扬州大学附属中学本部2020-2021学年高一年级第一学期期中考试
江苏扬州大学附属中学高一物理上学期期中考试试题卷 含解析

江苏扬州大学附属中学高一物理上学期期中考试试题卷含解析一、选择题1.一辆汽车由静止开始做匀速直线运动,从开始运动到驶过第一个100m距离时,速度增加了10m/s,汽车驶过第二个100m时,速度的增加量是A.4.1m/s B.8.2m/s C.10m/s D.20m/s2.下列各组物理量中,都属于矢量的是()A.位移和路程B.时间和时刻C.力和加速度D.平均速度和速率3.几个水球可以挡住子弹?实验证实:4 个水球就足够了!4个完全相同的水球紧挨在一起水平排列,如图所示,子弹(可视为质点)在水球中沿水平方向做匀变速直线运动,恰好穿出第 4 个水球,则以下说法正确的是()A.子弹在每个水球中速度变化相同B.由题干信息可以确定子弹穿过每个水球的时间C.由题干信息可以确定子弹在每个水球中运动的时间相同D.子弹穿出第 3 个水球的瞬间速度与全程的平均速度相等4.粗细均匀的电线架在A、B两根电线杆之间.由于热胀冷缩,电线在夏、冬两季呈现如图所示的两种形状,若电线杆始终处于竖直状态,下列说法中正确的是()A.冬季,电线对电线杆的拉力较大B.夏季,电线对电线杆的拉力较大C.夏季与冬季,电线对电线杆的拉力一样大D.夏季,电线杆对地的压力较大5.他是第一个把实验引进力学的科学家,并且利用实验和数学逻辑推理相结合的方法研究物理学基本问题,从而有力地推进了人类科学认识的发展,这位科学家是A.爱因斯坦B .亚里士多德C .伽利略D .牛顿6.课间休息时,负责擦黑板的同学为方便老师下节课使用,将磁性板擦吸附在磁性黑板上如图所示,下列说法正确的是( )A .板擦受到四个力作用,其中有三个力的施力物体是黑板B .板擦受到的摩擦力大于重力C .作用在板擦上的磁力和弹力是一对相互作用力D .若磁力突然消失,板擦仍能保持静止不动7.如图所示,质量为m 的等边三棱柱静止在水平放置的斜面上.已知三棱柱与斜面间的动摩擦因数为μ,斜面的倾角为30°,则斜面对三棱柱的支持力与摩擦力的大小分别为( )A 33mgB .12mg 3mgC .12mg 和12mg μ D 3和12mg 8.在某次检测国产某品牌汽车的刹车性能时,通过传感器发现踩下刹车后,汽车的位移与时间的关系满足2305x t t =-,则关于该次测试,下列说法正确的是A .汽车4s 内的位移是40mB .汽车的加速度是﹣5m/s 2C .汽车的初速度是20m/sD .汽车刹车2s 后的速度是10m/s9.一物体在地面以速度为 v 向上竖直上抛,不计空气阻力,经过 t 时间到最高点,上升高度为 h ,则A .物体通过前半程和后半程所用时间之比为 1:(21 )B .物体通过2h 处的速度为 2v C .物体经过 2t时的速度为2v D .物体经过前 2t 和后 2t 的位移之比为 1:310.近几年,在国家宏观政策调控下,我国房价上涨出现减缓趋势。
江苏省扬州市扬州大学附属中学东部分校2023-2024学年高一上学期期中地理试卷(含答案)

扬大附中东部分校2023—2024学年第一学期期中考试高一年级地理(总分100分时间75分钟)一、选择题(共34题,每小题2分,共计68分)2022年12月4日,神舟十四号返回舱安全降落在东风着陆场预定区域,媒体报道了相关事宜,有细心的网友发现,着陆的返回舱外表“烧焦”,看起来“颜值”不是那么高。
据此完成1~3题。
图1 图21.图1中的神舟十四号返回舱表面有类似“烧焦”的痕迹,与其成因类似的是()A.流星 B.极光 C.彗星 D.磁暴2.按照天体类型,图2中在太空运行的空间站属于()A.恒星 B.行星 C.流星 D.卫星3.下列有关天体的叙述,正确的是()A.环绕恒星运转的天体是卫星 B.宇宙中最基本的天体是行星C.晴朗的夜空,肉眼所见的多是恒星 D.天体都是由固态物质组成的火星探测是指人类对火星进行的科学探测活动。
2023年7月13日,国际某著名学术期刊发表论文称,通过“毅力号”火星车在火星上的陨击坑,探测到多种有机分子的证据。
下图示意火星上“奋进”陨石坑西缘“不屈不挠峡谷”的景象。
据此完成4~5题。
4.火星被认为是可能拥有生命的行星,主要是由于火星()A.温度比地球高 B.与太阳距离适中 C.拥有大量淡水 D.太阳光照稳定5.火星表面遍布大小不等的陨石坑的主要原因是()A.风化作用弱,陨击坑保存完好 B.火星地表松软,容易被撞击成坑C.邻近小行星带,受撞击的概率高 D.火星地表温度高,不断融化形成坑青海省是有名的阳光地带,有10万平方千米的荒漠化、半荒漠化土地适合集中建设光伏发电项目。
光热发电连续性差,例如青海中控德令哈50MW光热电站(见下图)在2021年8月26日至9月25日的31天中,实际运行25天。
据此完成6~8题。
6.青海大规模建设光伏电站的自然原因是()A.发展快,市场需求大 B.纬度低,日照时间长 C.距海远,太阳活动强 D.海拔高,晴天多7.大面积荒漠化、半荒漠化土地适合集中建设光伏电站,因为()A.交通便利 B.可以少占耕地、良田 C.太阳能分布集中 D.太阳能电站占地面积少8.青海中控德令哈光热电站在该时间段实际少运行了6天,原因最有可能是()A.电站故障 B.阴雨天气 C.常规保养 D.耀斑爆发湖南长沙马王堆一号汉墓中曾出土一副彩绘帛画,画的右上方是一轮红色的太阳,太阳中间有一只三趾的乌鸦。
江苏省扬州师范大学附中2020┄2021学年度高一上学期期中考试英语

江苏省扬州师范大学附中2020┄2021学年度高一上学期期中考试(英语)(11)本试卷分五部分满分120分考试时间120分钟。
第Ⅰ卷(共 81分)第一部分:听力(共两节,满分16分)第一节(共5小题;每小题1分,满分5分)听下面5段对话。
每段对话后有一个小题,从题中所给的A,B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1.Where are the speakers talking?A. In the living room.B. In the bedroom.C. At a furniture store.2.How many friends is the man waiting for?A. Three.B. Four.C. Five.3.How much should the man pay for the tax?A. 8 dollars.B. 5 dollars.C. 4 dollars.4.What will the woman bring to the party?A. Nothing.B. Some snacks.C. Some drinks.5.What does the woman want to be?A. A singer.B. A director.C. An actress.第二节(共11小题;每小题1分,满分11分)听下面4段对话或独白。
每段对话或独白后有几个小题,从每题所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。
每段对话或独白读两遍。
6.What does Mr. Green do?A. A manager.B. A professor.C. A doctor.7.When can the man meet Mr. Green?A. On Tuesday morning.B. On Tuesday afternoon.C. On Wednesday afternoon.听第7段材料,回答第8至10题。
2020-2021学年江苏省扬州市某校高一(上)期中考试数学试卷

2020-2021学年江苏省扬州市某校高一(上)期中考试数学试卷一、选择题)1. 设集合A={0,1,3},集合B={2,3,4},则A∪B( )A.{3}B.{0,1,3,3,4}C.{0,1,2,4}D.{0,1,2,3,4}2. 设a∈R,则“a>1”是“a2>a”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3. 函数f(x)=0的定义域为()√|x|−xA.(−∞, 0)B.(−∞, −1)C.(−∞, −1)∪(−1, 0)D.(−∞, 0)∪(0, +∞)4. 函数y=4x的图象大致为( )x2+1A. B.C. D.5. 已知命题p:“∃x0>0,x0+t−1=0”,若p为真命题,则实数t的取值范围是( )A.(1,+∞)B.(−∞,1)C.[1,+∞)D.(−∞,1]<0和不等式ax2+bx−2>0的解集相同,则a,b的值为( )6. 若不等式4x+1x+2A.a=−8,b=−10B.a=−4,b=−9C.a=−1,b=9D.a=−1,b=27. 下列命题中,正确的是( ) A.若a >b ,c >d ,则ac >bd B.若ac >bc ,则a >bC.若ac2<b c 2,则a <bD.若a >b ,c >d ,则a −c >b −d8. 已知函数f (x )的定义域为R ,f (x )是偶函数,f (4)=2,f (x )在(−∞,0)上是增函数,则不等式f (4x −1)>2的解集为( ) A.(−34,54) B.(−∞,−34)∪(54,+∞) C.(−∞,54) D.(−34,+∞)二、多选题)9. 已知函数f (x )是一次函数,满足f(f (x ))=9x +8,则f (x )的解析式可能为( ) A.f (x )=3x +2 B.f (x )=3x −2 C.f (x )=−3x +4 D.f (x )=−3x −410. 下列根式与分数指数幂的互化正确的是( ) A.−√x =(−x )12B.√y 26=y 12(y <0)C.x −13=√x3x ≠0) D.[√(−x )23]34=x 12(x >0)11. 若函数f(x)同时满足:(1)对于定义域内的任意x ,有f(x)+f(−x)=0;(2)对于定义域内的任意x 1,x 2,当x 1≠x 2时,有f(x 1)−f(x 2)x 1−x 2<0,则称函数f(x)为“理想函数”.给出下列四个函数是“理想函数”的是( ) A.f(x)=x 2 B.f(x)=−x 3C.f(x)=x −1x D.f(x)={−x 2,x ≥0,x 2,x <012. 若a >0,b >0,则下列结论正确的有( ) A.√a 2+b 2a+b≤√22B.若1a +4b =2,则a +b ≥92 C.若ab +b 2=2,则a +3b ≥4 D.若a >b >0,则a +1b >b +1a三、填空题)13. 集合A ={a −2,2a 2+5a,12},且−3∈A ,则a =________.14. 已知9a =3,ln x =a ,则x =________.15. 已知x 1,x 2是函数f (x )=x 2−(2k +1)x +k 2的两个零点且一个大于1,一个小于1,则实数k 的取值范围是________.16. 已知正实数a ,b 满足a +b =1,则(1)ab 的最大值是________;(2)1a+2+1b+2的最小值是________. 四、解答题)17. 已知A ={x|2≤x ≤4},B ={x|−m +1≤x ≤2m −1}. (1)若m =2,求A ∩(∁R B);(2)若A ∩B =⌀,求m 的取值范围.18. 计算: (1)1.5−13+80.25×√24+(√23×√3)6−√(−23)23;(2)lg 12−lg 58+lg 12.5−log 89⋅log 278.19. 已知p :A ={x|x 2−5x +6≤0},q :B ={x|x 2−(a +a 2)x +a 3≤0,a >1}. (1)若a =2,求集合B ;(2)如果q 是p 的必要条件,求实数a 的取值范围.20. 已知函数f(x)=xx 2+1. (1)判断并证明函数f(x)的奇偶性;(2)判断当x ∈(−1,1)时函数f(x)的单调性,并用定义证明;(3)若f(x)定义域为(−1,1),解不等式f(2x−1)+f(x)<0.21. 北京、张家口2022年冬奥会申办委员会在俄罗斯索契举办了发布会,某公司为了竞标配套活动的相关代言,决定对旗下的某商品进行一次评估,该商品原来每件售价为25元,年销售8万件.(1)据市场调查,若价格每提高1元,销售量将相应减少2000件,要使销售的总收入不低于原收入,该商品每件定价最多为多少元?(2)为了抓住申奥契机,扩大该商品的影响力,提高年销售量.公司决定立即对该商品(x2−600)万作进行全面技术革新和营销策略改革,并提高定价到x元.公司拟投入16万元作为浮动宣传费用.试问:当为技改费用,投入50万元作为固定宣传费用,投入x5该商品改革后的销售量a至少应达到多少万件时,才可能使改革后的销售收入不低于原收入与总投入之和?并求出此时商品的每件定价.22. 已知二次函数f(x)满足f(x+1)−f(x)=−2x+1且f(2)=15.(1)求函数f(x)的解析式;(2)令g(x)=(1−2m)x−f(x).①若函数g(x)在区间[0,2]上不是单调函数,求实数m的取值范围;②求函数g(x)在区间[0,2]上的最小值.参考答案与试题解析2020-2021学年江苏省扬州市某校高一(上)期中考试数学试卷一、选择题1.【答案】D【考点】并集及其运算【解析】根据并集的定义即可求解.【解答】解:由题意可知,集合A={0,1,3},集合B={2,3,4},则A∪B={0,1,2,3,4}.故选D.2.【答案】A【考点】必要条件、充分条件与充要条件的判断【解析】由不等式解得a的范围,根据充分条件和必要条件的定义,即可判断得出结论.【解答】解:由题意可知,不等式a2>a,解得a>1或a<0,则a>1是a2>a的充分不必要条件.故选A.3.【答案】C【考点】函数的定义域及其求法【解析】由0指数幂的底数不等于0,分母中根式内部的代数式大于0,联立不等式组求得x的取值集合得答案.【解答】解:要使原函数有意义,则{x+1≠0,|x|−x>0,解得x<0且x≠−1,∴函数f(x)=0√|x|−x的定义域是(−∞, −1)∪(−1, 0).故选C.4.【答案】A【考点】函数奇偶性的判断函数的图象【解析】根据函数的奇偶性和函数值的正负即可判断.【解答】解:设f(x)=y=4xx2+1,由题知定义域为实数集R,∵f(−x)=4(−x)(−x)2+1=−4xx2+1=−f(x),∴函数f(x)为奇函数,故排除CD;当x>0时,f(x)>0,故排除B.故选A.5.【答案】B【考点】全称命题与特称命题【解析】根据题目所给信息可得命题p为真命题,进而即可得到t的取值范围.【解答】解:由x0+t−1=0,得x0=1−t.已知命题p:“∃x0>0,x0+t−1=0”为真命题,即1−t>0,解得t<1,则实数t的取值范围为(−∞,1).故选B.6.【答案】B【考点】一元二次不等式的解法根与系数的关系【解析】先求出分式不等式的解集,进而即可得到另一个不等式的根的情况,利用韦达定理进行求解即可.【解答】解:已知不等式4x+1x+2<0,即(4x+1)(x+2)<0,解得−2<x<−14.