2016年高三第一轮复习 金版教程选3-5-2a
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UNIT 2HEALTHY LIFESTYLEⅠ.完形填空I was in Washington covering a conference. However,I came across a curious⁃looking 1shop with its door partly open.Citizens here said it had been like that 2as long as they could remember,but no one knew 3. “Maybe the owner is just lazy,”I4the shady entrance,eager to find out the reason. A(n)5“Welcome!”came from a man’s voice. And out walked Mr Smith,the 90⁃year⁃old6of this candy shop which I suddenly realised had7candies. Mr Smith explained the place was a candy store opened by him 50 years ago,but his energy 8in old age so he sold nothing for long.I admired Mr Smith’s long excellent service in the area. He had a9attitude to life. I asked Mr Smith why his store front was in such a state. Mr Smith,smiling,replied,“I 10 a bag by the door that a customer left two years ago. Honestly I want to just 11the store,but because the bag’s owner might come by12for it,I keep the door open just a little bit.”I was moved to tears by Mr Smith’s shockingly 13deeds. I then looked down,and sure enough there was a small bag on the counter with a paper that read,“14 Item.”I left the shop feeling better about humanity and was glad there are people like Mr Smith living the good life and keeping 15forever.1.A.clothing B.foodC.fruitD.candy2.A.every day B.occasionallyC.every nightD.recently3.A.when B.whereC.whyD.how4.A.approached B.leftC.brokeD.ignored5.A.official B.enthusiasticC.elegantD.humble6.A.supplier B.ownerC.customerD.employee7.A.many B.someC.variousD.no8.A.broke off B.calmed downC.went upD.faded away9.A.pessimistic B.positiveC.humourousD.cool10.A.buy B.sellC.keepD.exhibit11.A.manage B.exchangeC.openD.close12.A.looking B.payingC.chargingD.longing13.A.unbelievable B.sweetC.oddD.sensible14.A.Recommended B.FoundC.LostD.Deserted15.A.patience B.silenceC.faithD.touch【语篇解读】这是一篇夹叙夹议文。
