恒成立,·················································································· (4 分) ∴ ( )f x 在 (0, ) 单调递增;····················································································· (5 分) (Ⅱ)不等式  ( ) x af x e x x  等价于 ln ln lnx a x x a ae x x a x e e x x         ,····· (6 分) 设 ( ) lnk t t t  ,即    x ak e k x  (*) ∵ 1 1( ) 1 tk t t t    ································································································· (7 分) ∴当 (0,1)t  , ( ) 0k t  , ( )k t 在 (0,1) 是减函数 (1, )t   , ( ) 0k t  , ( )k t 在 (1, ) 是增函数 ·························································(8 分) ∵ (1, )x  , 10 1xe e    ···············································································(9 分) 当 0a  时, 0 1ax  ,且 ( )k t 在 (0,1) 是减函数 则(*)式 ln x a xe x a x      令 ( ) ln xh x x  ( 1x  ),则 2 ln 1( ) (ln ) xh x x   ,······························································ (10 分) 当 (1, )x e 时, ( ) 0h x  , ( )h x 单调递减, 当 ( , )x e  时, ( ) 0h x  , ( )h x 单调递增,····························································(11 分) min( ) ( )h x h e e a e     ∴ a e  又 0a  ∴ 0e a   ································· (12 分) 22.(Ⅰ)设某个时间段在需要开启 3 套系统就被确定需要检查污染源处理系统的事件为 A 2 3 3 3 2 2 3 3 2 3 2 3 3 3 3 3 3 3 1 1 1 1 1 1( ) (1 ) 2 2 2 2 2 2P A C p p C p C C C C                              ,···············(2 分) 设某个时间段在需要开启另外 2 套系统才能确定需要检查污染源处理系统的事件为 B 3 2 1 2 2 1 3 3 1 1 9( ) (1 ) 1 (1 ) 12 2 32P B C p p p C                        ···········································(5 分) ∴某个时间段需要检查污染源处理系统的概率为 1 9 25 2 32 32   ··········································· (6 分) (Ⅱ)设某个时间段环境监测系统的运行费用为 X 元,则 X 的可能取值为 60,100.···············(7 分) 1 2 3( 100) (1 )P X C p p   1 2 3( 60) 1 (1 )P X C p p    ···············································(8 分) 1 2 1 2 2 3 3( ) 60 1 (1 ) 100 (1 ) 60 120 (1 )E X C p p C p p p p           令 2( ) (1 )g p p p  , (0,1)p ,则 2( ) (1 ) 2 (1 ) (3 1)( 1)g p p p p p p        .··············(9 分) 当 10, 3p     时, ( ) 0g p  , ( )g p 在 10, 3      上单调递增, 当 1,13p     时, ( ) 0g p  , ( )g p 在 1,13      上单调递减,············································ (10 分) ∴ ( )g p 的最大值为 1 4 3 27g      ,············································································· (11 分) ∴实施此方案,最高费用为 446 9000 60 120 10 7627         (万元), ∵ 76 80 ,故不会超过预算.···················································································· (12 分)
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