又不等式4x+1x+2<0与不等式ax2+bx−2>0的解集相等,则不等式ax2+bx−2>0的解集为−2<x<−14,则方程ax2+bx−2=0的两根分别为x1=−2,x2=−14.由根与系数的关系,得x1x2=−2a =12,x1+x2=−ba=−94,解得a=−4,b=−9.故选B.7.【答案】C【考点】不等式的基本性质【解析】根据特殊值法判断A,D,根据不等式的性质判断B,C即可.【解答】解:令a=1,b=−1,c=−1,d=−5,显然A,D不成立,对于B:若c<0,显然不成立,对于C:由c2>0,得:a<b,故C正确,故选C.8.【答案】A【考点】函数奇偶性的性质函数单调性的性质【解析】根据函数的单调性和奇偶性以及不等式进行求解即可.【解答】解:已知函数f(x)是偶函数,即该函数图象关于y轴对称.又f(x)在(−∞,0)上是增函数,则f(x)在(0,+∞)是减函数.因为f(4)=2,所以f(4x−1)>2,即f(4x−1)>f(4),且x∈R,则|4x−1|<4,解得−34<x<54.故选A.二、多选题9.【答案】A,D【考点】函数解析式的求解及常用方法【解析】利用待定系数法求解,设f(x)=kx+b,由题意可知f(f(x))=k(kx+b)+b= k2x+kb+b=9x+8,从而得{k2=9kb+b=8,进而求出k和b的值【解答】解:由题意,设f (x )=kx +b ,则f(f (x ))=k (kx +b )+b =k 2x +kb +b =9x +8, 即{k 2=9,kb +b =8, 解得{k =3,b =2 或{k =−3,b =−4,所以f (x )=3x +2或f (x )=−3x −4. 故选AD . 10.【答案】 C,D【考点】根式与分数指数幂的互化及其化简运算 【解析】根据题目所给信息利用根式与分式指数幂互化的法则,逐一进行筛选即可. 【解答】解:对于选项A ,−√x =−x 12≠(−x )12,故选项A 错误; 对于选项B ,√y 26=−y 13(y <0),故选项B 错误;对于选项C ,x−13=√x3≠0)成立,故选项C 正确;对于选项D ,当x >0时,[√(−x)23]34=[|−x|23]34=x 12,故选项D 正确. 故选CD . 11.【答案】 B,D【考点】函数单调性的判断与证明 函数奇偶性的判断 函数新定义问题【解析】由“理想函数”的定义可知:若f(x)是“理想函数”,则f(x)为定义域上的单调递减的奇函数,将四个函数一一判断即可. 【解答】解:对于定义域上的任意x ,恒有f(x)+f(−x)=0,即f(−x)=−f(x), 故函数f(x)是奇函数.对于定义域上的任意x 1,x 2,当x 1≠x 2时,有f(x 1)−f(x 2)x 1−x 2<0,即(x 1−x 2)[f(x 1)−f(x 2)]<0,∴ 当x 1<x 2时,f(x 1)>f(x 2),即函数f(x)是单调递减函数,故f(x)为定义域上单调递减的奇函数.A,f(x)=x2在定义域R上是偶函数,所以不是“理想函数”,故选项A不符合题意;B,f(x)=−x3在定义域R上是奇函数,且在R上单调递减,所以是“理想函数”,故选项B符合题意;C,f(x)=x−1x在定义域(−∞, 0),(0, +∞)上分别单调递增,所以不是“理想函数”,故选项C不符合题意;D,f(x)={−x2,x≥0,x2,x<0在定义域R上既是奇函数,又是减函数,所以是“理想函数”,故选项D符合题意.故选BD.12.【答案】B,C,D【考点】基本不等式在最值问题中的应用不等式性质的应用【解析】根据基本不等式,对选项逐一分析即可.【解答】解:A,若a>0,b>0,由基本不等式,得a2+b2≥2ab,即2(a2+b2)≥(a+b)2,即√2(a2+b2)≥√(a+b)2=a+b,故√a2+b2a+b ≥√22,当且仅当a=b时取等号,故A选项错误;B,因为a>0,b>0,12(1a+4b)=1,所以a+b=12(a+b)(1a+4b)=12(5+ba+4ab)≥12(5+2√ba⋅4ab)=92,当且仅当1a +4b=2,ba=4ab,即a=32,b=3时取等号,故B选项正确;C,由a>0,b>0,ab+b2=(a+b)b=2,由基本不等式,得a+3b=(a+b)+2b≥2√2b(a+b)=4,当且仅当ab+b2=2,a+b=2b,即a=b=1时取等号,故C选项正确;D,若a>b>0,则1b >1a>0,此时a+1b >b+1a成立,故D选项正确.故选BCD.三、填空题13.【答案】−3 2【考点】元素与集合关系的判断【解析】利用−3∈A,求出a的值,推出结果即可.【解答】解:集合A={a−2,2a2+5a,12},且−3∈A,所以a−2=−3或2a2+5a=−3,解得a=−1或a=−32.当a=−1时,a−2=2a2+5a=−3,不符合题意,舍去.所以a=−32.故答案为:−32.14.【答案】√e【考点】对数的运算性质【解析】由指数的运算性质化简等式右边,等式两边化为同底数的对数后可得x的值.【解答】解:由9a=3,得a=12,∴ln x=12=ln√e,解得x=√e.故答案为:√e.15.【答案】{k|0<k<2}【考点】函数的零点【解析】(1)由已知,关于x的方程的两个根一个大于1,一个小于1,可得f(1)<0,由此构造关于k的不等式,解不等式,即可得到k的取值范围.【解答】解:∵ x1,x2是函数f(x)=x2−(2k+1)x+k2的两个零点且一个大于1,一个小于1,∴ f(1)<0,即1−(2k+1)+k2<0,解得0<k<2.故答案为:{k|0<k<2}.16.【答案】14,45【考点】基本不等式在最值问题中的应用【解析】(1)由基本不等式可求解本题;【解答】解:(1)因为a +b =1,所以由基本不等式,ab ≤(a+b 2)2=14, 当且仅当a =b 时等号成立,所以ab 的最大值是14;(2)因为a +b =1,所以a +2+b +2=5,所以1a+2+1b+2=15(a +2+b +2)(1a +2+1b +2) =15(2+b +2a +2+a +2b +2) ≥15(2+2√b+2a+2⋅a+2b+2)=45, 当且仅当b+2a+2=a+2b+2,即a =b =12时等号成立,所以1a+2+1b+2的最小值为45.故答案为:14;45.四、解答题17.【答案】解:(1)当m =2时,B ={x|−1≤x ≤3},所以∁R B ={x|x <−1或x >3}.又A ={x|2≤x ≤4},所以A ∩(∁R B)={x|3<x ≤4}.(2)当B =⌀时,2m −1<−m +1,解得m <23;当B ≠⌀时,则{2m −1≥−m +1,−m +1>4或 {2m −1≥−m +1,2m −1<2, 解得23≤m <32.综上所述,m 的取值范围是(−∞,32).【考点】交、并、补集的混合运算集合关系中的参数取值问题【解析】此题暂无解析【解答】解:(1)当m =2时,B ={x|−1≤x ≤3},所以∁R B ={x|x <−1或x >3}.又A ={x|2≤x ≤4},所以A ∩(∁R B)={x|3<x ≤4}.(2)当B =⌀时,2m −1<−m +1,解得m <23;当B ≠⌀时,则{2m −1≥−m +1,−m +1>4或 {2m −1≥−m +1,2m −1<2, 解得23≤m <32.综上所述,m 的取值范围是(−∞,32).18.【答案】解:(1)原式=(23)13+234×214+22×33−(23)13=2+4×27=2+108=110.(2)原式=−lg 2−lg 5+lg 8+lg 12.5−23log 23⋅log 32 =−(lg 2+lg 5)+(lg 8+lg 12.5)−23=−1+lg (8×12.5)−23=−1+lg 100−23=−1+2−23=13.【考点】根式与分数指数幂的互化及其化简运算对数的运算性质换底公式的应用【解析】(1)通过根式与分数指数幂的互化及其化简运算求解即可.(2)利用导数的运算法则直接求解即可.【解答】解:(1)原式=(23)13+234×214+22×33−(23)13 =2+4×27=2+108=110.(2)原式=−lg 2−lg 5+lg 8+lg 12.5−23log 23⋅log 32 =−(lg 2+lg 5)+(lg 8+lg 12.5)−23=−1+lg (8×12.5)−23=−1+lg 100−23=−1+2−23=13.19.【答案】解:(1)当a =2时,x 2−(a +a 2)x +a 3=x 2−6x +8.由x 2−6x +8≤0,解得2≤x ≤4,即B ={x|2≤x ≤4},故B =[2,4] .(2)由题意可知,A ={x|x 2−5x +6≤0},∴ A =[2,3].又B ={x|x 2−(a +a 2)x +a 3≤0,a >1},∴ B =[a,a 2].∵ q 是p 的必要条件,可得 {a ≤2,a 2≥3,解得√3≤a ≤2.【考点】一元二次不等式的解法根据充分必要条件求参数取值问题【解析】【解答】解:(1)当a =2时,x 2−(a +a 2)x +a 3=x 2−6x +8.由x 2−6x +8≤0,解得2≤x ≤4,即B ={x|2≤x ≤4},故B =[2,4] .(2)由题意可知,A ={x|x 2−5x +6≤0},∴ A =[2,3].又B ={x|x 2−(a +a 2)x +a 3≤0,a >1},∴ B =[a,a 2].∵ q 是p 的必要条件,可得 {a ≤2,a 2≥3,解得√3≤a ≤2.20.【答案】解:(1)函数f(x)为奇函数. 证明如下:∵ 函数定义域为R ,又f(−x)=−x (−x)2+1=−x x 2+1=−f(x),∴ f(x)=xx 2+1为奇函数.(2)函数f(x)在(−1, 1)上单调递增. 证明如下:任取x 1,x 2∈(−1, 1),且x 1<x 2,则f(x 1)−f(x 2)=x 1x 12+1−x2x 22+1 =x 1(x 22+1)−x 2(x 12+1)(x 12+1)(x 22+1)=(x 2−x 1)(x 1x 2−1)(x 12+1)(x 22+1).∵ x 1,x 2∈(−1, 1),且x 1<x 2,∴ x 2−x 1>0,x 1x 2−1<0,x 12+1>0,x 22+1>0,∴ f(x 1)−f(x 2)<0,即f(x 1)<f(x 2),∴ f(x)在(−1, 1)上单调递增.(3)由(1)可知,f(x)为奇函数,∴ f(2x −1)+f(x)<0等价于f(2x −1)<−f(x)=f(−x),由(2)可知,f(x)在(−1,1)上单调递增,∴ {2x −1<−x,−1<2x −1<1,−1<x <1,解得0<x <13,∴ 不等式的解集为{x|0<x <13}. 【考点】函数奇偶性的判断函数单调性的判断与证明不等式的基本性质函数奇偶性的性质【解析】(1)利用函数的奇偶性的定义即可判断;(2)任取x1,x2∈(−1, 1),且x1<x2,通过作差可判断f(x1)与f(x2)的大小,根据单调性的定义即可作出判断;(3)利用函数的奇偶性、单调性可去掉不等式中的符号“f”,从而转化为具体不等式,注意考虑函数的定义域;【解答】解:(1)函数f(x)为奇函数. 证明如下:∵函数定义域为R,又f(−x)=−x(−x)2+1=−xx2+1=−f(x),∴f(x)=xx2+1为奇函数.(2)函数f(x)在(−1, 1)上单调递增. 证明如下:任取x1,x2∈(−1, 1),且x1<x2,则f(x1)−f(x2)=x1x12+1−x2x22+1=x1(x22+1)−x2(x12+1) (x12+1)(x22+1)=(x2−x1)(x1x2−1)(x12+1)(x22+1).∵x1,x2∈(−1, 1),且x1<x2,∴x2−x1>0,x1x2−1<0,x12+1>0,x22+1>0,∴f(x1)−f(x2)<0,即f(x1)<f(x2),∴f(x)在(−1, 1)上单调递增.(3)由(1)可知,f(x)为奇函数,∴f(2x−1)+f(x)<0等价于f(2x−1)<−f(x)=f(−x),由(2)可知,f(x)在(−1,1)上单调递增,∴{2x−1<−x,−1<2x−1<1,−1<x<1,解得0<x<13,∴不等式的解集为{x|0<x<13}.21.【答案】解:(1)设每件定价最多为t元.由题意,得(8−t−251×0.2)t≥25×8,整理,得t2−65t+1 000≤0,解得25≤t≤40,所以要使销售的总收入不低于原收入,每件定价最多为40元.(2)由题意可知,当x>25时,不等式ax≥25×8+50+16(x2−600)+15x有解,即当x >25时,a ≥150x +16x +15有解. 由于150x +16x ≥2 √150x ⋅x 6=10, 当且仅当150x =x 6,即x =30时等号成立,所以a ≥10.2,所以,当该商品改革后的销售量a 至少达到10.2万件时,才可能使改革后的销售收入不低于原收入与总投入之和,此时商品的每件定价为30元.【考点】一元二次不等式的应用基本不等式在最值问题中的应用【解析】(1)设每件定价为x 元,可得提高价格后的销售量,根据销售的总收人不低于原收入,建立不等式,解不等式可得每件最高定价;(2)由题意,x >25时,不等式ax ≥25×8+50+16(x 2−600)+15x 有解,等价于x >25时,a ≥150x +16x +15有解,利用基本不等式,我们可以求得结论. 【解答】解:(1)设每件定价最多为t 元.由题意,得(8−t−251×0.2)t ≥25×8,整理,得t 2−65t +1 000≤0,解得25≤t ≤40,所以要使销售的总收入不低于原收入,每件定价最多为40元.(2)由题意可知,当x >25时,不等式ax ≥25×8+50+16(x 2−600)+15x 有解, 即当x >25时,a ≥150x +16x +15有解. 由于150x +16x ≥2 √150x ⋅x 6=10, 当且仅当150x =x 6,即x =30时等号成立,所以a ≥10.2,所以,当该商品改革后的销售量a 至少达到10.2万件时,才可能使改革后的销售收入不低于原收入与总投入之和,此时商品的每件定价为30元.22.【答案】解:(1)由题意,设f (x )=ax 2+bx +c (a ≠0).∵ f (x +1)−f (x )=−2x +1 ,即a(x +1)2+b(x +1)+c −ax 2−bx −c=2ax +a +b =−2x +1,∴ {2a =−2,a +b =1,解得a =−1,b =2.又f (2)=15,即4a +2b +c =15, 解得c =15,∴ f (x )=−x 2+2x +15.(2)①由(1)可知,f (x )=−x 2+2x +15, 则g(x)=(1−2m)x −f(x)=x 2−(2m +1)x −15, 故对称轴为x =m +12.∵ 函数g (x )在区间[0,2]上不是单调函数, ∴ 0<m +12<2, ∴ m ∈(−12,32).②由①可知,函数g (x )的对称轴为x =m +12. 当m +12≤0时,即m ≤−12时,g (x )min =g (0)=−15;当0<m +12<2,即−12<m <32时, g (x )min =g (m +12)=−m 2−m −614;当m +12≥2,即m ≥32时,g (x )min =g (2)=−4m −13.综上所述, g(x)min ={ −15,m ≤−12,−m 2−m −614,−12<m <32,−4m −13,m ≥32. 【考点】函数解析式的求解及常用方法二次函数的性质二次函数在闭区间上的最值【解析】【解答】解:(1)由题意,设f (x )=ax 2+bx +c (a ≠0). ∵ f (x +1)−f (x )=−2x +1 ,即a(x +1)2+b(x +1)+c −ax 2−bx −c =2ax +a +b =−2x +1,∴ {2a =−2,a +b =1,解得a =−1,b =2.又f (2)=15,即4a +2b +c =15,解得c =15,∴ f (x )=−x 2+2x +15.(2)①由(1)可知,f (x )=−x 2+2x +15, 则g(x)=(1−2m)x −f(x)=x 2−(2m +1)x −15, 故对称轴为x =m +12. ∵ 函数g (x )在区间[0,2]上不是单调函数, ∴ 0<m +12<2,∴ m ∈(−12,32). ②由①可知,函数g (x )的对称轴为x =m +12. 当m +12≤0时,即m ≤−12时, g (x )min =g (0)=−15;当0<m +12<2,即−12<m <32时, g (x )min =g (m +12)=−m 2−m −614; 当m +12≥2,即m ≥32时,g (x )min =g (2)=−4m −13.综上所述, g(x)min ={ −15,m ≤−12,−m 2−m −614,−12<m <32,−4m −13,m ≥32.。
