皖中名校联盟2019届高三10月联考1

皖中名校联盟2019届高三10月联考英语试题第一部分听力（共两节，满分30分）第一节（共5小题；每小题1.5分，满分7.5分）请听下面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 is Dr. Greenspan most probably?A. A dentist.B. A nurse.C. A professor.2. When should Peter and Sophia leave according to the man?A. This Tuesday.B. Next Tuesday.C. Next Thursday.3. What does the man suggest the woman do?A. Read the notice on the window.B. Get a new bus schedule.C. Go and ask the staff.4. What are the speakers mainly talking about?A. The design of the room.B. Smoking.C. The woman’s health.5. Where was the woman when she saw the fish?A. In the water.B. In a boat.C. On the shore.第二节（共15小题；每小题1.5分，满分22.5分）请听下面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，每小题将给出5秒钟的作答时间。

每段对话或独白读两遍。

请听第6段材料，回答第6至8题。

皖中名校联盟2019届高三10月联考数学试题卷（理科）第Ⅰ卷（选择题 共60分）一、选择题：本大题共12小题，每小题5分，共60分．每一小题给出的四个选项中只有一项是符合题目要求的．1.命题“4,0x R x x ∀∈+≥”的否定是（ ） A. 4,0x R x x ∀∈+< B. 4,0x R x x ∀∈+≤ C. 4000,0x R x x ∃∈+≥D. 4000,0x R x x ∃∈+<2.已知2{|430},{|P x x x Q y y =-+<==，则P Q =（ ）A. [0,1)B. [0,2)C. (1,2]D. (1,2)3.由曲线3,y x y == ）A.512B.13 C. 14 D. 12 4.已知向量AB 与AC的夹角为3π，()2,3,,AB AC AM AB AC R λμλμ===+∈，且A M B C ⊥，则λμ=（ ） A.16B. 6C.14D. 45.设函数21()1xxf x e e x -=+-+，则使得(2)(1)f x f x >+成立的x 的取值范围是（ ） A. (,1)-∞B. (1,)+∞C. 1(,1)3- D. 1(,)(1,)3-∞-+∞U6.“0a ≥”是“函数()(1)f x ax x =+在区间(0,)+∞上单调递增”的（ ） A. 充分不必要条件 B. 必要不充分条件C. 充分必要条件D. 既不充分也不必要条件7.已知数列{}n a 为等差数列，其前n 项和为n S ，且13623a a S +=，给出以下结论： ①100a =；②10S 最小；③712S S =；④190S =． 其中一定正确的结论是（ ）A. ①②B. ①③④C. ①③D. ①②④8.函数4lg x x y x=的图象大致是（ ）A.B.C.D.9.已知函数()()sin 0,0,2f x A x A πωφωφ⎛⎫=+>><⎪⎝⎭，其图象相邻两条对称轴之间的距离为2π，且()f x 的图象关于点,012π⎛⎫- ⎪⎝⎭对称，则下列判断正确的是（ ） A. 要得到函数()f x的图象只将2y x =的图象向右平移6π个单位B. 函数()f x 的图象关于直线512x π=对称 C. 当,66x ππ⎡∈-⎤⎢⎥⎣⎦时，函数()f x 的最小值为D. 函数()f x 在,63ππ⎡⎤⎢⎥⎣⎦上单调递增10.已知定义在R 上的奇函数()f x 满足(2)()f x f x +=-，当01x <<时，()21x f x =-，则2(log 9)f =（ ）A. 79-B. 8C. 10-D. 259-11.设函数121,1(),4,1x x f x x x +⎧-≤⎪=⎨->⎪⎩若互不相等的实数,,p q r 满足()()(),f p f q f r ==则222p q r ++的取值范围是（ ） A. (8,16)B. (9,17)C. (9,16)D. 1735(,)2212.已知2()f x x ax b =++，集合{|()0}A x f x =≤，集合{|[()]3}B x f f x =≤，若A B =≠∅，则实数a 的取值范围是（ ）A .[6,2]-B.C. [2,-D. [6,--第П卷（非选择题 共90分）二、填空题：本大题共4小题，每小题5分，共20分．请将答案填写在答题卷相应位置上．13.已知平面向量,a b 满足2,1,223a b a b ==+=，则a b r r与的夹角为___________．14.函数()y f x =的图象和函数log (0a y x a =>且1)a ≠的图象关于直线y x =-对称，且函数()(1)3g x f x=--，则函数()y g x =图象必过定点___________。

