重庆市第八中学2018届高考适应性月考(五)数学(理)试题 含答案

秘密★启用前重庆市第八中学2023届高考适应性月考卷（二）数学注意事项：1．答题前，考生务必用黑色碳素笔将自己的姓名、准考证号、考场号、座位号在答题卡上填写清楚．2．每小题选出答案后，用2B 铅笔把答题卡上对应题目的答案标号涂黑，如需改动，用橡皮擦干净后，再选涂其他答案标号.在试题卷上作答无效.3.考试结束后，请将本试卷和答题卡一并交回.满分150分，考试用时120分钟.一、单项选择题（本大题共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.设全集U =R ，集合{}=02A x x ≤≤，{}=240x B x -≥，则集合()UA B = ð（）A.()0,2 B.(]0,2 C.[)0,2 D.[]0,22.设x =R ，则“01x <<”是“2230x x --<”的（）A.充分而不必要条件B.必要而不充分条件C.充要条件D.既不充分也不必要条件3.在医学生物学试验中，经常以果蝇作为试验对象，一个关有6只果蝇的笼子里，不慎混入了两只苍蝇（此时笼内共有8只蝇子：6只果蝇和2只苍蝇），只好把笼子打开一个小孔，让蝇子一只一只地往外飞，直到两只苍蝇都飞出，再关闭小孔.记事件k A 表示“第k 只飞出笼的是苍蝇”，1,2,,8k =⋅⋅⋅，则()52|P A A 为（）A.15B.16C.17D.254.定义在R 上的函数()f x 满足()()121f x f x x +=-++，则下列是周期函数的是（）A.()y f x x=+ B.()y f x x=- C.()2y f x x=+ D.()2y f x x=-5.我国古代数学家僧一行应用“九服晷影算法”在《大衍历》中建立了晷影长l 与太阳天顶距θ（0180θ︒≤≤︒）的对应数表，这是世界数学史上较早的一张正切函数表，根据三角学知识可知，晷影长度l 等于表高h 与太阳天顶距θ正切值的乘积，即tan l h θ=.对同一“表高”两次测量，第一次和第二次太阳天顶距分别为α，β，若第一次的“晷影长”是“表高”的2倍，且()1tan 3αβ-=，则第二次的“晷影长”是“表高”的（）倍A.1B.2C.3D.46.已知81log 32a =，0.01b π=，sin1c =，则,,a b c 的大小关系是（）A.c b a <<B.c a b<< C.a b c<< D.a c b<<7.在ABC 中，π3A =，G 为ABC 的重心，若12AG AB AG AC ⋅=⋅= ，则ABC 外接圆的半径为（）A.B.2C.D.8.若函数()32f x ax bx cx d=+++()0a >有极值点1x ，2x ，且()22f x x =，则关于x 的方程()()2320a f x bf x c ++=⎡⎤⎣⎦的不同实数根个数是（）A.3B.4C.5D.6二、多项选择题（本大题共4小题，每小题5分，共20分.在每小题给出的选项中，有多项是符合题目要求的.全部选对的得5分，部分选对的得2分，有选错的得0分）9.已知0a b >>，且1ab =，则下列式子正确的有（）A.22log log 0a b ->B.22log log 1a b +>C.22log log 0a b ⋅< D.224a b +>10.设首项为1的数列{}n a 的前n 项和为n S ，已知121n n S S +=+，则下列结论正确的是（）A.数列{}1n S +为等比数列B.数列{}n a 不是等比数列C.21n n S a =-D.{}n a 中任意三项不能构成等差数列11.已知函数()4f x x πω⎛⎫=+ ⎪⎝⎭()0ω>，则下列说法正确的是（）A.若函数()f x 的最小正周期为π，则其图象关于直线π8=x 对称B.若函数()f x 的最小正周期为2π，则其图象关于点π,04⎛⎫ ⎪⎝⎭对称C.若函数()f x 在区间π0,8⎛⎫⎪⎝⎭上单调递增，则ω的最大值为2D.若函数()f x 在[]0,2π有且仅有4个零点，则ω的取值范围是151988ω≤<12.已知F 为椭圆C ：221168x y +=的左焦点，直线l ：=y kx ()0k ≠与椭圆C 交于A ，B 两点，AE x ⊥轴，垂足为E ，BE 与椭圆C 的另一个交点为P ，则（）A.8AF BF +=B.14AF BF+的最小值为2C.直线BE 的斜率为12k D.PAB ∠为钝角三、填空题（本大题共4小题，每小题5分，共20分）13.复数z 满足：23i z z +=+，则=z ______.14.定义在R 上的函数()f x 满足以下两个性质：①()() f x f x -=，②()()20f x f x +-=，满足①②的一个函数是______.15.已知M 是边长为1的正ABC 的边AC 上的动点，N 为AB 的中点，则BM MN ⋅的最大值是_____.16.已知函数()()2log 41xf x x =+-，数列{}n a 是公差为4的等差数列，若()()()()112233440a f a a f a a f a a f a +++=，则数列{}n a 的前n 项和=n S ______.四、解答题（共70分.解答应写出文字说明，证明过程或演算步骤）17.如图，在棱柱111ABC A B C -中，D 为棱BC 的中点.（1）证明：1//A B 平面1AC D ；（2）若该三棱柱为正三棱柱，且所有棱长均相等，求直线AC 与平面1AC D 所成角的正弦值.18.在ABC中，角A ，B ，C的对边分别为a ，b ，c ，已知()()()2sin sin sin sin sin sin cos cos C A B A B C A C B ----=-+.（1）求B ；（2）已知2a c -=，ABC S =△，求b .19.记n S 为数列{}n a 的前n 项和，已知1=2a ，{}32n n a S -是公差为2的等差数列.（1）求{}n a 的通项公式；（2）证明：121111na a a ++⋅⋅⋅+<.20.核电站某项具有高辐射危险的工作需要工作人员去完成，每次只派一人，每人只派一次，工作时长不超过15分钟，若某人15分钟内不能完成该工作，则撤出，再派下一人，现有小胡、小邱、小邓三人可派，且他们各自完成工作的概率分别为1p ，2p ，3p .假设1p ，2p ，3p 互不相等，且假定三人能否完成工作是相互独立.（1）任务能被完成的概率是否与三个人被派出的先后顺序有关？试说明理由；（2）若按某指定顺序派出，这三人各自能完成任务的概率依次为1q ，2q ，3q ，其中1q ，2q ，3q 是123,,p p p 的一个排列.①求所需派出人员数目X 的分布列和数学期望()E X ；②假定1231>>>p p p ，为使所需派出的人员数目的数学期望达到最小，应以怎么样的顺序派出？21.已知函数()()()ln 3f x x a x =++()a ∈R .（1）若函数()f x 在定义域内单调递增，求a 的取值范围；（2）若=2a ，()f x kx >在()1,x ∈+∞上恒成立，求整数k 的最大值.（参考数据：ln 20.69≈，ln 3 1.1≈）22.已知双曲线E ：22221x y a b-=0a >，0b >）一个顶点为()2,0A -，直线l 过点()3,0Q 交双曲线右支于M ，N 两点，记AMN ，AOM △，AON △的面积分别为S ，1S ，2S .当l 与x 轴垂直时，1S 的值为152.（1）求双曲线E 的标准方程；（2）若l 交y 轴于点P ，PM MQ λ= ，PN NQ μ=，求证：λμ+为定值；（3）在（2）的条件下，若121625S S mS μ=+，当58λ<≤时，求实数m 的取值范围.秘密★启用前重庆市第八中学2023届高考适应性月考卷（二）数学注意事项：1．答题前，考生务必用黑色碳素笔将自己的姓名、准考证号、考场号、座位号在答题卡上填写清楚．2．每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑，如需改动，用橡皮擦干净后，再选涂其他答案标号.在试题卷上作答无效.3.考试结束后，请将本试卷和答题卡一并交回.满分150分，考试用时120分钟.一、单项选择题（本大题共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题答案】【答案】C【2题答案】【答案】A【3题答案】【答案】C【4题答案】【答案】B【5题答案】【答案】A【6题答案】【答案】D【7题答案】【答案】C【8题答案】【答案】A二、多项选择题（本大题共4小题，每小题5分，共20分.在每小题给出的选项中，有多项是符合题目要求的.全部选对的得5分，部分选对的得2分，有选错的得0分）【9题答案】【答案】ACD 【10题答案】【答案】ACD 【11题答案】【答案】ACD 【12题答案】【答案】AC三、填空题（本大题共4小题，每小题5分，共20分）【13题答案】【答案】1+i ##i+1【14题答案】【答案】()πcos 2f x x =（答案不唯一）【15题答案】【答案】2364-【16题答案】【答案】228n n-四、解答题（共70分.【17题答案】【答案】（1）证明见解析（2）55【18题答案】【答案】（1）3π（2）b =【19题答案】【答案】（1）31nn a =-（2）证明见解析【20题答案】【答案】（1）无关；理由见解析（2）①分布列见解析；期望为121232q q q q --+；②完成任务概率大的人先派出【21题答案】【答案】（1）)5e ,-⎡+∞⎣（2）6【22题答案】【答案】（1）22143x y -=（2）证明见解析（3）1832,55⎛⎤⎥⎝⎦第8页/共8页。