高中江苏省扬州大学附属中学高一上学期期中数学试题

江苏省扬州大学附属中学【精品】高一上学期期中数学试题 学校:___________姓名:___________班级:___________考号:___________一、单选题1.已知集合{}{}0,1,2,3,02A B x x ==≤≤,则AB =( ) A .[]0,2 B .{}0,2C .{}0,1D .{}0,1,22.函数()f x = )A .(),2-∞B .(],2-∞C .()2,+∞D .[)2,+∞ 3.终边在直线y x =上的角α的取值集合是( )A .2,4k k Z πααπ⎧⎫=+∈⎨⎬⎩⎭B .2,4k k Z πααπ⎧⎫=-∈⎨⎬⎩⎭C .,4k k Z πααπ⎧⎫=-∈⎨⎬⎩⎭ D .,4k k Z πααπ⎧⎫=+∈⎨⎬⎩⎭4.已知一个扇形的圆心角为3弧度,半径为4,则这个扇形的面积等于( ). A .48 B .24 C .12 D .65.已知函数2log ,1,()(2),01,x x f x f x x ⎧=⎨<<⎩则f ⎝⎭的值是( ) A .0 B .1 C .12 D .-12 6.设()f x 为偶函数,且当0x ≥时,()101x f x =-,则当0x <时,()f x =( )A .101x --B .101x -+C .101x ---D .101x --+ 7.给定函数:①12y x =;②12log (1)y x =+;③|1|y x =-;④12x y +=,其中在区间(0,1)上单调递减的函数序号是( )A .①②B .②③C .③④D .①④ 8.函数26()log f x x x =-的零点所在区间是( ) A .()0,1 B .()1,2 C .()3,4 D .()4,+∞ 9.已知奇函数()f x 在0x ≥时的图象如图所示,则不等式()0xf x <的解集为( )A .(1,2)B .(2,1)--C .(2,1)(1,2)--⋃D .(1,1)- 10.若方程()()21210x k x k +--+=有两个不相等的实数根,且仅有一个根在区间(2,3)内,则实数k 的取值范围是( )A .(3,4)B .(2,3)C .(1,3)D .(1,2) 11.已知函数()ln f x x =,若()()()0f m f n m n =>>,则1111m n +=++( ) A .12 B .1 C .2D .4 12.若一系列函数的解析式相同,值域相同,但定义域不同,则称这些函数为“孪生函数”,那么函数解析式为221y x =-,值域为{}1,7的“孪生函数”共有( ) A .10个B .9个C .8个D .4个二、填空题13.1lg lg 707+的值为______. 14.幂函数()f x 的图象过点(4,2),则()2f =______.15.当0a >且1a ≠时,函数1()1x f x a +=-的图象一定过点______. 16.若函数()()12,2,{log ,2a a x a x f x x x --<=≥在R 上单调递减,则实数a 的取值范围是__________.三、解答题17.已知集合{}{}{}37,210,5A x x B x x C x a x a =≤≤=≤≤=-≤≤.(1)求A R ;(2)若()C A B ⊆⋃,求实数a 的取值范围.18.已知函数()31log 1x f x x+=-. (1)判断函数()y f x =的奇偶性并证明;(2)解方程()210x f -=.19.已知二次函数()f x 的最大值为-2,且()()023f f ==-.(1)求()f x 的解析式;(2)若()f x 在区间[],1a a +上的最大值为-6,求实数a 的值.20.某市今年出现百年不遇的旱情,市自来水厂观察某蓄水池供水情况以制定未来12小时的供水措施.现发现某蓄水池中有水450吨,水厂每小时可向蓄水池中注水80吨,同时蓄水池又向居民小区供水,t 小时内供水量为现在开始向水池注水并向居民小区供水.(1)请将蓄水池中存水量S 表示为时间t 的函数;(2)根据蓄水池使用要求,当蓄水池水量低于60吨时,蓄水池必须停止供水.请你判断该居民小区是否会停水,阐述你的理由.21.已知函数()22x x f x -=+. (1)试判断并证明函数()f x 在区间[)0,+∞上的单调性;(2)若()()20f x t f x +⋅≥对任意[]1,2x ∈-恒成立,求实数t 的取值范围.22.已知函数()y f x =,若对于给定的正整数k ,()f x 在其定义域内存在实数0x ,使得()()()00f x k f x f k +=+,则称此函数()f x 为“保k 值函数”.(1)若函数()2xf x =为“保1值函数”,求0x ; (2)①试判断函数()1f x x x =+是否是“保k 值函数”,若是,请求出k ;若不是,请说明理由;②试判断函数()ln1x a f x e =+是否是“保2值函数”,若是,求实数a 的取值范围;若不是,请说明理由.参考答案1.D【解析】【分析】由交集的定义,结合集合A,B ,即可写出A B . 【详解】 因为{}02B x x =≤≤,所以B 中整数有0,1,2,又{}0,1,2,3A =,所以{}0,1,2A B =,故选:D.【点睛】本题考查集合的运算,掌握集合交集的定义是解题的关键,属于简单题.2.D【分析】开偶次方根,被开方数要非负,求函数()f x 的定义域,只需要解不等式20x -≥即可.【详解】要使函数()f x 有意义,只需20x -≥,2x ≥,故选:D.【点睛】本题考查求已知函数的定义域,难度较易.常见函数求定义域需要注意:分式分母不为零、偶次根式被开方数大于等于零、对数的真数大于零、0y x =中{}|0x x ≠.3.D【分析】在π-到π内终边在直线y x =上的角是,44ππ-,由终边相同的角的表示方法可得出终边在直线y x =上的角的集合,可得解.【详解】当的终边在直线y x =(0x >)时, 24k παπ=+,k Z ∈,当的终边在直线y x =(0x <)时,24k παππ=++,k Z ∈,所以角α的取值集合是2,2,44k k Z k k Z ππααπααππ⎧⎫⎧⎫=+∈⋃=++∈⎨⎬⎨⎬⎩⎭⎩⎭=,4k k Z πααπ⎧⎫=+∈⎨⎬⎩⎭, 故选:D.【点睛】本题考查终边相同的角的表示方法,掌握终边相同的角的表示是解题的关键,属于基础题. 4.B【解析】因为扇形的弧长l =3×4=12,则面积S =12×12×4=24,选B. 5.C【分析】 先确定函数自变量的取值范围再代入分段函数解析式求解.【详解】∵2log ,1(),01(2),01x x f x f x x ⎧⎪=<<⎨<<⎪⎩.∴21log 22f f ⎛=== ⎝⎭, 故选C.【点睛】本题主要考查分段函数求值,意在考查学生对该知识的理解掌握水平,属于基础题. 6.A【分析】由()f x 为偶函数,则()()f x f x -=,结合已知,即可求出0x <时函数的解析式.【详解】因为()f x 为偶函数,所以()()f x f x -=,因为0x ≥时,()101x f x =-,所以0x <时,()()101x f x f x -=-=-,故选:A.【点睛】本题主要考查函数解析式的求法,属于中档题.求函数的解析式常见题型有以下几种:(1)根据实际应用求函数解析式;(2)换元法求函数解析式,利用换元法一定要注意,换元后参数的范围;(3)待定系数法求函数解析式,这种方法适合求已知函数名称的函数解析式;(4)消元法求函数解析式;(5)由函数的奇偶性,求分段函数的解析式.7.B【分析】 ①12y x =,(0)x 为幂函数,且x 的指数102α=>,在[0,)+∞上为增函数;②12log (1)y x =+,(1)x >-,为对数型函数,且底数1(0,1)2a =∈,在(1,)-+∞上为减函数;③|1|y x =-,在(,1)-∞上为减函数,④12x y +=为指数型函数,底数21a =>在(,)-∞+∞上为增函数,可得解.【详解】 ①12y x =,(0)x 为幂函数,且x 的指数102α=>,在[0,)+∞上为增函数,故①不可选; ②12log (1)y x =+,(1)x >-,为对数型函数,且底数1(0,1)2a =∈,在(1,)-+∞上为减函数,故②可选;③|1|y x =-,在(,1)-∞上为减函数,在(1,)+∞上为增函数,故③可选;④12x y +=为指数型函数,底数21a =>在(,)-∞+∞上为增函数,故④不可选; 综上所述,可选的序号为②③,故选B.【点睛】本题考查基本初等函数的单调性,熟悉基本初等函数的解析式、图像和性质是解决此类问题的关键,属于基础题.8.C【分析】根据连续函数()26f x log x x=-,可得f (3),f (4)的函数值的符号,由此得到函数()26f x log x x=-的零点所在的区间. 【详解】 ∵连续减函数()26f x log x x =-,∴f (3)=2﹣log 23>0,f (4)=64﹣log 24<0, ∴函数()26f x log x x=-的零点所在的区间是 (3,4), 故选C .【点睛】 本题主要考查函数的零点的定义,判断函数的零点所在的区间的方法,属于基础题. 9.C【分析】根据图象及奇函数的性质判断()y f x =在各个区间的正负,再结合x 与()y f x =异号,即得解.【详解】由图像可知在0x ≥时,在()()012+∞,,,()0f x >;在(1,2),()0f x <; 由()f x 为奇函数,图象关于原点对称,在0x <时,在()(),21,0∞-⋃--,()0f x <;在(2,1)--,()0f x >;又()y xf x =,在0x ≥时与()y f x =同号,在0x <时与()y f x =异号故不等式()0xf x <的解集为:(2,1)(1,2)--⋃故选:C【点睛】本题考查了函数的奇偶性在解不等式中的应用,考查了学生数形结合,转化划归的能力,属于中档题.10.D【分析】根据二次函数图像列不等式,通过解一元二次不等式可解得结果.【详解】因为方程()f x =()()21210x k x k +--+=有两个不相等的实数根,且仅有一个根在区间(2,3)内,所以①当(2)(3)0<f f 时,(44)(105)0k k --<,(1)(2)0k k --<,12k <<; ②令(2)0f =,1k =,方程240x -=另一解为2x =-,不适合;③令(3)0f =,2k =,方程260x x --=另一解为3x =-,不适合.综上k 的取值范围是(1,2),故选:D.【点睛】本题考查根据二次函数零点分布求参数,考查基本分析求解能力,属中档题.11.B【分析】通过讨论x 和1的关系,即可去绝对值,再结合等式即可得到1mn =,代入即可求值.【详解】因为()ln f x x =,若()()()0f m f n m n =>>,所以ln ln n m -=,10m n >>>,即1n m =,所以1111111111m n m m+=+=++++, 故选:B.【点睛】本小题主要考查对数函数的图像,考查函数的图像和单调性,考查化归与转化的数学思想方法,属于中档题.12.B【分析】由值域可求得所有x 可能的取值;则定义域中元素分别为2个,3个和4个,列举出所有可能的结果即可求得个数.【详解】由2211x -=得:1x =±;由2217x -=得:2x =±∴所求“孪生函数”的定义域分别为:{}1,2,{}1,2-,{}1,2-,{}1,2--,{}1,1,2-,{}1,1,2--,{}1,2,2-,{}1,2,2--,{}1,1,2,2--∴共有9个“孪生函数”故选B【点睛】本题考查新定义的问题,涉及到函数定义域的求解;易错点是将值域误认为是无限集,造成求解错误.13.1【分析】直接利用对数指数运算法则得到答案.【详解】11lg lg 70lg(70)lg10177+=⋅==, 故答案为:1.【点睛】本题考查了指数对数的计算,意在考查学生的计算能力.14【分析】首先设出幂函数的解析式,代入点(4,2),进而求出解析式,即可求得结果.【详解】设()f x x α=,因为()f x 的图象过点(4,2),所以42α=,222α=,12α= 12()f x x =,所以(2)f =.【点睛】本题考查函数的求值,形如y x α=的函数是幂函数,注意幂函数的系数为1,考查了运算求解能力.15.()1,0-【分析】根据指数函数的性质可知(1)0f -=,从而求得结果.【详解】因为110(1)110f a a -+-=-=-=,所以函数()f x 的图象一定过点()1,0-.故答案为:()1,0-.【点睛】本题考查指数函数的概念和性质,注意到01(0)a a =≠是解本题的关键,属基础题. 16.⎫⎪⎪⎣⎭【分析】根据题意,由函数的单调性的性质可得1001log 22(1)2aa a a a -<⎧⎪<<⎨⎪≤--⎩,解可得a 的取值范围,即可得答案.【详解】由题意得,因为函数()()12,2,{log ,2a a x a x f x x x --<=≥在R 上单调递减,则1001log 22(1)2aa a a a -<⎧⎪<<⎨⎪≤--⎩.∴12a ≤< ∴实数a的取值范围是⎫⎪⎪⎣⎭.故答案为,12⎫⎪⎪⎣⎭. 【点睛】本题主要考查分段函数的解析式及单调性,属于中档题.