皖中名校联盟2019届高三10月联考数学试题卷（理科）第Ⅰ卷（选择题 共60分）一、选择题：本大题共12小题，每小题5分，共60分．每一小题给出的四个选项中只有一项是符合题目要求的．1．命题“0||,4≥+∈∀x x R x ”的否定是（ ）A ．0||,4<+∈∀x x R xB ．0||,4≤+∈∀x x R xC ．0||,4000≥+∈∃x x R xD ．0||,4000<+∈∃x x R x2．已知}24|{},034|{2x y y Q x x x P -==<+-=，则=Q P （ ）A ．)1,0[B ．)2,0[C ．]2,1(D ．)2,1(3．由曲线x y x y ==,3围成的封闭图形的面积为（ ）A ．125 B ．31C ．41 D ．21 4．已知向量AB 与AC 的夹角为3π，),(,3||,2||R AC AB AM AC AB ∈+===μλμλ，且⊥，则=μλ（ ） A ．61B ．6C ．41 D ．45．设函数11)(2+-+=-x e e x f xx ，则使得)1()2(+>x f x f 成立的x 的取值范围是（ ） A ．)1,(-∞ B ．),1(+∞ C ．)1,31(-D ．),1()31,(+∞--∞6．“0≥a ”是“函数|)1(|)(x ax x f +=在区间),0(+∞上单调递增”的（ ）A ．充分不必要条件B ．必要不充分条件C ．充分必要条件D ．既不充分也不必要条件7．已知数列}{n a 为等差数列，其前n 项和为n S ，且63132S a a =+，给出以下结论： ①010=a ；②最小10S ；③127S S =；④019=S ． 其中一定正确的结论是（ ） A ．①②B ．①③④C ．①③D ．①②④8．函数||||ln 4x x x y =的图象大致是（ ）9．已知函数)2||,0,0)(sin()(πϕωϕω<>>+=A x A x f 的最大值为2，其图象相邻两条对称轴之间的距离为2π且)(x f 的图象关于点)0,12(π-对称，则下列判断正确的是（ ）A ．要得到函数)(x f 的图象，只需将x y 2cos 2=的图象向右平移6π个单位B ．函数)(x f 的图象关于直线π125=x 对称 C ．当]6,6[ππ-∈x 时，函数)(x f 的最小值为2- D ．函数)(x f 在]3,6[ππ上单调递增 10．已知定义在R 上的奇函数)(x f 满足)()2(x f x f -=+，当10<<x 时，12)(-=x x f ，则=)9(log 2f （ ） A ．97-B ．8C ．10-D ．925-11．设函数,1,41|,12|)(1⎩⎨⎧>-≤-=+x x x x f x 若互不相等的实数r q p ,,满足),()()(r f q f p f ==则r q p 222++的取值范围是（ ）A ．)16,8(B ．)17,9(C ．)16,9(D ．)235,217(12．已知b ax x x f ++=2)(，集合}0)(|{≤=x f x A ，集合}3)]([|{≤=x f f x B ，若∅≠=B A ，则实数a 的取值范围是（ ）A ．]2,6[-B ．]6,32[C ．]32,2[-D ．]32,6[--第П卷（非选择题 共90分）二、填空题：本大题共4小题，每小题5分，共20分．请将答案填写在答题卷相应位置上．13．已知平面向量b a ,满足32|2|,1||,2||=+==b a b a，则b a 与的夹角为___________．14．函数)(x f y =的图象和函数0(log >=a x y a 且)1≠a 的图象关于直线x y -=对称，且函数3)1()(--=x f x g ，则函数)(x g y =的图象必过定点___________．15．=︒-︒+︒-︒︒⋅︒)5.7cos 5.7(sin 35.7tan 15tan 15tan 5.7tan 22___________．16．若直线b kx y +=是曲线2ln +=x y 的切线，也是曲线xe y =的切线，则=b ___________．三、解答题：本大题共6小题，共70分．解答应写出文字说明，演算步骤或证明过程．解答写在答题卡上的指定区域内． 17．（本小题满分10分）已知0>m ，命题:p 函数)2(log )(mx x f m -=在]1,0[上单调递减，命题:q 不等式1||>-+m x x 的解集为R ，若q p ∧为假命题，q p ∨为真命题，求m 的取值范围．18．（本小题满分12分）已知等差数列}{n a 的公差为2，且1,1,1421---a a a 成等比数列． （1）求数列}{n a 的通项公式； （2）设11+=n n n a a b (*N n ∈)，n S 是数列}{n b 的前n 项和，求使152<n S 成立的最大正整数n ．19．（本小题满分12分）ABC ∆中，c b a ,,分别是内角C B A ,,所对的边，且满足02cos cos =+-+cb a C B ． （1）求角C 的值；（2）若2=b ，AB 边上的中线3=CD ，求ABC ∆的面积．20．（本小题满分12分）已知函数a xax x f 21)(-+=，当]3,1[∈x 时，)(x f 的最小值为0． （1）求a 的值；（2）若0>a ，不等式02)2(≥⋅-x x k f 在区间]1,1[-上有解，求k 的取值范围．21．（本小题满分12分）已知函数x axxx f ln 1)(+-=． （1）若0)(≥x f 对0>∀x 恒成立，求a 的值； （2）求证：2221...3221)1ln(nn n -+++>+（*N n ∈）．22．（本小题满分12分）已知函数2)1()(ax e x x f x --=． （1）讨论)(x f 的单调性；（2）若函数)(x f 有两个零点分别记为21,x x ． ①求a 的取值范围； ②求证：0)2(21<+'x x f ． 皖中名校联盟2019届高三10月联考数学试题卷（理科）参考答案一、选择题二、填空题 13．3π 14．)4,1(- 15．2- 16．0或1三、解答题17．解：命题:p 令mx x u -=2)( )(x u 在]1,0[∈x 上单减 1>∴m又0)(>x u 02)1()(m i n>-==∴m u x u 21<<∴m ………3分命题:q ⎩⎨⎧<≥-=-+m x m mx m x m x x ,,2||1||>-+m x x 的解集为R ∴只需1|)|(min >=-+m m x x ………6分q p ∧ 为假命题，q p ∨为真命题 ∴p 、q 一真一假（1）若p 真q 假，则无解m m m m ∴⎪⎩⎪⎨⎧>≤<<0121 （2）若p 假q 真，则20121≥∴⎪⎩⎪⎨⎧>>≥≤m m m m m 或 综上所述，),2[+∞∈m . ………10分18．