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河北省衡水市2019-2020学年七年级上学期期中数学试题(教师版)

河北省衡水市2019-2020学年七年级上学期期中数学试题(教师版)

河北省衡水市2019-2020学年七年级上学期期中数学试题一、选择题1.中国是世界上最早认识和应用负数的国家,比西方早一千多年.在我国古代著名的数学专著《九章算术》中,首次引入负数,如果收入100元记作+100元,则﹣80元表示( )A. 支出20元B. 收入20元C. 支出80元D. 收入80元【答案】C【解析】【分析】在一对具有相反意义的量中,先规定其中一个为正,则另一个就用负表示.【详解】解:根据题意,收入100元记作+100元,则-80表示支出80元.故选:C . 【点睛】本题考查了正数和负数,解题关键是理解“正”和“负”的相对性,确定一对具有相反意义的量.2.多项式2332x y x -的次数为( )A. 8B. 5C. 3D. 2 【答案】B【解析】【分析】 利用多项式中次数最高的项的次数叫做多项式的次数,进而得出答案.【详解】解:多项式2332x y x -的次数为:5.故选:B . 【点睛】此题主要考查了多项式,正确把握多项式的次数确定方法是解题关键.3.若||a a =-,则有理数a 可以是( ) A. 113 B. -1 C. 0.345 D. 9【答案】B【解析】【分析】根据负数的绝对值是它的相反数进行解答即可.【详解】解:∵||a a =-,∴0a <,故选:B.【点睛】此题考查绝对值,关键是根据绝对值的非负性进行解答.4.与﹣(a ﹣b )相等的式子是( )A. ﹣a+bB. ﹣a ﹣bC. a ﹣bD. ﹣(b ﹣a ) 【答案】A【解析】【分析】根据去括号的法则进行解答.【详解】解:﹣(a ﹣b )=﹣a+b .故选:A .【点睛】本题考查去括号的方法:去括号时,运用乘法的分配律,先把括号前的数字与括号里各项相乘,再运用括号前是“+”,去括号后,括号里的各项都不改变符号;括号前是“﹣”,去括号后,括号里的各项都改变符号.顺序为先大后小.5.3+3+3+3+3+3+3+3+3可以表示为( )A. 93B. 39C. 33D. 99 【答案】C【解析】【分析】根据有理数的乘方的定义解答即可.【详解】解:3333333333393333++++++++=⨯=⨯⨯=.故选:C .【点睛】本题考查了有理数的乘方,解题的关键是理解乘方的含义.6.2019年4月10日,人类首次公布了拍摄到的黑洞照片,这个黑洞位于代号为M87的星系中,距离地球大约55000000光年.数据55000000科学记数法表示为( )A. 75.510⨯B. 75510⨯C. 65510⨯D. 85.510⨯ 【答案】A【解析】【分析】科学记数法的表示形式为a×10n 的形式,其中1≤|a|<10,n 为整数.确定n 的值时,要看把原数变成a 时,小数点移动了多少位,n 的绝对值与小数点移动的位数相同.当原数绝对值>10时,n 是正数;当原数的绝对值<1时,n 是负数.【详解】解:将55000000科学记数法表示为:75.510 .故选:A .【点睛】此题考查科学记数法的表示方法.科学记数法的表示形式为a×10n 的形式,其中1≤|a|<10,n 为整数,表示时关键要正确确定a 的值以及n 的值.7.-2的相反数可以是( )A. +(-2)B. -(+2)C. -12D. -(-2) 【答案】D【解析】【分析】根据求一个数的相反数就是在这个数前面添上“-”号,即可得出答案.【详解】解:-2的相反数是2,可以是-(-2);故选:D .【点睛】本题考查了相反数的意义,一个数的相反数就是在这个数前面添上“-”号.一个正数的相反数是负数,一个负数的相反数是正数,0的相反数是0.学生易把相反数的意义与倒数的意义混淆.8.单项式4x 2y 的系数与一个单项式的次数相同,则这个单项式可以是( )A. 24xyB. 4xyC. 33xyD. 23x y 【答案】C【解析】分析】直接利用单项式的次数与系数确定方法分析得出答案.【详解】解:∵单项式4x 2y 的系数是4,与一个单项式的次数相同,A. 24xy ,次数是3;B. 4xy ,次数是2;C. 33xy ,次数是4;D. 23x y ,次数是3;∴这个单项式可以是3xy 3.故选:C .【点睛】此题主要考查了单项式,正确把握单项式的次数与系数确定方法是解题关键.9.下列四个单项式中,①214x y -;②214a b -;③24y x ⋅;④214xy -;是同类项的是( ) A. ①②B. ①③C. ①④D. ③④ 【答案】B【解析】【分析】所含字母相同,并且相同字母的指数也相同,这样的项叫做同类项,由此进行判断即可. 【详解】解:四个单项式中,①214x y -;②214a b -;③24y x ⋅;④214xy -;只有214x y -与24y x ⋅是同类项,即是同类项的是①③.故选:B .【点睛】本题主要考查了同类项,解题的关键是熟记同类项的定义.10.阅读下列解题过程: 22427382x x x x +++--=22482372x x x x -+++- (1)=()2248(23)(72)x x x x -+++- (2)=2(48)(23)(72)x x -⋅++⋅+- (3)=2455x x -++ (4)上面这个题计算过程中,(3)的依据是( )A. 有理数加法法则B. 加法交换律C. 加法结合律D. 分配律. 【答案】D【解析】【分析】利用乘法分配律判断即可.【详解】解:在计算()2248(23)(72)x x x x -+++- (2)=2(48)(23)(72)x x -⋅++⋅+- (3)运用了乘法分配律,故选:D .【点睛】此题考查了有理数的混合运算,熟练掌握运算法则是解本题的关键.11.所有整数组成整数集合,所有负数组成负数集合,阴影部分也表示一个集合,则这个集合可以包含的有理数为( )A. 3B. -2019C. 227D. 0【答案】B【解析】【分析】 直接利用负整数的定义分析得出答案.详解】解:阴影部分表示负整数,选项中只有-2019符合题意.故选:B .【点睛】此题主要考查了有理数,正确把握相关定义是解题关键.12.若(2018)63p -⨯=,则(2018)62-⨯的值可表示为( ),A. p-1B. p+2018C. p-2018D. 6263p 【答案】B【解析】【分析】直接利用已知将原式变形进而得出答案.【详解】解:∵(2018)63p -⨯=,∴()()()()()201862201863120186320182018p -⨯=-⨯-=-⨯--=+.故选:B .【点睛】此题主要考查了有理数的减法和乘法,正确将原式变形是解题关键.13.对于|1|3a --及2(3)2b -++,佳佳和音音提出了两个观点:佳佳的观点: |1|3a --有最小值,最小值为3.音音的观点:2(3)2b -++有最大值,最大值为2.对于以上观点,则( )A. 佳佳和音音均正确B. 佳佳正确,音音不正确C. 佳佳不正确.音音正确D. 佳佳和音音均不正确【答案】C【解析】【分析】根据有理数的平方、绝对值的定义解答即可.【详解】解:因为|a-1|≥0,所以|a-1|-3有最小值,最小值为-3;因为(b+3)2≥0,所以-(b+3)2≤0,所以-(b+3)2+2有最大值,最大值为2,所以佳佳不正确,音音正确,故选:C .【点睛】本题考查了非负数的性质,明确有理数的平方、绝对值是非负数是解题的关键.14.如图,将四张边长各不相同的正方形纸片按如图方式放入矩形ABCD 内(相邻纸片之间互不重叠也无缝隙),未被四张正方形纸片覆盖的部分用阴影表示,设右上角与左下角阴影部分的周长的差为l .若知道l 的值,则不需要测量就能知道周长的正方形的标号为( )A. ①B. ②C. ③D. ④【答案】D【解析】【分析】 设①、②、③、④四个正方形的边长分别为a 、b 、c 、d ,用a 、b 、c 、d 表示出右上角、左下角阴影部分的周长,利用整式的加减混合运算法则计算,得到答案.【详解】设①、②、③、④四个正方形的边长分别为a 、b 、c 、d ,由题意得,(a+d-b-c+b+a+d-b+b-c+c+c )-(a-d+a-d+d+d )=l ,整理得,2d=l ,则知道l 的值,则不需测量就能知道正方形④的周长,故选D .【点睛】本题考查的是整式加减运算的应用,根据图形正确表示出右上角、左下角阴影部分的周长是解题的关键.二、填空题15.-0.1的倒数为_____.【答案】-10【解析】【分析】先将-0.1转化成分数,然后根据倒数的定义解答即可. 【详解】解:因为10.110-=-, 所以-0.1的倒数为-10.故答案为:-10.【点睛】本题考查了倒数,解题的关键是熟练掌握倒数的定义.16.若一支圆珠笔的笔芯的价格为0.9元,买一些笔芯需付款0.9x 元,则x 表示的实际意义是______.【答案】这些笔芯共x 支【解析】【分析】根据题意解答即可.【详解】解:一支圆珠笔的笔芯的价格为0.9元,买一些笔芯需付款0.9x 元,则x 表示的是笔的支数,即:x表示的实际意义是:这些笔芯共x支【点睛】本题考查了对未知数的含义的理解.17.若a+b=2019,c+d=-5,则代数式(a-2c)-(2d-b)=______.【答案】2029【解析】【分析】根据去括号、添括号法则把原式变形,代入计算,得到答案.【详解】(a-2c)-(2d-b)=a-2c-2d+b=(a+b)-2(c+d)=2019+10=2029,故答案为:2029.【点睛】本题考查的是整式的加减混合运算,掌握去括号、添括号法则是解题的关键.注意整体思想的应用.18.某校园学子餐厅把WIFI密码做成了数学题,小亮在餐厅就餐时,思索了一会,输入密码,顺利地连接到了学子餐厅的网络,那么他输入的密码是______.【答案】143549【解析】【分析】根据题中密码规律确定所求即可.【详解】5⊗3⊗2=5×3×10000+5×2×100+5×(2+3)=1510259⊗2⊗4=9×2×10000+9×4×100+9×(2+4)=183654,8⊗6⊗3=8×6×10000+8×3×100+8×(3+6)=482472,∴7⊗2⊗5=7×2×10000+7×5×100+7×(2+5)=143549.故答案为:143549【点睛】本题考查有理数的混合运算,根据题意得出规律并熟练掌握运算法则是解题关键.三、解答题19.比较大小:78-和89-【答案】78->89-【解析】【分析】根据负数的大小比较法则:两个负数比较大小,绝对值大的反而小,进行比较即可.【详解】解:因为7788-=,8899-=所以763648 =872729<=所以78->89-【点睛】本题考查了实数的大小比较法则和有理数的大小比较法则的应用,能熟记法则的内容是解此题的关键.20.蜜蜂是自然界神奇的“建筑师“,它能用最少的材料造成最牢固的建筑物“蜂窝“,观察下列的“蜂窝图(1)若“”中每条边看成1个建筑单位,则第1个图形中共有19个建筑单位,第2个图案中共有_____个建筑单位;第3个图案中共有_____个建筑单位.(2)第n个图案中共有多少个建筑单位.【答案】(1)30,41;(2)8+11n.【解析】【分析】(1)根据题目中的图形,可以发现建筑单位的变化规律,从而可以解答本题;(2)根据发现的建筑单位的变化规律,可以写出第n个图案中共有多少个建筑单位.【详解】解:(1)第1个图形中有:8+11=19个建筑单位,第2个图形中有:8+11×2=30个建筑单位,第3个图形中有:8+11×3=41个建筑单位,第4个图形中有:8+11×4=52个建筑单位,故答案为:30,41;(2)由(1)可知:第n个图形中共有:(8+11n)个建筑单位.【点睛】本题考查图形的变化类,解答本题的关键是明确题意,发现题目中建筑单位的变化规律,利用数形结合的思想解答.21.计算:221411 (3)264332⎛⎫⎛⎫--⨯-+-⨯⎪ ⎪⎝⎭⎝⎭【答案】4【解析】【分析】根据有理数的运算法则计算即可. 【详解】解:()2214113264332⎛⎫⎛⎫--⨯-+-⨯ ⎪ ⎪⎝⎭⎝⎭916119+664932=-⨯⨯-⨯9423=-+-4=【点睛】本题考查了有理数的混合计算,熟悉运算法则是解题的关键.22.已知在数轴上原点处有一点A,将点A先向左移动3个单位长度,再向右移动5个单位长度.(1)移动后点A在数轴上所表示的数为;(2)若数轴上有一点B与移动后点A相距4个单位长度,求点B表示的数;(3)在(2)的条件下,若将点B移动3个单位长度后与点C重合,求点C所表示的数.【答案】(1)2;(2)﹣2或6;(3)点C所表示的数±1,±3,±5,7,9.【解析】【分析】(1)根据有理数的加法运算进行计算即可;(2)分两种情况分别计算,一是点B在A的左侧,二是点B在A的右侧;(3)由(2)得B有两种可能,而每种中又有四种情况,因此点C有8种情况,对应8种结果.【详解】(1)0﹣3+5=2.故答案为:2.(2)2﹣4=﹣2或2+4=6.答:点B表示的数为﹣2或6.(3)①右移3个单位长度:-2+3=1,6+3=9;②左移1个单位长度,右移2个单位长度:-2-1+2=-1,6-1+2=7;③左移2个单位长度,右移1个单位长度:-2-2+1=-3,6-2+1=5;④左移3个单位长度:-2-3=-5,6-3=3.综上所述:点C所表示的数为±1,±3,±5,7,9.答:点C所表示的数±1,±3,±5,7,9.【点睛】本题考查了有理数、绝对值的意义,分类讨论是常用的方法.23.老师设计了一个数学实验,给甲、乙、丙三名同学各一张写有已化为最简(没有同类项)的代数式的卡片,规则是两位同学的代数式相减等于第三位同学的代数式,则实验成功,甲、乙、丙的卡片如下,丙的卡片有一部分看不清楚了.(1)计算出甲减乙的结果,并判断甲减乙能否使实验成功;(2)嘉琪发现丙减甲可以使实验成功,请求出丙的代数式.【答案】(1)甲减乙不能是实验成功;(2)3x2-5x+2.【解析】【分析】(1)根据题意列出关系式,去括号合并后即可作出判断;(2)根据题意列出关系式,去括号合并即可确定出丙.【详解】解:(1)根据题意得:()()2223123x x x x ----+ 2223123x x x x =--+--24x x =--,由于丙卡片的常数项为2,结果与题意不符,因此甲减乙,实验不成功;(2)根据题意得:丙表示的代数式为甲加乙.即:()()2223123x x x x -+-+- 2352x x -=+.【点睛】此题考查了整式的加减,熟练掌握运算法则是解本题的关键.24.育才中学为提高学生的身体素质,经常在课间开展学生跳绳比赛,下表为该校6(1)班50名学生参加某次跳绳比赛的情况,规定标准数量为每人每分钟100个.(1)6(1)班50人中跳绳最多的同学一分钟跳的次数是多少个,跳绳最少的同学一分钟跳的次数是多少个;(2)跳绳比赛的计分方式如下:①若每分钟跳绳个数是规定标准数量,不计分;②若每分钟跳绳个数超过规定标准数量,每多跳1个绳加2分;③若每分钟跳绳个数没有达到规定标准数量,每少跳1个绳扣1分.如果班级跳绳总积分超过200分,便可得到学校的奖励,请你通过计算说明6(1)班能否得到学校奖励?【答案】(1)106个;98个;(2)能得到学校奖励,理由见解析;【解析】【分析】见解析.【详解】(1)6(1)班50人中跳绳最多的同学一分钟跳的次数是:100+6=106(个),跳绳最少的同学一分钟跳的次数是:100﹣2=98(个).答:6(1)班50人中跳绳最多的同学一分钟跳的次数是106个,跳绳最少的同学一分钟跳的次数是98个;(2)依题意得:(4×6+5×11+6×8)×2﹣(﹣2×6﹣1×12)×(﹣1)=230>200.所以6(1)班能得到学校奖励.【点睛】理解正负数的意义是解题的关键.25.桐城市发起了“保护龙眠河”行动,某学校七年级两个班的115名学生积极参与,踊跃捐款,已知甲班有13的学生每人捐了10元,乙班有的学生每人捐了10元,两个班其余学生每人捐了5元,设甲班有学生x人。