阶段示范性金考卷 (十一 )本卷测试内容:选修3-3本试卷分为第Ⅰ卷 (选择题 )和第Ⅱ卷 (非选择题 )两部分,共 110 分。
测试时间 90 分钟。
第Ⅰ卷(选择题,共 60 分)一、选择题 (此题共 12 小题,每题 5 分,共 60 分。
在每题给出的四个选项中,第 1、3、 5、6、7、9、12 小题,只有一个选项正确;第2、4、8、10、11 小题,有多个选项正确,所有选对的得 5 分,选对但不全的得 3 分,有选错的得 0 分。
)1. [2014 上·海市十校高三联考 ]以下说法中正确的选项是 ()A.温度低的物体内能小B.外界对物体做功时,物体的内能必定增添C.温度低的物体分子运动的均匀动能小D.做加快运动的物体,因为速度愈来愈大,所以物体分子的均匀动能愈来愈大分析:物体的内能跟物体所含的分子数、物体的温度和体积等因素相关,所以温度低的物体内能不必定小,选项 A 错误;做功和热传达均能改变物体的内能,当外界对物体做功,而物体放热时,物体的内能可能减小,选项 B 错误;物体的温度低表示物体分子运动的均匀动能小,选项 C 正确;物体做机械运动时的动能与物体分子做热运动时的动能不同,明显,选项 D 错误。
答案: C2.以下图,两个同样的玻璃瓶分别装有质量同样的热水和冷水,以下说法正确的选项是()A.热水中每个分子的动能都比冷水中每个分子的动能大B.热水的内能大于冷水的内能C.同样的固体悬浮颗粒,在热水中的布朗运动比在冷水中明显D.热水和冷水的分子个数不同分析:温度是分子均匀动能的标记,分子均匀动能大,其实不意味着每个分子的动能都大, A 错;热水的温度高,内能大, B 正确;温度越高,布朗运动越明显, C 正确;热水和冷水的质量同样,物质的量同样,分子个数同样, D 错误。
答案: BC3. 以下图,甲分子固定于坐标原点O,乙分子从无量远处由静止开释,在分子力的作用下凑近甲。
图中 b 点是引力最大处, d 点是分子靠得近来处,则乙分子速度最大处是()A. a 点B. b 点C. c 点D. d 点分析:由分子力与分子之间距离的图象能够看出,乙分子从无量远处到 c 点过程中,分子力做正功,分子动能增大,从 c 到 d 过程中,分子力做负功,动能减小,所以经过地点 c 时速度最大。
第一章 集合与常用逻辑用语考点测试1 集合高考 概览本考点在高考中是必考知识点,常考题型为选择题,分值为5分,低难度考纲 研读1.了解集合的含义,体会元素与集合的属于关系2.能用自然语言、图形语言、集合语言(列举法或描述法)描述不同的具体问题3.理解集合之间包含与相等的含义,能识别给定集合的子集 4.在具体情境中,了解全集与空集的含义5.理解两个集合的并集与交集的含义,会求两个简单集合的并集与交集 6.理解在给定集合中一个子集的补集的含义,会求给定子集的补集 7.能使用Venn 图表达集合的关系及运算一、基础小题1.已知集合A ={x |x 2-x -6<0},B ={x |2<x <5},则A ∪B =( ) A .(1,6) B .(-2,5) C .(2,3) D .(3,5) 答案 B解析 A ={x |-2<x <3},A ∪B =(-2,5).故选B.2.满足M ⊆{a 1,a 2,a 3,a 4},且M ∩{a 1,a 2,a 3}={a 1,a 2}的集合M 的个数是( )A .1B .2C .3D .4 答案 B解析 集合M ={a 1,a 2}或{a 1,a 2,a 4},有2个.故选B. 3.已知集合P =⎩⎨⎧⎭⎬⎫x ⎪⎪⎪1x <13,则(∁R P )∩N =()A .{x |0<x <3}B .{x |0<x ≤3}C .{0,1,2,3}D .