解：（1）由题意知，)1)(1()1(4122--=-a a a 即)5)(1()1(1121+-=+a a a 解得31=a故12+=n a n ，*N n ∈………5分 （2）由)321121(21)32)(12(1+-+=++=n n n n b n得n n a a a a S ++++=...321 )321121...71515131(21+-+++-+-=n n )32131(21+-=n )32(3+=n n 由152)32(3<+n n解得6<n故所求的最大正整数n 为5. ………12分 19．解：（1）02cos cos =+-+cba C B由正弦定理得0s i n s i n s i n 2c o s c o s =+-+CB AC B 即0)sin sin 2(cos sin cos =+-⋅+⋅B A C C B 从而0cos sin 2)sin(=⋅-+C A C B 即0cos sin 2sin =⋅-C A A 又ABC ∆中，0sin >A 故21cos =C 得3π=C .………6分（2）由)(21CB CA CD +=得)60cos 222(41322︒⋅⋅⋅++=a a从而2=a 或a=)(4舍-故360sin 2221sin 21=︒⨯⨯⨯=⋅=∆C ab S ABC .………12分 20．解：（1）22211)(xax x a x f -=-=' 31≤≤x 912≤≤∴x①当91≤a 时，0)(≤'x f 恒成立，)(x f 在[]3,1上单减 02313)3()(min =-+==∴a a f x f 31-=∴a②当191<<a 时，)(,0)(,)1,1[x f x f a x <'∈时单减；)(,0)(,]3,1(x f x f ax >'∈时单增02)1()(m i n =-+==∴a a a af x f 0=-∴a a 0=∴a （舍）或1=a （舍）③当1≥a 时，0)(≥'x f 恒成立，)(x f 在]3,1[上单增021)1()(min =-+==∴a a f x f 1=∴a综上所述：131=-=a a 或. ………6分（2）由（1）可知：21)(,1-+==xx x f a 0222122)2(≥⋅--+=⋅-xx x x x k k f ，122)2(2+-1≤x x k 要使不等式在]1,1[-上有解，则只需max 2]122)2[(+-1≤x x k令12122)21(),221(2122+-=+-≤≤=t t t t x x x ，其最大值为1，1≤∴k . (12)分21. 解：（1）22222211)1()(xa x x a ax a x x a x a ax x f -=-=+---='①当0<a 时，0)(>'x f 恒成立，)(x f 在),0(+∞上单增 0)1()(,)1,0(=<∈∴f x f x 时当，不满足题意 ②当0>a 时，单减时)(,0)(,)1,0(x f x f ax <'∈；单增时)(,0)(,),1(x f x f ax >'+∞∈0ln 11)1()(min ≥--==∴a a a f x f令a a a g ln 11)(--=，则22111)(aaa a a g -=-='单增时)(,0)(,)1,0(a g a g a >'∈；单减时)(,0)(,),1(a g a g a <'+∞∈0)1()(=≤∴g a g ∴由0ln 11≥--a a解得1=a . ………6分（2）由（1）：取等号）当且仅当1(11ln =-≥x xx令)(11*∈>+=N n nn x ，则有111ln +>+n n n122->n n 21111ln n n n n n ->+>+∴ 21ln )1ln(nn n n ->-+∴ ⎪⎪⎪⎩⎪⎪⎪⎨⎧->-+>->-∴2221ln )1ln(212ln 3ln 101ln 2ln n n n n 累加得*)(1...3221)1ln(222N n n n n ∈-+++>+，原命题得证. ………12分 22. 解：（1）)2(2)1()(a e x ax e x e x f x x x -=--+='（i ）当0≤a 时，02>-a e x单减时)(,0)(,)0,(x f x f x <'-∞∈；单增时)(,0)(,),0(x f x f x >'+∞∈（ii ）当210<<a 时， 单增时)(,0)(,))2ln(,(x f x f a x >'-∞∈；单减时)(,0)(,)0),2(ln(x f x f a x <'∈单增时)(,0)(,),0(x f x f x >'+∞∈（iii ）当21=a 时，0)(≥'x f 恒成立，)(x f 在R 上单增 （iv ）当21>a 时，单增时)(,0)(,)0,(x f x f x >'-∞∈；单减时)(,0)(,))2ln(,0(x f x f a x <'∈单增时)(,0)(,)),2(ln(x f x f a x >'+∞∈综上所述：0≤a 时，)(x f 在上单调递增上单调递减),0(,)0,(+∞-∞；210<<a 时，)(x f 在),0())2ln(,(,)0),2(ln(+∞-∞和上单调递减a a 上单调递增；21=a 时，)(x f 在R 上单调递增； 21>a 时，)(x f 在)),2(ln()0,(,))2ln(,0(+∞-∞a a 和上单调递减上单调递增.………4分（2）①1)0(-=f（i ）当0=a 时，x e x x f )1()(-=，只有一个零点，舍去（ii ）当0<a 时，单增单减在),0(,)0,()(+∞-∞x f 01)0()(m i n <-==∴f x f 又0)1(>-=a f ，取)2ln(1a b b -<-<且则2)1()(ab e b b f b --=2)1(2ab b a --->)12(22-+-=b b a0)12)(1(2>-+-=b b a )(x f ∴存在两个零点 （iii ）当210<<a 时， )(x f 在),0(+∞上单调递增，0≤x 时，0)(<x f )(x f ∴不可能有两个零点，舍去（iv ）当21=a 时，)(x f 在)(,x f R 上单增不可能有两个零点，舍去 （v ）当21>a 时，)(x f 在上单增）（上单减）（),2(ln ,)2ln,0(+∞a a 0≤x 时，0)(<x f )(x f ∴不可能有两个零点，舍去综上所述：0<a （本题也可用分离参数法）………8分②由①知：0<a ，)(x f 在上单增上单减),0(,)0,(+∞-∞ ∴要证0)2(21<+'x x f11 即证0221<+x x ，即证021<+x x 令)()()(x f x f x g --=，则)()()(x f x f x g -'+'=')2)(()2(a e x a e x x x --+-=-)(x x e e x --= 当0>x 时，单增)(,0)(x g x g >'不妨设210x x >>，则)0()(1g x g >，即0)()(11>--x f x f 又)()(21x f x f = )()(12x f x f ->∴ )(x f 在上单减)0,(-∞ 12x x -<∴ 021<+∴x x ，原命题得证. ………12分。