2019-2020学年河北省衡水中学实验学校七年级(上)期中数学试卷 (有答案)

2019-2020学年河北省衡水中学实验学校七年级(上)期中数学试卷 (有答案)

2019-2020学年河北省衡水中学实验学校七年级(上)期中数学试卷一、选择题(本大题共16小题,共42.0分)1.若()×12=−1,则括号内应填的数是()A. 2B. −2C. 12D. −122.一个数的平方是16,则这个数的3次方是()A. 48B. 64C. −64D. 64或−643.下列图形中,属于立体图形的是()A. B. C. D.4.有理数(−1)2,(−1)3,−12,|−1|,−(−1),−1−1中,等于1的有()个.A. 3B. 4C. 5D. 65.下列说法:①0是绝对值最小的有理数;②相反数大于自身的数是负数;③数轴上原点两侧的数互为相反数;④两个数相互比较绝对值大的反而小.其中正确的是()A. ①②B. ①③C. ①②③D. ②③④6.下列说法中不正确的是()①过两点有且只有一条直线②连接两点的线段叫两点的距离③两点之间线段最短④点B在线段AC上,如果AB=BC,则点B是线段AC的中点A. ①B. ②C. ③D. ④7.如图,下列说法正确的是()A. ∠1与∠BOC表示同一个角.B. ∠β表示的是∠AOCC. ∠1+∠β=∠AOC.D. ∠β>∠BOA.8.钟表在8:20时,时针与分针的夹角是()度.A. 101.5B. 130C. 120D. 1259.如果A、B、C三点在同一直线上,且线段AB=6cm,BC=4cm,若M,N分别为AB,BC的中点,那么M,N两点之间的距离为()A. 5 cmB. 1 cmC. 5或1 cmD. 无法确定10.如图,B、C是线段AD上任意两点,M是AB的中点,N是CD的中点.若MN=a,BC=b,则线段AD等于()A. 2(a−b)B. 2a−bC. a+bD. a−b11.已知a+b<0,且b<0<a,则数a、b在数轴上距离原点较近的是()A. aB. bC. a、b一样远近D. 无法判断12.已知a、b为有理数,下列说法①若a、b互为相反数,则ab=−1;②若a+b<0,ab>0,则|3a+4b|=−3a−4b;③若|a−b|+a−b=0,则b>a;④若|a|>|b|,则(a+b)⋅(a−b)是正数,其中正确的有()个.A. 1B. 2C. 3D. 413.有理数a,b在数轴上的位置如图所示,则下列关系时中正确的有()①a−b>0;②a+b>0;③1a >1b;④b−a>0.A. 1个B. 2个C. 3个D. 4个14.已知有理数a≠1,我们把11−a 称为a的差倒数,如:2的差倒数是11−2=−1,−1的差倒数是11−(−1)=12.如果a1=−2,a2是a1的差倒数,a3是a2的差倒数,a4是a3的差倒数……依此类推,那么a1+a2+⋯+a100的值是()A. −7.5B. 7.5C. 5.5D. −5.515.点A,B,C在一条直线上,AB=6,BC=2,点M是AC的中点,则AM的长度为()A. 4B. 6C. 2或6D. 2或416.如图是由●按照一定规律组成的图形,其中第①个图中共有3个●,第②个图中共有8个●,第③个图中共有15个●,第④个图中共有24个●……照此规律排列下去,则第⑩个图中●的个数为()A. 105B. 110C. 120D. 140二、填空题(本大题共3小题,共12.0分)17.43°29′7″+36°30′53″=______ .18.m,n是有理数,它们在数轴上的对应点的位置如图所示,化简|n−m|的结果是______.19.已知一个有理数为x,则|x+3|+|x−2|的最小值是___________.三、解答题(本大题共7小题,共66.0分)20.计算:(1)−14−8÷(−2)3+22×(−3);(2)[45−(79−1112+56)×36]÷5.21.已知|a|=5,|b|=2.(1)若a<0,b>0,求3a−2b的值;(2)若a>0,b<0,|c−2|=1,求2ab c+|b−c|的值.22.点A,B,C在同一直线上,若AB=8,AC:BC=3:1,求线段AC的长度;23.如图所示,如果直线l上依次有3个点A,B,C,那么:(1)在直线l上共有多少射线?多少条线段?(2)在直线l上增加一个点,共增加了多少条射线?多少条线段?(3)在直线l上增加到n个点,共有多少条射线?多少条线段?24.已知m,n满足(m−6)2+|n−2|=0.(1)求m,n的值;(2)已知线段AB=m,在直线AB上取一点P,使AP=nPB,Q为PB的中点,求线段AQ的长.25.一点A从数轴上表示+2的点开始移动,第一次先向左移动1个单位,再向右移动2个单位;第二次先向左移动3个单位,再向右移动4个单位;第三次先向左移动5个单位,再向右移动6个单位…(1)写出第一次移动后这个点在数轴上表示的数为______;(2)写出第二次移动结果这个点在数轴上表示的数为______;(3)写出第五次移动后这个点在数轴上表示的数为______;(4)写出第n次移动结果这个点在数轴上表示的数为______;(5)如果第m次移动后这个点在数轴上表示的数为56,求m的值.26.有一科技小组进行了机器人行走性能试验,在试验场地有A、B、C三点顺次在同一笔直的赛道上,A、B两点之间的距离是70米.甲、乙两机器人分别从A、B两点同时同向出发,乙机器人始终以60米/分的速度行走,乙行走7分钟到达C点.设两机器人出发时间为t(分钟),当t=2分钟时,甲追上乙.前3分钟甲机器人的速度保持不变,3分钟后甲的速度变为另一数值.已知在3≤t≤4分钟时,甲、乙两机器人之间的距离保持不变.请解答下面问题:(1)B、C两点之间的距离是______ 米,3分钟后甲机器人的速度为______ 米/分.(2)求甲机器人前2分钟的速度为多少米/分?(3)求两机器人前4分钟内出发多长时间相距28米?-------- 答案与解析 --------1.答案:B=−1×2=−2,解析:解:根据题意得:−1÷12故选:B.此题考查了有理数的乘除法,熟练掌握运算法则是解本题的关键.根据积除以一个因式得到另一个因式即可.2.答案:D解析:解:这个数是4或−4.当这个数是4时,它的三次方是64;当这个数是−4时,它的立方是−64.故选D.根据平方根的定义求得这个数,然后利用乘方定义求解.本题考查了平方根的定义,正确理解正数的平方根有两个,这两个平方互为相反数是解题的关键.3.答案:C解析:【分析】本题主要考查的是立体图形的认识,掌握相关概念是解题的关键.依据立体图形的定义回答即可.【解答】解:长方形、圆、三角形是平面图形,圆锥体是立体图形.故选C.4.答案:B解析:【分析】本题考查了乘方的性质,即−1的偶次幂是1,−1的奇次幂是−1;绝对值的性质,即负数的绝对值是它的相反数;相反数的概念,即−1的相反数是1.注意:−12表示12的相反数.根据乘方的性质、绝对值的性质、相反数的概念等分别化简各个数,进而判断.【解答】=1,解:∵(−1)2=1,(−1)3=−1,−12=−1,|−1|=1,−(−1)=1,−1−1∴等于1的有4个.故选B.5.答案:A解析:【分析】根据相反数和绝对值的概念进行判断.理解相反数和绝对值的概念是解答此题的关键.相反数:符号不同,绝对值相等的两个数互为相反数;绝对值:数轴上,一个数到原点的距离叫做这个数的绝对值.【解答】解:①正确;②若−a>a,则2a<0,即a是负数,故②正确;③数轴上原点两侧,且到原点距离相等的数互为相反数;故③错误;④两个负数相互比较,绝对值大的反而小;故④错误;所以正确的结论是①②.故选:A.6.答案:B解析:解:①过两点有且只有一条直线,正确;②连接两点的线段的长度叫两点间的距离,错误③两点之间线段最短,正确;④点B在线段AC上,如果AB=BC,则点B是线段AC的中点,正确;故选:B.依据直线的性质、两点间的距离、线段的性质以及中点的定义进行判断即可.本题主要考查了直线的性质、两点间的距离、线段的性质以及中点的定义,平面上任意两点间都有一定距离,它指的是连接这两点的线段的长度,注意强调最后的两个字“长度”.7.答案:C解析:【分析】此题考查了角的表示方法以及角的大小比较,根据图形特点将每个角用合适的方法表示出来是解题的关键.【解答】解:A.∠1与∠AOB表示的是同一个角,故A说法错误;B.∠β表示的是∠BOC,故B说法错误;C.∠1+∠β=∠AOC,故C说法正确;D.∠AOC>∠BOA,故D说法错误.故选C.8.答案:B解析:【分析】本题考查了钟面角:钟面被分成12大格,每大格为30°;分针每分钟转6°,时针每分钟转0.5°.因为钟表上的刻度是把一个圆平均分成了12等份,每一份是30°,借助图形,找出时针和分针之间相差的大格数,用大格数乘30°即可.【解答】解:∵时针在钟面上每分钟转0.5°,分针每分钟转6°,钟表12个数字,每相邻两个数字之间的夹角为30°,∴8:20时分针与时针的夹角4×30°+20×0.5°=130°.故选B.9.答案:C解析:解:如图1,当点B在线段AC上时,∵AB=6cm,BC=4cm,M,N分别为AB,BC的中点,∴MB=12AB=3,BN=12BC=2,∴MN=MB+NB=5cm,如图2,当点C在线段AB上时,∵AB=6cm,BC=4cm,M,N分别为AB,BC的中点,∴MB=12AB=3,BN=12BC=2,∴MN=MB−NB=1cm,故选:C.分点B在线段AC上和点C在线段AB上两种情况,根据线段中点的性质进行计算即可.本题考查的是两点间的距离,掌握线段中点的性质、灵活运用数形结合思想、分情况讨论思想是解题的关键.10.答案:B解析:【分析】本题考查了比较线段长短的知识,利用中点性质转化线段之间的关系是解题的关键,在不同的情况下灵活选用它的不同表示方法,有利于解题的简洁性.同时,灵活运用线段的和、差、倍、分转化线段之间的数量关系也是十分关键的一点.【解答】解:∵MN=MB+CN+BC=a,BC=b,∴MB+CN=a−b,∵M是AB的中点,N是CD中点∴AB+CD=2(MB+CN)=2(a−b),∴AD=2(a−b)+b=2a−b.故选B.11.