江淮十校2020届高三第三次联考

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2020年5月安徽省江淮十校2020届高三毕业班第三次联考理科综合试题及答案解析

2020年5月安徽省江淮十校2020届高三毕业班第三次联考理科综合试题及答案解析

绝密★启用前安徽省江淮十校联盟2020届高三毕业班第三次联考质量检测理科综合试题2020年5月考生注意:1.本试卷分第Ⅰ卷和第Ⅱ卷两部分。

满分300分,考试时间150分钟。

2.考生作答时,请将答案答在答题卡上。

必须在题号所指示的答题区城作答,超出..答.题.区域书写的...................无效,在试题卷、草稿纸上答题无效。

.....答案3.做选考题时,考生须按照题目要求作答,并用2B铅笔在答题卡上把所选题目的题号涂黑。

可能用到的相对原子质量:H 1 C 12 O 16 S 32 Ni 59 Cu 64第Ⅰ卷(共126分)一、选择题:本题共13小题,每小题6分,共78分。

在每小题给出的四个选项中,只有一项是符合题目要求的。

1.在适宜的条件下.将丽藻培养在含NH4NO3的完全营养液中,一段时间后,发现营养液中NH4+和NO3-的含量不变,下列叙述合理的是A.NH4+和NO3-可通过自由扩散进入根细胞B.NH4NO3必须溶解在水中才能被根吸收C.植物需要N元素,但NH4+和NO3-没有被吸收D.温度升高会促进丽藻对NH4+和NO3-的吸收2.关于细胞结构和功能的叙述,错误的是A.当细胞衰老时,其细胞膜的通透性会发生改变B.膜蛋白的形成与核糖体、内质网、高尔基体有关C.成人心肌细胞中线粒体数量比腹肌细胞的多D.在光镜的高倍镜下观察新鲜菠菜叶装片,可见叶绿体的结构3.从二倍体哺乳动物精巢中取细胞分析其分裂图像,其中甲、乙两类细胞的染色体组数和同源染色体对数如图所示。

下列叙述正确的是A.甲类细胞是精原细胞或初级精母细胞B.甲类细胞处于减数第一次分裂的前、中、后期C.乙类细胞中,性染色体只有一条X染色体或Y染色体D.乙类细胞的分裂会因同源染色体分离导致染色体组数减半。

2020届安徽省江淮十校高三下学期第三次联考英语试题及解析

2020届安徽省江淮十校高三下学期第三次联考英语试题及解析
“Yellow face” actors
Charlie Chan and Fu Manchu may have been Chinese characters, but the actors were usually white men made up to look like Asian. Actors Sidney Toler, Roland Winters and Ross Martin all played Charlie Chan. Yellow face meant they actually yellowed up their skin. White actors just played the lead characters in The Good Earth, a 1937 film about Chinese farmers. Asian actors had parts in the film, but they needed bankable actors , however , there were no Asian American bankable actors.
C. The old library is preferred.
听第9段材料,回答第14至16题。
14. What does the man want to know?
A. The price of a training course.
B. Information about a training course.
听第8段材料,回答第11至13题。
11. Who is sad about the news?
A. Baker. B. Grandpa. C. Alex.
12. What is the most serious problem with the community center on Cranberry Street?

2020届安徽省江淮十校高三第三次联考数学(理)试题(解析版)

2020届安徽省江淮十校高三第三次联考数学(理)试题(解析版)

2020届安徽省江淮十校高三第三次联考数学(理)试题一、单选题1.已知复数满足(其中为虚数单位),则()A.B.C.D.【答案】B【解析】根据复数除法法则化简即可.【详解】由知:,,故选.【点睛】本题考查复数除法法则,考查基本求解能力,属基础题.2.已知命题,,如果命题是命题的充分不必要条件,则实数的取值范围是()A.B.C.D.【答案】B【解析】先解不等式,,再根据命题是命题的充分不必要条件即得。

【详解】记,对于命题,即为,由是的充分不必要条件知:是的真子集,,故选.【点睛】本题主要考察一元二次不等式的解法,充分不必要条件的概念,属于基础题。

3.如图所示,程序框图的输出结果是()A.B.C.D.【答案】C【解析】读懂流程图,其功能是求四项的和,计算求值即可.【详解】计算结果是:,故选.【点睛】本题考查循环结构流程图,考查基本分析求解能力,属基础题.4.已知数列满足,,则的最小值为()A.B.C.D.【答案】C【解析】对式子中的n赋值,依次得到,,…,,进行累加得到和,进而得到的最小值。

【详解】由知:,,…,,相加得:,,又,且,故选.【点睛】本题考查由数列的递推关系得到数列的通项公式,赋值法,累加法,属于基础题。

5.已知一个四棱锥的正视图、侧视图如图所示,其底面梯形的斜二测画法的直观图是一个如图所示的等腰梯形,且该梯形的面积为,则该四棱锥的体积是()A.4 B.C.D.【答案】A【解析】根据三视图以及斜二测画法确定四棱锥的高以及底面面积,再根据锥体体积公式求结果.【详解】由三视图可知,该四棱锥的高是3,记斜二测画法中的等腰梯形的上底为,高为,则斜二测中等腰梯形的腰为,而积,由斜二测画法的特点知直观图中底面梯形的高为,面积,,故四棱锥的体积,故选.(也可用结论直接得出:,,)【点睛】本题考查三视图、斜二测画法以及四棱锥体积,考查基本分析求解能力,属中档题. 6.已知,,,则下列结论成立的是()A.B.C.D.【答案】A【解析】利用幂函数和指数的单调性判断大小。