地理试卷注意事项：1.答题前，考生务必用黑色碳素笔将自己的姓名、准考证号、考场号、座位号在答题卡上填写清楚。

2.每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑，如需改动，用橡皮擦干净后，再选涂其他答案标号。

在试题卷上作答无效。

3.考试结束后，请将本试卷和答题卡一并交回。

满分150分，考试用时75分钟。

一、选择题：本题共15小题，每小题3分，共45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。

2024年第14号台风“普拉桑”于9月19日21时登陆上海，第21号台风“康妮”于10月31日17时登陆台湾。

图1为“普拉桑”和“康妮”路径示意图。

据此完成1~3题。

图11.图2中最能反映“普拉桑”气流运动特征的是B.②A.①C.③D.④2.“普拉桑”从江苏移入黄海南部海面后强度发生变化，主要原因是①海域提供热量补给②水汽补给增多③地转偏向力增大④地表摩擦力减小A.①②③B.①②④C.①③④D.②③④3.“康妮”比“普拉桑”登陆位置更靠南，是因为A.秋季冷空气南下B.“副高”南移C.“康妮”移动速度快D.东北信风的影响第七次全国人口普查数据显示，重庆2023年常住人口3191.43万，户籍人口3413.8万。

随着重庆社会经济的良好发展和一系列的政策措施的施行，重庆市来渝人口规模一直稳步增长。

在重庆常住人口中，外省来渝人口为219.36万，约占常住人口的6.8%。

图3为“2023年来渝人口前6名省统计图”。

据此完成4~6题。

qwe4.图中甲省最可能为A.陕西B.四川C.甘肃D.广东5.与重庆相邻的贵州、湖北来渝人口数量存在差异，主要影响因素是A.空间位置B.人口总量C.产业结构D.对外交通6.大量外来人口对重庆的作用是A.增加环境人口容量B.促进人口结构年轻化C.增加第一产业从业人员D.促进社会结构多元化茶树喜光怕寒、喜湿怕涝。