答案:A解析:解:∵a+b<0,且b<0<a,∴|a|<|b|,∴数a在数轴上距离原点较近,故选:A.根据已知条件判断出a,b的符号及绝对值的大小即可.本题考查的是数轴,熟知数轴上右边的数总比左边的大的特点是解答此题的关键.12.答案:B没有意义,本选项错误;解析:解:①0与0互为相反数,但是ab②由a+b<0,ab>0,得到a与b同时为负数,即3a+4b<0,∴|3a+4b|=−3a−4b,本选项正确;③∵|a−b|+a−b=0,即|a−b|=−(a−b),∴a−b≤0,即a≤b,本选项错误;④若|a|>|b|,当a>0,b>0时,可得a>b,即a−b>0,a+b>0,∴(a+b)⋅(a−b)为正数;当a>0,b<0时,a−b>0,a+b>0,∴(a+b)⋅(a−b)为正数;当a <0,b >0时,a −b <0,a +b <0,∴(a +b)⋅(a −b)为正数; 当a <0,b <0时,a −b <0,a +b <0,∴(a +b)⋅(a −b)为正数, 本选项正确, 则其中正确的有2个. 故选B①0的相反数为0,而ab 没有意义;②由两数之和小于0,两数之积大于0,得到a 与b 都为负数,即3a +4b 小于0,利用负数的绝对值等于它的相反数化简得到结果,即可作出判断;③由a −b 的绝对值等于它的相反数,得到a −b 为非正数,得到a 与b 的大小,即可作出判断; ④由a 绝对值大于b 绝对值,分情况讨论,即可作出判断.此题考查了有理数的混合运算,熟练掌握各种运算法则是解本题的关键.13.答案:B解析: 【分析】本题主要考查数轴及有理数的加减法则及不等式的基本性质,熟练掌握有理数的加减法则、不等式的基本性质是关键.由图可知,a <b <0,再根据有理数的加减法则、不等式的基本性质逐一判断即可. 【解答】解:由数轴可知,a <b <0, a −b <0,故①错误 a +b <0,故②错误 ∵a <b <0, ∴1a >1b ,故③正确 ∵a <b ,∴b −a >0,故④正确. 故选B .14.答案:A解析: 【分析】本题考查了规律型:数字的变化类:通过从一些特殊的数字变化中发现不变的因素或按规律变化的因素,然后推广到一般情况.求出数列的前4个数,从而得出这个数列以−2,13,32依次循环,且−2+13+32=−16,再求出这100个数中有多少个周期,从而得出答案. 【解答】 解:∵a 1=−2,∴a 2=11−(−2)=13,a 3=11−13=32,a 4=11−32=−2,…… ∴这个数列以−2,13,32依次循环,且−2+13+32=−16,∵100÷3=33…1,∴a 1+a 2+⋯+a 100=33×(−16)−2=−152=−7.5,故选:A .15.答案:D解析:【分析】根据题意画出符合条件的两种情况,求出AC 的值,根据线段中点定义得出AM =12AC ,代入求出即可.本题考查了求两点间的距离和线段中点的定义,主要考查学生的计算能力.【解答】解:分为两种情况:①当C 在线段AB 上时,AC =AB −BC =6−2=4,∵M 是AC 的中点,∴AM =12AC =2; ②当C 在线段AB 的延长线上时,AC =AB +BC =6+2=8,∵M 是AC 的中点,∴AM =12AC =4.∴AM 的长度为2或4.故选:D .解析:解:∵第①个图中●有3=1×3个,第②个图中●有8=2×4个,第③个图中●有15=3×5个,第④个图中●有24=4×6个,……∴第⑩个图中●的个数为10×12=120个,故选:C.根据已知条件得出第n个图中●的个数为n(n+2),据此可得.本题主要考查图形的变化规律,解题的关键是根据题意得出第n个图中●的个数为n(n+2).17.答案:80°解析:解:43°29′7″+36°30′53″=79°59′60″=80°,故答案为:80°.根据度、分、秒的换算,即可解答.本题考查了度、分、秒的换算,解决本题的关键是熟记度、分、秒的换算.18.答案:m−n解析:解:观察数轴可知n<m,∴n−m<0∴|n−m|=−(n−m)=m−n故答案为m−n.根据数轴可判断n<m,可得n−m<0,再进一步去掉绝对值符号即可得到化简结果.本题考查的是绝对值的相关化简,先判断绝对值内代数式的正负,再按法则去掉绝对值符号是化简的主要过程.19.答案:5解析:解:|x+3|+|x−2|表示数轴上x和−3的两点之间与x和2的两点之间距离和,即当−3≤x≤2时有最小值,这个最小值就是2到−3的距离,故|x+3|+|x−2|最小值是5.故答案为:5.|x+3|+|x−2|表示数轴上x和−3的两点之间与x和2的两点之间距离和.本题主要考查的是数轴、绝对值,理解绝对值的几何意义是解题的关键.20.答案:解:(1)原式=−1+1−12(2)原式=(45−28+33−30)÷5=4.解析:(1)直接利用有理数的混合运算法则计算得出答案;(2)直接利用乘法分配律进而计算得出答案.此题主要考查了有理数的混合运算,正确掌握相关运算法则是解题关键.21.答案:解:∵|a|=5,|b|=2,∴a=±5,b=±2,(1)∵a<0,b>0,∴a=−5,b=2,∴3a−2b=3×(−5)−2×2=−19;(2)∵a>0,b<0,|c−2|=1,∴a=5,b=−2,c=3或c=1,当c=3时,2ab c+|b−c|=2×5×(−2)3+|−2−3|=−80+5=−75;当c=1时,2ab c+|b−c|=2×5×(−2)+|−2−3|=−20+5=−15,综上所述,2ab c+|b−c|的值为−75或−15.解析:本题主要考查了代数式求值,利用绝对值的定义解得a,b,c是解答此题的关键.(1)根据绝对值的性质可得a=±5,b=±2,根据(1)的条件取得合适的a,b,再代入计算即可;(2)根据(2)的条件取得合适的a,b,再由绝对值的性质求得c,再代入计算即可.22.答案:解:如图:当点C在线段AB上时,∵AB=8,AC:BC=3:1,∴AC=6;当点B在线段AC上时,∵AB=8,AC:BC=3:1,∴AC=AB+BC=12.解析:本题考查的是两点间的距离的计算,灵活运用数形结合思想和分情况讨论思想是解题的关键.分点C在线段AB上和点B在线段AC上两种情况,结合图形计算即可.23.答案:解:(1)以A,B,C为端点的射线各自有2条,因而共有射线6条,线段有:AB,AC,BC,共有线段3条;(2)由分析得:增加一个点增加2条射线,增加3条线段;(3)由分析(1)可得共有2n条射线,n(n−1)条.线段的总条数是12解析:本题考查直线射线及线段的知识,难度不大,注意基本概念的掌握及规律的总结.(1)一个直线上的每一个点对应两条射线,可求出射线的条数,分别以A、B为起点可查找出线段的条数;(2)根据分析(1)可得出答案;(3)根据(1)(2)可得出增加一个点后增加的射线条数及线段条数,由特殊到一般总结即可得出答案.24.答案:解:(1)由题意得(m−6)2=0,|n−2|=0,所以m=6,n=2;(2)当点P在线段AB上时,AP=2PB,所以AP=4,PB=2.而Q为PB的中点,所以PQ=1,故A Q=AP+PQ=5;当点P在线段AB的延长线上时,AP−PB=AB,即2PB−PB=6,所以PB=6.而Q为PB的中点,所以BQ=3,AQ=AB+BQ=6+3=9.故线段AQ的长为5或9.解析:本题考查了线段的中点,线段的和差,利用非负数的和为零得出每个非负数同时为零是解题关键.(1)根据非负数的和为零,可得每个非负数同时为零,可得答案;(2)分点P在线段AB上和点P在线段AB的延长线上两种情况讨论,根据AP=nPB和线段中点的性质,即可得答案.25.答案:(1)3;(2)4;(3)7;(4)n+2;(5)m+2=56,解得m=54.解析:解:(1)第一次移动后这个点在数轴上表示的数是3;(2)第二次移动后这个点在数轴上表示的数是4;(3)第五次移动后这个点在数轴上表示的数是7;(4)第n次移动后这个点在数轴上表示的数是n+2;(5)见答案.故答案为3,4,7,n+2,54.【分析】(1)一点A从数轴上表示+2的点开始移动,第一次先向左移动1个单位,再向右移动2个单位,实际上点A最后向左移动了1个单位,则第一次后这个点表示的数为1+2=3;(2)第二次先向左移动3个单位,再向右移动4个单位,实际上点A最后向左移动了1个单位,则第二次后这个点表示的数为2+2=4;(3)根据前面的规律得到第五次移动后这个点在数轴上表示的数是5+2=7;(4)第n次移动后这个点在数轴上表示的数是n+2;(5)由(4)得到第m次移动后这个点在数轴上表示的数为m+2,则m+2=56,然后解方程即可.本题考查了规律型:数字的变化类:通过从一些特殊的数字变化中发现不变的因素或按规律变化的因素,然后推广到一般情况.26.答案:解:(1)420;60(2)设甲机器人前2分钟的速度为x米/分,根据题意,得2x−2×60=70,解得x=95.答:甲机器人前2分钟的速度为95米/分;(3)设两机器人前4分钟内出发y分相距28米.分两种情况:①甲没有追上乙,根据题意,得95y−60y=70−28,解得y=1.2;②甲追上乙后,根据题意,得95y−60y=70+28,解得y=2.8.答:两机器人前4分钟内出发1.2或2.8分时相距28米.解析:【分析】本题考查了一元一次方程的运用,解题关键是理解题意,找到等量关系列出方程.(1)根据路程=速度×时间求出B、C两点之间的距离;根据在3≤t≤4分钟时,甲、乙两机器人之间的距离保持不变,可得3分钟后甲机器人的速度=乙机器人的速度=60米/分;(2)设甲机器人前2分钟的速度为x米/分,根据当t=2分钟时,甲追上乙得出方程2x−2×60=70,解方程即可;(3)设两机器人前4分钟内出发y分相距28米.分两种情况进行讨论:①甲没有追上乙;②甲追上乙.分别根据两机器人相距28米列出方程,解方程即可.【解答】解:(1)∵乙机器人从B出发,以60米/分的速度行走,乙行走7分钟到达C点,∴B、C两点之间的距离是60×7=420(米);∵在3≤t≤4分钟时,甲、乙两机器人之间的距离保持不变,∴3分钟后甲机器人的速度=乙机器人的速度=60米/分.故答案为420;60;(2)见答案;(3)见答案.。