2020年5月安徽省江淮十校2020届高三毕业班第三次联考英语试题

2020年5月安徽省江淮十校2020届高三毕业班第三次联考英语试题

绝密★启用前安徽省江淮十校联盟2020届高三毕业班第三次联考质量检测英语试题2020年5月注意事项:1.本试卷由四个部分组成。

其中第一、二部分和第三部分的第一节为选择题。

第三部分的第二节和第四部分为非选择题。

共150分。

2.全部答案在答题卡上相应区域内完成,在本试卷上作答无效。

选择题请使用2B铅笔填涂,非选择题请使用0.5毫米黑色签字笔作答。

要求字体工整、笔迹清晰。

3. 请在答题卡规定的地方填写好个人信息,并认真核对答题卡上所粘贴的条形码是否与本人的信息一致。

4. 考试结束后,将本试卷和答题卡一并交回。

第一部分听力(共两节,满分30分)做题时先将答案标在试卷上,录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节(共5小题;每小题1.5分,满分7.5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. What does the woman want to do?A. Take a bus to Brooklyn.B. Go to the 12th street.C. Put up a sign at the bus stop.2. When is Sara's car supposed to arrive?A. At 4:20 pm.B. At 4:35 pm.C. At 6: 20 pm.3. What is the probable relationship between the speakers?A. Customer and clerk.B. Teacher and student.C. Manager and employee.4. Where does the conversation take place?A. At the man's office.B. At a clothing store.C. At a travel agency.5. Who plays tennis best in the woman's opinion?A. David.B. Steven.C. Mike.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白,每段对话或独白后有几个小题,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。

安徽省江淮十校2020届高三第三次联考化学试卷(5月)带详细解析

安徽省江淮十校2020届高三第三次联考化学试卷(5月)带详细解析

“江淮十校”2020届高三第三次联考化学试卷2020.5考生注意:1.本试卷分第Ⅰ卷和第Ⅱ卷两部分。

满分300分,考试时间150分钟。

2.考生作答时,请将答案答在答题卡上。

必须在题号所指示的答题区城作答,超出.....答案..无...答题..区域书写的效,在试题卷、草稿纸上答题无效。

................3.做选考题时,考生须按照题目要求作答,并用2B铅笔在答题卡上把所选题目的题号涂黑。

可能用到的相对原子质量:H 1 C 12 O 16 S 32 Ni 59 Cu 64第Ⅰ卷一、选择题:本在每小题给出的四个选项中,只有一项是符合题目要求的。

7.化学与生活密切相关,下列观点错误的是A.硝酸铵、液氨可用作制冷剂B.苏打、小苏打可用作食用面碱C.二氧化氯、漂白粉可用作自来水的杀菌消毒剂D.氧化铝、二氧化硅可用作制坩埚8.BASF高压法制备醋酸,所采用钴碘催化循环过程如图所示,则下列观点错误的是A.CH3OH转化为CH3I的有机反应类型属于取代反应B.从总反应看,循环过程中需不断补充CH3OH、H2O、CO等C.与乙酸乙酯互为同分异构体且与CH3COOH互为同系物的物质有2种结构D.工业上以淀粉为原料也可以制备醋酸9.设N A为阿伏加德罗常数的值。

下列说法正确的是A.52 g苯乙烯含碳碳双键数目为2N AB.1 L 0.5 mol·L-1醋酸钠溶液中阴阳离子总数目小于N AC.标准状况下,22.4L一氯甲烷含共用电子对数为4N AD.有铁粉参加的反应若生成3 mol Fe2+,则转移电子数一定为6N A10.已知A、B、C、D、E是原子序数依次增大的五种短周期主族元素,其中B和D同主族,中学阶段常见物质X、Y、Z为其中的三种元素组成的化合物,且均为强电解质,当X、Y按物质的量之比为1:2反应时,有如图转化关系。

下列说法正确的是A.C、D、E三种元素一定位于同一周期B.物质X、Y 、Z既含有离子键又含有共价键C.C和D形成的化合物溶于水,其溶液显中性D.简单氢化物的热稳定性:D>E11.次磷酸钴[Co( H 2PO 2)2]广泛应用于化学电镀,工业上利用电渗析法制取次磷酸钴的原理图如右图所示。

2020届安徽省江淮十校高三第三次联考(5月)数学(理)试题 PDF版

2020届安徽省江淮十校高三第三次联考(5月)数学(理)试题 PDF版
19.(12 分) 2019 新型冠状病毒感染的肺炎的传播有飞沫、气溶胶、接触等途径,为了有效抗击疫情,隔离性防护是一 项具体有效措施.某市为有效防护疫情,宣传居民尽可能不外出,鼓励居民的生活必需品可在网上下单,
商品由快递业务公司统一配送(配送费由政府补贴).快递业务主要由甲公司与乙公司两家快递公司承接: “快递员”的工资是“底薪+送件提成”.这两家公司对“快递员”的日工资方案为:甲公司规定快递员每天 底薪为 70 元,每送件一次提成 1 元;乙公司规定快递员每天底薪为 120 元,每日前 83 件没有提成,超过 83 件部分每件提成 5 元,假设同一公司的快递员每天送件数相同,现从这两家公司往年忙季各随机抽取一 名快递员并调取其 100 天的送件数,得到如下条形图:
B. 20
C. 44
D. 40
7.某多面体的三视图如图所示,该多面体的各个面中有若干个是三角形,这些三角形的面积之和为
A.16
B.12
C.8 4 2
D. 8 4 6
8.执行如右图所示的程序框图,若输出的结果为1 1 1 1 ,
35
2019
则判断框内应填人的条件是
A. i 1008?
B. i 1008?