皖中名校联盟2019届高三第一次联考（阜阳一中、六安一中）化学试题卷考试说明:1.考查范围：金属及其化合物，非金属及其化合物.2.试卷结构：分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）；试卷分值:100分，考试时间:90分钟。

3.所有答案均要答在答题卷上，否则无效。

考试结束后只交答题卷.可能用到的相对原子质量:H 1 C 12 O 16 Na 23 Al 27 Cr 52 Ca 40 Si 28第Ⅰ卷（选择题共48分）本卷共16小题，每小题3分。

每小题只有一个选项符合题意.答案填涂到答题卡上。

1。

下列对有关文献的理解错误的是（）A。

“所在山洋，冬月地上有霜，扫取以水淋汁后，乃煎炼而成”过程包括了溶解、蒸发、结晶等操作。

B．《物理小识》记载“青矾(绿矾）厂气熏人，衣服当之易烂，栽木不没，“气”凝即得“矾油”。

青矾厂气是NO 和NO2。

C．《开宝本草》中记载了中药材铁华粉的制作方法:“取钢煅作时如笏或团，平面磨错令光净，以盐水洒之，于醋瓮中阴处埋之一百日，铁上衣生，铁华成矣。

”中药材铁华粉是醋酸亚铁。

D．唐代《真元妙道要略》中有云“以硫磺、雄黄合硝石并蜜烧之，焰起烧手、面及屋舍者"，描述了黑火药制作过程。

2。

下列说法正确的是（）A．液态HCl、固态AgCl均不导电，所以HCl、AgCl 是非电解质B．NH3、CO2的水溶液均能导电,所以NH3、CO2均是电解质C．非金属氧化物不一定是酸性氧化物，酸性氧化物不一定都是非金属氧化物D．碱性氧化物一定是金属氧化物，金属氧化物一定是碱性氧化物3。