河北省衡水中学2019-2020学年高三上学期期中考试政治试题 附详细解析

河北省衡水中学2019-2020学年高三上学期期中考试政治试题 附详细解析

2019�2020学年度高三年级期中考试试题政治试卷命题人:李文静审核人:位维维本试卷分I 卷(选择题)和I I 卷(非选择题)两部分。

选择题共48题,非选择题共5题。

试卷共8页。

考试时间110分钟。

满分100分。

卷I <选择题共48分)一、选择题(本大题共48小题。

每小题1分。

满分48分。

在四个选项中,只有一项是最符合题意的。

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下列推导能正确反映企业提高劳动生产率意义的是@商品价值总量增加@单位商品价值量降低@个别劳动时间缩短@提高企业经济效益@缩短社会必要劳动时间@增加商品的附加值A.@-®-@B.@一CD-@-:.C. @-®-CDD. ® 一@一@高三政治期中试卷 3.人民币汇率中间价,不考虑其他因素,对图表信息认识正确的是品种2018年3月20日I 201s年12月20曰美元/人民币I :s. 324s I s. a 93s@人民币贬值,有孩子在美国读书的家庭支出增加@人民币贬值,海外代购和海淘成本明显降低@美元升值,我国原材料依赖进口的企业成本增加@美元贬值,我国出口型企业价格竞争优势增强A .(D@ B.CD@ C.®@ D.®@4.最新数据显示,当前世界上90%的二维码个人用户在中国,我国已成为名副其实的二维码大国。

河北省衡水市2019-2020学年九年级(上)期中数学试卷

河北省衡水市2019-2020学年九年级(上)期中数学试卷

2019-2020学年九年级(上)期中数学试卷一、选择题(本大题有16个小题共42分.1-10小题每小题3分,11~16小题每小题3分,在每个小题给出的四个选项中只有一项是符合题目要求的)1.在图示的四个汽车标志图案中,能用平移变换来分析其形成过程的图案是()A.B.C.D.2.一元二次方程x2=4x的解是()A.x=0 B.x=4 C.x1=2,x2=﹣2 D.x1=0,x2=4 3.用配方法解方程x2﹣2x﹣8=0,下列配方结果正确的是()A.(x+1)2=9 B.(x+1)2=7 C.(x﹣1)2=9 D.(x﹣1)2=7 4.一学生推铅球,铅球行进的高度y(m)与水平距离x(m)之间的关系为y=﹣x2+ x+,则学生推铅球的距离为()A.B.3m C.10m D.12m5.下列函数中,满足y的值随x的值增大而增大的是()A.y=﹣2x B.y=3x﹣1 C.y=D.y=x26.如图,在△ABC中,∠ACB=90°,∠A=20°.将△ABC绕点C按逆时针方向旋转得△A′B′C,且点B在A′B′上,CA′交AB于点D,则∠BDC的度数为()A.40°B.50°C.60°D.70°7.如图,在平面直角坐标系中,点A的坐标为(﹣1,),以原点O为中心,将点A顺时针旋转150°得到点A′,则点A′的坐标为()A.(0,﹣2)B.(1,﹣)C.(2,0)D.(,﹣1)8.如图,⊙O中,弦AB、CD相交于点P,若∠A=30°,∠APD=70°,则∠B等于()A.30°B.35°C.40°D.50°9.如图,⊙O的半径是3,点P是弦AB延长线上的一点,连接OP,若OP=4,∠APO=30°,则弦AB的长为()A.2B.C.2D.10.“若a是实数,则|a|≥0”这一事件是()A.必然事件B.不可能事件C.不确定事件D.随机事件11.某居委会组织两个检查组,分别对“垃圾分类”和“违规停车”的情况进行抽查.各组随机抽取辖区内某三个小区中的一个进行检查,则两个组恰好抽到同一个小区的概率是()A.B.C.D.12.设α,β是方程x2+x+2012=0的两个实数根,则α2+2α+β的值为()A.﹣2014 B.2014 C.2013 D.﹣201313.某校九年级一班共有学生50人,现在对他们的生日(可以不同年)进行统计,则正确的说法是()A.至少有两名学生生日相同B.不可能有两名学生生日相同C.可能有两名学生生日相同,但可能性不大D.可能有两名学生生日相同,且可能性很大14.如图,在△ABC中,CA=CB,∠ACB=90°,AB=2,点D为AB的中点,以点D为圆心作圆心角为90°的扇形DEF,点C恰在弧EF上,则图中阴影部分的面积为()A.B.C.D.15.如图,将矩形ABCD绕其右下角的顶点按顺时针方向旋转90°至图①位置,继续绕右下角的顶点按顺时针方向旋转90°至图②位置,以此类推,这样连续旋转2017次.若AB =4,AD=3,则顶点A在整个旋转过程中所经过的路径总长为()A.2017πB.2034πC.3024πD.3026π16.二次函数y=ax2+bx+c(a≠0)的图象如图所示,下列结论:①ac>0;②当x≥1时,y随x的增大而减小;③2a+b=0;④b2﹣4ac<0;⑤4a﹣2b+c>0,其中正确的个数是()A.1 B.2 C.3 D.4二、填空题(本大题共6个小题每小题3分,共18分把答案写在题中横线上)17.如图,△ABC中,∠ACB=90°,∠ABC=25°,以点C为旋转中心顺时针旋转后得到△A′B′C′,且点A在A′B′上,则旋转角为.18.将二次函数y=2x2﹣4x+3的图象先向左平移3个单位长度,再向下平移1个单位长度,得到函数的图象的表达式是.19.如图,MN是⊙O的直径,MN=2,点A在⊙O上,∠AMN=30°,B为弧AN的中点,P 是直径MN上一动点,则PA+PB的最小值为.20.关于x的方程kx2﹣4x﹣=0有实数根,则k的取值范围是.21.小强与小红两人下军棋,小强获胜的概率为46%,小红获胜的概率是30%,那么两人下一盘棋小红不输的概率是%.22.如图,正方形ABCD和正方形CEFG边长分别为a和b,正方形CEFG绕点C旋转,给出下列结论:①BE=DG;②BE⊥DG;③DE2+BG2=2a2+b2,其中正确结论是(填序号)三、解答题(本大题共6个小题,共ω0分,解答应写出文字说明,证明或演算过程)23.已知关于x的方程x2+ax+a﹣2=0.(1)证明:不论a取任何实数,该方程都有两个不相等的实数根;(2)当a=1时,求该方程的根.24.荆岗中学决定在本校学生中,开展足球、篮球、羽毛球、乒乓球四种活动,为了了解学生对这四种活动的喜爱情况,学校随机调查了该校m名学生,看他们喜爱哪一种活动(每名学生必选一种且只能从这四种活动中选择一种),现将调查的结果绘制成如下不完整的统计图.(1)m=,n=;(2)请补全图中的条形图;(3)根据抽样调查的结果,请估算全校1800名学生中,大约有多少人喜爱踢足球;(4)在抽查的m名学生中,喜爱打乒乓球的有10名同学(其中有4名女生,包括小红、小梅),现将喜爱打乒乓球的同学平均分成两组进行训练,且女生每组分两人,求小红、小梅能分在同一组的概率.25.如图,在△ABC中,∠C=60°,⊙O是△ABC的外接圆,点P在直径BD的延长线上,且AB=AP.(1)求证:PA是⊙O的切线;(2)若AB=2,求图中阴影部分的面积.(结果保留π和根号)26.如图,已知等腰直角三角形ABC,点P是斜边BC上一点(不与B,C重合),PE是△ABP 的外接圆⊙O的直径.(1)求证:△APE是等腰直角三角形;(2)若⊙O的直径为2,求PC2+PB2的值.27.某商店经营儿童益智玩具,已知成批购进时的单价是20元.调查发现:销售单价是30元时,月销售量是230件,而销售单价每上涨1元,月销售量就减少10件,但每件玩具售价不能高于40元.设每件玩具的销售单价上涨了x元时(x为正整数),月销售利润为y元.(1)求y与x的函数关系式并直接写出自变量x的取值范围.(2)每件玩具的售价定为多少元时,月销售利润恰为2520元?(3)每件玩具的售价定为多少元时可使月销售利润最大?最大的月利润是多少?28.如图,点O是等边△ABC内一点,∠AOB=110°,∠BOC=α.将△BOC绕点C按顺时针方向旋转60°得△ADC,连接OD.(1)求证:△COD是等边三角形;(2)当α=150°时,试判断△AOD的形状,并说明理由;(3)探究:当α为多少度时,△AOD是等腰三角形?。

河北省衡水中学2019-2020学年高一上学期期中考试(化学)

河北省衡水中学2019-2020学年高一上学期期中考试(化学)

河北省衡水中学2019-2020学年高一上学期期中考试高一化学试卷满分:100分考试时间:110分钟可能用到的相对原子质量:H~1 C~12 N~14 O~16 S~32 Zn~65 Na~23 K~39 Cu~64 Fe~56 Mg~24 Al~27 Cl~35.5 Ca~40 Ag~108第Ⅰ卷(选择题共50分)一、选择题(下列每小题所给选项只有1项符合题意,请将正确答案的序号填涂在答题卡上,每小题1分,共10分)1、在下列反应中,属于氧化还原反应,且水既不作氧化剂又不作还原剂的是()A、2Na+2H2O=2NaOH+H2↑B、2F2+H2O=4HF+O2↑C、CaO+H2O=Ca(OH)2D、Cl2+H2O=HCl+HClO2.下列各组离子能在溶液中大量共存,且溶液无色的是( )A、Fe3+、K+、Cl-、NO3-B、Cu2+、SO42-、Na+、CO32-C、K+、Cl-、NO3-、CO32-D、Ba2+、OH-、Mg2+、NO3-3.下列变化中,需加入氧化剂才能实现的是( )①MnO4-→Mn2+②S2-→S ③HCO3-→CO2④Cl2→Cl-⑤NH4+→NH3⑥H2S→SO2A、①②④⑥B、②④⑥C、②⑥D、①④⑤4.下列说法正确的是()A、铜、石墨均能导电,所以它们均是电解质B、NH3、CO2的水溶液均能导电,所以NH3、CO2均是电解质C、液态HCl、固体NaCl均不能导电,所以HCl、NaCl均是非电解质D、蔗糖、酒精在水溶液或熔融状态下均不能导电,所以它们均是非电解质5.胶体区别于其他分散系的本质特征是()A.胶体的分散质能通过滤纸空隙,而浊液的分散质不能B.产生丁达尔现象C.分散质粒子直径在1nm~100nm之间D.胶体在一定条件下能稳定存在6.在强酸性溶液中,可以大量共存的离子组是( )A、Na+、K+、S2-、Cl-B、Al3+、Mg2+、SO42-、Cl-C、CO32-、Cl-、NO3-、K+D、Ba2+、OH-、Ca2+、Cl-7.下列操作过程中一定有氧化还原反应发生的是()A B C D8.下列物质中,属于电解质但不能导电的是( )A、稀硝酸B、液态HClC、NH3D、熔融的氢氧化钠9.在下列反应中,钾元素失去电子被氧化的是()A、2K+2H2O=2KOH+H2↑B、2K2O2+2H2O=4KOH+O2↑C、2KCl 2K+Cl2↑D、2KClO32KCl+3O2↑10.下列化学实验操作中均要用到玻璃棒,其中玻璃棒作用相同的是()①过滤②蒸发③溶解④向容量瓶转移液体⑤测溶液的pH 值A.①②④B.②③⑤C.③和④D.①和④二、不定项选择题(下列每小题所给选项有1个或2个符合题意,请将正确答案的序号填涂在答题卡上,每个2分,部分分1分,错选不得分。

2019-2020学年河北省衡水中学高三(上)期中物理试卷 (含答案解析)

2019-2020学年河北省衡水中学高三(上)期中物理试卷 (含答案解析)

2019-2020学年河北省衡水中学高三(上)期中物理试卷一、单选题(本大题共5小题,共20.0分)1.如图是蹦床运动员落在弹簧床面的示意图,忽略空气阻力,下面说法正确的是()A. 运动员下落到刚接触蹦床时,速度最大B. 运动到最低点时,床对运动员的作用力大于运动员对床的作用力C. 从刚接触蹦床到运动至最低点的过程中,运动员一直处于超重状态,运动员的加速度先减小后增大D. 在下落过程中,重力对运动员所做的功等于其重力势能的减小2.在平直公路上行驶的a车和b车,其位移−时间(x−t)图像分别为图中直线a和曲线b,已知b车的加速度恒定且等于−2m/s2,当t=3s时,直线a和曲线b刚好相切,则()m/sA. a车做匀速运动且其速度为v a=83B. t=3s时a车和b车相遇但此时速度不相等C. t=1s时b车的速度为10m/sD. t=0时a车和b车间的距离x0=9m3.如图所示,细绳一端与光滑小球连接,另一端系在竖直墙壁上的A点,在缩短细绳小球缓慢上移的过程中,细绳对小球的拉力F、墙壁对小球的弹力N的变化情况为()A. F、N都不变B. F变大、N变小C. F、N都变大D. F变小、N变大4.质量为2kg的物块放在粗糙水平面上,在水平拉力的作用下由静止开始运动,物块动能E k与其发生位移x之间的关系如图所示.已知物块与水平面间的动摩擦因数μ=0.2,重力加速度g取10m/s2,则下列说法正确的是()A. x=1m时速度大小为2m/sB. x=3m时物块的加速度大小为2.5m/s2C. 在前4m位移过程中拉力对物块做的功为9JD. 在前4m位移过程中物块所经历的时间为2.8s5.如图所示,有A、B两个质量均为m的小车,在光滑的水平地面上以相等的速率v0在同一直线上相向运动,A车上有一质量也为m的人,他现在从A车跳到B车上,为了避免两车相撞,他跳离A车时的速率(相对地面)最小为()A. 3v0B. v0C. 53v0 D. 35v0二、多选题(本大题共10小题,共40.0分)6.如图甲所示,A,B两物体静止叠放在光滑水平面上。