印度尼西亚是世界重要产茶国，爪哇岛山区是其重要产茶区（如图4），全年可采茶，但由嫩叶制成的茶叶品质最佳。

2022届重庆市巴蜀中学高三下学期高考适应性月考（八）数学试题一、单选题1．已知集合{}5,8A =，{}23100B x x x =--≤，则()R A B ⋂=（ ）A ．{}5B ．{}8C ．{}2,5,8-D ．{}2-【答案】B【分析】求出集合B ，利用交集和补集的定义可求得结果.【详解】因为{}{}2310025B x x x x x =--≤=-≤≤，则{R 2B x x =<-或}5x >，因此，(){}R 8A B ⋂=. 故选：B.2．已知变量x 与y 正相关，且由观测数据算得样本平均数2x =，10y =，则由观测的数据得到的线性回归方程可能为（ ） A ．1511y x =-+.B ．0.511y x =-+C ．0.59y x =+D ． 1.58y x =+【答案】C【分析】根据变量正相关排除AB ，再利用线性回归方程过点(,)x y 判断CD 即可. 【详解】因为变量x 与y 正相关，所以0b >，故排除AB ； 又线性回归方程过点(2,10)，代入CD 检验，可知C 正确，D 错误. 故选：C3．将函数()y f x =的图象向右平移2π个单位长度得到函数()sin3g x x =的图象，则()f x =（ ）A ．cos3xB ．cos3x -C ．sin3xD ．sin3x -【答案】B【分析】利用三角函数的图象变换可得出()2f x g x π⎛⎫=+ ⎪⎝⎭，结合诱导公式化简可得出函数()f x 的解析式.【详解】由题意可知，将函数()sin3g x x =的图象向左平移2π个单位长度，可得到函数()f x 的图象，则()3sin 3sin 3cos322f x x x x ππ⎡⎤⎛⎫⎛⎫=+=+=- ⎪ ⎪⎢⎥⎝⎭⎝⎭⎣⎦. 故选：B.4．已知O 为坐标原点，直线():22l y kx k =+-上存在一点P ，使得2OP =，则k 的取值范围为（ ） A ．32,32⎡⎤-+⎣⎦B ． (][,2323,)-∞-⋃++∞C ．23,23⎡⎤-+⎣⎦D ． ](,3232,) [-∞-⋃++∞【答案】C【分析】根据题意得坐标原点到直线l 距离d OP ≤，利用点到直线的距离公式即可求解.【详解】点()0,0O 到直线():22l y kx k =+-的距离为 222|0022||22|(1)1k k k d k k ⨯-+--==+-+，由题意得坐标原点到直线l 距离d OP ≤，2OP =， 所以2|22|21k k -≤+，解得2323k -≤≤+所以k 的取值范围为23,23⎡⎤-+⎣⎦.故选：C.5．如图甲所示，古代中国的太极八卦图是以同圆内的圆心为界，画出相等的两个阴阳鱼，阳鱼的头部有眼，阴鱼的头部有个阳殿，表示万物都在相互转化，互相涉透，阴中有阳，阳中有阴，阴阳相合，相生相克，蕴含现代哲学中的矛盾对立统一规律.其平面图形记为图乙中的正八边形ABCDEFGH ，其中2OA =，则以下结论错误的是（ ）A 20OB OE OG ++= B ．22OA OD =-C ．4AH EH +=D ．422+=+AH GH 【答案】D【分析】根据题意，建立平面直角坐标系，写出需要点的坐标，然后利用向量加法的坐标运算、向量的数量积坐标运算及向量的坐标运算即可求解. 【详解】由题意可知，建立如图所示的平面直角坐标系，因为正八边形ABCDEFGH ，所以AOH HOG AOB EOF FOG ∠∠∠∠∠==== DOE COB COD ∠∠∠︒=====︒360458作AM HD ⊥，则OM AM =, 因为2OA =，所以2OM AM == 所以(,A --22，同理可得其余各点坐标， (,),(,(,),(,),(,)B E G D H ---022******* ，对于A ((),)OB OE OG ++=-=202222220，故A 正确； 对于B ，((222022OA OD =-⨯+-⨯=-，故B 正确；对于C ，(22,2AH =-，(22,2EH =-，()4,0AH EH +=- 所以()22404AH EH +=-+=，故C 正确；对于D ,(22,2AH =-，(22,2GH =-，()422,0AH GH =-++()224220422AH GH +=-++=-D 不正确.故选：D.6．已知正实数a ，b 满足220ab a +-=，则4a b +的最小值是（ ） A ．2 B ．422 C ．32 D ．6【答案】B【分析】根据220ab a +-=变形得22a b =+，进而转化为a b b b +=++842， 用凑配方式得出()b b ++-+8222，再利用基本不等式即可求解.【详解】由220ab a +-=，得22a b =+，所以()a b b b b b b +=+=++-⋅=+++888422222222，当且仅当,a b b b ==+++28222，即a b ==22取等号. 故选：B.7．从编号分别为1、2、3、4、5、6、7的七个大小完全相同的小球中，随机取出三个小球，则至少有两个小球编号相邻的概率为（ ） A ．57B ．35C ．25D ．13【答案】A【分析】先用组合数算出基本事件的总数，再用排异法求出任意两个小球编号都不相邻的事件的个数，进而得出至少有两个小球编号相邻的事件的个数，然后用古典概型的计算公式即可求解.【详解】随机取出三个小球共有3735C =种情况，任意两个小球编号都不相邻的基本事件有：()()()()()()()()()(),,,,,,,,,,,,,,,,,,,,,,,,,.,,135136137146147157246247257357共有10种，故所求概率为-=35105357. 故选：A.8．已知点(A B ，若曲线0(0,0)⎛⎫⎛⎫+-=>> ⎪⎪⎝⎭⎝⎭x y x y a b a b a b 上存在点P 满2PA PB -=，则下列选项一定正确的是（ ）A ．1b a <+B ．b >C ．1b a >+D ．b <【答案】D【分析】根据双曲线的定义知，点P 在双曲线的右支上，求得双曲线的方程为221(0)2y x x -=>和渐近线方程为y =，转化为曲线b y x a =±与双曲线相交，得到ba.【详解】由题意，点(A B 且2PA PB -=，根据双曲线的定义知，点P 在双曲线的右支上，且1122,a c =b ==所以双曲线的方程为221(0)2y x x -=>，其渐近线方程为2y x =±，又点P 在曲线0x y x y a b a b ⎛⎫⎛⎫+-= ⎪⎪⎝⎭⎝⎭，即22220x y a b-=，即点P 在曲线b y x a =±，若曲线0x y x y a b a b ⎛⎫⎛⎫+-= ⎪⎪⎝⎭⎝⎭上存在点P 满2PA PB -=，即曲线b y x a =±与双曲线221(0)2y x x -=>相交，所以2b a <，即2b a <.故选：D. 二、多选题9．设α，β为两个平面，下列选项中是“//αβ”的充分条件的是（ ） A ．异面直线a ，b 满足a ∥α，b ∥β B ．α内有两条相交直线与平面β均无交点 C ．α，β与直线l 都垂直D ．α内有无数个点到β的距离相等 【答案】BC【分析】AD 选项可举出反例，BC 选项可以通过，面面平行的判定及线面垂直的性质进行证明.