分段函数的单调性是分段函数性质中的难点,也是高考命题热点,要正确解答这种题型,必须熟悉各段函数本身的性质,在此基础上,不但要求各段函数的单调性一致,最主要的也是最容易遗忘的是,要使分界点两函数的单调性与整体保持一致.17.(1){3R C A x x =<,或}7x >;(2)(,3]-∞.【分析】(1)由补集的定义和集合A ,即可求出和R C A ;(2)由()C A B ⊆⋃,可知集合C 是A B 的子集,分两种情况:C =∅和C ≠∅,分别讨论即可.【详解】(1)因为{}37A x x =≤≤,所以{3R C A x x =<,或}7x > ;(2)因为{}37A x x =≤≤,{}=210B x x ≤≤,所以{}210A B x x ⋃=≤≤, 因为()C A B ⊆⋃,所以C φ≠时,55210a a a a -≤⎧⎪-≥⎨⎪≤⎩,得532a ≤≤; C φ=时5a a ->,52a <, 综上a 的取值范围是(,3]-∞.故答案为:(,3]-∞.【点睛】本题考查了集合的并集和补集,考查了集合间的包含关系,考查了不等式的解法,属于基础题.18.(1)()f x 为奇函数;(2)0x =【分析】(1)根据题意,求函数定义域结合对数函数真数大于零得到关于x 的不等式组,求解即可得函数的定义域关于原点对称,由函数的解析式和奇偶性的定义即可确定函数的奇偶性; (2) 根据题意结合对数函数的单调性,解方程进行求解,即可得出方程的解.【详解】(1)()f x 为奇函数.使函数()f x 有意义,只需101x x +>-,101x x +<-,11x -<<, 由()31log 1x f x x +=-,得13311()log log ()()11x x f x f x x x --+-===-+-,所以()f x 为奇函数. (2)(21)0xf -=,32log 022x x =-,2122xx =-,21x =,0x =,检验知适合1211x -<-<,所以原方程的解为0x =.【点睛】本题主要考查函数的奇偶性以及对数函数的相关知识,掌握对数函数真数大于零以及对数函数的单调性,考查了运算能力,属于中档题.19.(1)2()23f x x x =-+-;(2)2a =-或3a =【分析】(1)由等式可得出函数的对称轴,设出二次函数的解析式,由最大值为-2,即可求得解析式;(2)由(1)的结论,讨论对称轴和a,a+1的关系,结合最大值为-6,即可求得实数a 的值.【详解】(1)由()()023f f ==-,可知函数的对称轴为1x =,设2()(1)2f x m x =--,0m <,因为(0)3f =-,所以23m -=-,1m =-,所以22()(1)223f x x x x =---=-+-;(2)因为()f x 在区间[],1a a +上的最大值为-6,最大值没有在顶点处取到,所以①1a ≥时,()f x 在区间[],1a a +上递减,2max ()()23f x f a a a ==-+-, 所以2236a a -+-=-,3a =,1a =-(舍),得3a =;②11a +≤时即0a ≤时,()f x 在区间[],1a a +上递增,2max ()(1)2f x f a a =+=--,所以226a --=-,2a =-,2a =(舍),得2a =-;01a <<时max ()(1)2f x f ==-,不适合条件.综上2a =-或3a =.【点睛】本题考查二次函数的解析式以及二次函数在闭区间上的最值,考查了分类讨论思想和运算求解能力,属于中档题.20.(1)45080S t =+-[0,12]t ∈.(2) 小区在t ∈要停水 【分析】(1)设t 小时候水池中存水量为S 吨,利用题设条件能将S 表示为时间t 的函数;(2)令60S <,解不等式4508060t +-<,即可求出结果.【详解】(1)由开始时蓄水池中有水450吨,又水厂每小时可向蓄水池中注水80吨,同时蓄水池又向居民小区供水,t 小时内供水量为t 小时蓄水池中存水量45080S t =+-[0,12]t ∈.(2)由(1)令60S <,4508060t +-,8390t -<,<,又012t ≤≤,所以414188t -+<<,所以小区在t ∈要停水. 【点睛】 本题考查函数的应用,考查了建模能力和一元二次不等式的解法,属于中档题.21.(1) 函数()f x 在区间[)0,+∞上是增函数(2) [1,)-+∞【分析】(1)根据函数单调性的定义,利用作差法,即可证得函数的单调性;(2)利用换元法,将函数()g x 转化为二次函数,利用二次函数的性质,即可求得t 的取值范围.【详解】(1)函数()f x 在区间[)0,+∞上是增函数.设1x ,2x ∈[0,)+∞,120x x ≤<,由()22x x f x -=+, 得12121211()()2(2)22x x x x f x f x -=+-+121212(22)(221)22x x x x x x --=, 因为120x x ≤<,所以12122x x ≤<,得12())0(f x f x -<,12()()f x f x <, 所以函数()f x 在区间[)0,+∞上是增函数.(2)由(1)知()f x 在区间[0,2]上是增函数,(0)()(2)f f x f ≤≤,172()4f x ≤≤, 又()22()x x f x f x --=+=,所以()f x 为偶函数,所以在[1,2]-的值域为17[2,]4.因为()()20f x t f x +⋅≥对任意[]1,2x ∈-恒成立,2222(22)0x x x x t --+++≥,2(22)2(22)0x x x x t --+-++≥,令22x x s -=+,所以不等式220s ts -+≥在17[2,]4s ∈恒成立,max 2()t s s ≥-, 由2()g s s s =-在17[2,]4s ∈递减,所以max ()(2)1g s g ==-,所以1t ≥-,故t 的取值范围为[1,)-+∞.【点睛】本题考查了函数单调性的判断与证明,注意一般单调性的证明选用定义法证明,证明的步骤是:设值,作差,化简,定号,下结论.同时考查了二次函数的最值,解题的关键是确定函数的单调性,从而确定参数的范围,属于中档题.22.(1)01x =(2)①函数()1f x x x=+不是“保k 值函数” ②当2221(,1)e a e e+∈+时函数()ln 1x a f x e =+是“保2值函数”; 当2221(0,][ 1.)e a e e+∈++∞时函数()ln 1x a f x e =+不是“保2值函数”. 【分析】(1函数()2xf x =为“保1值函数”,列方程即可求解;(2)①由“保k 值函数”的定义,转化为二次函数是否有解问题,即可进行判断;②由题意可得()022111x e a e a e -+=--,再由00x e >,解不等式即可进行判断.【详解】(1)因为函数()2x f x =为“保1值函数”,所以存在0x 使00(1)()(1)f x f x f +=+,001222x x +=+,022x =,01x =.(2) ①若函数()1f x x x=+是“保k 值函数”,则存在实数00x ≠,使得()()()00f x k f x f k +=+,0000111x k x k x k x k ++=++++,22000x kx k ++=,0k ≠时23k ∆=-0<,方程无解;0k =时00x =,与00x ≠不符. 综上,函数()1f x x x =+不是“保k 值函数”. ②若函数()ln 1x a f x e =+是否是“保2值函数”,则()f x 在其定义域内存在实数0x ,使得()()()0022f x f x f +=+,即0022ln ln ln 111x x aa a e e e +=++++,即0022111x x a a a e e e +=⋅+++, 可得()()0022111x x e e a e +++=+,化简可得()022111x e a e a e -+=--,由00x e >,解得22211e a e e+<<+, 故当22211e a e e+<<+时,函数是“保2值函数”,又0a >,所以当2221(0,][ 1.)e a e e+∈++∞时函数()ln 1x a f x e =+不是“保2值函数”. 【点睛】本题考查了函数的新定义等综合知识,考查了二次函数有解问题,考查指数非负,求解一元二次不等式问题,考查了分类讨论思想的运用,属于中档题.。
江苏省扬州大学附属中学东部分校2020-2021学年高一上学期期中数学试卷及解析

江苏省扬州大学附属中学东部分校2020-2021学年高一上学期期中数学试卷注意事项:1.答题前填写好自己的姓名、班级、考号等信息 2.请将答案正确填写在答题卡上第I 卷(选择题)一、选择题1.已知幂函数y f x =的图象过点(,则()81f =( ) A.3B.13C.9D.192.已知集合{}{}20,1,4A B x x ==≤,则A B =( )A.{}0,1B.{}0,1,2C.{}02x x ≤<D.{}02x x ≤≤3.已知10x y -<<<,比较2211,,,x y x y的大小关系得( ) A.2211x y y x<<< B.2211y x x y<<< C.2211y x y x <<< D.2211y x y x<<< 4.下列图形中,表示函数图象的个数是( )A.1个B.2个C.3个D.4个5.已知函数y =f(x),部分x 与y 的对应关系如表:则f(f(4))=( )A. −1B. −2C. −3D. 36.已知偶函数f(x)在[0,+∞)上单调递增,则对实数a 、b ,“a>|b|”是“f(a)>f(b)”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7.下列命题为真命题的是( ) A.若ac bc >,则a b > B.若22a b >,则a b >C.若11a b>,则a b < D.<a b <8.已知函数()3122xx f x x =+-,若()()2120f a f a -+≤,则实数a 的取值范围为( ) A.(]1,1,2⎡⎫-∞-+∞⎪⎢⎣⎭ B.1,12⎡⎤-⎢⎥⎣⎦ C.[)1,1,2⎛⎤-∞-+∞ ⎥⎝⎦D.11,2⎡⎤-⎢⎥⎣⎦第II 卷(非选择题)二、填空题210x >的解”的否定________,并判断其真假_________(填“真命题”或“假命题”). 10.若0,0x y >>,化简:21113333243x yx y ---⎛⎫÷- ⎪⎝⎭得__________. 11.设lg 6,lg12a b ==,用,a b 表示lg 75得__________. 12.下列几个命题:①下列函数中2y =;y 2log 2x y =;2log 2xy =,与函数y x =相同的函数有2个;②函数()f x x x bx c =++的图象关于点()0,c 对称;③函数y =是偶函数,但不是奇函数;④()f x 是定义在R 上的奇函数,当0x <时,()221f x x x =+-,则当0x ≥时,()221f x x x =-++;⑤函数3222xx y -=+的值域是31,2⎛⎫- ⎪⎝⎭.其中正确命题的序号有__________.三、解答题13.设集合1{|()8}22xA x =<<,{|||1}B x x a =+<. (1)若3a =,求AB ;(2)设命题:p x A ∈,命题:q x B ∈,若p 是q 成立的必要不充分条件,求实数a 的取值范围.14.已知一次函数()y f x =满足()12f x x a -=+, . 在所给的三个条件中,任选一个补充到题目中,并解答. ①()5f a =,②142a f ⎛⎫=⎪⎝⎭,③()()41226f f -=. (1)求函数()y f x =的解析式;(2)若()()()g x x f x f x x λ=⋅++在[]0,2上的最大值为2,求实数λ的值. 15.已知a ∈R ,且a ≠1,比较a +2与31a-的大小. 16.已知函数2()f x x x m =-+.(1)当2m =-时,解不等式()0f x >; (2)若0m >, ()0f x <的解集为(,)a b ,求14a b+的最小値. 17.某渔场鱼群的最大养殖量为m 吨,为保证鱼群的生长空间,实际的养殖量x 要小于m ,留出适当的空闲量,空闲量与最大养殖量的比值叫空闲率,已知鱼群的年增加量y (吨)和实际养殖量x (吨)与空闲率的乘积成正比(设比例系数k >0). (1)写出y 与x 的函数关系式,并指出定义域; (2)求鱼群年增长量的最大值;(3)当鱼群年增长量达到最大值时,求k 的取值范围.18.已知函数()f x 为偶函数,当0x ≥时,()11x xe f x e -=+. (1)求当0x <时,函数()f x 的解析式; (2)判断函数()f x 在(),0-∞上的单调性并证明;(3)设函数()()()2g x f ax f x a =--+,使函数()g x 有唯一零点的所有a 构成的集合记为M ,求集合M .四、新添加的题型) A.若1x >,则21x > B.=x y =C.若220x x +-=,则1x =D.若x AB ∈,则x A B ∈20.对任意实数,,a b c ,下列命题中正确的是( ) A.“5a <”是“3a <”的必要条件 B.“5a +是无理数”是“a 是无理数”的充要条件C.“a b =”是“ac bc =”的充要条件D.“a b >”是“22a b >”的充分条件21.已知函数()225,1,1x ax x f x a x x⎧++<⎪=⎨-≥⎪⎩在区间(),-∞+∞上是减函数,则整数a 的取值不可以为( ) A.-2B.-1C.0D.122.关于定义在R 上的偶函数()f x ,当0x ≥时,()22f x x x =+,则下列说法正确的是( )A.0x <时,函数解析式为()22f x x x =- B.函数在定义域R 上为增函数C.不等式()328f x -<的解集为40,3⎛⎫ ⎪⎝⎭D.不等式()210f x x x ---<恒成立参考答案1.C【解析】1.设幂函数解析式,代入点的坐标,求出幂函数解析式,即可求得结果. 由题意设()y f x x α==,图象过点(,得3α=解得12α=, ∴()12f x x=,()1281819f ==;故选:C. 2.A【解析】2.利用一元二次不等式的解法化简集合B ,再利用交集的定义求解即可. 因为{}{}20,1,{|4}22A B x x x x ==≤=-≤≤,所以{}0,1AB =,故选:A. 3.C【解析】3.利用不等式的性质求解即可. 由10x y -<<<, 得22110y x y x<<<<, 故选:C. 4.B【解析】4.利用函数的定义判断即可.利用函数的定义,在定义域内的任一个x ,都有唯一确定的y 与之对应, 观察图像得第一个图和第二个图正确,第三个图和第四个图不正确; 故选:B. 5.D【解析】5. 先求f(4)=−3,再求f(−3)=3通过表格可以得到f(4)=−3,f(f(4))=f(−3)=3故选:D 6.A【解析】6.