2x y 1 0
14.设
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x
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x 3y 3 0
15.如图所示,点 F 是抛物线 y2=4x 的焦点,点 A,B 分别在抛物线 y2=4x 及圆 x2+y2-2x-8=0 的实线部
分上运动,且 AB 总是平行于 x 轴,则△FAB 的周长的取值范围是
1.已知集合 A x y ln( x 1) , B x 2x 1) ,则 A B =

安徽省江淮十校2020届高三第三次联考(5月)数学(理)

安徽省江淮十校2020届高三第三次联考(5月)数学(理)

“江淮十校”2020届高三第三次联考数学(理科)2020.5一、选择题:本题共12小题,每小题5分,共60分。

在每小题给出的四个选项中,只有一个选项是符合题目要求的。

1.已知集合{})1ln(-==x y x A ,{})12>=x x B ,则B A I =A .),1[+∞B .),1(+∞C .),0(+∞D .)1,0(2.复数z 满足1)2321(=+-z i ,则z 的共轭复数为 A .i 2321+B .i 2321-C .i 2321+-D .i 2321-- 3.已知双曲线)0,0(12222>>=-b a bx a y 的离心率为2.则其渐近线的方程为A .03=±y xB .03=±y xC .02=±y xD .0=±y x 4.如图,点A 的坐标为(1,0),点C 的坐标为(2,4).函数2)(x x f =,若在矩形ABCD 内随机取一点.则该点取自阴影部分的概率为 A .31 B .21 C .32 D .1255.等差数列{}n a 的首项为5.公差不等于零.若542a a a ,,成等比数列,则2020a =A .21B .23C .23-D .2014-6.34)21()1(x x -+展开式中6x 的系数为A .20B .20-C .44D .407.某多面体的三视图如图所示,该多面体的各个面中有若干个是三角形,这些三角形的面积之和为A .16B .12C .248+D .648+8.执行如右图所示的程序框图,若输出的结果为2019151311++++Λ,则判断框内应填人的条件是 A .?1008>i B .?1008≤i C .?1010≤i D .?1009>i9.已知函数2)6(sin 2)6(cos )(22++-+=ππx x x f .则关于它有关性质的说法中,正确的是A .周期为π2B .将其图象向右平移6π个单位,所得图象关于y 轴对称C .对称中心为))(0,212(Z k k ∈+ππD .]20[π,上单调递减 10.为推进长三角一体化战略,长三角区域内5个大型企业举办了一次协作论坛.在这5个企业董事长A ,B ,C ,D ,E 集体会晤之前,除B 与E ,D 与E 不单独会晤外,其他企业董事长两两之间都要单独会晤.现安排他们在正式会晤的前两天的上午、下午单独会晤(每人每个半天最多只进行一次会晤),那么安排他们单独会晤的不同方法共有 A .48种 B .36种 C .24种 D .8种11.已知函数f (x )的定义域为R .其图象关于原点成中心对称,且当x >0时1)(--=x e x f x,则不等式2ln )1(ex f ≤-的解集为A .]12ln ,12ln [++-B .]12ln ,12ln [---C .),2(ln )2ln ,(+∞--∞YD .),()0,(+∞-∞e Y12.侧棱长为32的正四棱锥ABCD V -内,有一半球,其大圆面落在正四棱锥底面上,且与正四棱锥的四个侧面相切,当正四棱锥的体积最大时,该半球的半径为 A .1 B .2 C .22D .2 二、填空题:本题共4小题,每小题5分,共20分。

安徽省江淮十校2020届高三第三次联考(文数)

安徽省江淮十校2020届高三第三次联考(文数)