设N A表示阿伏加德罗常数的值，下列说法正确的是( )A．6g SiO2晶体中含有Si-O键的数目为0.2N AB．1mol乙酸与足量的乙醇发生酯化反应，生成乙酸乙酯分子数为N A个C．某温度下，pH=2的H2SO4溶液中，硫酸和水电离的H+总数为0.01N AD．常温下，1L 0.1mol·L-1醋酸钠溶液中加入醋酸至溶液为中性，则溶液含醋酸根离子数0.1N A个4.下列离子方程式书写正确的是（）A．Na2S2O3溶液与稀H2SO4反应的离子方程式：S2O32—+ 6H+ =2S↓+3H2OB．向饱和Na2CO3溶液中通入过量CO2，CO32-+CO2+H2O═2HCO3-C．CuSO4与过量浓氨水反应的离子方程式：Cu2++2NH3·H2O= Cu(OH)2↓+2NH4+D．Fe与稀硝酸反应，当n（Fe)∶n(HNO3）=1∶2时,3Fe+2NO3-+8H+═3Fe2++2NO↑+4H2O5。

【题文】阅读下面的材料，根据自己的感悟和联想，写一篇不少于800字的文章。

尼康退出中国并不是被同行打败，居然是毫不相关的行业：智能手机。

康师傅和统一方便面销量直线下滑，并不是因为其它品牌方便面崛起，而是因为美团、饿了吗等外卖。

打败卖单车店铺或修自行车铺的也不是同行，而是共享单车。

有人曾开玩笑说：“打败小偷的是移动支付。

”这一切都很像科幻小说《三体》中所说的：“我消灭你，与你无关。

”“敌人”在哪里？你想过吗？要求：①自选角度，自拟标题；②文体自选（诗歌除外），文体特征鲜明；③不得抄袭，不得套作。

【答案】居安思危方能不败康可能从没想过它们会在时代的潮流中被智能手机打败，方便面巨头也可能没想过越来越懒散的人会有一天竟然拒绝它们。

时代并没有给这些处在鼎盛时期的企业任何提醒就抛弃了它们，它们也没能在时代之中看到自己的发展方向而走向下坡。

所以，任何人都应该站在时代的潮流之上，居安思危方能立于不败之地。

代打败你，可能并不是你违背了历史潮流，而只是因为你只是在时代潮流之中不能立于潮头。

每一个“弄涛儿”只能向涛头而立，方能“手把红旗旗不湿”。

马从不曾忘记奔跑，因为它知道只有跑得快才能逃避敌害，获得生存；蚂蚁一入秋就开始不停忙碌，因为它知道只有积累得多才能穿越寒冬；小树在还是苗的时候就不停地向深处扎根，因为它知道只有根扎得牢才能顶得住暴雨狂风。

动植物尚且知道在安逸的条件下为以后的艰难做准备，我们人类呢，不是更应该懂得，居安思危方能立于不败之地吗？经济发展迅猛、生活环境不断改善、生活水平不断提高的今天，人们往往会过度沉浸于自我的优越感之中，而忽视了来自周围的危机和挑战。

要知道，糖衣炮弹有时比冷兵器的杀伤力来得更为强大。

你的对手可能是“智能手机”“外卖”等你意想不到的对象。

而作为一个人，一个社会，甚至一个民族，一个国家，如果失去了赖以生存的精神家园，失去了前进的精神动力、智力支持和思想保证，又怎么能够脚踏实地地向未来发起挑战，又怎么能够如初生的朝阳般充满生机和活力，不断向上攀登，直至达到人生的辉煌点？因此，摒弃盲目的优越感，树立必要的危机意识，我们才能在发展路上越探越深，越行越远。

皖中名校联盟2019届高三10月联考英语试题第一部分听力（共两节，满分30分）第一节（共5小题；每小题1.5分，满分7.5分）请听下面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秒钟的时间回答有关小题和阅读下一小题。