本道题结合偶函数满足f (x )=f (−x )以及单调递增关系,前后推导,即可.结合偶函数的性质可得f (x )=f (−x ),而当a >|b |,−a <b <a ,所以结合f (x )在 [0,+∞)单调递增,得到f (a )=f (−a )>f (b ),故a >|b |可以推出f (a )>f (b ).举特殊例子,f (−3)=f (3)>f (1),但是−3<|1|,故由f (a )>f (b )无法得到a >|b |,故a >|b |是f (a )>f (b )的充分不必要条件,故选A.7.D【解析】7.根据不等式的性质判断各个命题.A 中若0c <,则得不出a b >,错误;B 中,若0,0a b <<,则有a b <,错误;C 中若0,0a b ><,则仍然是a b >,错误;由不等式的性质知D 正确.故选:D. 8.D【解析】8.先求出函数的定义域,再利用函数奇偶性的定义判断奇偶性,最后利用幂函数和指数函数的单调性判断函数的单调性,即可解不等式. 由()3122xxf x x =+-定义域为R ,()()33112222x xx xf x x x f x ---=-+-=--+=-, 所以函数()f x 为奇函数,利用幂函数和指数函数的单调性易知:函数()f x 为R 上的增函数,()()()()()()2221201212f a f a f a f a f a f a -+≤⇒-≤-⇒-≤-,则211212a a a -≤-⇒-≤≤, 故选:D.9.存在大于3的自然数不是不等式210x >的解 假命题【解析】9.利用“改量词,否结论.”求命题的否定,判断原命题的真假即可判断. 由命题:大于3的自然数是不等式210x >的解,得命题的否定为:存在大于3的自然数不是不等式210x >的解, 因为大于3的自然数有4,5,6,它们的平方一定大于10,即大于3的自然数都是不等式210x >的解, 故该否定为假命题.故答案为:存在大于3的自然数不是不等式210x >的解;假命题. 10.6x -【解析】10.利用指数幂的运算法则求解即可. 由0,0x y >>, 得2111213333113333234432x yx y x y ---+-+⎛⎫÷-=-⨯ ⎪⎝⎭066xy x =-=-;故答案为:6x -. 11.432a b -+【解析】11.由题意条件得出lg 2lg3lg32lg 2ab+=⎧⎨+=⎩,解出lg 2和lg 3,由此可得出lg 75lg32lg 22=-+,代入即可得出答案.lg6lg 2lg3a =+=, lg12lg32lg 2b =+=,即lg 2lg3lg32lg 2ab +=⎧⎨+=⎩,解得lg 2lg32b aa b=-⎧⎨=-⎩,753lg 75lg2lg 2lg32lg 224321004a b ∴=+=+=-+=-+, 故答案为:432a b -+. 12.②⑤【解析】12.对于选项①:判断函数的定义域与对应关系是否相等即可判断;对于选项②:求解()()2f x f x c +-=即可判断;对于选项③:先求函数的定义域,写出函数解析式即可判断;对于选项④:利用函数为定义在R 上的奇函数,则()00f =,即可判断;对于选项⑤:令()20xt t =>,原函数变为:()25351222t t y t t t -++-===-++++,利用t 的范围求解即可判断.对于选项①:由2y =定义域为{}0x x ≥,y x ==,2log 2x y x ==,2log 2x y =定义域为{}0x x >,得与函数y x =相同的函数只有1个;故①不正确; 对于选项②:由()f x x x bx c =++,得()()2f x f x x x bx c x x bx c c +-=++--+=, 则函数()f x x x bx c =++的图象关于点()0,c 对称;故②正确;对于选项③:由函数y ,得2210110x x x ⎧-≥⇒=±⎨-≥⎩,所以()01y x ===±即是偶函数,也是奇函数;故③不正确;对于选项④:因为()f x 是定义在R 上的奇函数, 则()00f =,而题干中,当0x ≥时,()221f x x x =-++;此时()01f =,故不满足题意, 故④不正确;对于选项⑤:令()20xt t =>,原函数变为:()25351222t t y t t t -++-===-++++ 因为5522,022t t +><<+, 则531122t -<-<+, 所以函数3222xx y -=+的值域是31,2⎛⎫- ⎪⎝⎭,故⑤正确; 故答案为:②⑤. 13.(1)(4,1)A B =-(2)[0,2]【解析】13.(1)将3a =代入B ,求得B ,再求得AB ;(2)将问题转化为集合B 是集合A 的真子集,再根据真子集关系列式可得. (1)由已知可得(3,1)A =-,(4,2)B =--,∴(4,1)A B =-.(2)由题意可得集合B 是集合A 的真子集,∵(1,1)B a a =---+,∴1311a a ---⎧⎨-+<⎩或1311a a -->-⎧⎨-+⎩,∴02a ,∴实数a 的取值范围是[0,2]. 14.(1)()23f x x =+(2)2λ=-【解析】14.利用待定系数法求出()22f x x a =++,(1)根据所选条件,都能求出1a =,可得()23f x x =+;(2)根据对称轴与区间中点值的大小分两种情况讨论求出最大值,结合已知最大值可求得λ的值.设()f x kx b =+(0)k ≠,则(1)2k x b x a -+=+,即2kx k b x a -+=+, 所以2k =,2b a ,所以()22f x x a =++,若选①,(1)由()5f a =得225a a ++=,得1a =,所以()23f x x =+.(2)()(23)(23)g x x x x x λ=++++=22(42)3x x λλ+++,区间[]0,2的中点值为1,对称轴为()22x λ+=-,当()212λ+-≤,即4λ≥-时,max()(2)8843716f x f λλλ==+++=+,所以7162λ+=,解得2λ=-;当()212λ+->,即4λ<-时,max()(0)3f x f λ==,所以32λ=,解得23λ=(舍), 综上所述:2λ=-. 若选②, (1)由142a f ⎛⎫=⎪⎝⎭得14222a a =⨯++,解得1a =,所以()23f x x =+; (2)()(23)(23)g x x x x x λ=++++=22(42)3x x λλ+++,区间[]0,2的中点值为1,对称轴为()22x λ+=-,当()212λ+-≤,即4λ≥-时,max()(2)8843716f x f λλλ==+++=+,所以7162λ+=,解得2λ=-;当()212λ+->,即4λ<-时,max()(0)3f x f λ==,所以32λ=,解得23λ=(舍), 综上所述:2λ=-. 若选③,(1)由()()41226f f -=得4(22)2(42)6a a ++-++=,解得1a =,所以()23f x x =+;(2)()(23)(23)g x x x x x λ=++++=22(42)3x x λλ+++,区间[]0,2的中点值为1,对称轴为()22x λ+=-,当()212λ+-≤,即4λ≥-时,max()(2)8843716f x f λλλ==+++=+,所以7162λ+=,解得2λ=-;当()212λ+->,即4λ<-时,max()(0)3f x f λ==,所以32λ=,解得23λ=(舍), 综上所述:2λ=-. 15.当1a <时,321a a +<-;当1a >时,321a a+>-【解析】15.利用作差的方法比较数值的大小关系22213()3(2)(1)31124(2)11111a a a a a a a a a a a a a +++-----+++-====----- 我们不难发现:分式中分子始终为正值,所以:1a <时3(2)01a a+-<- 当1a >时,3(2)01a a+->-; 故:当1a <时,321a a +<-;当1a >时,321a a+>- 16.(1){2x x >或}1x <-;(2)最小值为9.【解析】16.(1)由一元二次不等式的解法即可求得结果;(2)由题()0f x =的根即为a ,b ,根据韦达定理可判断a ,b 同为正,且1a b +=,从而利用基本不等式的常数代换求出14a b+的最小值.(1)当2m =-时,不等式0f x >(),即为220x x -->, 可得()()210x x -+>,即不等式()0f x >的解集为{2x x >或}1x <-.(2)由题()0f x =的根即为a ,b ,故1a b +=,0ab m =>,故a ,b 同为正,则14a b +=144()559a b a b a b b a ⎛⎫⎛⎫++=++≥+= ⎪ ⎪⎝⎭⎝⎭, 当且仅当13a =,23b =等号成立,所以14a b+的最小值为9.17.(1).定义域为;(2)当时,;(3)的取值范围是.【解析】17.试题分析:(1)由题意求出空闲率,然后利用正比例关系得与的函数关系式,并确定函数的定义域;(2)利用配方法求二次函数的最值;(3)鱼群年增长量达到最大值时,应保证实际养殖量和增加量的和在0到之间,由此列不等式求解的取值范围即可. 试题解析:(1)空闲率为,由已知得:. (2)因为,所以当时,.(3)由题意得:,即,解得.又因为,所以,所以的取值范围是.18.(1)()11xxe f x e-=+;(2)函数()f x 在(),0-∞上单调递减,证明见详解;(3){}1,0,1,2M =-.【解析】18.(1)当0x <时,0x ->,()1111x xx xe ef x e e-----==++,利用函数的奇偶性求解即可;(2)函数()f x 在(),0-∞上单调递减,利用定义证明函数的单调性即可;(3)把函数()g x 有唯一零点的问题转化为方程()()20f ax f x a --+=有唯一的解的问题,利用函数的奇偶性和单调性得到2ax x a =-+,两边平方,利用方程有唯一的解即可得出结果. (1)当0x <时,0x ->, 又函数()f x 为偶函数,则()()1111x xx xe ef x f x e e-----===++,所以函数()f x 的解析式为()11xxe f x e-=+; (2)函数()f x 在(),0-∞上单调递减, 设任意120x x <<,则()()()()()12212112212111111x x x x x x x x e e e e f x f x e e e e ----=-=++++, 因为xy e =在R 上单调递增, 所以12x x e e <,即120x x e e -<, 所以()()21f x f x <,所以函数()f x 在(),0-∞上单调递减; (3)因为函数()f x 为偶函数, 所以函数()f x 在()0,∞+上单调递减,函数()()()2g x f ax f x a =--+的零点就是方程()()20f ax f x a --+=的解, 因为函数()g x 有唯一零点,所以方程()()20f ax f x a --+=有唯一的解, 因为函数()f x 为偶函数, 所以方程变形为:()()2fax f x a =-+,因为函数()f x 在()0,∞+上单调递减, 所以2ax x a =-+, 平方得:()()()22212220a x a x a -+-+-=,当210a -=时,即1a =±,经检验方程有唯一解; 当210a -≠时,()()()222424120a a a ∆=----=,得()22200a a a -=⇒=或2a =, 综上可得:集合{}1,0,1,2M =-. 19.AD【解析】19.对于选项A :利用不等式的性质判断即可;对于选项B =x y =即可判断;对于选项C :解一元二次方程即可判断;对于选项D :利用元素与集合的关系判断即可.对于选项A :若1x >,则21x >,故选项A 正确;对于选项B =x y =或y x =-,故选项B 不正确;对于选项C :若220x x +-=,则1x =或2x =-,故选项C 不正确; 对于选项D :若x A B ∈,则x A B ∈,故选项D 正确;故选:AD. 20.AB【解析】20.利用充分与必要条件的定义,判定各选项中的充分性与必要性是否成立,从而选出正确答案.A 中,∵a <3时,得出a <5, ∴a <5是a <3的必要条件; ∴A 是正确的;B 中,5a +是无理数,得出a 是无理数,充分性成立;a 是无理数,得出5a +是无理数,必要性成立;∴B 是正确的;C 中,由a b =,得出ac bc =,充分性成立; 由ac bc =,不能得出a b =, 例如:c =0时,2×0=3×0,2≠3, ∴必要性不成立; ∴C 是不正确的;;D 中,∵a >b 不能得出22a b >, 例如:1,2a b =-=得22a b <, ∴充分条件不成立; D 不正确. 故选:AB . 21.CD【解析】21.根据题意,讨论1x <时,()f x 是二次函数,在对称轴对称轴左侧单调递减,1x 时,()f x 是反比例函数,在0a <时单调递减;再利用端点处的函数值即可得出满足条件的a的取值范围.解:由函数()225,1,1x ax x f x a x x⎧++<⎪=⎨-≥⎪⎩在区间(,)-∞+∞上是减函数,当1x <时,2(2)5f x x ax +=+,二次函数的对称轴为x a =-, 在对称轴左侧单调递减,1a ∴-,解得1a ≤-;当1x 时,()a f x x=-, 在0a <时单调递减; 又2152a a +≥-+, 即2a ≥-;综上,a 的取值范围是21a -≤≤-, 则整数a 的取值不可以为0或1; 故选:C D. 22.AC【解析】22.对于A ,利用偶函数定义求0x <时,函数解析式为()22f x x x =-;对于B ,研究当0x ≥时,()f x 的单调性,结合偶函数图像关于y 轴对称,知()f x 在R 上的单调性;对于C ,求出(2)8f =,不等式(32)8f x -<,转化为(32)(2)f x f -<,利用单调性解不等式;对于D ,分类讨论(0,)x ∈+∞与(,0)x ∈-∞两种情况是否恒成立. 对于A ,设0x <,0x ->, 则2()2f x x x -=-, 又()f x 是偶函数,所以()2()2f x f x x x =-=-,即0x <时,函数解析式为2()2f x x x =-,故A 正确; 对于B ,2()2f x x x =+,对称轴为1x =-, 所以当0x ≥时,()f x 单调递增, 由偶函数图像关于y 轴对称,所以()f x 在(),0-∞上为减函数,故B 不正确; 对于C ,当(0,)x ∈+∞时,2()28f x x x =+=, 解得12x =,24x =-(舍去), 即(2)8f =,所以不等式(32)8f x -<, 转化为(32)(2)f x f -<, 又()f x 在R 上为偶函数, 得432203x x -<⇒<<, 所以不等式的解集为40,3⎛⎫ ⎪⎝⎭,故C 对; 对于D ,当(,0)x ∈-∞时,2()2f x x x =-,222()12131f x x x x x x x x --=--=-----,不恒小于0;当0x ≥时,2()2f x x x =+,222()1211f x x x x x x x x --=+---=--不恒小于0,故D 错;故选:AC.。