安徽省江淮十校2020届高三第三次联考数 学(文科)考生注意:1.本试卷满分150分,考试时间120分钟.2.考生作答时,请将答案答在答题卡上.选择题每小题选出答案后,用2B 铅笔把答题卡上对应题目的答案标号涂黑;非选择题请用直径0.5毫米黑色墨水签字笔在答题卡上各题的答题区域内作答,超出答题区域书写的答案无效.............,在试题卷....、草稿纸上作答无效......... 3.做选考题时,考生须按照题目要求作答,并用2B 铅笔在答题卡上把所选题目的题号涂黑. 4.本卷命题范围:高考范围.一、选择题:本题共12小题,每小题5分,共60分.在每小题给出的四个选项中,只有一项是符合题目要求的.1.已知集合{|14}A x x =剟,{}2|23B x x x =-…,则A B ⋂=( ) A .{|14}x x -剟B .{|13}x x 剟C .{|13}x x -剟D .{|14}x x 剟 2.已知复数z 满足262z z i +=-(i 是虚数单位),则复数z 在复平面内对应的点位于( )A .第一象限B .第二象限C .第三象限D .第四象限3.已知双曲线2222:1(0,0)x y C a b a b-=>>的渐近线方程为0x ±=,则双曲线C 的离心率为( )A .3BC .D4.已知直线m ,n ,平面α,β,则m α∥的充分条件是( )A .n α⊂,m n ∥B .αβ⊥,m β⊥C .n α∥,m n ∥D .αβ∥,m β⊂ 5.已知等差数列{}n a 的前n 项和为n S ,若888S a ==,则公差d 等于( ) A .14B .12C .1D .26.新高考方案规定,普通高中学业水平考试分为合格性考试(合格考)和选择性考试(选择考).其中“选择考”成绩将计入高考总成绩,即“选择考”成绩根据学生考试时的原始卷面分数,由高到低进行排序,评定为,,,,A B C D E 五个等级.某试点高中2019年参加“选择考”总人数是2017年参加“选择考”总人数的2倍,为了更好地分析该校学生“选择考”的水平情况,统计了该校2017年和2019年“选择考”成绩等级结果,得到如下图表:针对该校“选择考”情况.2019年与2017年比较,下列说法正确的是( ) A .获得A 等级的人数不变 B .获得B 等级的人数增加了1倍 C .获得C 等级的人数减少了 D .获得E 等级的人数不变 7.函数()cos x xy e ex -=-的部分图象大致是( )A .B .C .D .8.在ABC V 中,5AC AD =u u u r u u u r ,E 是直线BD 上一点,且2BE BD =u u u r u u u r ,若AE mAB nAC =+u u u r u u u r u u u r,则m n +=( )A .25B .25-C .35D .35-9.已知等比数列{}n a 的前n 项和为n S ,若247a a =,423S S =,则5a =( ) A .2B .22C .4D .4210.已知2()2()3f x f x x x =-++,则函数()f x 图象在点(1,(1))f 处的切线方程为( ) A .1y x =-+ B .1y x =+ C .1y x =-- D .1y x =- 11.若函数()3cos f x x x =+在区间[,]a b 上是增函数,且()2,()2f a f b =-=,则函数()3sin g x x x =-在区间[,]a b 上( )A .是增函数B .是减函数C .可以取得最大值2D .可以取得最小值2- 12.在三棱锥P ABC -中,已知4APC π∠=,3BPC π∠=,PA AC ⊥,PB BC ⊥,且平PAC ⊥平面PBC ,三棱锥P ABC -3,若点,,,P A B C 都在球O 的球面上,则球O 的表面积为( )A .4πB .8πC .12πD .16π 二、填空题:本题共4小题,每小题5分,共20分.13.设,x y 满足约束条件1,1,33,x y x y x y --⎧⎪+⎨⎪-≥⎩……则2z x y =-的最小值为___________. 14.在平面直角坐标系中,若角α的始边是x 轴非负半轴,终边经过点22sin,cos 33P ππ⎛⎫ ⎪⎝⎭,则cos()a π+=_________.15.已知函数()f x 是定义域为R 的偶函数,x R ∀∈,都有(2)()f x f x +=-,当01x <≤时,213log ,02()11,12x x f x x x ⎧-<<⎪⎪=-≤≤,则9(11)4f f ⎛⎫-+= ⎪⎝⎭_____.16.已知抛物线2:2(0)C y px p =>,其焦点为F ,准线为l ,过焦点F 的直线交抛物线C 于点A 、B (其中A 在x 轴上方),A ,B 两点在抛物线的准线上的投影分别为M ,N ,若||3MF =,||2NF =,则||||AF BF =____________.三、解答题:共70分.解答应写出文字说明、证明过程或演算步骤.第17~21题为必考题,每个试题考生都必须作答,第22.23题为选考题,考生根据要求作答. (一)必考题:共60分. 17.(12分)在ABC V 中,内角,,A B C 的对边分别为,,a b c ,满足2cos cos cos a A b C c B =+. (1)求A ;(2)若ABC V 的面积为637a =,求ABC V 的周长. 18.(12分)如图,在四棱锥P ABCD -中,底面ABCD 为长方形,PA ⊥底面ABCD ,4PA AB ==,3BC =,E 为PB 的中点,F 为线段BC 上靠近B 点的三等分点.(1)求证:AE ⊥平面PBC ; (2)求点B 到平而AEF 的距离. 19.(12分)2019新型冠状病译(2019-nCoV )于2020年1月12日被世界卫生组织命名.冠状病毒是一个大型病毒家族,可引起感冒以及中东呼吸综合征(MERS )和严重急性呼吸综合征(SARS )等较严重疾 戴口罩 未戴口罩 总计 未感染 30 10 40 感染 4 6 10 总计341650(1)根据上表,判断是否有95%的把握认为未感染与戴口罩有关;(2)在上述感染者中,用分层抽样的方法抽取5人,再在这5人中随机抽取2人,求这2人都未戴口罩的概率.参考公式:22()()()()()n ad bc K a b c d a c b d -=++++,其中n a b c d =+++.参考数据:()20P K k … 0.150.10 0.05 0.025 0.010 0.005 0.0010k2.072 2.7063.841 5.024 6.635 7.879 10.82820.(12分)已知点1F ,2F 是椭圆2222:1(0)x y C a b a b+=>>的左、右焦点,椭圆上一点P 满足1PF x ⊥轴,125PF PF =,1222F F =(1)求椭圆C 的标准方程;(2)过2F 的直线l 交椭圆C 于,A B 两点,当1ABF V 的内切圆面积最大时,求直线l 的方程.21.(12分)已知函数2()()xf x e ax x R =-∈.(1)若函数()y f x =有两个极值点,试求实数a 的取值范围;(2)若02ea 剟且0x >,求证:()1f x >.(二)选考题:共10分.请考生在第22、23题中任选一题作答.如果多做,则按所做的第一题计分.22.[选修4-4:坐标系与参数方程](10分)在平面直角坐标系中,直线l 的参数方程为415315x t y t ⎧=+⎪⎪⎨⎪=+⎪⎩(t 为参数),以直角坐标系的原点为极点,以x 轴的非负半轴为极轴建立极坐标系,曲线C的极坐标方程为4πρθ⎛⎫=- ⎪⎝⎭.(1)求直线l 的极坐标方程和曲线C 的直角坐标方程;(2)已知直线l 与曲线C 交于,A B 两点,试求,A B 两点间的距离.23.[选修4-5:不等式选讲](10分) 已知0a >,0b >,1a b +=. (1(2)若不等式11|||1|x m x a b+-++…对任意x R ∈及条件中的任意,a b 恒成立,求实数m 的取值范围.数学(文科)参考答案1.B 集合{|13}B x x =-剟,∴{|13}A B x x ⋂=剟. 2.A 设(,)z a bi a b R =+∈,则2()2()362z z a bi a bi a bi i +=++-=-=-,362a b =⎧⎨-=-⎩,22a b =⎧⎨=⎩,即22z i =+,对应点为(2,2),在第一象限.3.A 由题知b a =,又222a b c +=,解得c e a ==.4.D ∵n α⊂,m n ∥,有可能m α⊂,A 错误;,m αββ⊥⊥,有可能m α⊂,B 错误;,n m n α∥∥,有可能m α⊂.5.D ∵888S a ==,∴1288a a a a +++=L ,∴70S =,∵747S a =,∴40a =. ∵480,8a a ==,∴84284a a d -==-. 