{1,2,3}答案 C 解析 由题意,得P =⎩⎨⎧⎭⎬⎫x ⎪⎪⎪1x <13=⎩⎪⎨⎪⎧⎭⎪⎬⎪⎫x ⎪⎪⎪x -33x >0={x |x >3或x <0},则(∁R P )∩N ={x |0≤x ≤3}∩N ={0,1,2,3}.故选C.4.已知集合A ={1,2},B ={(x ,y )|x ∈A ,y ∈A ,x -y ∈A },则B 的子集共有( )A .2个B .4个C .6个D .8个 答案 A解析 由已知得B ={(2,1)},所以B 的子集有2个.故选A.5.已知集合A ={x |(x -2)(x +2)≤0},B ={y |x 2+y 2=16},则A ∩B =( ) A .[-3,3] B .[-2,2] C .[-4,4] D .∅ 答案 B解析 由题意,得A ={x |-2≤x ≤2},B ={y |-4≤y ≤4},所以A ∩B ={x |-2≤x ≤2}.故选B.6.已知集合A ,B 均为全集U ={1,2,3,4}的子集,且∁U (A ∪B )={4},A ∩(∁U B )={3},则B =( )A .{1,2}B .{2,4}C .{1,2,4}D .∅答案 A解析 由∁U (A ∪B )={4},得A ∪B ={1,2,3}.由A ∩(∁U B )={3},得3∈A 且3∉B .现假设1∉B ,∵A ∪B ={1,2,3},∴1∈A .又1∉A ∩(∁U B )={3},∴1∉∁U B ,即1∈B ,矛盾.故1∈B .同理2∈B .故选A.7.已知集合A ={x |y =x 2-2},集合B ={y |y =x 2-2},则有( ) A .A =B B .A ∩B =∅ C .A ∪B =AD .A ∩B =A答案 C解析 A ={x |y =x 2-2}=R ,B ={y |y =x 2-2}=[-2,+∞),所以B ⊆A ,故A ∪B =A .故选C.8.已知集合M 是函数y =11-2x的定义域,集合N 是函数y =x 2-4的值域,则M ∩N =( )A .⎩⎨⎧⎭⎬⎫x ⎪⎪⎪x ≤12B .⎩⎨⎧⎭⎬⎫x ⎪⎪⎪-4≤x <12C .⎩⎨⎧⎭⎬⎫(x ,y )⎪⎪⎪x <12且y ≥-4D .∅ 答案 B解析 由题意,得M =⎝ ⎛⎭⎪⎫-∞,12,N =[-4,+∞),所以M ∩N =⎣⎢⎡⎭⎪⎫-4,12.故选B.9.若集合U =R ,A ={1,2,3,4,5},集合B ={x |0<x <4},则图中阴影部分表示( )A .{1,2,3,4}B .{1,2,3}C .{4,5}D .{1,4}答案 C解析 集合A ={1,2,3,4,5},B ={x |0<x <4},图中阴影部分表示A ∩(∁U B ),又∁U B ={x |x ≥4或x ≤0},所以A ∩(∁U B )={4,5}.故选C.10.已知集合A ={(x ,y )|y =2x },B ={(x ,y )|y =x +1},则A ∩B 中元素的个数为( )A .3B .2C.1 D.0答案 B解析由y=2x与y=x+1的图象可知,两函数图象有两个交点,如图所示.∴A∩B中元素的个数为2.故选B.11.(多选)已知全集U=R,函数y=ln (1-x)的定义域为M,集合N={x|x2-x<0},则下列结论正确的是()A.M∩N=N B.M∩(∁U N)≠∅C.M∪N=U D.M⊆(∁U N)答案AB解析由题意知M={x|x<1},N={x|0<x<1},所以M∩N=N.又∁U N={x|x≤0或x≥1},所以M∩(∁U N)={x|x≤0}≠∅,M∪N={x|x<1}=M,M⊆/(∁U N).故选AB.12.(多选)已知集合A={0,1,2},若A∩(∁Z B)≠∅(Z是整数集合),则集合B可以为()A.{x|x=2a,a∈A}B.{x|x=2a,a∈A}C.{x|x=a-1,a∈N}D.{x|x=a2,a∈N}答案ABD解析由题意知,集合A={0,1,2}.