每段对话仅读一遍。

1. What is Dr. Greenspan most probably?A. A dentist.B. A nurse.C. A professor.2. When should Peter and Sophia leave according to the man?A. This Tuesday.B. Net Tuesday.C. Net Thursday.3. What does the man suggest the woman do?A. Read the notice on the window.B. Get a new bus schedule.C. Go and as the staff.4. What are the speaers mainly taling about?A. The design of the room.B. Smoing.C. The woman’s health.5. Where was the woman when she saw the fish?A. In the water.B. In a boat.C. On the shore.第二节（共15小题；每小题1.5分，满分22.5分）请听下面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，每小题将给出5秒钟的作答时间。

每段对话或独白读两遍。

请听第6段材料，回答第6至8题。

皖中名校联盟2019届高三10月联考语文试题卷命题学校：合肥八中命题人：刘从良、许婧审题人：阮晋豹叶守义考试说明：1.考查范围：2019年高考大纲。

2.试卷分值:150分。

考试时间:150分钟。

3.所有答案均要答在答题卷上，否则无效。

考试结束后只交答题卷。

一、现代文阅读（36分）（一）论述类文本阅读（本题共3小题，9分）阅读下面的文字，完成1-3题。

人工智能安全性问题的根本问题，并不在于它能否真正超越人类，而在于它是否是一种安全可靠的工具和人类是否对其拥有充分的控制权。

就像高铁、飞机等交通工具那样，虽然它们的速度远远超过了人类，但人类拥有绝对控制权，所以人们相信它们是安全的。

为了实现对其控制的目标，首先需要对人工智能的自主程度进行限定。

虽然人工智能发展迅速，但人类智能也有自己的优势，比如目前人工智能的认知能力还远不如人类智能。

我们可以充分发挥人工智能在信息存储、处理等方面的优势，让它在一些重大事件上做人类的高级智囊，但最终的决定权仍在人类。

比如，当我们把人工智能应用于军事领域时，我们可以利用人工智能来评估危险程度，以及可以采取的措施，但是否应该发动战争、如何作战等重大决策，还是需要掌握在人类自己手里。