【详解】A 选项，如图，直线BC 为a ，直线EF 为b ，平面ADHE 为α，平面CDHG 为β，满足a ∥α，b ∥β，而α，β为相交平面，A 错误；B 选项，设α内两条相交直线为,m n 均与平面β无交点，即m ∥β，n ∥β，又,m n α⊂，且,m n 为相交直线，故α∥β，B 正确；C 选项，α，β与直线l 都垂直，不妨设l 与α内有两条相交直线,a b 均垂直，则在平面β内存在相交直线,m n 与l 都垂直，且m ∥a ，n ∥b ，因为m ⊄α，所以m ∥α，同理可知n ∥α，由于,m n 为相交直线，故可知α∥β，C 正确；D 选项，α内有无数个点到β的距离相等，这无数个点可能来自于同一条直线，此时不能推导出α∥β，D 错误. 故选：BC10．已知()4cos cos 25αβα+==-，其中,αβ为锐角，则以下命题正确的是（ ） A ．3sin 25α=B ．()cos αβ-=C．cos cos αβ=D ．1tan tan 3αβ=【答案】AB【分析】利用凑角的方式2()αβααβ-=-+，将角看成整体，但要注意角的范围， 根据同角三角函数的关系，两角和差的余弦公式及解方程即可求解.【详解】因为4cos 25α=-，π0,02π2αα<<∴<<，所以3sin 25α=，故A 正确；因为()cos αβ+=ππ0,0,0π22αβαβ<<<<∴<+<，所以()sin αβ+= 所以cos()cos[2()]cos2cos()sin 2sin()αβααβααβααβ-=-+=+++⎛⎛⎫=-⨯+= ⎪ ⎝⎭⎝⎭4355，故B正确； cos()cos cos sin sin αβαβαβ-=+=，cos()cos cos sin sin αβαβαβ+=-=②， 由+①②得，2cos co s αβ=，解得cos cos αβ=C 不正确； 由①-②得，2sin sin αβ=，解得sin sin αβ=sin sin tan tan 3cos c os αβαβαβ===，故D 不正确.故选：AB.11．202212a +能被7整除，则整数a 的值可以是（ ） A ．4 B ．6C ．11D ．13【答案】BD【分析】将202212a +化成2022(142)a -+，问题转化为20222a +能被7整除，再利用二项式定理计算推理作答.【详解】依题意202120222022202220222022012(142)[C 14(2)]2i i i i a a a -=+=-+=⨯⨯-++∑，显然2021202220220[C 14(2)]iii i -=⨯⨯-∑能被7整除，因此202212a +能被7整除，当且仅当20222a+能被7整除， 而()67367420226746746742871(C7)1k kk a a a a -=+=+=++=⨯++∑，又673674674(C 7)kk k -=⨯∑能被7整除，从而得1a +能被7整除，则整数a 的值可以是6或13. 故选：BD12．如图甲，在边长为2的正方形ABCD 中，点E ，F 分别是AB ，BC 的中点，点M 是AD 上靠近A 的四等分点. 现将△AED ，△DCF 分别沿DE ，DF 折起，使A ，C 两点重合于P ，连接PB ，如图乙所示，则下列说法正确的是（ ）A ．PB //平面EFM B ．PD PB ⊥C ．平面EFM 与平面BFDE 所成角的余弦值为63D ．点P 到平面BFDE 的距离为23【答案】ACD【分析】根据线面平行的判定可判断A ，证明PD ⊥平面PEF 可判断B ，由∠MGD 是二面角的平面角，利用余弦定理可判断C ，根据点P ，M 到面的距离的关系求解可判断D.【详解】如图，连接BD 与EF 相交于G ;3//4MD DG MG PB PD DB ==⇒⇒ PB //平面EFM ，故A 正确； 由PD ⊥PE ，PD ⊥PF ，PE∩PF=P 知PD ⊥平面PEF ，PB 平面PEF=P ，故PD 与PB 不垂直，故B 错误;平面EFM 与平面BFDE 所成角即为∠MGD ，在△MGD中，32GD MD ==，MG由余弦定理得222cos 2MG GD MD MGD MG GD +-∠==⋅故C 正确；由cos MGD ∠=知sin MGD ∠=MN BD ⊥，则点P 到平面BFDE 的距离等于点M 到平面BFDE 的距离的43，又442sin 333MN MG MGD =⋅∠=，故D 正确.故选：ACD 三、填空题13．设复数1z ，2z 是共轭复数，且12229i,-=-+z z ，则1z =___________.【分析】设复数1i z a b =+，进而得出共轭复数2z ，根据复数相等求出,a b ， 再利用复数的摸公式即可求解. 【详解】设1i z a b =+,则2i z a b =-，i (i)i i i z z a b a b a b a b a b -=+--=+-+=-+1222223因为12229i z z -=-+，所以i=i a b -+-+329即239a b -=-=⎧⎨⎩，解得2,3a b ==，所以1=2+3i z ，所以1z =14．函数13,0()32,02x x x f x x ---≤⎧⎪=⎨->⎪⎩的值域为___________.【答案】[3,)-+∞【分析】利用函数的单调性分别求出3y x =--在(,0]-∞上、1322x y -=-在(0,)+∞上函数值集合，再求并集作答.【详解】依题意，3y x =--在(,0]-∞上单调递减，则当0x ≤时，3y ≥-， 1322x y -=-在(0,)+∞上单调递增，则当0x >时，1y >-，所以函数13,0()32,02x x x f x x ---≤⎧⎪=⎨->⎪⎩的值域为[3,)-+∞.故答案为：[3,)-+∞15．已知P 为抛物线24y x =上任意一点，则点P 到y 轴的距离与点P 到直线:125130l x y -+=的距离之和的最小值为___________.【答案】1213【分析】将点P 到y 轴的距离与点P 到直线l 的距离之和转化为点P 到准线的距离与点P 到直线l 的距离之和，再借助抛物线定义求解作答.【详解】抛物线24y x =的焦点(1,0)F ，准线:1l x '=-，抛物线24y x =上的点P 到y 轴的距离等于它到准线l '距离d 减去1的差，由抛物线定义知，||d PF =，令点P 到直线:125130l x y -+=的距离为d '， 于是得点P 到y 轴的距离与点P 到直线:125130l x y -+=的距离之和为1||1d d PF d ''-+=+-，过P 作PM l ⊥于M ，连PF ，MF ，过点F 作FQ l ⊥于Q ，交抛物线24y x =于点P '，如图，显然，||PM d '=，当点P 与点P '不重合时，有：||1||||1||1||1||||1PF d PF PM MF FQ P F P Q '''+-=+->->-=+-，则当点P 是过焦点F 作直线l 的垂线与抛物线交点时，点P 到y 轴的距离与点P 到直线:125130l x y -+=的距离之和取得最小值，此最小值为2212||111312(5)FQ -==+-.故答案为：121316．