{x|x=2a,a∈A}={0,2,4},则A∩(∁Z B)={1}≠∅,A满足题意;{x|x=2a,a∈A}={1,2,4},则A∩(∁Z B)={0}≠∅,B 满足题意;{x|x=a-1,a∈N}={-1,0,1,2,3,…},则A∩(∁Z B)=∅,C 不满足题意;{x|x=a2,a∈N}={0,1,4,9,16,…},则A∩(∁Z B)={2}≠∅,D 满足题意.故选ABD.二、高考小题13.(2021·新高考Ⅰ卷)设集合A ={x |-2<x <4},B ={2,3,4,5},则A ∩B =( )A .{2}B .{2,3}C .{3,4}D .{2,3,4} 答案 B解析 因为A ={x |-2<x <4},B ={2,3,4,5},所以A ∩B ={2,3}.故选B.14.(2021·新高考Ⅱ卷)设集合U ={1,2,3,4,5,6},A ={1,3,6},B ={2,3,4},则A ∩(∁U B )=( )A .{3}B .{1,6}C .{5,6}D .{1,3}答案 B解析 由题意可得∁U B ={1,5,6},故A ∩(∁U B )={1,6}.故选B. 15.(2021·全国甲卷)设集合M ={x |0<x <4},N =⎩⎨⎧⎭⎬⎫x ⎪⎪⎪13≤x ≤5,则M ∩N =( )A .⎩⎨⎧⎭⎬⎫x ⎪⎪⎪0<x ≤13 B .⎩⎨⎧⎭⎬⎫x ⎪⎪⎪13≤x <4C .{x |4≤x <5}D .{x |0<x ≤5} 答案 B解析 由已知得M ∩N =⎩⎨⎧⎭⎬⎫x ⎪⎪⎪13≤x <4.故选B.16.(2021·全国乙卷)已知集合S ={s |s =2n +1,n ∈Z },T ={t |t =4n +1,n ∈Z },则S ∩T =( )A .∅B .SC .TD .Z 答案 C解析 因为s =2n +1,n ∈Z ,当n =2k ,k ∈Z 时,s =4k +1,k ∈Z ;当n =2k +1,k ∈Z 时,s =4k +3,k ∈Z ,所以TS ,S ∩T =T .故选C.17.(2021·天津高考)设集合A ={-1,0,1},B ={1,3,5},C ={0,2,4},则(A ∩B )∪C =( )A .{0}B .{0,1,3,5}C .{0,1,2,4}D .{0,2,3,4}答案 C解析 ∵A ={-1,0,1},B ={1,3,5},C ={0,2,4},∴A ∩B ={1},∴(A ∩B )∪C ={0,1,2,4}.故选C.18.(2020·新高考Ⅰ卷)设集合A ={x |1≤x ≤3},B ={x |2<x <4},则A ∪B =( )A .{x |2<x ≤3}B .{x |2≤x ≤3}C .{x |1≤x <4}D .{x |1<x <4} 答案 C解析 A ∪B =[1,3]∪(2,4)=[1,4).故选C.19.(2020·全国Ⅰ卷)设集合A ={x |x 2-4≤0},B ={x |2x +a ≤0},且A ∩B ={x |-2≤x ≤1},则a =( )A .-4B .-2C .2D .4答案 B 解析 ∵A ={x |x2-4≤0}={x |-2≤x ≤2},B ={x |2x +a ≤0}=⎩⎨⎧⎭⎬⎫x ⎪⎪⎪x ≤-a 2,A ∩B ={x |-2≤x ≤1},∴-a2=1,解得a =-2.故选B.20.(2020·全国Ⅲ卷)已知集合A ={(x ,y )|x ,y ∈N *,y ≥x },B ={(x ,y )|x +y =8},则A ∩B 中元素的个数为( )A .2B .3C .4D .6 答案 C解析 由题意,A ∩B 中的元素满足⎩⎨⎧y ≥x ,x +y =8,且x ,y ∈N *,由x +y =8≥2x ,得x ≤4,所以A ∩B 中的元素有(1,7),(2,6),(3,5),(4,4),共4个.故选C.三、模拟小题21.(2022·江苏镇江市第一中学高三上学期期初考试)已知集合A={x||x|≤2,x∈N},集合B={x|x2+x-6=0},则A∩B=()A.{2} B.{-3,2}C.{-3,1} D.{-3,0,1,2}答案 A解析集合A={x||x|≤2,x∈N}={0,1,2},集合B={x|x2+x-6=0}={-3,2},所以A∩B={2}.