江苏省扬州市扬州大学附属中学2020-2021学年第一学期第一次月考高一数学

江苏省扬州市扬州大学附属中学2020-2021学年第一学期第一次月考高一数学(本卷满分:150分 考试时间:120分钟)一、单选题(本大题共8小题,每小题5分,共40分)1、集合{}Z x x x A ∈<<-=,12中的元素个数为( )A 、1B 、2C 、3D 、42、已知集合{}{}3,1,4,3,2,1==A U ,则U A =( )A 、{}4,2B 、{}2,1 C 、{}3,2 D 、{}4,2,1 3、“1>x ”是“2>x ”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、下列命题中,是假命题的是( )A 、0,=∈∃x R xB 、1102,=-∈∃x R xC 、0,3>∈∀x R x D 、01,2>+∈∀x R x5、函数1322+-=x x y 的零点是( )A 、()0,1,0,21-⎪⎭⎫ ⎝⎛-B 、1,21-C 、()0,1,0,21⎪⎭⎫ ⎝⎛ D 、1,21 6、已知1,22,22-+=+=∈x x B x x A R x ,则A ,B 的大小关系是( )A 、B A = B 、B A >C 、B A <D 、无法判定7、如果0<<b a ,那么下列不等式成立的是( )A 、b a 11<B 、2b ab <C 、2a ab -<-D 、ba 11-<- 8、若不等式012≥--bx ax 的解集是⎥⎦⎤⎢⎣⎡21,31,则不等式02<--a bx x 的解集是( ) A 、()3,2 B 、⎪⎭⎫ ⎝⎛21,31 C 、()2,3-- D 、⎪⎭⎫ ⎝⎛+∞⎪⎭⎫ ⎝⎛∞-,2131,二、多选题(本大题共4小题,每小题5分,共20分)9、若C C B A B A == ,,则集合A ,B ,C 之间的关系必有( )A 、C A ⊆B 、C A = C 、B A ⊆D 、B A =10、已知q p ,都是r 的充分条件,s 是r 的必要条件,q 是s 的必要条件,则( )A 、p 是q 的既不充分也不必要条件B 、p 是s 的充分条件C 、r 是q 的必要不充分条件D 、s 是q 的充要条件11、下列说法正确的是( )A 、xx 1+的最小值是2 B 、223x x +的最小值是32 C 、2322++x x 的最小值是2 D 、x x 1+的最小值是2 12、已知函数()02>++=a b ax x y 有且只有一个零点,则( )A 、422≤-b aB 、412≥+b a C 、若不等式02<-+b ax x 的解集为()21,x x ,则021<x xD 、若不等式c b ax x <++2的解集为()21,x x ,且421=-x x ,则4=c 三、填空题(本大题共4小题,每小题5分,共20分)13、命题“01,2>++∈∀x x R x ”的否定是 .14、某班共30人,其中15人喜爱篮球,10人喜爱乒乓球,8人对这两项运动都不喜爱,则喜爱篮球但不喜爱乒乓球的人数是 .15、设P ,Q 为两个非空实数集合,定义集合{}Q b P a ab z z Q P ∈∈==*,,,若{}{}2,2,1,0,1-=-=Q P ,则集合Q P *有 个子集.16、已知0,0>>y x ,且114=+yx ,则y x +的最小值为 . 四、解答题(本大题共6小题,共70分)17、(本题满分10分)已知全集{}{}{}22,3,23,21,2,5U U a a A a A =+-=-=,求实数a 的值.18、(本题满分12分)求下列不等式的解集.(1)0432≤--x x (2)1342>-+x x19、(本题满分12分)已知命题:p “关于x 的方程012=++mx x 有两个不相等的实数根”是真命题.(1)求实数m 的取值集合M ;(2)若{}2+<<=a m a m N ,且“N m ∈”是“M m ∈”的充分不必要条件,求实数a 的取值范围.20、(本题满分12分)已知集合{}(){}0112,04222=-+++==+=a x a x x B x x x A .(1)若B A B A =,求实数a 的值;(2)若A B A = ,求实数a 的取值范围.21、(本题满分12分)为了改善居民的居住条件,某城建公司承包了棚户区改造工程,按合同规定: 若提前完成,则每提前一天可获2万元奖金,但要追加投入费用;若延期完成,则将被罚款. 追加投入的费用按以下关系计算:11837846-++x x (万元),其中x 表示提前完工的天数(附加效益=所获奖金-追加费用). (1)求附加效益y (万元)与x 的函数关系式;(2)提前多少天,能使公司获得最大的附加效益? 并说明理由.22、(本题满分12分)已知二次函数()()m x m x y -+-+-=222. (1)若“0,<∈∀y R x ”为真命题,求实数m 的取值范围;(2)是否存在小于4的整数a ,使得关于x 的不等式()()4222≤-+-+-≤m x m x a 的解集恰好为[]4,a ? 若存在,求出所有可能的a 的取值集合;若不存在,说明理由.。
江苏省扬州市扬州大学附属中学东部分校高一第一学期期中考试化学试题

江苏省扬州市扬州大学附属中学东部分校2020-2021学年高一第一学期期中考试化学试题学校:___________姓名:___________班级:___________考号:___________一、单选题1.我校本月提出“节约用水用电,倡导绿色简约生活”。
下列做法不应该提倡的是A.路远的小伙伴们拼车上学B.少用修正带、涂改液C.将化学实验室的废水直接排入水池D.将食堂产生的“地沟油”制成肥皂2.小苏打是焙制糕点常用的发酵粉,小苏打属于A.碱B.氧化物C.盐D.有机物3.下列过程中涉及化学变化的是A.海水淡化B.海水晒盐C.风力发电D.海带提碘I的说法正确的是( )4.下列关于具有放射性的12553A.是一种新元素B.中子数是125I有很大区别D.质子数是53C.其化学性质与131535.下列常见物质的俗名或主要成份与化学式相对应的是A.苏打——NaHCO3B.生石灰 Ca(OH)2C.漂白粉——Ca(ClO)2和CaCl2D.酒精 CH3COOH6.下列溶液中,跟2mol/L的K2CO3溶液所含的K+ 物质的量浓度相同的是A.4mol/LKCl溶液B.1mol/LKCl溶液C.5mol/LKOH溶液D.1mol/LK2SO3溶液7.下列物既能导电,又属于电解质的是A.固体NaCl B.食盐水C.酒精D.熔融的NaCl8.关于反应方程式Fe2O3+2O3+2Fe下列叙述正确的是A.Al是氧化剂B.Fe2O3被还原C.Fe2O3失去电子D.Al2O3是还原产物9.下列电离方程式,书写正确的是A.KMnO4═K++Mn7++4O2-B.NaHCO3= Na++ H++ CO32-C.FeCl3═Fe3++Cl3—D.Al2(SO4)3═2Al3++3SO42-10.下列实验操作或装置错误的是A.蒸馏B.过滤C.萃取D.转移溶液11.胶体与溶液的本质差异在于( )A.是否稳定B.粒子直径大小C.是否透明D.是否有颜色12.将绿豆大小的方形金属钠投入足量且含酚酞的水中,根据相应现象不能得出的结论是A.A B.B C.C D.D13.标准状况下,32g的甲烷(CH4)的体积为( )A.5.6L B.22.4 L C.11.2 L D.44.8 L 14.下列关于氯水的叙述,正确的是()A.氯水中只含Cl2和H2O分子B.新制氯水可使蓝色石蕊试纸先变红后褪色C.光照氯水有气泡逸出,该气体是Cl2D.氯水放置数天后其酸性减弱15.在实验室用下列装置制取氯气并验证氯气的某些化学性质,其中不能达到实验目的是A.甲闻氯气的气味B.实验室用装置乙制取氯气C.用装置丙验证氯气能支持H2燃烧D.用装置丁除去尾气中的氯气16.下列除杂所选用的试剂及操作方法均正确的一组是(括号内为杂质)A.A B.B C.C D.D17.把V L含有MgSO4和K2SO4的混合溶液分成两等份,一份加入含a mol NaOH的溶液,恰好使镁离子完全沉淀为氢氧化镁;另一份加入含b mol BaCl2的溶液,恰好使硫酸根离子完全沉淀为硫酸钡。
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第二部分阅读理解(共两节,满分50分)ACNK DigitalCNK Digital is a leader in making learning fun. Since first offering its gaming platform early in the present decade, CNK Digital has become a leader in game-based learning, with a variety of leveled games that work with students from pre-school through the beginning of middle school. The company’s games are fun, but also instructive, with lessons in phonics, sight words, the alphabet, sentence structure, and much more.Book AdventureFit for students in grades K-8, Book Adventure has a dual mission in the world of online reading education. First, the service is designed to help early learners master the basic concepts of reading comprehension and the principles of literacy. This is accomplished through a series of games and “adventures” that require input from students, corresponding with literacy lessons, to progress through the “adventure” and reach its conclusion.K5 Learning for ReadingOne of the things that originally made K5 Learning unique was its founding members: parents. When the company was founded, virtually all literacy tools were marketed exclusively to schools and sold on a “volume license basis,” making it nearly impossible for parents and homeschoolers to take advantage. The company’s original mission was to democratize this process, bringing high-end, highly effective phonics and literacy tools to early learners across the countryReading BearThere is one problem that has almost always stopped homeschoolers and budget-weary parents in their tracks: licensing fees for literacy programs. Many of the most popular literacy programs.Reading Bear was founded to put a stop to this problem. All of its literacy and reading comprehension lessons are completely free for home use, making this tool the best choice for some poor students.21.what do CNK Digital and Book Adventure have in common?A.They are both fit for any middle school student.B.They both encourage reading through games.C.They are both leaders in online reading.D.They both have two different goals.22.What makes K5 learning for Reading special?A.Its method of readingB.Its usersC.Its foundersD.Its literacy tools23.Which program is meant for a kid from a poor family?K DigitalB.Book AdventureC.K5 Learning for ReadingD.Reading BearBAs the President of Ariel Investments, I am often asked how to cultivate (培养)green hands to be qualified for a new job within a short time. My memories take me straight back to the days when I worked for John W. Rogers.While in college, I worked as an intern (实习生)at Ariel Investments. As the youngest girl of six kids from a single mom, I had very little knowledge of the business world. When I graduated, I decided to return to the company to improve myself.At the