设y R ∈，[x ]表示不超过x 的最大整数，设正项数列{n a }满足284n n n S a a n N +=+∈（），设数列{bn }的前n 项和为n T ，且n b ，则[35T ]=___________. 【答案】5【分析】由,n n a S 的关系可推出{n a }为等差数列，求出通项公式代入n b ，利用放缩法及裂项相消法可求出35T 的范围，即可得解.【详解】由284n n n S a a =+可得2111842)(n n n S a a n ---=+≥，两式相减得：2211844n n n n n a a a a a --=-+- (2)n ≥，化简得1114()()()n n n n n n a a a a a a ---+=-+，又由正项数列{n a }可知，10n n a a ->+， 所以14n n a a --= (2)n ≥，又211184S a a =+，解得14a =所以{n a }是以4为首项，4为公差的等差数列， 故44(1)4n a n n =+-=，n b ==== 3512351213243353435T b b b ∴=+++<+-+-+-++-=，又n b ===> 35123521324336353615T b b b ∴=+++>-+-+-++-=-=，355T ∴<<35[]5T ∴=.故答案为：5 四、解答题17．如图，在ABC 中，点D 在线段BC 上，且2BD DC =，2AD =.(1)若ABD △是正三角形，求AC 的长； (2)若310cos B =135ADC ∠=，求ABCS 的值.【答案】(1)7AC =(2)6ABCS=【分析】（1）利用余弦定理可求得AC 的长；（2）利用两角差的正弦公式可求得sin BAD ∠的值，利用正弦定理可求得BD 的长，可得出CD 的长，再利用三角形的面积公式及3ABC ACD S S =△△可求得结果. 【详解】(1)解：因为120ADC ∠=，112CD BD ==， 由余弦定理可得222cos 7AC AD CD AD CD ADC =+-⋅∠(2)解：因为310cos B =B 为锐角，则210sin 1cos B B =- 则())225sin sin 135cos sin BAD B B B ︒∠=-=+=由正弦定理sin sin BD ADBAD B=∠得252sin 542sin 10AD BAD BD B ∠===22CD =，因此，133sin13562ABC ACD S S AD CD ==⨯⋅⋅=△△.18．美国职业篮球联赛（NBA 联赛）分为常规赛和季后赛，常规赛共82场比赛，以全明星假期为界，分为前半赛季和后半赛季，东、西部排名前8的球队进入季后赛，季后赛共四轮，最后一轮总决赛采用七场四胜制（“七场四胜制”是指在七场比赛中先胜四场者获得比赛胜利，胜者成为本赛季的总冠军）.下表是A 队在常规赛的前80场比赛中的比赛结果记录表 阶段 比赛场数 主场场数 获胜场数 主场获胜场数 前半赛季 52 25 43 23 后半赛季28151712(1)根据表中信息完成列联表，并判断是否有95%的把握认为比赛的“主客场”与“胜负”之间有关？(2)已知A 队与B 队在季后赛的总决赛中相遇，假设每场比赛结果相互独立，A 队每场比赛获胜的概率等于A 队常规赛前80场比赛获胜的频率，求总决赛五场结束的概率. 附∶()()()()()22n ad bc K a b c d a c b d -=++++【答案】(1)见解析 (2)2164【分析】（1）根据题设信息填写列联表，计算2K 进行独立性检验；（2）总决赛五场结束意味着前4场中A 队（或B 队）胜3场败1场，第五场A 队（或B 队）获胜，再由概率公式计算即可. 【详解】(1)列联表如下表所示：2280(3515255)20 6.667 3.841602040403K ⨯-⨯==≈>⨯⨯⨯故有95%的把握认为比赛的“主客场”与“胜负”之间有关.(2)由题意可知，A 队常规赛前80场比赛获胜的频率603804=，即A 队每场比赛获胜的概率为34总决赛五场结束意味着前4场中A 队（或B 队）胜3场败1场，第五场A 队（或B 队）获胜.即总决赛五场结束的概率为3333443131312144444464C C ⎛⎫⎛⎫⋅⨯⨯+⋅⨯⨯= ⎪ ⎪⎝⎭⎝⎭.19．已知数列{}n a 的首项10a =，()134N n n a a n n *+=+∈.(1)证明：数列{}21n a n ++是等比数列； (2)求数列{}100n a -的前n 项和n S 的最小值. 【答案】(1)证明见解析 (2)304-【分析】（1）由已知等式变形得出()()1211321n n a n a n ++++=++，结合等比数列的定义可证得结论成立；（2）分析数列{}n b 的单调性，确定{}n b 的符号，由此可求得n S 的最小值.【详解】(1)解：因为()134N n n a a n n *+=+∈，则()()1211321n n a n a n ++++=++，且133a +=，所以，数列{}21n a n ++是以3为首项，3为公比的等比数列. (2)解：由（1）知，121333n n n a n -++=⋅=，则321n n a n =--.所以，10032101nn n b a n =-=--，所以，113322320n n nn n b b ++-=--=⋅->，故数列{}n b 为递增数列，1100b =-，296b =-，380b =-，428b =-，5132b =，，故当14n ≤≤时，0n b <；当5n ≥时，0n b >. 所以，n S 的最小值为4304S =-.20．如图所示，在正三棱台111ABC A B C -中，O 为BC 的中点，M 为11B C 的中点，平面ABC 与平面11BCC B 所成角的余弦值为13.(1)证明：平面ABC ⊥平面AOM ；(2)求直线1CC 与平面11ABB A 所成角的正弦值. 【答案】(1)证明见解析 6【分析】（1）延长111,,BB AA CC 相交于点P ，则可得三棱锥P ABC -为正三棱锥，可得延长OM 也交于点P ，由等腰三角形的性质可得PO BC ⊥，AO BC ⊥，由线面垂直的判定可得BC ⊥平面AOH ，再由面面垂直的判定可证得结论，（2）由（1）可知AOP ∠为平面ABC 与平面11BCC B 所成角，设2AB =，侧棱为a ，然后在POA 中利用余弦定理列方程可求出2a =，从而可得正三棱锥P ABC -为正四面体，过P 作1PO 垂直于平面ABC ，则1PBO ∠为PB 与平面ABC 所成角，然后计算即可 【详解】(1)延长111,,BB AA CC 相交于点P ， 因为棱台111ABC A B C -为正三棱台， 所以三棱锥P ABC -为正三棱锥， 所以延长OM 也交于点P ，所以PBC 为等腰三角形，O 为BC 的中点， 所以PO BC ⊥，因为ABC 为正三角形，O 为BC 的中点， 所以AO BC ⊥，因为AO PO O =，所以BC ⊥平面POA ，即BC ⊥平面AOH ， 因为BC ⊂平面ABC ，所以平面ABC ⊥平面AOM ； (2)如图，,PO BC AO BC ⊥⊥，所以AOP ∠为平面ABC 与平面PBC 所成角，即平面ABC 与平面11BCC B 所成角，设2AB =，侧棱为a ，所以3AO =，21PO a =-，所以在POA 中，222222131cos 23213PO OA PA a a POA PO OA a +--+-∠===⋅-⋅，解得2a =， 所以正三棱锥P ABC -为正四面体，所以1CC 与平面11ABB A 所成角，即为PC 与平面PBA 所成角， 即PB 与平面ABC 所成角，过P 作1PO 垂直于平面ABC ，则1O 为正ABC 的中心， 连接1O B ，则1PBO ∠为PB 与平面ABC 所成角， 因为122,33PB O B ==，所以22111226493PO PB O B =-=-=, 所以112663sin 23PO PBO PB ∠===， 所以直线1CC 与平面11ABB A 所成角的正弦值为6321．