故选A.22.(2022·广东广州荔湾区高三上调研考试)已知全集U=R,设集合A={x|x2-x-6≤0},B={x|x-1<0},则图中阴影部分表示的集合是()A.{x|x≤3} B.{x|-3≤x<1}C.{x|-2≤x<-1} D.{x|1≤x≤3}答案 D解析由题意得,A={x|-2≤x≤3},B={x|x<1},∴∁U B={x|x≥1},∴A ∩(∁U B)={x|1≤x≤3}.故选D.23.(2021·新高考八省联考)已知M,N均为R的子集,且∁R M⊆N,则M∪(∁R N)=()A.∅B.M C.N D.R答案 B解析解法一:∵∁R M⊆N,∴M⊇∁R N,据此可得M∪(∁R N)=M.故选B.解法二:如图所示,设矩形区域ABCD 表示全集R ,矩形区域ABHE 表示集合M ,则矩形区域CDEH 表示集合∁R M ,矩形区域CDFG 表示集合N ,满足∁R M ⊆N ,结合图形可得M ∪(∁R N )=M .故选B.24.(2021·河南南阳模拟)设集合P ={3,log 2a },Q ={a ,b },若P ∩Q ={0},则P ∪Q =( )A .{3,0}B .{3,0,1}C .{3,0,2}D .{3,0,1,2}答案 B解析 ∵P ∩Q ={0},∴log 2a =0,∴a =1,从而b =0,∴P ∪Q ={3,0,1}.故选B.25.(2022·河北沧州第一中学等十五校高三上摸底考试)已知集合A =⎩⎨⎧⎭⎬⎫x ∈R ⎪⎪⎪y = x -4x -7,集合B ={3,4,5,6,7},则A ∩B =( ) A .(3,4) B .{3,4} C .[3,4] D .{3,4,7} 答案 B解析 由x -4x -7≥0得⎩⎨⎧(x -4)(x -7)≥0,x ≠7,得x ≤4或x >7,所以A ={x |x ≤4或x >7},因为B ={3,4,5,6,7},所以A ∩B ={x |x ≤4或x >7}∩{3,4,5,6,7}={3,4}.故选B.26.(2022·湖北襄阳五中高三开学考试)已知集合M ={x |1-a <x <2a },N =(1,4),且M ⊆N ,则实数a 的取值范围是( )A .(-∞,2]B .(-∞,0]C .⎝ ⎛⎦⎥⎤-∞,13D .⎣⎢⎡⎭⎪⎫13,2答案 C解析 因为M ⊆N ,而∅⊆N ,所以当M =∅时,2a ≤1-a ,则a ≤13;当M ≠∅时,M ⊆N ,则⎩⎪⎨⎪⎧1-a <2a ,1-a ≥1,2a ≤4⇒⎩⎪⎨⎪⎧a >13,a ≤0,a ≤2,无解.综上得a ≤13,即实数a 的取值范围是⎝ ⎛⎦⎥⎤-∞,13.故选C. 27.(2022·湖南长沙长郡中学高三上开学考试)已知集合A =⎩⎨⎧⎭⎬⎫x ∈N ⎪⎪⎪12<2x +1<16,B ={x |x 2-4x +m =0},若1∈A ∩B ,则A ∪B =( )A .{1,2,3}B .{1,2,3,4}C .{0,1,2}D .{0,1,2,3} 答案 D 解析由题可知,A =⎩⎨⎧⎭⎬⎫x ∈N ⎪⎪⎪12<2x +1<16,即2-1<2x +1<24,解得-2<x <3,又x ∈N ,所以A ={0,1,2}.因为1∈A ∩B ,则1∈B ,所以1-4+m =0,解得m =3,所以B ={x |x 2-4x +3=0}={1,3},所以A ∪B ={0,1,2,3}.故选D.28.(多选)(2021·江苏沭阳如东中学测试)设A ={x |x 2-8x +15=0},B ={x |ax -1=0},若A ∩B =B ,则实数a 的值可以为( )A .15B .0C .3D .13 答案 ABD解析 ∵x 2-8x +15=0的两个根为3和5,∴A ={3,5},∵A ∩B =B ,∴B ⊆A ,∴B =∅或B ={3}或B ={5}或B ={3,5},当B =∅时,满足a =0即可,当B ={3}时,满足3a -1=0,∴a =13,当B ={5}时,满足5a -1=0,∴a =15,当B ={3,5}时,显然不符合条件,∴实数a 的值可以是0,13,15.