6.D7.B 易判断函数()cos x xy x e e-=-为奇函数,由此排除A ,C ,又1x =时,()cos 0x x y x e e -=->,排除D ,故选B .8.D 222()5AE AB BE AB BD AB AD AB AB AC =+=+=+-=-+u u u r u u u r u u u r u u u r u u u r u u u r u u u r u u u r u u u r u u u r ,∴35m n +=-.9.C10.A ∵2()2()3f x f x x x =-++,∴2()2()3f x f x x x -=+-.∴2()f x x x =-.∴(1)0f =,()12f x x '=-.∴(1)1f '=-,∴过(1,(1))f 切线方程:1y x =-+.11.C ()cos 2sin 6f x x x x π⎛⎫=+=+ ⎪⎝⎭,()sin 2cos 2sin 662g x x x x x πππ⎛⎫⎛⎫=-=+=++ ⎪ ⎪⎝⎭⎝⎭,()g x 的图象由()f x 得图象向左平移2π个单位长度所得.()f x 在区间[,]a b 上是增函数,且()2,()2f a f b =-=,令6x t π+=,可取,22t ππ⎡⎤∈-⎢⎥⎣⎦,向左平移2π个单位长度,即14个周期,可得,22t ππ⎡⎤∈-⎢⎥⎣⎦时()g x 可取得最大值为2.12.A 取PC 中点O ,连接AO ,BO ,设球半径为R ,因为3BPC π∠=,PA AC⊥,PB BC ⊥,所以AO BO R ==,2PC R =,PB R =,3BC R =,由平面PAC ⊥平面PBC ,4APC π∠=得,AO ⊥平面PBC ,因为三棱锥P ABC -的体积为36.所以33366R =,∴1R =,∴球的表面积为4π.13.1 由约束条件1,1,33,x y x y x y --⎧⎪+⎨⎪-⎩………作出可行域如图,由图可知,最优解为A ,联立133x y x y -=-⎧⎨-=⎩,解得(2,3)A .∴2z x y =-的最小值为2231⨯-=.14.32-3122P ⎛⎫- ⎪⎝⎭,∴3cos 2α=,3cos()2πα+=-.15.5 由题知,函数()f x 为偶函数且周期为2,∴91(11)(1)50544f f f f ⎛⎫⎛⎫-+=+=+= ⎪ ⎪⎝⎭⎝⎭. 16.3 由抛物线的定义得:||||AF AM =,||||BF BN =,易证2MFN π∠=,∴222||||||16MN NF MF =+=,∴||4MN =∵11||||||2322MNF S p MN MF NF =⋅=⋅=V ,∴3p =.∴3MFO π∠=,∵||||AF AM =,∴AMF V 为等边三角形.∴直线AB 的倾斜角3πθ=.∴||1cos p AF θ=-,||1cos p BF θ=+.∴||3||AF BF =.17.解:(1)由2cos cos cos a A b C c B =+及正弦定理得,2sin cos sin cos sin cos A A B C C B =+, ∴2sin cos sin A A A =. ∵0A π<<,∴1cos ,23A A π==.(2)∵1sin 2ABC S bc A ==V ,∴24bc =. 由余弦定理2222cos 28a b c bc A =+-=,可得2252b c +=, ∴2()252,10b c bc b c +-=+=,∴ABC V 的周长为10+18.(1)证明:∵PA AB =,E 为线段PB 中点,∴AE PB ⊥. ∵PA ⊥平面ABCD ,BC ⊂平面ABCD ,∴BC PA ⊥. 又∵底面ABCD 是长方形,∴BC AB ⊥.又PA AB A ⋂=, ∴BC ⊥平面PAB .∵AE ⊂平面PAB ,∴AE BC ⊥. 又PB BC B ⋂=,∴AE ⊥平面PBC .(2)由(1)知,AE ⊥平面PBC ,又EF ⊂平面PBC ,∴AE EF ⊥,∴3EF ==.由题知PA ⊥平面ABCD ,E 为PB 中点, ∴点E 到平面ABCD 的距离为122PA =, 设点B 平面AEF 的距离为h ,则B AEF E ABF V V --=,即111134123232h ⨯⨯⨯=⨯⨯⨯⨯,解得h =∴点B 到平面AEF 的距离为319.解:(1)2250(306410) 4.504 3.84134164010K ⨯⨯-⨯=≈>⨯⨯⨯. 所以有95%的把握认为未感染与戴口罩有关.(2)由(1)知,感染者中有4人戴口罩,6人未戴口罩,用分层抽样的方法抽取5人,则2人戴口罩记为,A B ,3人未戴口罩记为1,2,3,从中随机抽取2人,共有AB ,1A ,2A ,3A ,1B ,2B ,3B ,12,13,23共10种可能,其中2人都未戴口罩的有12,13,23共3种, ∴这2人都未戴口罩的概率310P =.20.解:(1)由题知215PF PF =,12F F =2221122PF F F PF +=,解得23PF =,13PF =233a =+=∴a =,c =1b =,∴椭圆C 的标准方程为2213x y +=. (2)要使1AF B V 的内切圆的面积最大,需且仅需其1AF B V 的内切圆的半径r 最大.因为12(F F ,设()()1122,,,A x y B x y ,易知,直线l 的斜率不为0,设直线:l x ty =+221,3x ty x y ⎧=+⎪⎨+=⎪⎩故()22310t y ++-=,故1223y y t +=-+,12213y y t =-+; 故11212121212AF B F F A F F B S S S F F y y =+=-=V V V==, 又()11111||22AF B S AF F B AB r r =++⋅=⋅=V ,故23t =+,即12r ==;=,即1t =±时等号成立,∴直线l 的方程为y x =-或y x =-+21.解:∵2()xf x e ax =-,∴()2xf x e ax '=-.∴()20xf x e ax '=-=,2x e a x =,不妨令()(0)xe g x x x=≠.2(1)()x e x g x x -'=,∴()g x 在(,0)-∞、(0,1]递减,在[1,)+∞递增,(1)g e =,且0x <时,()0g x <.∵函数()y f x =有两个极值点,∴2,2e a e a >>. (2)方法一:()2xf x e ax '=-, 令()(),()2x h x f x h x e a ''==-,(ⅰ)当102a 剟时,()0,()h x f x ''>单调递增,()(0)1f x f ''>=, ()f x 单调递增,()(0)1f x f >=,满足题意.(ⅱ)当122ea <≤时,()20x h x e a '=-=,解得ln2x a =. 当(0,ln2)x a ∈,()0h x '<,()f x '单调递减; 当(ln2,)x a ∈+∞,()0h x '>,()f x 单调递增, 此时ln 2min ()(ln 2)2ln 22(1ln 2)af x f a e a a a a ''==-=-,∵2ea „,1ln20a -…,即min ()0f x '…, ∴()f x 单调递增,()(0)1f x f >=,满足题意.综上02ea 剟且0x >时,()1f x >成立. 方法2:不妨令2()(0)xG a e ax a e =-剟, ∴2()xG a x a e =-+在0,2e a ⎡⎤∈⎢⎥⎣⎦递减.2min ()22x e e G a G e x ⎛⎫==- ⎪⎝⎭,不妨令:2()2x e g x e x =-,∴()x g x e ex '=-.令()()xx g x e ex ϕ'==-, 则()xx e e ϕ'=-, 由()0x ϕ'>得1x >, 由()0x ϕ'<得1x <,∴()()x g x ϕ'=在(,1]-∞递减,在[1,)+∞递增. ∴min ()(1)0g x g ''==,11 ∴()0g x '…,∴()g x 在[0,)+∞递增.∴min ()(0)1g x g ==, 当02e a 剟且0x >时,()1f x >. 22.解:(1)直线:3410l x y -+=,即3cos 4sin 10ρθρθ-+=;曲线:4C πρθ⎛⎫=- ⎪⎝⎭,即2cos sin ρρθρθ=+,曲线C 的普通方程为220x y x y +--=.(2)将直线l 的参数方程代入220x y x y +--=得2705t t +=,即75t =-或0t =, ∴,A B 两点间的距离127||5AB t t =-=.23.解:(1)21111116a b a b a b +=+++++++++++=, 当且仅当12a b ==+. (2)1111()24b a a b a b a b a b ⎛⎫+=++=++ ⎪⎝⎭…,当且仅当a b =时取等号, ∴11a b+的最小值为4. 又|||1||1|x m x m +-+-„,∴不等式11|||1|x m x a b +-++„对任意x R ∈恒成立,只需|1|4m -„即可,解得35m -≤≤,即m 的取值范围为[3,5]-.。