正如霍金斯所说的那样：“对于智能机器我们也要谨慎，不要太过于依赖它们。

”与限定人工智能的自主程度类似，我们也需要对人工智能的智能水平进行某种程度的限定。

从长远来看，人工智能是有可能全面超越人类智能的。

从人工智能的发展历程来看，尽管它的发展并非一帆风顺，但短短六十年取得的巨大进步让我们完全有理由相信将来它会取得更大的突破。

从世界各国对人工智能高度重视的现实情况来看，想要阻止人工智能的发展步伐是不现实的，但为了安全起见，限定人工智能的智能程度却是完全可以做到的。

我们应当还需要成立“人工智能安全工程”学科，建立人工智能安全标准与规范，确保人工智能不能自我复制，以及在人工智能出现错误时能够有相应的保护措施以保证安全。

皖中名校联盟2019届高三10月联考英语试题第二部分阅读理解（共两节，满分40分）ATerrific New TechnologiesA fashionable smart speakerThe new Amazon Echo smart speaker has the same popular feature as the original---an always listening voice assistant ready to play music and news, set timers, and use third-party apps. Only now it’s actually stylish. The $120 speaker comes in different finishes, including fabric and wood.A helping hand for parentsSuzy Snooze helps babies and kids sleep so parents can catch up on their own sleep. It’s a sound machine and a nightlight. It connects with an app over Wi-Fi and turns into an audio monitor. If a kid cries at midnight, it’ll detect the sound and try to calm him back to sleep. It can also let early risers know when it’s OK to get out of bed. Made by BleepBleeps, the $249 “smart nightlight” has an adorable little face.Safety for ding dongsDoorbells used to just ring, ding, dong. But thanks to modern technology, doorbells can now double as Internet-connected surveillance (监视) devices. Ring’s $249 Vid eo Doorbell Pro can record HD video of your front steps, even at night. It includes motion sensors and two-way audio, so you can yell “get off my lawn” from anywhere. You can also use the app to see what’s going on in front of your house while you’re at wo rk or in the backyard.The must-have kitchen gadgetThe Instant Pot makes food fast in one pot. The surprise kitchen hit is a combination of pressure cooker, rice cooker, slow cooker and even yogurt maker. It has sensors and settings to make cooking as safe as possible. Starting at $80, an Instant Pot can make dishes in less time than it typically takes to cook them.21. What can Suzy Snooze do for parents?A. Provide sunlight.B. Play with babiesC. Comfort crying babies.D. Remind parents to wake up kids.22. What is the advantage of the Video Doorbell Pro compared with traditional doorbells?A. It can play HD videos.B. It can make a sound of ding-dong.C. It can communicate with the users.D. It can take on the work of a monitor.23. Who are most probably interested in the Instant Pot?A. Those who love fast food.B. Those who enjoy cooking.C. Those who are busy working.D. Those who like going camping.BBike sharing have become popular words in cities from Cape Town to Shanghai to Melbourne. Planners, politicians and media keep showing off their benefits: reducing pollution, congestion, travel costs and oil dependence, while improving public health. Bike sharing also helps make cities appear modern, dynamic and world-wide-----qualities much sought after by the creative class.But what makes for a successful public bike-sharing program? This is an important question because installing one requires significant public and private investment and adjustment to the built environment.While many programs have been launched among much praise, often their popularity has soon declined. Many end up operating at a financial loss and depend on other profitable enterprises to cross-subsidize (交叉补贴) them. Some have resulted in thrown-away bikes becoming an eyesore.Understanding which factors enhance or stop public bike sharing is critical in helping cities decide whether such a program is workable, before considering what design and sitting will work best.Drawing on current knowledge, we discuss the importance of the local landscape, climate, cycling infrastructure (基础设施) and land use. We also touch on other factors, such as the legal environment and the characteristics of the bike-sharing program itself.Take natural environment for example. Two natural environment factors are known to affect participation: hilliness and weather. Hilliness discourages balanced bike-sharing use, as users avoid returning bicycles to stations on hilltops. Those stations end up being empty, while stations on flat areas are often full, so users cannot find a station to return their bikes.As for weather, ideal temperature ranges vary by the climate zone. Case studies show warmand dry weather encourages public bike-sharing use. Rain and strong wind reduce the frequency of trips. However, some approaches, such as providing sheltered, shaded, or even heated or cooled cycling infrastructure, could prove useful.24.What can we infer about bike sharing from Paragraph 1?A.It has no disadvantage.B.It is welcomed worldwide.C.It can solve every problem.D.It is the symbol of modern cities.25.What is the current situation of bike-sharing programs?A.Many of them run at a loss.B.Most of them make huge profits.C.They are the most promising business.D.Their development is determined by public investment.26.Why does the author mention the natural environment?A.To show user’s interest in various natural environment.B.To show the decisive function of natural environment.C.To show the necessity of bike-sharing programs.D.To show the significance of some factors.27.Which of the following might increase bike-sharing use?A.Hilliness.B.Attractive bikes.C.Rain and strong windD.Perfect cycling infrastructure.CWater is precious especially in dry areas. A new device can produce drinkable water from desert air using nothing but sunlight.“With this device, you can harvest a Coke can’s worth of water in an hour,” says co-creator Omar Yaghi, a chemist at the University of California. “That’s about how much water a person needs to survive in the desert.”The current device is just a sample. But the technology could be used to supply fresh water to some remote regions of the globe, like the Middle East and North Africa, Yaghi says.“Previous attempts at low-energy water collection functioned below 50 percent relative humidity, while the new device pulled water from air with as low as 20 percent humidity,” Yaghi said.Getting that water out is easy when the air is wet. But drawing water from the drier air in parched areas is a greater challenge. Spongy (柔软吸水的) materials can take water out from the dry air. Those materials, however, either collect water too slowly or require lots of energy to gather the water.The new device uses a material that avoids both problems. MIT mechanical engineer Evelyn Wang, and her colleagues remade an existing material composed of electrically charged metal atoms. This metal-organic framework MOF-801, creates a network of sponge like pores (气孔) to trap water vapor. At room temperature, water vapor is collected in the pores. As temperatures rise, the water escapes into a box beneath the pores. A condenser (冷凝器) in the box cools the vapor, converting it into a drinkable liquid. This entire process takes around two hours.Laboratory tests of the device harvested 2. 8 liters of water per day. The device could be used as a personal water source in dry regions, Yaghi says, or scaled up (按比例增加) to produce enough water for a whole community.28.How does MOF-801 work?A. By taking water out in wet areas.B. By collecting and cooling water vapor.C. By using spongy materials to store water.D. By creating metal pores and using a condenser.29.What does the underlined word “parched” in Paragraph 5 mean?A. Dry.B. Cool.C. Deserted.D. Exposed.30.Which of the following is an advantage of MOF-801?A. The small size.B. The low price.C. Making water fresher.D. Saving time and energy.31.What can be inferred from the text?A. Evelyn Wang created MOF-801 on her own.B. People are curious about the creation of MOF-801.C. MOF-801 isn’t available for customers now.D. The problem of water shortage will be solved by MOF-801.DScientists have found living organisms (有机体) that could be 50,000 years old. The organisms were found in ahot, but beautiful cave system in Naica, Mexico. These ancient life forms can be seen only with a microscope.The ancient microbes (微生物) were able to exist by eating minerals such as iron and manganese, said Penelope Boston, head of NASA’s Astrobiology Institute. If the findings are confirmed, they will show how microbes can survive in extreme conditions.Researchers have been studying the microbes for nine years. Forty different kinds of microbes and some viruses were found in the underground area. The genetic structures of these organisms are different from those of their nearest relatives. That means their closest relative is about as genetically different as humans are from mushrooms growing in the forest. The caves in Naica are 800 kilometers deep. Before the miners began working in the caves, the area was isolated from the surface and the outside world.Some of the caves are as big as the large religious centers built during Europe’s Middle Ages. There are crystals (结晶体) covering the walls. Scientists must wear special clothing to keep cool. The clothing keeps the crystals safe from human germs or other damage. The researchers could only work for about 20 minutes at a time before they had to go to a room that was 38 degrees Celsius to cool down.NASA officials would not let Boston share her findings with scientific experts before last week’s announcement. So scientists could not say much about the findings. But Norine Noonan, a biologist with the University of South Florida, said she believed them. “Why are we surprised?” Noonan said. “As a biologist, I would say life on Earth is extremely tough and extremely adaptable.” The microbes are not the oldest life forms on Earth. Several years ago, scientists published studies about microbes that may be 500,000 years old and still alive.32. What do the ancient microbes survive on?A. Minerals.B. Germs.C. Viruses.D. Crystals.33. Why are mushrooms mentioned in Paragraph 3?A. To show the caves are extremely hot.B. To tell us the microbes are a kind of mushroom.C. To explain how the structures of the microbes are unusual.D. To say the microbes are as useful to humans as mushrooms.34. What can we learn from the last paragraph about the microbes?A. They are the same microbes found several years ago.B. They live longer than scientists predicted.C. They didn’t surprise Norine Noonan at all.D. They are the oldest life forms on Earth.35. What does the text mainly tell us?A. A complex cave system was discovered in Naica.B. 50,000-year-old “super life” was found in Mexico.C. Researchers never stop studying living organisms.D. Some life forms can live in extreme conditions.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，并在答题卡上将该项涂黑。