已知椭圆C ∶22221(0)x y a b a b+=>>经过点P 332），O 为坐标原点，若直线l与椭圆C 交于A ，B 两点，线段AB 的中点为M ，直线l 与直线OM 的斜率乘积为-14.(1)求椭圆C 的标准方程；(2)若3OM =AOB 面积的最大值. 【答案】(1)221123x y += (2)3【分析】（1）根据椭圆经过点P 332），得到223914a b+=，再利用点差法，根据直线l 与直线OM 的斜率乘积为-14，得到 2214b a -=-求解；（2）当AB x ⊥轴时，易得12AOBSOM AB =⋅AB 与x 轴不垂直时，设直线AB 的方程为y kx t =+，联立221123x y y kx t ⎧+=⎪⎨⎪=+⎩，根据OM =k ，t 的关系，再求得AB 和点O 到直线AB 的距离为d ，由12AOB S AB d =⋅⋅求解.【详解】(1)解：因为椭圆经过点P32）， 所以223914a b +=， 设()()1122,,,A x y B x y ，因为直线l 与椭圆C 交于A ，B 两点，所以22112222222211x y a b x y ab ⎧+=⎪⎪⎨⎪+=⎪⎩，两式相减得2121221212y y x x b x x a y y -+=-⋅-+， 因为线段AB 的中点为M ，且直线l 与直线OM 的斜率乘积为-14，所以 2214b a -=-，解得223,12b a ==，所以椭圆方程为：221123x y +=； (2)当AB x ⊥轴时，点M 在x 轴上，且OM AB ⊥，由OM =3AB =，所以12AOBSOM AB =⋅ 当直线AB 与x 轴不垂直时，设直线AB 的方程为y kx t =+，由221123x y y kx t ⎧+=⎪⎨⎪=+⎩，消去y 得()2221484120k x ktx t +++-=， 则21212228412,1414kt t x x x x k k -+=-⋅=++，224,1414kt t M k k ⎛⎫- ⎪++⎝⎭，由OM =()2222314116k t k +=+，因为AB =，点O 到直线AB 的距离为d所以12AOBSAB d =⋅⋅=3≤=，当且仅当221214k k =+，即218k =时，等号成立，综上 AOB 面积的最大值是3.22．已知函数()()ln e x x f x x g x x+==421，（e为自然对数的底数）. (1)求()f x 的极值；(2)（i ）证明∶()e ln xm x x x =+342与()e x n x x=-21有相同的零点； （ii ）若()()()Z f x g x x ax a -->∈210恒成立，求整数a 的最大值. 【答案】(1)()f x 的极小值为e f ⎛⎫-=- ⎪⎝⎭1144，无极大值；(2)（i ）证明见解析，（ii ）1【分析】（1）根据导数法求函数的极值的步骤即可求解；（2）（i ）首先利用导数法判断函数()m x ，()n x 的单调性，然后利用 函数零点的存在性定理即可证明；（ii ）先将不等式()()()Z f x g x x ax a -->∈210恒成立等价转化为 min ln e x x a x +⎛⎫+<- ⎪⎝⎭42211012在()0,+∞恒成立，然后利用导数法求函数()ln e x x h x x +=-42212的最小值即可，进而能够求出整数a 的最大值. 【详解】(1)由题意可知，()()e xf x x '=+441，令0fx，即()e x x +=4410，解得14x =-；当14x >-时，0fx，所以()f x 在,⎛⎫-+∞ ⎪⎝⎭14单调的递增；当14x <-时，0f x ，所以()f x 在,⎛⎫-∞- ⎪⎝⎭14单调的递减；当14x =-时，()f x 取得极小值为e f ⎛⎫-=- ⎪⎝⎭1144，无极大值；(2)（i ）由()e ln xm x x x =+342知()()e x m x x x x'=++>3241860，所以()m x 在()0,+∞上单调递增；由()e xn x x =-21知()()e xn x x x'=+>>221200，所以()n x 在()0,+∞上单调递增；又,e n n ⎛⎫⎛⎫=<=-> ⎪ ⎪⎝⎭⎝⎭11402042，故必存在唯一,x ⎛⎫∈ ⎪⎝⎭01142使得()00n x =，即有e ,ln ln x x x x x ===-020000112， 故()eln x m x x x x x x ⎛⎫=+=⋅-= ⎪⎝⎭243300000012220，所以()e ln x m x x x =+342与()e xn x x=-21有相同的唯一零点0x x =； （ii ）由()()f x g x x ax -->210，得()()ln e x f x g x x a x x -++<=-422211012恒成立，minln e x x a x +⎛⎫+<- ⎪⎝⎭42211012在()0,+∞恒成立，令()ln e xx h x x +=-42212，()0,x ∞∈+，则 ()ln ()e e ln x x x x h x x x x x'=+=+434434482， 由（i ）知()()e xm x x x x'=++>3241860单调递增且()e ln x m x x x =+342存在唯一零点0x x =；则当()00,x x ∈时，()0h x '<，()h x 单调递减； 当()0,x x ∈+∞时，()0h x '>，()h x 单调递增； 故()()min h x h x =0； 由（i ）知e,ln ln x x x x x ===-020000112；又e ,e n n ⎛⎫⎛⎫=-<=-> ⎪ ⎪⎝⎭⎝⎭2311302032， 故进一步确定011,32x ⎛⎫∈ ⎪⎝⎭；故()min ln ()e (,)x x x h x h x x x x x x +-+==-=-=+∈04000222200000214121421221， 即a +≤10112，解得.a ≤11，又Z a ∈； 所以整数a 的最大值为1.【点睛】求解不等式问题的关键：适当变形，灵活转化，结合题设条件，有时需要对不等式进行“除法”变形， 从而分离参数，有时需要进行移项变形，可使不等式两边具有相同的结构特点； 构造函数，利用导数求解，若分离参数，则直接构造函数，并借助导数加以 求解，若转化为不等式两边具有相同的结构特点，则可根据该结构特点构造函数，并借助导数加以求解.。