安徽省江淮十校2020届高三第三次联考数学(理科)试卷 含解析

安徽省江淮十校2020届高三第三次联考数学(理科)试卷 含解析
5
19.(12 分) 快递业务主要由甲公司与乙公司两家快递公司承接:“快递员”的工资是“底薪+送件提成”.这两家公司对“快 递员”的日工资方案为:甲公司规定快递员每天底薪为 70 元,每送件一次提成 1 元;乙公司规定快递员每 天底薪为 120 元,每日前 83 件没有提成,超过 83 件部分每件提成 5 元,假设同一公司的快递员每天送件 数相同,现从这两家公司往年忙季各随机抽取一名快递员并调取其 100 天的送件数,得到如下条形图:
B. i 1008 ?
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2
9.已知函数 f (x) = cos2 (x + ) − 2sin 2 (x + ) + 2.则关于它有关性质的说法中,正确的是
6
6
A.周期为 2
B.将其图象向右平移 个单位,所得图象关于 y 轴对称 6
C.对称中心为 ( + k ,0)(k Z ) 12 2
13.已知向量|a|=3,|b|=2,|2a+b|= 2 3 ,则 a,b 的夹角为
D.2 .
3
2x − y −1 0
14.设
x,y
满足约束条件
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x − 3y − 3 0
15.如图所示,点 F 是抛物线 y2=4x 的焦点,点 A,B 分别在抛物线 y2=4x 及圆 x2+y2-2x-8=0 的实线部
A. 20
B. − 20
C. 44
D. 40
7.某多面体的三视图如图所示,该多面体的各个面中有若干个是三角形,这些三角形的面积之和为

江淮十校2020届高三第三次联考语文试题(含答案和解析)

江淮十校2020届高三第三次联考语文试题(含答案和解析)
本试卷为江淮十校2020届高三第三次联考的语文试题,包含了丰富的题目类型,旨在全面考察学生的语文知识和应用能力。试卷由多个部分组成,包括阅读理解、古文翻译、词语运用等,每个部分都精选了具有代表性的题目,以检验学生在不同语文领域的掌握情况。同时,本试卷还附带了详细的答案和解析,方便生在完成测试后对照检查,及时了解自己的错误和不足,从而进行有针对性的提升。然而,请注意,本试卷并非用户所搜索的2021年江淮十校第二次联考的语文试卷,内容可能存在一定的差异。
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“江淮十校”2020届高三第三次联考英语2020.5注意事项:1.本试卷由四个部分组成。