2019-2020学年河北省衡水中学高三(上)期中数学试卷(理科)

2019-2020学年河北省衡水中学高三(上)期中数学试卷(理科)

2019-2020学年河北省衡水中学高三(上)期中数学试卷(理科)一、选择题(每小题5分,共60分.下列每小题所给选项只有-项符合题意.请将正确答案的序号填涂在答题卡上)1. 已知曲线f(x)=xcosx+3x在点(0, f(0))处的切线与直线ax+4y+1=0垂直,则实数a的值为()A.−4B.−1C.1D.4【答案】C【考点】利用导数研究曲线上某点切线方程【解析】求得f(x)的导数,可得切线的斜率,由两直线垂直的条件可得a的方程,解方程可得所求值.【解答】f(x)=xcosx+3x的导数为f′(x)=cosx−xsinx+3,可得在点(0, f(0))处的切线斜率为cos0−0+3=4,由切线与直线ax+4y+1=0垂直,可得−a4=−14,即a=1.2. 已知各项不为0的等差数列{a n}满足a5−2a72+2a8=0,数列{b n}是等比数列且b7=a7,则b2b12等于()A.49B.32C.94D.23【答案】C【考点】等差数列与等比数列的综合【解析】由条件利用等差数列的性质可得3a7=2a72,求得a7的值,再根据b2b12=b72计算.【解答】由a5−2a72+2a8=0,得a5+2a8=2a72,即3(a1+6d)=2a72,即3a7=2a72,∵a7≠0,∴a7=32=b7,则b2b12=b72=94.3. 对于函数f(x),若存在区间A=[m, n]使得{y|yf(x), x∈A}=A则称函数f(x)为“同域函数”,区间A为函数f(x)的一个“同城区间”.给出下列四个函数:①f(x)=cosπ2x;②f(x)=x2−1;③f(x)=|x2−1|;④f(x)=log2(x−1).存在“同域区间”的“同域函数”的序号是()A.①②B.①②③C.②③D.①②④【答案】B【考点】命题的真假判断与应用 【解析】解题思路:对于每一个选项找到其“同域区间”就判定为“同域函数”.逐项寻找就可以了! 【解答】对于函数①f(x)=cosπx 2,当x ∈[0, 1],则有f(x)∈[0, 1],符合题意;对于函数②f(x)=x 2−1,当x ∈[−1, 0]时,则有f(x)∈[−1, 0],符合题意; 对于函数③f(x)=||,当x ∈[0, 1]时,则有f(x)∈[0, 1],符合题意; 由选项可知,应选B ,4. 设θ为两个非零向量a →,b →的夹角,已知对任意实数t ,|b →+ta →|的最小值为1,则( )A.若|a →|确定,则 θ唯一确定 B.若|b →|确定,则θ唯一确定 C.若θ确定,则|a →|唯一确定 D.若θ确定,则|b →|唯一确定 【答案】 D【考点】平面向量数量积的性质及其运算 【解析】由题意可得,(b →+ta →)2=b →2+2ta →⋅b →+t 2a →2,则令g(t)=b →2+2ta →⋅b →+t 2a →2,可得判别式△<0,运用二次函数的性质,求出最小值,结合向量的数量积的性质,即可得到答案. 【解答】(b →+ta →)2=b →2+2ta →⋅b →+t 2a →2,则令g(t)=b →2+2ta →⋅b →+t 2a →2, 可得判别式△=4(a →⋅b →)2−4a →2b →2=4a →2b →2cos 2θ−4a →2b →2=−4a →2b →2sin 2θ<0,由二次函数的性质,可得g(t)>0恒成立. 且当t =−2a →⋅b →2a →2=−|b →||a →|cosθ时,g(t)最小,且为1.即g(−|b →||a →|cosθ)=−|b →|2cos 2θ+|b →|2=|b →|2sin 2θ=1,故当θ唯一确定时,|b →|唯一确定.5. 已知点P(x, y)是直线y =2√2x −4上一动点,PM 与PN 是圆C:x 2+(y −1)2=1的两条切线,M,N为切点,则四边形PMCN的最小面积为()A.4 3B.23C.53D.56【答案】A【考点】直线与圆的位置关系【解析】四边形PMCN的面积是两个三角形的面积的和,因为CM⊥PM,CM=1,显然PM最小时,四边形面积最小,此时PC最小,由此可得结论.【解答】圆C:x2+(y−1)2=1圆心坐标为(0, 1),半径为1;由题意过点P作圆C的两条切线,切点分别为M,N,可知四边形PMCN的面积是两个三角形的面积的和,因为CM⊥PM,CM=1,显然PM最小时,四边形面积最小,此时PC最小.∵P是直线y=2√2x−4上的动点,∴PC最小值=√8+1=53,∴PM最小值=√(53)2−12=43,∴四边形PMCN面积的最小值为:2×12×43×1=43.6. 已知函数f(x)=Asin(ωx+φ)(A>0, ω>0, 0<φ<π2)的部分图象如图所示,则f(3π4)=()A.−√22B.−12C.−1D.√22【答案】C【考点】由y=Asin(ωx+φ)的部分图象确定其解析式【解析】由函数的图象的顶点坐标求出A,由周期求出ω,由五点法作图求出φ的值,可得函数的解析式,从而求得f(3π4)的值.【解答】由函数f(x)=Asin(ωx+φ)(A>0, ω>0, 0<φ<π2)的部分图象,可得A=2,由2sinφ=√3,求得φ=π3.再根据五点法作图,可得ω⋅7π12+π3=3π2,∴ω=2,∴f(x)=2sin(2x+π3),∴f(3π4)=2sin(3π2+π3)=−2cosπ3=−1,7. 已知函数f(x)=12−4sinxcosx,若f(x−a)=−f(x+a)恒成立,则实数a的最小正值为()A.2πB.πC.π2D.π4【答案】D【考点】函数恒成立问题【解析】将函数式f(x)进行化简求出最小正周期,并将恒成立问题转化为周期问题即可.【解答】∵f(x)=12−4sinxcosx=12−2sin2x∴f(x)的最小正周期为T=π;又∵f(x−a)=−f(x+a)恒成立,∴f(x)=−f(x+2a)⇒−f(x)=f(x+2a),而−f(x)=f(x−2a),∴f(x+2a)=f(x−2a)⇒f(x)=f(x+4a),∴f(x)是以4a为周期的函数,∴4a=π,⇒a=π4;8. 设S n为数列{a n}的前n项和,a1=1,a n+1=2S n,则数列{1a n}的前20项和为()A.3 2−12×319B.74−14×319C.3 2−12×3D.74−14×3【答案】A【考点】数列的求和【解析】根据数列的递推公式可得数列{a n}是以1为首项,以3为公比的等比数列,即可得到1 a n =(13)n−1,再根据等比数列的求和公式即可求出.【解答】设S n为数列{a n}的前n项和,a1=1,a n+1=2S n,∴a n=2S n−1,∴a n+1−a n=2a n,∴a n+1=3a n,∴数列{a n}是以1为首项,以3为公比的等比数列,∴a n=3n−1,当n=1时也满足,∴1a n =(13)n−1,∴数列{1a n }的前20项和为1−13201−13=32−12×3199. 椭圆x2a2+y2b2=1(a>b>0)的左右焦点分别是F1、F2,以F2为圆心的圆过椭圆的中心,且与椭圆交于点P,若直线PF1恰好与圆F2相切于点P,则椭圆的离心率为()A.√3−1B.√3+12C.√22D.√5−12【答案】A【考点】椭圆的离心率【解析】利用已知条件以及椭圆的性质列出关系式,求解椭圆的离心率即可.【解答】椭圆x2a2+y2b2=1(a>b>0)的左右焦点分别是F1、F2,以F2为圆心的圆过椭圆的中心,且与椭圆交于点P,若直线PF1恰好与圆F2相切于点P,可得(2a−c)2+c2=4c2,可得2a2−2ac=c2,所以e2+2e−2=0,e∈(0, 1),解得e=−2+√122=√3−1.10. 已知函数f(x)=asinx−√3cosx图象的一条对称轴为直线x=5π6,且f(x1)f(x2)=−4,则|x1+x2|的最小值为()A.−π3B.0 C.π3D.2π3【答案】D【考点】三角函数的恒等变换及化简求值【解析】首先通过三角函数的恒等变换把函数关系式变性成正弦型函数,进一步利用对称轴确定函数的解析式,再利用正弦型函数的最值确定结果.【解答】解:函数f(x)=asinx−√3cosx=√a2+3sin(x+θ)的图象的一条对称轴为直线x=5π6,∴f(5π6)=a2+32=±√a2+3,解得a=1.则f(x)=sinx−√3cosx=2sin(x−π3),∵f(x1)f(x2)=−4,则f(x1)和f(x2)一个为−2,另一个为2,可设x1=2kπ−π6,x2=2kπ+5π6,则|x1+x2|=|4kπ+2π3|,k∈Z.故当k=0时,|x1+x2|取得最小值为2π3.故选D.11. 若函数f(x)=e x(x−3)−13kx3+kx2只有一个极值点,则k的取值范围为()A.(−∞, e) B.[0, e] C.(−∞, 2) D.(0, 2]【答案】B【考点】利用导数研究函数的极值【解析】利用函数求导函数f′(x)=e x(x−2)−kx2+2kx=(x−2)(e x−kx),只有一个极值点时f′(x)=0只有一个实数解有e x−kx≥0,设新函数设u(x)=e x,v(x)=kx,等价转化数形结合法即可得出结论,【解答】函数f(x)=e x(x−3)−13kx3+kx2只有一个极值点,f′(x)=e x(x−2)−kx2+2kx=(x−2)(e x−kx),若函数f(x)=e x(x−3)−13kx3+kx2只有一个极值点,f′(x)=0只有一个实数解,则:e x−kx≥0,从而得到:e x≥kx,当k=0时,成立.当k≠0时,设u(x)=e x,v(x)=kx如图:当两函数相切时,k=e,此时得到k的最大值,但k<0时不成立.故k的取值范围为:(0, e]综上:k的取值范围为:[0, e]12. 双曲线x2a2−y2b2=1(a>0,b>0)的左右焦点分别为F1,F2,过F1的直线交曲线左支于A,B两点,△F2AB是以A为直角顶点的直角三角形,且∠AF2B=30∘.若该双曲线的离心率为e,则e2=()A.11+4√3B.13+5√3C.16−6√3D.19−10√3【答案】D【考点】双曲线的离心率【解析】设|BF2|=2m,根据△F2AB是以A为直角顶点的直角三角形,且∠AF2B=30∘,以及双曲线的性质可得|AF2|=2a(3−√3),|AF1|=2a(2−√3),再根据勾股定理即可求出【解答】设|BF2|=2m,∵△F2AB是以A为直角顶点的直角三角形,且∠AF2B=30∘,∴|AB|=12|BF2|=m,|AF2|=√32|BF2|=√3m,由|AF2|−|AF1|=2a,∴|AF1|=√3m−2a,由|BF2|−|BF1|=2a,∴|BF1|=2m−2a,∴|AF1|+|BF1|=AB,∴√3m−2a+2m−2a=m,∴m=2a(√3−1),∴|AF|=√3⋅2a(√3−1)=2a(3−√3)2|AF1|=2a(3−√3)−2a=2a(2−√3)又在Rt△F1AF2中|AF1|2+|AF2|2=4c2,即4a2(3−√3)2+4a2(2−√3)2=4c2,即(19−10√3)a2=c2,∴e2=19−10√3,二、填空题(每题5分,共20分.