重庆市第八中学校2024届高三下学期高考适应性月考（五）英语试卷学校：___________姓名：___________班级：___________考号：___________一、阅读理解DAY TRIP ITINERARY (行程)After a warm reception from your tour manager at your dedicated pick-up point in London, settle in your comfortable coach as we set off on our journey towards Stonehenge. The world's most famous prehistoric monument has inspired people to study and interpret it for centuries, yet many questions remain to be answered-about who built it, when, and why.After visiting Stonehenge, we drive to Windsor Castle, which is home to royalty and 1,000 years of royal history. The stunning 13-acre site is the largest and oldest occupied, working castle in the world. There are many famous, must-see moments within these spectacular rooms in the castle, like the grand Waterloo Chamber and the magnificent Crimson Drawing Room.In the early evening, we make our way towards London and proceed towards your respective drop off points and bid farewell to all friends you have made on the tour. ESSENTIALS TO CARRY WITH YOUWe recommend you wear comfortable clothing and carry essentials such as a jacket or jumper, dry snacks, water, tissues, chargers, power bank, ete., in your handbag as access to the luggage hold is only possible until a comfort stop or arrival at your destination. Hot foods are not allowed to be carried or consumed inside the coach.PICK-UP POINTSand travelling.1．How many tourist spots will the participants visit?A.Two.B.Three.C.Four.D.Five.2．It is advised to put your essentials in the handbag because _______.A.the luggage hold is inaccessible when the bus is in motionB.they are not allowed to be carried inside the coachC.it's convenient for you to enjoy hot foodsD.drivers are likely to access them3．Where is the text most probably taken from?A.A journal.B.A webpage.C.A travel brochure.D.A magazine.It all started with a simple question："Can I paint your portrait? "One day in the summer of 2015, Peterson was relaxing in his living room, reading the book Love Does, about the power of love in action, when his quiet was disturbed by a homeless man outside his apartment. Inspired by the book's compassionate message, Peterson made a decision：He was going to go outside and introduce himself.In that first conversation, Peterson learned that the man's name was Matt Faris. He'd moved to Southern California from Kentucky to pursue a career in music, but he soon fell on hard times and ended up living on the street for more than a decade. "I saw beauty on the face of a man who hadn't shaved in probably a year, because his story, the life inside of him, inspired me," Peterson recalled. Therefore, Peterson asked if he could paint Faris's portrait. Faris agreed.Peterson's connection with Faris led him to form Faces of Santa Ana, a nonprofit organization focused on befriending and painting portraits of members of the community who are unhoused. Peterson sells the paintings for money, splitting the proceeds and putting half into a "love account" for his model. He then helps people use the money to get back on their feet.Many of Peterson's new friends use the donations to secure immediate necessities medical care, hotel rooms, food. Faris used the funds from his portrait to record an album, fulfilling his musical dreams. Another subject, Kimberly Sondoval, had never been able to financially support her daughter. She asked, "Can I use the money to pay my daughter's rent? " When the check was delivered, "They both wept in my arms, "Peterson recalls.Peterson has painted 41 of these portraits himself. But there's more to the finished products than the money they bring to someone who's down and out. He's discovered that the buyers tend to connect to the story of the person in the painting, finding similarities and often friendship with someone they might have otherwise overlooked or stereotyped."People often tell me, 'I was the one that would cross the street, but I see homeless people differently now, ' " Peterson says. "I didn't know that would happen."4．What made Peterson start a conversation?A.The curiosity about strangers.B.The touching story of Matt Faris.C.The disturbance by a homeless man.D.The sympathetic message in Love Does.5．What do we know about Faces of Santa Ana?A.It pays the homeless salaries.B.It is an official nonprofit organization.C.It tries to restore the lives of Peterson's models.D.It spends all the money on helping the unhoused.6．After buying a portrait, a buyer might view the homeless as ______?zy and poorB.odd but inspiringC.disturbing and untidyD.pitiful but respectable7．Which of the following is the best title of this text?A.Art with Purpose: Love Account MattersB.Hope in Paintings: Help Knows No RaceC.Faces of Compassion: Painting a New PerspectiveD.Encounter with Strangers: Embracing New FriendsFilm Puts Justifiable Defense in SpotlightThe film, Article 20, directed by Zhang Yimou, draws its name from Article 20 of the Criminal Law, which focuses on the sometimes controversial legal concept of justifiable defense. Drawing inspiration from real-life cases of justifiable defense, the film gained widespread attention and struck a chord with the public during the Spring Festival holiday. The movie calls for a detailed interpretation of legal terms and urges against compromising on unlawful conduct.According to the Criminal Law, when a person, faced with an unlawful attack, takes action to protect his or her own rights or interests or those of others, and the attacker is thereby harmed, the defender will be considered to have acted in justifiable defense and will not bear criminal responsibility. For some time, justifiable defense has been regarded as a "dormant clause" (沉寂条款) , due to the influence of misconceptions, such as "whoever is injured or killed is right".But the true meaning of the law should be to increase the responsibility for wrongdoers, rather than burdening good people. Recent cases have shown that, for ordinary citizens, justifiable defense is no longer a pipe dream. A high-profile case in Kunshan, Jiangsu province, in 2018 served as a wake-up call and caught the attention of authorities regarding such cases. In that case, a traffic argument led to a motorist who took a knife with him confronting another man.The motorist was killed, and police and judges determined that the defender's actions constituted justifiable defense. Since then, the justifiable defense clause has been applied in several places across the nation. The concern over justifiable defense reflects the public's demand for fairness, justice, security and rule of law.Luo Xiang, a renowned professor of criminal law at China University of Political Science and Law, said in a recent comment about the film Article20 that the public and judges should avoid taking a "godlike" perspective. Instead, they should consider the situation in which the defender was involved, empathize with the defender's position, and stop themselves from making excessive demands on the defender, Luo said.8．Why did the film Article 20 attract the audience's attention?A.It was released during the Spring Festival holiday.B.It is named after one article in the Criminal Law.C.It explores real-life cases of justifiable defense.D.It was directed by Zhang Yimou.9．According to the Criminal Law, the victim will be free from criminal responsibility when ______.A.the victim gave up his legal rightsB.the robber kept silent about the robberyC.the robber was hurt worse than the victimD.the victim knifed the armed robber to stop the ongoing crime10．