其中第一、二部分和第三部分的第一节为选择题。

第三部分的第二节和第四部分为非选择题。

共150分。

2.全部答案在答题卡上相应区域内完成,在本试卷上作答无效。

选择题请使用2B铅笔填涂,非选择题请使用0.5毫米黑色签字笔作答。

要求字体工整、笔迹清晰。

3. 请在答题卡规定的地方填写好个人信息,并认真核对答题卡上所粘贴的条形码是否与本人的信息一致。

4. 考试结束后,将本试卷和答题卡一并交回。

第一部分听力(共两节,满分30分)做题时先将答案标在试卷上,录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节(共5小题;每小题1.5分,满分7.5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. What does the woman want to do?A. Take a bus to Brooklyn.B. Go to the 12th street.C. Put up a sign at the bus stop.2. When is Sara's car supposed to arrive?A. At 4:20 pm.B. At 4:35 pm.C. At 6: 20 pm.3. What is the probable relationship between the speakers?A. Customer and clerk.B. Teacher and student.C. Manager and employee.4. Where does the conversation take place?A. At the man's office.B. At a clothing store.C. At a travel agency.5. Who plays tennis best in the woman's opinion?A. David.B. Steven.C. Mike.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白,每段对话或独白后有几个小题,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。

听第6段材料,回答第6、7题。

6. Who is the man?A. An actor.B. A journalist.C. A bodyguard.7. Why does the woman want to wear sunglasses?A. She likes being photographed.B. She doesn't want to be recognized.C. It's too bright outside.听第7段材料,回答第8至10题。

8. When will Fiona go to Florida?A. On July 10th.B. On July 13th.C. On July 15th.9. What is Fiona going to do in Everglades?A. Go to a nature reserve.B. Visit theme parks.C. Go shopping.10. Where is the Florida International University?A. Orlando.B. Everglades.C. Miami.听第8段材料,回答第11至13题。

11 . Who is sad about the news?A. Baker.B. Grandpa.C. Alex.12. What is the most serious problem with the community center on Cranberry Street?A. It's too noisy.B. It's too small.C. It's too far.13 . Why has the plan for the community center been put off?A. Many people are against it.B. There is short of money.C. The old library is preferred .听第9段材料,回答第14至16题。

14. What does the man want to know?A. The price of a training course.B. Information about a training course.C. The trainer of a course.15 . How long does the course last?A. About 10 days.B. About 5 days.C. About 3 days.16. What will the man finally do during the training course?A. Attend some lectures.B. Work in a company.C. Take a test for computer science.听第10段材料,回答第17至20题。

17. Who is the speaker probably talking to?A. Visitors.B. New employees.C. Directors・18. What comes first for the company?A. Interpersonal relationship.B. Knowledge about products.C. Being punctual.19. What can they do in the afternoon?A. Getting to know the customers.B. Visiting the workshops.C. Reading the book about the company.20. What's the speaker's advice in the end?A. Leaving before 5:30.B. Having lunch in the canteen.C. Keeping learning.第二部分阅读理解(共两节,满分40分)第一节(共15小题;每小题2分,满分30分)阅读下列短文,从所给的四个选项(A、B、C、D)中选出最佳选项,并在答题卡上将选项涂黑。