把答案填在答题纸的横线上)已知向量a→,b→,|a→|=1,|b→|=2,且|2a→+b→|=√10,则a→⋅b→=________.【答案】12【考点】平面向量数量积的性质及其运算【解析】根据|a→|=1,|b→|=2,对|2a→+b→|=√10两边平方即可得出4+4a→⋅b→+4=10,从而可求出a→⋅b→.【解答】∵|a→|=1,|b→|=2,且|2a→+b→|=√10,∴(2a→+b→)2=4a→2+4a→⋅b→+b→2=4+4a→⋅b→+4=10,∴a→⋅b→=1.2已知抛物线E:y2=12x的焦点为F,准线为l,过F的直线m与E交于A,B两点,过A作AM⊥l,垂足为M,AM的中点为N,若AM⊥FN,则|AB|=________.【答案】16【考点】抛物线的性质【解析】由题意画出图形,得到直线AB的斜率,进一步求得直线AB的方程,与抛物线方程联立求解即可得答案.【解答】由题意画出图形如图,∵AF=AM,N为AM的中点,且FN⊥AM,∴∠AFN=30∘,则直线AB的倾斜角为60∘,斜率为√3.由抛物线y2=12x,得F(3, 0),则直线AB的方程为y=√3(x−3).联立{y=√3(x−3)y2=12x,得x2−10x+9=0.则x A+x B=10,∴|AB|=x A+x B+p=16.已知函数f(x)=(x2−2x)e x−1,若当x>1时,f(x)−mx+l+m≤0有解,则m的取值范围为________.【答案】(−1, +∞)【考点】利用导数研究函数的单调性【解析】先求导,判断出函数的单调性,可得函数值的情况,即可求出m的取值范围.【解答】∵f(x)−mx+1+m≤0,∴f(x)≤m(x−1)−1,∵y=m(x−1)−1且过定点(1, −1),∵当x>1时,f(x)−mx+1+m≤0有解,∴当x>1时,存在y=f(x)在y=m(x−1)−1的下方,∵f′(x)=(x2−2)e x−1,令f′(x)=0,解得x=√2,当1<x<√2时,f′(x)<0,当x>√2时,f′(x)>0,∴f(x)在(1, √2)上递减,在(√2,+∞)上递增,∵当x>2时,f(x)>0,又f(1)=−1,f(√2)<−1,f(2)=0,∴m>−1,故答案为:(−1, +∞)数列{a n}为1,1,2,1,1,2,3,1,1,2,1,1,2,3,4,…,首先给出a1=1,接着复制该项后,再添加其后继数2,于是a2=1,a3=2,然后再复制前面所有的项1,1,2,再添加2的后继数3,于是a4=1,a5=1,a6=2,a7=3,接下来再复制前面所有的项1,1,2,1,1,2,3,再添加4,…,如此继续,则a2019=________.【答案】1【考点】数列的求和【解析】由数列{a n}的构造方法可知a1=1,a3=2,a7=3,a15=4,可得a2n−1=n,即a2n−1+k=a k(1≤k<2n−1),进而得出结论.【解答】由数列{a n}的构造方法可知a1=1,a3=2,a7=3,a15=4,可得a2n−1=n,即a2n−1+k=a k(1≤k<2n−1),故a2019=a996=a485=a230=a103=a40=a9=a2=1.三、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)如图为一块边长为2km的等边三角形地块ABC,为响应国家号召,现对这块地进行绿化改造,计划从BC的中点D出发引出两条成60∘角的线段DE和DF,与AB和AC围成四边形区域AEDF,在该区域内种上草坪,其余区域修建成停车场,设∠BDE=α.(1)当α=60∘时,求绿化面积;(2)试求地块的绿化面积S(α)的取值范围.【答案】解:(1)当α=60∘时,DE // AC,DF // AB,则四边形AEDF为平行四边形,所以△BDE和△CDF都为边长为1km的等边三角形,且面积都为√34km2,故绿化面积为√34×22−2×√34=√32(km2).(2)由题意可得,30∘<α<90∘,在△BDE中,∠BED=120∘−α,由正弦定理可得BEsinα=1sin(120∘−α),∴BE=sinαsin(120∘−α),在△CDF中,∠CDF=120∘−α,∠CFD=α,由正弦定理可得1sinα=CFsin(120∘−α),∴CF=sin(120∘−α)sinα,∴BE+CF=sin(120∘−α)sinα+sinαsin(120∘−α)=sin2(120∘−α)+sin2αsinα⋅sin(120∘−α)=(√32cosα+12sinα)2+sin2αsinα⋅(12sinα+√32cosα)=134√32sinαcosα+12sin2α=1+34√34sin2α−14cos2α+14=1+32×1sin(2α−30∘)+12,∴S(α)=S△ABC −S△BDE−S△CDF=√3−√34(BE+CF)=3√34−3√38⋅112+sin(2α−30∘)(30∘<α<90∘).∵12<sin(2α−30∘)≤1,1<sin(2α−30∘)+12≤32,∴23≤112+sin(2α−30∘)<1,∴3√38<S(α)≤√32.故地块的绿化面积S(α)的取值范围(3√38,√3 2].【考点】两角和与差的正弦公式三角形的面积公式解三角形正弦函数的定义域和值域【解析】(1)当α=60<em>o</em>时,DE // AC,DF // AB,四边形AEDF为平行四边形,△BDE和△CDF都为边长为1km的等边三角形,结合已知即可求解;(2)由题意可得,30∘<α<90∘,在△BDE中,由正弦定理可表示BE,同理可得CF,然后结合和差角公式及同角平方关系对BE+CF进行化简,而s(α)=s△<em><em>ABC</em></em>−s△<em><em>BDE</em></em>−s<em>CDF</em>=√3−√34(BE+CF),代入结合正弦函数的性质可求.【解答】解:(1)当α=60∘时,DE // AC,DF // AB,则四边形AEDF为平行四边形,所以△BDE和△CDF都为边长为1km的等边三角形,且面积都为√34km2,故绿化面积为√34×22−2×√34=√32(km2).(2)由题意可得,30∘<α<90∘,在△BDE中,∠BED=120∘−α,由正弦定理可得BEsinα=1sin(120−α),∴BE=sinαsin(120∘−α),在△CDF中,∠CDF=120∘−α,∠CFD=α,由正弦定理可得1sinα=CFsin(120−α),∴ CF =sin(120∘−α)sinα,∴ BE +CF =sin(120∘−α)sinα+sinαsin(120∘−α)=sin 2(120∘−α)+sin 2αsinα⋅sin(120∘−α)=(√32cosα+12sinα)2+sin 2αsinα⋅(12sinα+√32cosα)=134√32sinαcosα+12sin 2α=1+34√34sin2α−14cos2α+14=1+32×1sin(2α−30∘)+12,∴ S(α)=S △ABC −S △BDE −S △CDF =√3−√34(BE +CF)=3√34−3√38⋅112+sin(2α−30∘)(30∘<α<90∘).∵ 12<sin(2α−30∘)≤1,1<sin(2α−30∘)+12≤32, ∴ 23≤112+sin(2α−30∘)<1,∴3√38<S(α)≤√32. 故地块的绿化面积S(α)的取值范围(3√38,√32].已知等差数列{a n }前n 项和S n ,等比数列{b n }前n 项和为T n ,a 1=1,b 1=1,a 2+b 2=4.(1)若a 3+b 3=7,求数列{b n }的通项公式;(2)若T 3=13,求S 5. 【答案】设等差数列{a n }的公差为d ,等比数列{b n }的公比为q , 由a 1=1,b 1=1,a 2+b 2=4,a 3+b 3=7,得 {1+d +q =41+2d +q 2=7,解得q =2. ∴ b n =b 1q n−1=2n−1;由b 1=1,T 3=13,得1+q +q 2=13,即q =−4或q =3.当q =−4时,b 2=−4,此时a 2=4−b 2=8,d =a 2−a 1=7,S 5=5+5×42×7=75;当q =3时,b 2=3,此时a 2=4−b 2=1,d =a 2−a 1=0,S 5=5a 1=5. 综上,S 5=75或5. 【考点】等差数列的通项公式 等差数列的前n 项和 【解析】(1)设等差数列{a n }的公差为d ,等比数列{b n }的公比为q ,由已知列关于d 和q 的方程组,求得q ,可得数列{b n }的通项公式;(2)由b 1=1,T 3=13列式求得q ,然后分类求解S 5. 【解答】设等差数列{a n }的公差为d ,等比数列{b n }的公比为q , 由a 1=1,b 1=1,a 2+b 2=4,a 3+b 3=7,得 {1+d +q =41+2d +q 2=7,解得q =2. ∴ b n =b 1q n−1=2n−1;由b 1=1,T 3=13,得1+q +q 2=13,即q =−4或q =3.当q =−4时,b 2=−4,此时a 2=4−b 2=8,d =a 2−a 1=7,S 5=5+5×42×7=75;当q =3时,b 2=3,此时a 2=4−b 2=1,d =a 2−a 1=0,S 5=5a 1=5. 综上,S 5=75或5.已知圆D :(x −2)2+(y −1)2=1,点A 在抛物线C:y 2=4x 上,O 为坐标原点,直线OA 与圆D 有公共点.(1)求点A 横坐标的取值范围;(2)如图,当直线OA 过圆心D 时,过点A 作抛物线的切线交y 轴于点B ,过点B 引直线l 交抛物线C 于P 、Q 两点,过点P 作x 轴的垂线分别与直线OA 、OQ 交于M 、N ,求证:M 为PN 中点. 【答案】由题意直线OA 斜率存在且不为零,设l OA :y =kx ,则 由{y =kx y 2=4x ′解得x A =4k 2, 又D(2, 1)到l OA :kx −y =0的距离为√k 2+1≤1,即0≤k ≤43, 所以x A ∈[94,+∞).证明:当直线OA 过圆心D(2, 1)时,k =12,x A =4k 2=16,A(16, 8), 由y 2=4x(y >0)可得y =2√x , 所以y ′=√x ,所以k AB =y ′|x=16=14,所以l AB :y −8=14(x −16),即y =14x +4,所以B(0, 4), 设l:y =mx +4,P(y 124,y 1),Q(y 224,y 2),由l OA :y =12x ,l OQ :y =4y 2x ,得y M =y128,y N =y 12y 2,由{y =mx +4y 2=4x ,解得my 2−4y +16=0, 所以y 1+y 2=4m ,y 1y 2=16m,所以y P +y N =y 1+y 12y 2=y 12(y 1+y 2)y 1y 2=y 12⋅4m16m =y 124=2y M ,即M 为PN 中点. 