故选ABD.29.(多选)(2021·山东滨州模拟)设S 为复数集C 的非空子集.若对任意x ,y ∈S ,都有x +y ,x -y ,xy ∈S ,则称S 为封闭集.下列命题中的真命题有( )A .集合S ={a +b i|a ,b 为整数,i 为虚数单位}为封闭集B .若S 为封闭集,则一定有0∈SC .封闭集一定是无限集D .若S 为封闭集,则满足S ⊆T ⊆C 的任意集合T 也是封闭集 答案 AB解析 因为两个复数的和是复数,两个复数的差是复数,两个复数的积也是复数,所以集合S ={a +b i|a ,b 为整数,i 为虚数单位}为封闭集,A 正确;当S 为封闭集时,因为x -y ∈S ,取x =y ,得0∈S ,B 正确;集合S ={0}显然是封闭集,但S 是有限集,C 错误;取S ={0},T ={0,1},满足S ⊆T ⊆C ,但由于0-1=-1不属于T ,故T 不是封闭集,D 错误.故选AB.30.(多选)(2022·湖南衡阳模拟)对于集合M ,定义函数f M (x )=⎩⎨⎧-1,x ∈M ,1,x ∉M .对于两个集合M ,N ,定义集合M ⊗N ={x |f M (x )·f N (x )=-1}.已知集合A ={2,4,6},B ={1,2,4},则下列结论正确的是( )A .1∈A ⊗B B .2∈A ⊗BC .4∉A ⊗BD .A ⊗B =B ⊗A答案 ACD解析 由题意知,f A (x )=⎩⎨⎧-1,x ∈{2,4,6},1,x ∉{2,4,6},f B (x )=⎩⎨⎧-1,x ∈{1,2,4},1,x ∉{1,2,4}.当x =1时,f A (1)=1,f B (1)=-1,所以f A (1)f B (1)=1×(-1)=-1,故1∈A ⊗B ,A 正确;当x =2时,f A (2)=-1,f B (2)=-1,所以f A (2)f B (2)=(-1)×(-1)=1,故2∉A ⊗B ,B 错误;当x =4时,f A (4)=-1,f B (4)=-1,所以f A (4)f B (4)=(-1)×(-1)=1,故4∉A ⊗B ,C 正确;由定义及乘法的交换律可知,D 正确.一、高考大题本考点在近三年高考中未涉及此题型.二、模拟大题1.(2021·江西南昌高三模拟)已知全集U =R ,集合A ={x |x 2-4x -5≤0},B ={x |2≤x ≤4}.(1)求A ∩(∁U B );(2)若集合C ={x |a ≤x ≤4a ,a >0},满足C ∪A =A ,C ∩B =B ,求实数a 的取值范围.解 (1)由题意,得A ={x |-1≤x ≤5},∁U B ={x |x <2或x >4}, ∴A ∩(∁U B )={x |-1≤x <2或4<x ≤5}.(2)由C ∪A =A 得C ⊆A ,则⎩⎨⎧a ≥-1,4a ≤5,解得-1≤a ≤54.由C ∩B =B 得B ⊆C ,则⎩⎨⎧a ≤2,4a ≥4,解得1≤a ≤2. 从而实数a 的取值范围为⎩⎨⎧⎭⎬⎫a ⎪⎪⎪1≤a ≤54.2.(2022·云南师大附中月考)设集合A =⎩⎨⎧⎭⎬⎫x ⎪⎪⎪12≤2x ≤4,B ={x |x 2+(b -a )x -ab ≤0}.(1)若A =B 且a +b <0,求实数a ,b 的值;(2)若B 是A 的子集,且a +b =2,求实数b 的取值范围.解 (1)A =⎩⎨⎧⎭⎬⎫x ⎪⎪⎪12≤2x ≤4={x |-1≤x ≤2},∵a +b <0,∴a <-b ,∴B ={x |(x -a )(x +b )≤0}={x |a ≤x ≤-b }, ∵A =B ,∴a =-1,b =-2.(2)∵a +b =2,∴B ={-b ≤x ≤2-b }, ∵B 是A 的子集,∴-b ≥-1且2-b ≤2, 解得0≤b ≤1,即实数b 的取值范围为[0,1].考点测试2 充分条件与必要条件、全称量词与存在量词高考概览高考在本考点的常考题型为选择题,分值为5分,低难度考纲研读1.理解命题的概念2.理解充分条件、必要条件与充要条件的含义3.