What is the function of the case in Kunshan in Paragraph 3?A.Making comparisons.B.Listing reasons.C.Explaining misconceptions.D.Providing evidence.11．What does Luo Xiang suggest judges do?A.Take a “godlike” viewpoint with the public.B.Put oneself in the defender's shoes.C.Demand more from the defender.D.Side with the attacker.Researchers have discovered the oldest black hole ever observed, dating from the dawn of the universe, and found that it is "eating" its host galaxy to death.The international team, led by the University of Cambridge, used the James Webb Space Telescope (JWST) to detect the black hole, which dates from 400 million years after the big bang. The results, which lead author Professor Roberto Maiolino says are "a giant leap forward", are reported in the journal Nature.This surprisingly massive black hole even exists so early in the universe challenges our assumptions about how black holes form and grow. The size of this newly-discovered black hole suggests that they might form in other ways: they might be 'born big' or they can eat matter at a rate that's five times higher than had been thought possible. Like all black holes,According to standard models, supermassive black holes form from the remains of dead stars, which collapse and may form a black hole about a hundred times the mass of the Sun. If it grew in an expected way, this newly-detected black hole would take about a billion years to grow to its observed size. However, the universe was not yet a billion years old when this black hole was detected.The young host galaxy, called GN-z11, is a compact galaxy, about one hundred times smaller than the Milky Way, but the black hole is likely harming its development. When black holes consume too much gas, it pushes the gas away like an ultra-fast wind. This "wind" could stop the process of star formation, slowly killing the galaxy, but it will also kill the black hole itself, as it would cut off the black hole's source of "food".Maiolino says that the gigantic leap forward provided by JWST makes this the most exciting time in his career. "It's a new era: the giant leap is like upgrading from Galileo's telescope to a modern telescope overnight," he said. "The universe has been quite generous in what it's showing us, and this is just the beginning."12．What does the underlined word "devours" mean in Paragraph 3?A.Changes.B.Swallows.C.Observes.D.Forms.13．According to Paragraph 5, why is GN-z11 likely to be harmed?A.Because the host galaxy is too small.B.Because the black hole is killing itself.C.Because the “wind” ceases star formation.D.Because black holes produce too much gas.14．What is Maiolino's attitude to the new discovery about the black hole?A.Favorable.B.Intolerant.C.Doubtful.D.Ambiguous.15．What can we learn from this passage?A.The black hole took a billion years to achieve its size.B.Supermassive black holes are assumed to form overnight.C.The new discovery of the host galaxy is a giant leap forward.D.The host galaxy and the black hole can be destroyed by the "wind".二、七选五16．The Failed New Year's Resolution: Three Tips to Get on Track January is officially over, and many people are taking stock of their progress towards New Year's resolutions. The fact is that you probably haven't kept up with them as much as you hoped. But that's not your fault. ①______.If you feel like you have already failed, here are three tips before you let go.Practice self-compassionMany people talk to themselves in harsh ways when struggling with new habits, believing self-criticism will help them reach their goals. Research shows, however, that the opposite is true. Self-compassion is more effective for personal improvement, especially when facing failure. ②______, try to be kind and gentle with yourself, just as you would with a loved one.③______Resolutions are often phrased as definitive goals. I will exercise daily. I will kick desserts. ④______. But setting all-or-nothing goals can lead to all-or-nothing decisions that one gives up when faced with challenges. In contrast, intentions focus more on your values than specific actions. For example, the resolution "I will exercise daily" may become an intention of "I want to move my body because it feels good." This approach allows for more flexibility when unexpected stress arises.Solve problems by overcoming barriersIf you are struggling to maintain your desired habits, there are evidence-based techniques available to help you. ⑤______. This involves identifying the specific barriers that lead to your quitting something that you want to do. Perhaps you keep forgetting the new habit, or perhaps you don't understand how to do it. Whatever it is, identify the barrier and cope with it specifically.A. Set all-or-nothing goalsB. Change your resolutions into intentionsC. One such skill is called missing links analysisD. Setting specific behavioral goals can be effectiveE. If you're persuaded to give up on your resolutionsF. Old habits tend to die hard, and new habits tend to die easyG. When you are upset about yourself for not keeping resolutions三、完形填空（15空）My son asked me months ago if he could switch from his mainstream high school to aanother learning environment to complete his study. But my heart hurt a little that he wouldThen, we decided the vocational school (职业学校) was an option with the same generalcarry over, and he could graduate early.17．pulsory B.technical C.unprofessional D.academic18．A.heartbroken B.nervous C.angry D.confused19．A.apply B.adapt C.transfer D.talk20．A.ceremony B.moment C.dilemma D.stage21．A.curriculum B.wellbeing C.friendship D.decision22．A.bald B.generous C.straightforward D.shy23．A.anxiety B.conduct C.responsibility D.awareness24．A.studied B.canceled C.required D.banned25．A.credits B.subjects C.medals D.reputations26．A.graduate B.worsen C.thrive D.suffer27．A.formed B.understood C.abandoned D.remembered28．A.got away B.made up C.lined up D.got along29．A.crying B.murmuring plaining D.smiling30．A.oversell B.learn C.share D.predict31．A.affordable B.suitable C.available D.sensible四、短文填空32．While there is growing consumer ①______ (realize) of the environmental impact of fast fashion, can the same be said about fast furniture? The chairs and tables that fill many of our homes and everyday spaces are manufactured on a mass scale, and the cheaper items often end up ②______ landfills.According to the Environmental Protection Agency (EPA) , in 2020 over 12 million tons of furniture ③______ (throw) out in America and some into the forest illegally. Buying furniture can be ④______ (incredible) expensive. Many of us switch over to cheaper, instant brands like IKEA, ⑤______ manufacturers use materials harder to recycle, which is likely to damage the environment.With growing calls for sustainability, many brands are announcing to change though it remains ⑥______ (see) whether they will keep these promises. In ⑦______ (it) current sustainability strategy, IKEA commits itself to using only recyclable materials in all its products in an effort to practice "circular" design and cut emissions to net-zero. The concept of circular design ⑧______ (win) increasing concern up to now. In ⑨______ circular system, furniture products would be designed to last longer and be fully recyclable, thus forming a ⑩______ (close) loop (环) .五、书面表达33．假定你是高三学生李华。