皖中名校联盟2019届高三10月联考物理试题卷第I 卷（选择题共40分）一、本题共10小题，每小题4分。

1－7为单选題，8－10为多选题。

答案填涂到答题卡上1．如图所示，用轻绳AO 和OB 将重为C 的重物悬挂在水平天花板和竖直墙壁之 间处于静止状态，AO 绳水平，OB 绳与竖直方向的夹角为θ。

AO 绳的拉力T 1、OB 绳的拉力T 2的大小与G 之间的关系为（）A. T 1=Gsin θB. T 1tan G θ=C. T 2= cos G θD. T 2=Gcos θ 2．轻绳固定在A 点，下端系着铁球，在A 点的正下方B 处有一人用一根可伴缩的轻杆顶住铁球，人沿着轻杆方向施力、铁球和人都可以看做质点，逐渐增加杆子的长度，则（ ）A ．绳子的拉力大小不变B ．绳子的拉力变小C ．杆子的支持力大小不变D ．杆子的支持力变小3．如图所示，物块A 放在倾斜的木板上，木板的倾角α为30°和45°时物块所受摩擦力的大小恰好相等，则物块和木板间的滑动摩擦因数为（ ）A. 12B. 2C. 2D. 24．从水平地面上以20m ／s 的速度竖直上抛一个物体，空气阻力不计，重力加速度＝10/s 2，当速度大小变为10m ／s 时，（）A ．平均速度大小一定是15m ／sB ．平均速度的大小可能小于10m ／sC ．平均速率一定是15m ／sD ．平均速率可能等于5m ／s5．一条质量分布均匀软绳AB 上端固定在B 点，用水平力F 拉动绳子的下端A 稳定后绳子在B 点的切线方向与竖直方向的夹角为θ，绳子中点的切线方向与竖直方向的夹角为α，下列说法正确的是（ ）A ．sin sin 2αθ=，越靠近A 端绳子的张力越大 B ．sin sin 2αθ=，越靠近B 端绳子的张力越大 C ．n n 2ta ta αθ=，越靠近A 端绳子的张力越大 D ．n n 2ta ta αθ=，越靠近B 端子的张力越大 6．水平地面上有一个三角形斜面C ，斜面上有两个滑块A 、B ，A 的质量小于B 的质量，它们与斜面的动降擦因数相同，A 紧在起沿着斜面减速下滑的过程中，斜面保持不动，（ ）A ．斜面对地面有向右的摩擦力作用B ．斜面对地面没有摩擦力作用C ．滑块A 对滑块B 有弹力作用D ．滑块A 与滑块B 之间没有弹力作用7．两块形状相同的板A、B固定并排在一起，一颗子弾以水平速度垂直射向板，由于两块板的材质不同，子弹在A板内的阻力恒为f，在B板中的阻力恒为2f。