理解全称量词与存在量词的意义4.能正确地对含有一个量词的命题进行否定一、基础小题1.下面四个条件中,使a>b成立的必要不充分条件是()A.a-1>b B.a+1>bC.|a|>|b| D.a3>b3答案 B解析寻找使a>b成立的必要不充分条件,若a>b,则a+1>b一定成立,a3>b3也一定成立,但是当a3>b3成立时,a>b也一定成立.故选B.2.命题“所有实数的平方都是正数”的否定为()A.所有实数的平方都不是正数B.有的实数的平方是正数C.至少有一个实数的平方是正数D.至少有一个实数的平方不是正数答案 D解析根据全称量词命题的否定为存在量词命题知,把“所有”改为“至少有一个”,“是”的否定为“不是”,故命题“所有实数的平方都是正数”的否定为“至少有一个实数的平方不是正数”.故选D.3.命题“∃x∈(0,+∞),ln x=x-1”的否定是()A .∀x ∈(0,+∞),ln x ≠x -1B .∀x ∉(0,+∞),ln x =x -1C .∃x ∈(0,+∞),ln x ≠x -1D .∃x ∉(0,+∞),ln x =x -1 答案 A解析 存在量词命题的否定为全称量词命题,所以∃x ∈(0,+∞),ln x =x -1的否定是∀x ∈(0,+∞),ln x ≠x -1.故选A.4.已知0<α<π,则“α=π6”是“sin α=12”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 答案 A解析 ∵0<α<π,则α=π6⇒sin α=12,sin α=12⇒α=π6或α=5π6,∴已知0<α<π,则“α=π6”是“sin α=12”的充分不必要条件.故选A.5.“直线l 与曲线C 只有一个交点”是“直线l 与曲线C 相切”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 答案 D解析 若直线l 与曲线C 只有一个交点,直线l 与曲线C 不一定相切,比如当直线l 与双曲线的渐近线平行时,直线l 与该双曲线只有一个交点,但不相切;反之,若直线l 与曲线C 相切,直线l 与曲线C 也不一定只有一个交点.6.已知命题“∃x ∈R ,使2x 2+(a -1)x +12≤0”是假命题,则实数a 的取值范围是( )A.(-∞,-1) B.(-1,3)C.(-3,+∞) D.(-3,1)答案 B解析因为命题“∃x∈R,使2x2+(a-1)x+12≤0”是假命题,所以2x2+(a-1)x+12>0在R上恒成立为真命题,所以Δ=(a-1)2-4×2×12<0,解得-1<a<3,故实数a的取值范围是(-1,3).故选B.7.(多选)下列命题中,假命题是()A.∃x∈R,使得e x≤0B.a>1,b>1是ab>1的充分不必要条件C.∀x∈R,2x>x2D.sin x+1sin x≥2(x≠kπ,k∈Z)答案ACD解析对于A,由指数函数的性质可得e x>0,所以命题“∃x∈R,使得e x ≤0”为假命题;对于B,由a>1,b>1,可得ab>1成立,即充分性成立.反之,例如a=12,b=4时,ab>1,所以必要性不成立,所以命题“a>1,b>1是ab>1的充分不必要条件”为真命题;对于C,例如当x=2时,2x=x2,所以命题“∀x∈R,2x>x2”为假命题;对于D,当sin x<0时,sin x+1sin x≥2不成立,所以是假命题.故选ACD.8.(多选)下列叙述中正确的是()A.“a<1”是“方程x2+x+a=0有一个正根和一个负根”的必要不充分条件B.若a,b,c∈R,则“ab2>cb2”的充要条件是“a>c”C.“a>1”是“1a<1”的充分不必要条件D.若a,b,c∈R,则“ax2+bx+c≥0”的充要条件是“b2-4ac≤0”答案AC解析 令f (x )=x 2+x +a ,方程x 2+x +a =0有一个正根和一个负根,则f (0)<0,则有a <0,∴“a <1”是“方程x 2+x +a =0有一个正根和一个负根”的必要不充分条件,A 正确;当b =0时,若a >c 成立,则ab 2=0=cb 2,充分性不成立,B 错误;a >1⇒1a <1,1a <1⇒/a >1,∴“a >1”是“1a <1”的充分不必要条件,C 正确;由ax 2+bx +c ≥0可得a >0,b 2-4ac ≤0或a =b =0,c ≥0,∴“b 2-4ac ≤0”是“ax 2+bx +c ≥0”的必要不充分条件,D 错误.