【考点】直线与抛物线的位置关系 【解析】(1)根据题意设出直线OA 的方程,联立抛物线方程可表示出交点A 的坐标,再根据圆心到直线的距离小于半径可以求得OA 斜率范围,继而算出A 点横坐标的范围;(2)对抛物线求导,可求出AB 的斜率,继而写出AB 的方程,可以求得B 点坐标,设出直线l 及交点坐标,联立直线与抛物线方程可以推得y P +y N =2y M ,得出结论. 【解答】由题意直线OA 斜率存在且不为零,设l OA :y =kx ,则 由{y =kx y 2=4x′解得x A =4k 2, 又D(2, 1)到l OA :kx −y =0的距离为√k 2+1≤1, 即0≤k ≤43, 所以x A ∈[94,+∞).证明:当直线OA 过圆心D(2, 1)时,k =12,x A =4k 2=16,A(16, 8), 由y 2=4x(y >0)可得y =2√x , 所以y ′=√x ,所以k AB =y ′|x=16=14,所以l AB :y −8=14(x −16),即y =14x +4, 所以B(0, 4), 设l:y =mx +4,P(y 124,y 1),Q(y 224,y 2),由l OA :y =12x ,l OQ :y =4y 2x ,得y M =y128,y N =y 12y 2,由{y =mx +4y 2=4x ,解得my 2−4y +16=0, 所以y 1+y 2=4m ,y 1y 2=16m,所以y P+y N=y1+y12y2=y12(y1+y2)y1y2=y12⋅4m16m=y124=2y M,即M为PN中点.已知等差数列{a n}的公差d∈(0, π],数列{b n}满足b n=sin(a n),集合S={x|xb n, n∈N∗}.(1)若a1=0,d=2π3,求集合S;(2)若a1=π2,求d使得集合S恰有两个元素;(3)若集合S恰有三个元素,b n+T=b n,T是不超过5的正整数,求T的所有可能值,并写出与之相应的一个等差数列{a n}的通项公式及集合S.【答案】∵等差数列{a n}的公差d∈(0, π],数列{b n}满足b n=sin(a n),集合S={x|xb n, n∈N∗},∴a1=0,d=2π3,a n=2π3(n−1),∴b n=sin(a n)=0,−√32,√3 2,故S={0, −√32,√32};a1=π2,a n=π2+(n−1)d,d∈(0, π],根据题意,集合S恰有两个元素;当d=π时,sin(π2+(n−1)π)={1,n−1,n,故成立,因为a1=π2,要使a n(n≥2)的值唯一,在一个周期内,角的终边关于y轴对称,且值相等如图3d=2π,d=2π3,故d=π或2π3;①当T=3时,b n+3=b n,集合S={b1, b2, b3},符合题意.与之相应的一个等差数列a n的通项公式为a n=2π3n,此时S={−√32,√32,0}.②当T=4时,b n+4=b n,sin(a n+4d)=sina n,a n+4d=a n+2kπ或a n+4d=2kπ−a n,等差数列a n的公差d∈(0, π],故a n+4d=a n+2kπ,d=kπ2,又k=1或2,∴当k=1时满足条件,此时S={0, 1, −1}与之相应的一个等差数列a n的通项公式为a n=π2n,此时S={0, 1, −1}【考点】数列与函数的综合【解析】(1)根据等差数列的通项公式写出a n,进而求出b n,再根据周期性求解;(2)由集合S的元素个数,分析数列b n的周期,进而可求得答案;(3)分别令T=1,2,3,4,5进行验证,判断T的可能取值,并写出与之相应的一个等差数列a n的通项公式及集合S.【解答】∵等差数列{a n}的公差d∈(0, π],数列{b n}满足b n=sin(a n),集合S={x|xb n, n∈N∗},∴a1=0,d=2π3,a n=2π3(n−1),∴b n=sin(a n)=0,−√32,√3 2,故S={0, −√32,√32};a1=π2,a n=π2+(n−1)d,d∈(0, π],根据题意,集合S恰有两个元素;当d=π时,sin(π2+(n−1)π)={1,n−1,n,故成立,因为a1=π2,要使a n(n≥2)的值唯一,在一个周期内,角的终边关于y轴对称,且值相等如图3d=2π,d=2π3,故d=π或2π3;①当T=3时,b n+3=b n,集合S={b1, b2, b3},符合题意.与之相应的一个等差数列a n的通项公式为a n=2π3n,此时S={−√32,√32,0}.②当T=4时,b n+4=b n,sin(a n+4d)=sina n,a n+4d=a n+2kπ或a n+4d=2kπ−a n,等差数列a n的公差d∈(0, π],故a n+4d=a n+2kπ,d=kπ2,又k=1或2,∴当k=1时满足条件,此时S={0, 1, −1}与之相应的一个等差数列a n的通项公式为a n=π2n,此时S={0, 1, −1}已知函数f(x)=(x −1)lnx ,g(x)=x −lnx −3e . (Ⅰ)求函数f(x)的单调区间;(Ⅱ)令ℎ(x)=mf(x)+g(x)(m >0)两个零点x 1,x 2(x 1<x 2),证明:x 1+e >x 2+1e . 【答案】(1)由题可知f ′(x)=lnx +1−1x ,f ′(x)单调递增,且f ′(1)=0, 当0<x <1时,f ′(x)<0,当x ≥1时,f ′(x)≥0; 因此f(x)在(0, 1)上单调递减,在[1, +∞)上单调递增.(2)证明:由ℎ(x)=m(x −1)lnx +x −lnx −3e 有两个零点可知 由ℎ(x)=m(1+lnx −1x )+1−1x 且m >0可知, 当0<x <1时,ℎ′(x)<0,当x ≥1时,ℎ′(x)≥0; 即ℎ(x)的最小值为ℎ(1)=1−3e <0,因此当x =1e 时,ℎ(1e )=m(1e −1)(−1)+1e −(−1)−3e =m(e−1)+e−2e>0,可知ℎ(x)在(1e ,1)上存在一个零点;当x =e 时,ℎ(e)=m(e −1)+e −1−3e >0, 可知ℎ(x)在(1, e)上也存在一个零点; 因此x 2−x 1<e −1e ,即x 1+e >x 2+1e .【考点】利用导数研究函数的单调性 利用导数研究函数的最值 【解析】(Ⅰ)求出函数f(x)的导数,利用导函数的符号判断函数的单调性,求出单调区间;(Ⅱ)求出ℎ(x)=mf(x)+g(x)(m >0)的导数,求解函数的最小值,通过零点判断定理,转化两个零点x 1,x 2(x 1<x 2),所在位置,即可证明:x 1+e >x 2+1e . 【解答】(1)由题可知f ′(x)=lnx +1−1x ,f ′(x)单调递增,且f ′(1)=0, 当0<x <1时,f ′(x)<0,当x ≥1时,f ′(x)≥0; 因此f(x)在(0, 1)上单调递减,在[1, +∞)上单调递增.(2)证明:由ℎ(x)=m(x −1)lnx +x −lnx −3e 有两个零点可知 由ℎ(x)=m(1+lnx −1x )+1−1x 且m >0可知, 当0<x <1时,ℎ′(x)<0,当x ≥1时,ℎ′(x)≥0; 即ℎ(x)的最小值为ℎ(1)=1−3e <0,因此当x =1e 时,ℎ(1e )=m(1e −1)(−1)+1e −(−1)−3e =m(e−1)+e−2e>0,可知ℎ(x)在(1e ,1)上存在一个零点;当x =e 时,ℎ(e)=m(e −1)+e −1−3e >0, 可知ℎ(x)在(1, e)上也存在一个零点; 因此x 2−x 1<e −1e ,即x 1+e >x 2+1e .已知椭圆C:y 2a 2+x 2b 2=1(a >b >0)的离心率为√22,且过定点M(1, √22). (1)求椭圆C 的方程;(2)已知直线l:y =kx −13(k ∈R)与椭圆C 交于A 、B 两点,试问在y 轴上是否存在定点P ,使得以弦AB 为直径的圆恒过P 点?若存在,求出P 点的坐标和△PAB 的面积的最大值,若不存在,说明理由. 【答案】由已知可得{ e =ca =√22b 2+c 2=a 212a 2+1b 2=1⇒{a 2=52b 2=54, ∴ 椭圆C 的方程为2y 25+4x 25=1;由{y =kx −132y 25+4x 25=1得:9(2k 2+4)x 2−12kx −43=0① 设A(x 1, y 1),B(x 2, y 2),则x 1、x 2是方程①的两根, ∴ x 1+x 2=12k9(2k 2+4),x 1x 2=−439(2k 2+4),设P(0, p),则PA →=(x 1,y 1−p),PB →=(x 2,y 2−p),PA →⋅PB →=x 1x 2+y 1y 2−p(y 1+y 2)+p 2=x 1x 2+(kx 1−13)(kx 2−13)−pk(x 1+x 2)+2p 3+p 2=(18p 2−45)k 2+36p 2+24p −399(2k 2+4)假设在y 轴上存在定点P ,使得以弦AB 为直径的圆恒过P 点, 则PA →⊥PB →,即PA →⋅PB →=0.即(18p 2−45)k 2+36p 2+24p −39=0对任意k ∈R 恒成立, ∴ {18p 2−45=036p 2+24p −39=0 ,此方程组无解,∴ 不存在定点满足条件. 【考点】直线与椭圆结合的最值问题椭圆的离心率 【解析】(1)运用离心率公式和点M 满足椭圆方程,解方程可得a ,b ,进而得到椭圆方程; (2)联立直线方程和椭圆方程,运用韦达定理,设P(0, p),求得向量PA ,PB 和数量积,再由直径所对的圆周角为直角,结合向量垂直的条件,即可得到结论. 【解答】由已知可得{ e =ca =√22b 2+c 2=a 212a 2+1b 2=1⇒{a 2=52b 2=54, ∴ 椭圆C 的方程为2y 25+4x 25=1;由{y =kx −132y 25+4x 25=1得:9(2k 2+4)x 2−12kx −43=0① 设A(x 1, y 1),B(x 2, y 2),则x 1、x 2是方程①的两根, ∴ x 1+x 2=12k9(2k 2+4),x 1x 2=−439(2k 2+4),设P(0, p),则PA →=(x 1,y 1−p),PB →=(x 2,y 2−p),PA →⋅PB →=x 1x 2+y 1y 2−p(y 1+y 2)+p 2=x 1x 2+(kx 1−13)(kx 2−13)−pk(x 1+x 2)+2p 3+p 2=(18p 2−45)k 2+36p 2+24p −399(2k 2+4)假设在y 轴上存在定点P ,使得以弦AB 为直径的圆恒过P 点, 则PA →⊥PB →,即PA →⋅PB →=0.即(18p 2−45)k 2+36p 2+24p −39=0对任意k ∈R 恒成立, ∴ {18p 2−45=036p 2+24p −39=0 ,此方程组无解,∴ 不存在定点满足条件.。