age of 22, my perspective was greatly influenced by lunching with the company's founder and then-president, John W. Rogers. He told me that I would often find myself in the company of rich and successful people with big titles, lots of degrees and lots of experience. And yet, my ideas could still be as good or even better, he said. This early advice became the guiding force that has helped me throughout my entire career. When I was starting out, I thought my inexperience was holding me back, but John said that it was a good thing. He told me that being inexperienced means that you have a fresh point of view.As my tutor, John encouraged me to be brave. As my boss, he took some giant leaps of faith to support me. Because of our first lunch and the years of support that followed, I never thought twice about speaking up, raising my hand or taking a seat at the table. Staying in the background was never a choice. But John 's advice came with one important warning. I needed to keep my attitude in check. It's one thing to be firm but quite another to be rude. Similarly, there is a fine line between confidence and overconfidence. With that in mind, I have had to learn to balance the way I act around others. As someone once told me, “the goal is to be non-threatened andnon-t hreatening at the same time. ”24.What can we learn about the author from the text?A.She went to Ariel Investments after graduation for practice.B.She made her voice heard with encouragement from her boss.C.Her miserable childhood had a negative effect on her work.D.Her performance during internship was highly praised in public.25.Due to John W. Rogers' advice, the author_______ .A.learned to strike a balance between work and lifeB.contributed her ideas only after listening to others'C.no longer felt lost and became aware of her own valueD.switched to a different company to display her talent26.What important lesson does the author learn from his experience?A.Encourage Your Employees to GrowB.Keep Your Attitude under ControlC.Let Your Success Be Your Noise.D.Live a Life with a High Goal.27.Who would probably benefit most from this article?pany bosses.B.Those new to a job.C.Business consultants.D.Experienced teammates.CDo you know some people have memories that are completely wrong? They remember things that never happened. Sometimes, many people have the same wrong memory. When this happens, it’s believed that it is because of the Mandela Effect.People first noticed the Mandela Effect in 2013. That Year,Nelson Mandela passed away. However, thousands of people remembered him dying in the 1980s. They described watching his funeral on TV. But in fact, Mandela lived to be 95 years old. How could that be? How could so many people have the same wrong memory?Most experts explain the Mandela Effect using brain science. When we make memories, our brains keep them in groups. Each group includes memories that are similar to each other. Sometimes, our brains mess up. They mix two similar memories together. That’s why people thought Nelson Mandela died in 1980s. They knew he was imprisoned during the 1980s. For some reason, they mixed his imprisonment with his death.That would explain why so many people in 2013 thought Nelson Mandela was already dead. But would it explain why all those people gave similar descriptions of the funeral? Could all those people have mixed up the same two memories?There are many other examples of the Mandela Effect. Have you ever seen the Mona Lisa? Many people remember the woman in the painting with a frown. Others say she was straight-faced. If you look at the Mona Lisa today, you’ll find she’s smiling. Did you ever read a series of children’s book about the Berenstein Bears? No, you didn’t. In our reality, the books are about the Berenstein Bears, spelled with an “a” instead of an “e”. Still, thousands of people believe they remember the name spelled differently.Are these wrong memories the result of mixed memories, or is there another explanation? Without further science research, we can’t say for sure!48.What is the Mandela Effect?A.Some people have worse memory than others.B.Some people may have the same wrong memory.C.People are better at remembering certain things.D.People remember unhappy experiences more easily.29.Why does the author give an example in Paragraph 2?A.To prove his point.B.To correct the view above.C.To introduce the topic.D.To provide the background.30.What is the reason for the Mandela Effect given in Paragraph 3?A.People are easily influenced by others.B.People pay little attention to the fact.C.People mix two similar things up.D.People have poor memory.31.What does the author think of the Mandela Effect?A.Its reason is unknown.B.It lasts for a long time.C.It is hard to avoid.D.Its effect is great.Scientist are MIT have developed a smart diaper that can call for change when it is wet.Though many single-use diapers can change color when the diapers are wet, it’s often hard to tell for sure without opening the diapers up. Some companies have offered some good products for recognizing wet diapers, but their products don’t come cheap, and need to be cleaned by hand before they can be used again.Scientists at MIT decided to explore the possibility of using cheap tags known as RFID tags. RFID is short for “radio frequency identification”. RFID tags are commonly used in stores for keeping track of products. They are also part of many bus or subway cards. They can be used to lock or unlock doors, and are also used to electronically collect fees on highways. The tags are cheap, so they can be used once and thrown away.Even better, they don’t need batteries. Instead, their antennas get power from radio waves sent by nearby RFID readers. When an RFID tag is made active by a reader, it sends back a simple code stored in a chip inside the tag.What’s unusual in the MIt project is that the RFID tags are not only being used for identification, but also as sensors. The scientist found a clever way to do that. They are using the part of the diaper that urine goes into as the antenna, but it only become an antenna when it is wet. That means a dry diaper doesn’t send out any signal. But as soon as it gets wet, the diaper antenna can send out a signal with the RFID chip’s code.The scientists say that adding the RFID tag will only cost about 2 cents per diaper. The diaper can send a signal to an RFID reader up to 1 meter away. An RFID reader