AWhen film was first invented in the late 1800s and early 1900s, movie-goers could actually see images offar-away places, like China, and that fueled interest in the area. Throughout history, many Chinese Americans are devoted in this area. Now, the back room of the Formosa Cafe looks like a museum that honors the works of Chinese Americans and their contributions to Hollywood. Chinese stereotypes (刻板印象)Stereotypes of the Chinese in America were strengthened by the otherness of U. S. China towns in the late 1800s and early 1900s. There was an idea that the Chinese was the " yellow peril" , who you couldn't trust. And that resulted in the character called Fu Manchu. Fu Manchu was an evil character who wanted to destroy the western world. He appeared in movies and in a television series.In 1926, Charlie Chan, a Chinese investigator from Hawaii, appeared for the first time in a movie. This created a different, yet still problematic Asian stereotype." Yellow face" actorsCharlie Chan and Fu Manchu may have been Chinese characters, but the actors were usually white men made up to look like Asian. Actors Sidney Toler, Roland Winters and Ross Martin all played Charlie Chan. Yellow face meant they actually yellowed up their skin. White actors just played the lead characters in The Good Earth, a 1937 film about Chinese farmers. Asian actors had parts in the film, but they needed bankable actors, however, there were no Asian American bankable actors.China factorsOver the years, Asian and Chinese Americans did find work in Hollywood, and a few earned a star on the Hollywood Walk for Fame. Hollywood is also changing the way it presents the Chinese culture. As the biggest market for movies outside the U. S, Hollywood has been making films that will not offend movie-goers in China or the country's government. The industry has been careful not to show the Chinese as evil. Co-productions between Hollywood and Chinese companies put Chinese characters and China in a favorable or satisfactory way.21. Why is the back room of the Formosa Caf6 mentioned?A. To show appreciation for Chinese American filmmakers.B. To display the richness of Chinese American films.C. To attract more customers to enjoy coffee in the Formosa Caf6.D. To recommend a place to mover-goers to learn Chinese American films.22. What can be known about Charlie Chan?A. He could be found in a TV series.B. He was an actor bearing evil reputation.C. He represented investigators from Hawaii.D. He wasn't much appreciated by movie-goers then.23. What can be concluded about Chinese American films?A. Chinese actors were preferred in casting Chinese roles.B. They focused on evil Chinese eager to ruin America.C. China factors are positively viewed in them.D. They have been introducing Chinese stereotypes.BThis is the colourful waste created by a Swedish city with a unique recycling system. Like many cities in Sweden, Eskilstuna has an impressive recycling record. It met the EU 2020 target of recycling 50% of waste many years ago. But almost everyone who lives here follows a strict recycling policy at home. People are expected to sort their household waste into seven separate categories, including food, textiles, cartons(纸板箱)and metal. But what really makes the system stand out is the bright colour code.The reason for this becomes clear at the city's recycling plant. The bags arrive all jumbled up because they're collected altogether in a rush, once a fortnight from outside people's houses. But thanks to those bright colors, scanners can select the bags and separate them efficiently. The food waste in green bags is processed on site into a certain thick liquid to make burnable gas, which powers the city's buses. One of the benefits of this method of recycling is that there is less crosspollution , so more of the recycled waste can actually be used to make new things. Like the rest of Sweden, Eskilstuna is committed to sending zero waste from its citizens to landfill. Waste that cannot be recycled is burnt at a local plant to generate electricity. This reduces reliance on fossil fuels, but does create greenhouse gases.As countries around the world try to improve their recycling rates, some may look to Eskilstuna as an example to follow as long as they think they can persuade their citizens to get busy sorting at home. 24. What can we learn about Eskilstuna?A. It is flooded with colorful waste.B. It is best-known for waste-recycling.C. Waste there is painted into seven colors.D. Rubbish recycling is handled seriously there.25. What does the underlined word "jumbled” in paragraph 2 mean?A. Mixed together.B. Collected orderly.C. Sealed cautiously.D. Piled purposefully.26. What may the new recycle-system bring about?A. The increase in the efficiency of waste recycling.B. The convenience to the citizens in waste dropping.C. The avoidance of rubbish related environmental problems.D. The decrease of profit for waste plants.27. Where is this text most likely from?A. Parents.B. In style.C. Fortune.D. The New York Times.CChocolates, syrup(糖浆),cream on top, many of the hot drinks we consume even more at this time of year certainly sound sugary, but just how much sugar they contain might come as a shock.The health campaign group Action on Sugar has found that Starbucks hot chocolate made with milk has almost 94 grams of sugar. That equals 23 teaspoons. At Caffe Nero, the salted hot chocolate packs in almost 60 grams of sugar, about 15 teaspoons. And Casta's cream latte(拿铁)has 32 grams of sugar, that's 8 teaspoons.The drink makers won't have those details listed next to the price. If known, it would be quite alarming. That's how capitalism works. The capitalists have to get people addicted to certain things to keep selling it more. If told the amount of sugar contained, many would take hot drinks just as a treat not on a daily basis. Some drinks have actually reduced in sugar over the past few years, but many more haven't. The most sugary seasonal drink was from Starbucks with more than 14 teaspoons of sugar. Other companies like breakfast cereal manufacturers and yogurt manufacturers are reducing sugar. And so is soft drinks, yet the milk based drinks, milk shakes, hot chocolates and lattes just don't seem to have changed in the same way.And while some companies are actually being really responsible reducing their sugar, some have actually increased that in the last two years which just seems ridiculous. Coffee chains do offer low- calorie alternatives. But Action on Sugar are still calling for more tax on hot coffee drinks. Not much of a Christmas present for the coffee chains, but maybe a gift in the long term for our health.28. What does the first two paragraphs mainly talk about?A. Consumers favor sugar contained hot drinks.B. The amount of sugar in some hot drinks is frightening.C. Famous hot drink makers prefer hot chocolate series.D. Cream latte is the signature hot drink for Starbucks.29. What can we learn about hot drink makers?A. Most of them are trying to reduce sugar in certain drinks.B. Chocolate and syrup are irreplaceable part of their products.C. They won't show the consumer how much sugar a drink contains.D. Four of them have been investigated by the health campaign group.30. What's the writer's attitude toward the hot drink makers?A. Supportive.B. Doubtful.C. Objective.D. Reserved.31 . How can the sugar-originated hot drink problem be eased?A. Taxes can be raised on hot drink makers like coffee chains.B・ More official groups like Action on Sugar could be founded.C. Sugar content can be labeled in place of the price for hot drinks.D. Warnings can be given to those enjoying weekly hot drinks.DDo some kinds of video games cause violence? Scientific studies do not suggest a link. But the idea that there is a link between violent video games and violent acts reappeared following the mass shooting in El Paso, Texas, last weekend. An online statement thought to be written by the El Paso gunman mentioned the video game "Call of Duty".On Monday, President Donald Trump said that " terrifying video games”contribute to a "glorification of violence''. American politicians have long made similar statements・ Benjamin Burroughs is a professor of media at the University of Las Vegas. He said that there is no linkage to gun violence, when mentioning video games.Burroughs pointed out that some studies show a short-term increase in aggressive thoughts and feelings after playing video games, but nothing that rises to the level of violence. "Plenty of gamers get upset when they lose or feel the game was 'cheating', but it doesn't lead to violent outputs," Burroughs stressed.In 2006, a small study by researchers at Indiana University found that teenagers who played violent video games showed higher levels of emotional arousal (激发)---strong emotions like anger or fear. The teenagers also showed less activity in the parts of the brain associated with the ability to plan, control and direct thoughts and behavior.Patrick Markey, the psychology professor, found in his research that men who commit severe acts of violence actually play violent video games less than the average male. Another study by Markey and other researchers showed that violence tends to go down when a new violent movie or video game comes out. One possible explanation is that people are at home playing the game or in theaters watching the movie. Markey believes that video games might excite people, but t hey do not change who people are. "It is like going to see a sad movie,” Markey said of playing video games. "It might make you cry but it doesn't make you clinically depressed," he said.32. Why is the mass shooting in El Paso, Texas mentioned in the first paragraph?A. To show the necessity of scientific studies.B. To support the writer's own viewpoint.C. To show the seriousness of violent acts.D. To serve as evidence for the assumption.33. What are the similarities between Benjamin Burroughs and Patrick Markey?A. Both are specialists in psychology in the USA.B. Both acknowledge video game aroused emotional change.C. Both present their ideas through doing research.D. Both worry about the potential dangers caused by video games.34. What can be learned about the research in 2006?A. Teenagers tested in it become more emotional.B. Its findings set alarm for young video game players.C. Teenagers mentioned in it mainly come from Indiana.D. Its researchers are strongly for banning video games.35. What is the best title for the text?A. Violence—a by-product of video games.B. Video games—the cause of violence or not.C. Video games—the promoter of the mass shoot.D. Violence—a threat for game players or not.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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