衡水中学2019-2020学年上学期期中考试编

衡水中学2019-2020学年上学期期中考试编

2019~2020 学年度高三年级上学期期中考试英语试卷ASmart Kids Festival Events21 it isS mart K ids is a collection of one hundred events scheduled in O ctober. T his year, experimenting with Pay What You Decide (PWYD). That is, you can decide to pay what you want to or can afford, after you have attended an event. You can pre-book events without paying for a ticket in advance. Here are some of the director's picks.Walk on the Wild SideNot ticketed, Free,Join storyteller Sarah Law to hear science stories about animals. Along the way you'll meetall sorts of beautiful creatures and discover life cycles and food chains. Best suited to children aged 5-9. Children under 8 must be accompanied by an adult.Introduction to WavesPre-book, PWYD (21 )Subjects range from sound waves to gravity waves, and from waves of light to crashing waves on the ocean. Mike Goldsmith explores the fundamental features shared by all waves in the natural world.Science in the Field Not ticketed, Free22T his storytelling night features a scientist sharing his favourite memories of gatheringfirst-hand data on various field trips. Come along for inspiring and informative stories straight from the scientist's mouth. Join Mark Samuels to find out more in this fun-filled workshop.Festival DinnerPre-book, 2邺er personWhether you want to explore more about food, or just fancy a talk over a meal, join us tomark the first science festival in London. Which foods should you eat to trick your brain into thinking that you are full? Find out more from Tom Crawford.21. In which event can you decide the payment?A. Walk on the Wild Side.B. Introduction to Waves.C. Science in the Field.D. Festival Dinner.22. Who will talk about experiences of collecting direct data?A. Sarah Law.B. Mike Goldsmith.C. Mark Samuels.D. Tom Crawford.23. What do the four events have in common? (关键词蓝色标注)A. Family-based.B. Science- themed.C. Picked by children.D. Filled with adventuresAlmost none of us have the time to read everything we'd like to read. Yet we lose countless hours to daily activities that bring us little joy like taking buses and waiting in line. 24What if we could turn these little blocks of unoccupied time into precious and rewarding moments for learningand reflection?Founded in 2012, iReader, a micro-learning app on mobile phones, brings the biggest ideas from best-selling books through 15-minute audio( 音频) and text. So far, more than 3,000 books have been included, ranging from psychology and parenting to management and economics, with new titles added every day.iReader is pioneering a new method of reading, with over 9 million users enjoying the benefits already. 25A ccording to the P ew R esearch C enter (PRC), the B ritish read just 4 paper books a year and over 25% haven't read a single paper book this year, but reading isn't dying. There are now more ways for the British to read than ever before, due to the widespread use of e-books and audio books.26T he books in i R eader are rewritten to ensure it is easy to remember the main content.The way the content is edited has been specifically designed to ensure it is useful in practice. Besides, the content is rewritten with relevant examples in real life, which means users are more likely toremember and apply what is helpful to them.Holger Seim, German co-founder of this app, declares,27 "iReader gives you the biggest ideas in the shortest possible time. It transforms great ideas into little packs you can listen to or read in just 15 minutes. "24. What does the author suggest people do in the unoccupied time?A. Read and think.B. Write and share.C. Avoid taking buses.D. Bring joy to daily routines.25. What can we infer from the PRC findings?A. The British benefit a lot from reading.B. Reading methods are more important than before.C. Digital technology is taking the place of paper books.D. New forms of books are changing the way the British read.26. How does iReader make the content easy to remember?A. By bringing fun to it.B. By making it useful.C. By using artistic designs.D. By taking users as examples.27. What is the best title for the text?A. iReader Prevents Reading from DyingB. iReader Unites Worldwide Book-lovers.C. iReader: The Best New App CreationD. iReader: Big Ideas in Small PackagesCI visited Elba last June, joining Mary and John on a bicycling vacation. They made the arrangements for the car, hotel and bicycles. 28 I studied the history of the island, which of course particularly features Napoleon.Napoleon (now I know) picked Elba as a place for peace when he was forced to give up thethrone (王权)as Emperor of France in 1814. Far from being a prison island, Elba is beautiful with towering mountains, tick forests and sweeping bays and beaches.It is also an island filled with treasure. Very early on this island, locals discovered rich deposits of iron. Soon outsiders, too, discovered the iron and 150 other valuable minerals on this little piece of land. 29Long before Etruscans and other Greeks set foot on it, Dorians had movedin by the tenth century B.C. and were mining the island. The Romans ruled next, obtaining the minerals and building grand houses overlooking the sea. From the twelfth century until the nineteenth, the island was traded back and forth and was passed to France in 1802. Then came Napoleon, the new ruler of Elba.I was eager to visit his house in Portoferraio. 30 The Emperor lived with his court and his mother, but his wife, Marie Louise had ensconced herself in the splendid Viennese palace of her father, Emperor of Austria. She lived safely there and showed lttle interest in Visiting her husband in his mini-kingdom. Apparently, Napoleon wasn't troubled much by this. He was too busy riding everywhere on horseback, building roads, modernizing agriculture and, above all, sharpening his tiny army and navy into readiness for his escape.31 In the formal gardens behind the house it seemed to me that I could imagine the exiled (流放的)conqueror's anxious thoughts. He might gaze over where 1 stood now, toward the lighthouse of the Stella fort, the sandy bay, and across it, the green mountains of the Tuscan coast. Napoleon spent only ten months here before making his victorious return to France and the throne.28. What did the author do for the visit to EIba?A. He did research on its past,B. He arranged transportation.C. He planned bicycling routes.D. He booked accommodation.29. Who might be the earliest outsiders to Elba according to the text?A. Napoleon and his army.B. Etruscans and other Greeks.C. DoriansD. Romans.30. What does the underlined word “ ensconced" probably mean?A. SettledB. LockedC. CuredD. Controlled.31. What came to the author's mind during his visit to Napoleon's gardens?A. Beautiful views on Elba.B. Terrible living conditions on Elba.C. Napoleon's ambition to regain power.D. Hardship of Napoleon's return to France.D32/35DNA analysis has revealed family relationships between more than 10 generations of Stone Age people at megalithic (巨石的)tombs in Ireland and Sweden.The evidence suggests that megaliths, prehistoric large stone structures, sometimes acted as graves for family groups in northwestern Europe thousands of years ago. The latest findings throw new light on the origins and social structure of the groups that built megaliths in this region--- a history that has long been hidden in mystery.33For their study, the international team of researchers analyzed the genomes(基因组)--- the complete set of genetic material in a cell- - of 24 Stone Age individuals from five megalithic burial sites in Ireland, Scotland and Gotland, a large Swedish island in the Baltic Sea.32This analysis showed that many of the individuals buried at each megalith, who alllived between 3,800 B.C. and 2,600 B.C, according to radiocarbon-dating of their remains, were closely related via family ties.34The results also showed that the individuals buried at the megaliths were related to Neolithic farmers in northern and western Europe but genetically distinct from other hunter-gatherers. This was particularly noticeable at the Ansarve site on the island of Gotland.“The people buried in the Ansarve tomb are remarkably different on a genetic level compared to the individuals dug out from hunter-gather contexts, showing that the burial tradition in this megalithic tomb, which lasted for over 700 years, was performed by distinct groups with roots in the European Neolithic expansion," Magdalena Fraser, co-first author from Uppsala University, said in the statement.32. What's the significance of the new findings?A. It reveals the family ties between people in Ireland and Sweden.B. It implies that many people buried in the tombs were closely related.C. It indicates the long-hidden mystery concerning DNA analysis.D. It suggests that the megaliths became tombs thousands of years ago.33. How did the researchers reach their findings?A. By interviewing individuals.B. By traveling to different regions.C. By analyzing genes.D. By studying the burial sites.34. Which is true according to the latest findings?A. Some people buried in the tombs were related to farmers.B. Few people buried in the tombs were genetically different.C. All the people buried at megaliths "had family relationships.D. People buried in the Ansarve tomb were dug out 700 years later.35. What's the main idea of this passage?A. Megaliths served as tombs thousands of years ago.B. People buried at the megaliths were recently analyzed.C. Latest findings shed light on a mystery about burials.D. Stone-Age people in Ireland and Sweden had close ties.第二节(共5小题:每小题2分,满分10分)根据短文内容,从短文后的选项中选出能填入空白处的最佳选项。

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2019~2020学年度高三年级上学期期中考试英语试卷ASmart Kids Festival EventsSmart Kids is a collection of one hundred events scheduled in October. This year,21it is experimenting with Pay What You Decide (PWYD). That is, you can decide to pay what you want to or can afford, after you have attended an event. You can pre-book events without paying for a ticket in advance. Here are some of the director's picks.Walk on the Wild SideNot ticketed, Free,Join storyteller Sarah Law to hear science stories about animals. Along the way you'll meet all sorts of beautiful creatures and discover life cycles and food chains. Best suited to children aged 5-9. Children under 8 must be accompanied by an adult.Introduction to WavesPre-book, PWYD(21)Subjects range from sound waves to gravity waves, and from waves of light to crashing waves on the ocean. Mike Goldsmith explores the fundamental features shared by all waves in the natural world.Science in the FieldNot ticketed, Free22This storytelling night features a scientist sharing his favourite memories of gatheringfirst-hand data on various field e along for inspiring and informative stories straight from the scientist's mouth. Join Mark Samuels to find out more in this fun-filled workshop. Festival DinnerPre-book, £25 per personWhether you want to explore more about food, or just fancy a talk over a meal, join us to mark the first science festival in London. Which foods should you eat to trick your brain into thinking that you are full? Find out more from Tom Crawford.21. In which event can you decide the payment?A. Walk on the Wild Side.B. Introduction to Waves.C. Science in the Field.D. Festival Dinner.22. Who will talk about experiences of collecting direct data?A. Sarah Law.B. Mike Goldsmith.C. Mark Samuels.D. Tom Crawford.23. What do the four events have in common?(关键词蓝色标注)A. Family-based.B. Science- themed.C. Picked by children.D. Filled with adventuresBAlmost none of us have the time to read everything we'd like to read. Yet we lose countlesshours to daily activities that bring us little joy like taking buses and waiting in line. 24What if we could turn these little blocks of unoccupied time into precious and rewarding moments forlearning and reflection?Founded in 2012, iReader, a micro-learning app on mobile phones, brings the biggest ideas from best-selling books through 15-minute audio(音频) and text. So far, more than 3,000 books have been included, ranging from psychology and parenting to management and economics, with new titles added every day.iReader is pioneering a new method of reading, with over 9 million users enjoying thebenefits already. 25According to the Pew Research Center (PRC), the British read just 4 paper books a year and over 25% haven't read a single paper book this year, but reading isn't dying. There are now more ways for the British to read than ever before, due to the widespread use of e-books and audio books.26The books in iReader are rewritten to ensure it is easy to remember the main content.The way the content is edited has been specifically designed to ensure it is useful in practice. Besides, the content is rewritten with relevant examples in real life, which means users are more likely toremember and apply what is helpful to them.Holger Seim, German co-founder of this app, declares,27"iReader gives you the biggest ideas in the shortest possible time. It transforms great ideas into little packs you can listen to or read in just 15 minutes."24. What does the author suggest people do in the unoccupied time?A. Read and think.B. Write and share.C. Avoid taking buses.D. Bring joy to daily routines.25. What can we infer from the PRC findings?A. The British benefit a lot from reading.B. Reading methods are more important than before.C. Digital technology is taking the place of paper books.D. New forms of books are changing the way the British read.26. How does iReader make the content easy to remember?A. By bringing fun to it.B. By making it useful.C. By using artistic designs.D. By taking users as examples.27. What is the best title for the text?A. iReader Prevents Reading from DyingB. iReader Unites Worldwide Book-lovers.C. iReader: The Best New App CreationD. iReader: Big Ideas in Small PackagesCI visited Elba last June, joining Mary and John on a bicycling vacation. They made thearrangements for the car, hotel and bicycles.28I studied the history of the island, which of course particularly features Napoleon.Napoleon (now I know) picked Elba as a place for peace when he was forced to give up the throne (王权) as Emperor of France in 1814. Far from being a prison island, Elba is beautiful with towering mountains, tick forests and sweeping bays and beaches.It is also an island filled with treasure. Very early on this island, locals discovered rich deposits of iron. Soon outsiders, too, discovered the iron and 150 other valuable minerals on thislittle piece of land. 29Long before Etruscans and other Greeks set foot on it, Dorians had moved in by the tenth century B.C. and were mining the island. The Romans ruled next, obtaining the minerals and building grand houses overlooking the sea. From the twelfth century until the nineteenth, the island was traded back and forth and was passed to France in 1802. Then came Napoleon, the new ruler of Elba.I was eager to visit his house in Portoferraio.30The Emperor lived with his court and his mother, but his wife, Marie Louise had ensconced herself in the splendid Viennese palace of her father, Emperor of Austria.She lived safely there and showed lttle interest in Visiting her husband in his mini-kingdom. Apparently, Napoleon wasn't troubled much by this. He was too busy riding everywhere on horseback, building roads, modernizing agriculture and, above all, sharpening his tiny army and navy into readiness for his escape.31In the formal gardens behind the house it seemed to me that I could imagine the exiled(流放的) conqueror's anxious thoughts. He might gaze over where 1 stood now, toward the lighthouse of the Stella fort, the sandy bay, and across it, the green mountains of the Tuscan coast. Napoleon spent only ten months here before making his victorious return to France and the throne.28. What did the author do for the visit to EIba?A. He did research on its past,B. He arranged transportation.C. He planned bicycling routes.D. He booked accommodation.29. Who might be the earliest outsiders to Elba according to the text?A. Napoleon and his army.B. Etruscans and other Greeks.C. DoriansD. Romans.30. What does the underlined word“ensconced" probably mean?A. SettledB. LockedC. CuredD. Controlled.31. What came to the author's mind during his visit to Napoleon's gardens?A. Beautiful views on Elba.B. Terrible living conditions on Elba.C. Napoleon's ambition to regain power.D. Hardship of Napoleon's return to France.D32/35DNA analysis has revealed family relationships between more than 10 generationsof Stone Age people at megalithic (巨石的) tombs in Ireland and Sweden.The evidence suggests that megaliths, prehistoric large stone structures, sometimes acted as graves for family groups in northwestern Europe thousands of years ago. The latest findings thrownew light on the origins and social structure of the groups that built megaliths in this region--- a history that has long been hidden in mystery.33For their study, the international team of researchers analyzed the genomes(基因组)---the complete set of genetic material in a cell-- of 24 Stone Age individuals from five megalithic burial sites in Ireland, Scotland and Gotland, a large Swedish island in the Baltic Sea.32This analysis showed that many of the individuals buried at each megalith, who alllived between 3,800 B.C. and 2,600 B.C, according to radiocarbon-dating of their remains, were closely related via family ties.34The results also showed that the individuals buried at the megaliths were related toNeolithic farmers in northern and western Europe but genetically distinct from other hunter-gatherers. This was particularly noticeable at the Ansarve site on the island of Gotland.“The people buried in the Ansarve tomb are remarkably different on a genetic level compared to the individuals dug out from hunter-gather contexts, showing that the burial tradition in this megalithic tomb, which lasted for over 700 years, was performed by distinct groups with roots in the European Neolithic expansion," Magdalena Fraser, co-first author from Uppsala University, said in the statement.32. What's the significance of the new findings?A. It reveals the family ties between people in Ireland and Sweden.B. It implies that many people buried in the tombs were closely related.C. It indicates the long-hidden mystery concerning DNA analysis.D. It suggests that the megaliths became tombs thousands of years ago.33. How did the researchers reach their findings?A. By interviewing individuals.B. By traveling to different regions.C. By analyzing genes.D. By studying the burial sites.34. Which is true according to the latest findings?A. Some people buried in the tombs were related to farmers.B. Few people buried in the tombs were genetically different.C. All the people buried at megaliths "had family relationships.D. People buried in the Ansarve tomb were dug out 700 years later.35. What's the main idea of this passage?A. Megaliths served as tombs thousands of years ago.B. People buried at the megaliths were recently analyzed.C. Latest findings shed light on a mystery about burials.D. Stone-Age people in Ireland and Sweden had close ties.第二节(共5小题:每小题2分,满分10分)根据短文内容,从短文后的选项中选出能填入空白处的最佳选项。

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