connected to the Internet could announce the wet diaper in many different ways, such as by a message or through an app. And if the smart diaper even fails, there’s always the backup system---a crying baby.In fact, the new technology wouldn’t just help babies, the scientists point out. Many old people and people who have trouble moving also wear diapers. Changing wet diapers quickly can help circumvent some health problems. The MIT scientists believe the technology will be very useful in a hospital where a nurse needs to care for many patients or babies.32.What’s the author’s purpose in writing Paragraph 2?A.To show recognizing wet diapers is not an easy thing.B.To suggest the advantages the new smart diapers have.C.To show many single-use diapers don’t work very well.D.To suggest the big effort made to develop better diapers.33.What can we learn about RFID tags?A.They have to be thrown away once used.B.They are useful but not cheap to make.C.They give signals by changing color.D.They are widely used in our lives.34.What do Paragraph 4-6 mainly talked about?A.How the new smart diapers work.B.How RFId reader can be made active.C.Why dry diapers don’t sent out signals.,D.Why antennas are needed instead of batteries35.What does the underlined word “circumvent” in the last paragraph mean?A.AnalyzeB.DescribeC.AvoidD.Face第二节Today’s families are busy. Mom and Dad are both working, the kids are involved in a variety of activities, but there is little time to enjoy being a family. What can you do to build some family unity? How can you prevent those you love from not knowing each other? ________36______It takes time and patience, but it is definitely possible._______37_______If it’s not possible to have them every night, make sure you eat together as a family at the dinner table at least twice a week. If you need to cut something out of your schedule to achieve this, then do it.Make a rule that all members of the family are treated with respect. _____38_____Speaking disrespectfully with others in the family should not be allowed.Talk to your kids, and really listen. Busy parents are often guilty at not listening to their kids.______39______During your family dinners and game nights, let everyone talk and make the others listen.Forgive when necessary.______40_______If someone has done something wrong,get it our in the open, and then forgive that person. Teach your children how to do this with each other.A.Have family dinnersB.Plan a family night now and thenC.Keep physical contact with your family membersD.Nothing breaks up family unity faster than a grudgeE.Start by modeling this as a parent, and then apply the rule to your childrenF.With creativity and effort, you can build a closely united family although you are busyG.When children know their thoughts or opinions are valued, they will feel closer to their family members.第三部分语言运用Clara Daly was seated on an Alaska Airlines flight from Boston to Los Angeles when a flight attendant asked an urgent question over the loudspeaker:“Does anyone on board___41____American Sign language?” She knew she needed to help.Clara,15 at the time,pressed the call button. The flight attendant came by and __42___ the situation.“We have a passenger on the plane who’s blind and deaf,”she said. The passenger seemed to___43____something,but he was traveling alone and the flight attendants couldn’t__44____ what he needed,according to PEOPLE magazine.Clara had been studying ASL for the past year to help with her dyslexia and knew she’d be__45___to finger spell into the man’s palm. So she ___46____ her seat belt,walked toward the front of the plane,and knelt by the aisle seat of Tim Cook,then 64.__47___ taking his hand,she signed,“How are you? Are you OK/” Cook ___48___ some water. When it arrived,Clara___49___ to her seat. She came by again a bit later because he wanted to know the time. On her ___50___ visit,she stopped and stayed for a while.“He didn’t need anything. He was __51____ and wanted to talk,” Clara said. So for the next hour,that’s what they did. She __52____ about her family and her plans for the future (she wants to be a politician).Cook told Clara how he had gradually become blind over time and shared stories of his days as a traveling salesman. Even though he couldn’t ___53___ her,she “looked attentively at his face with such kindness,”a passenger reported.“Clara was __54___.” a flight attendant told Alaska Airlines in a blog interview.“You could tell Tim was very __55____ to have someone he could speak to,and she was such an angel.41.A.know B.teach C.prefer D. choose42.A.studied B.changed C.wondered D.explained43.A.affect B.carry C.want D.lose44.A.offer B.show C.express D.understand45.A.able B.ashamed C.afraid D.anxious46.A.kept B.fixed C.opened D.measured47.A.roughly B.gently C.wildly D.excitedly48.A.put away B.ask for C.turned down D.gave away49.A.tell B.waved C.returned D.rushed50.A.first B.second C.third D.fourth51.A.regretful B.lonely C.injured D.ill52.A.chatted B.doubted C.argued D.thought53.A.see B.contact C.accept D.connect54.A.brave B.interesting C.creative D.amazing55.A.frightened B.disappointed C.excited D.sad第二节George Washington was born in 1732 in Virginia, __(1)__ father died when he was 11 years old. In his __(2)__(young), Washington was not educated in normal schools. As __(3)__ matter of fact, he went to war rather than going to college. He __(4)__(fight) against the French and Indians as a British Army officer.He also led the army during the American Revolution War and became a national hero after they won the war. In 1783, with a peace treaty (条约) signed between Great Britain and the U.S.,Washington considered __(5)__(give) up his command of the army and returned to his quiet and __(6)__(peace) life in Mount Vernon with his family. However, because __(7)__ his good qualityto be a leader, he was first elected (选举) President in 1789. At first Washington wanted__(8)__(refuse), but public opinion was so strong that finally he gave in. The United States was a small nation when he took office, having 13 states and about 4 million people, but Washington devoted __(9)__(he) to setting a good example to American people.In 1797, Washington finally retired and returned to his family life. __(10)__(sad), he died on the night of December 14 in 1799, at the age of 67.第四部分写作假如你是李华,2022年冬奥会即将在北京召开,请根据以下要点给志愿者主治的负责人Mr.Brown写一封邮件,内容包括:1.表达自己想要成为志愿者2.介绍自己的优势3.期待回复注意:词数80词左右,可增加细节,使行文连贯。