故选AC.9.已知全集U =R ,A ⊆U ,B ⊆U ,如果命题p :x ∈(A ∩B ),那么¬p 是________. 答案 x ∉A 或x ∉B解析 x ∈(A ∩B )即x ∈A 且x ∈B ,所以其否定为x ∉A 或x ∉B .10.设p :ln (2x -1)≤0,q :(x -a )[x -(a +1)]≤0,若q 是p 的必要不充分条件,则实数a 的取值范围是________.答案 ⎣⎢⎡⎦⎥⎤0,12解析 由p 得,12<x ≤1,由q 得,a ≤x ≤a +1,因为q 是p 的必要不充分条件,所以a ≤12且a +1≥1,所以0≤a ≤12.11.已知“p :(x -m )2>3(x -m )”是“q :x 2+3x -4<0”的必要不充分条件,则实数m 的取值范围为________.答案 (-∞,-7]∪[1,+∞)解析 由p 中的不等式(x -m )2>3(x -m ),得(x -m )(x -m -3)>0,解得x >m +3或x <m .由q 中的不等式x 2+3x -4<0,得(x -1)(x +4)<0,解得-4<x <1.因为p 是q 的必要不充分条件,所以q ⇒p 且p ⇒/ q ,即m +3≤-4或m ≥1,解得m ≤-7或m ≥1.所以实数m 的取值范围为(-∞,-7]∪[1,+∞).12.设p ,r 都是q 的充分条件,s 是q 的充要条件,t 是s 的必要条件,t 是r 的充分条件,那么p 是t 的________条件,r 是t 的________条件.(用“充分”“必要”或“充要”填空)答案 充分 充要解析 由题知p ⇒q ⇔s ⇒t ,又t ⇒r ,r ⇒q ,q ⇒s ⇒t ,故p 是t 的充分条件,r 是t 的充要条件.二、高考小题13.(2021·天津高考)已知a ∈R ,则“a >6”是“a 2>36”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 答案 A解析 若a >6,则a 2>36,故充分性成立;若a 2>36,则a >6或a <-6,推不出a >6,故必要性不成立.所以“a >6”是“a 2>36”的充分不必要条件.故选A.14.(2021·北京高考)已知f (x )是定义在[0,1]上的函数,那么“函数f (x )在[0,1]上单调递增”是“函数f (x )在[0,1]上的最大值为f (1)”的( )A .充分而不必要条件B .必要而不充分条件C .充分必要条件D .既不充分也不必要条件 答案 A解析 若函数f (x )在[0,1]上单调递增,则f (x )在[0,1]上的最大值为f (1),若f (x )在[0,1]上的最大值为f (1),比如f (x )=⎝ ⎛⎭⎪⎫x -132,但f (x )=⎝ ⎛⎭⎪⎫x -132在⎣⎢⎡⎦⎥⎤0,13上单调递减,在⎣⎢⎡⎦⎥⎤13,1上单调递增,故f (x )在[0,1]上的最大值为f (1)推不出f (x )在[0,1]上单调递增,故“函数f (x )在[0,1]上单调递增”是“函数f (x )在[0,1]上的最大值为f (1)”的充分而不必要条件.故选A.15.(2021·浙江高考)已知非零向量a ,b ,c ,则“a ·c =b ·c ”是“a =b ”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答案 B解析由a·c=b·c可得(a-b)·c=0,所以(a-b)⊥c或a=b,所以“a·c=b·c”是“a=b”的必要不充分条件.故选B.16.(2021·全国甲卷)等比数列{a n}的公比为q,前n项和为S n.设甲:q>0,乙:{S n}是递增数列,则()A.甲是乙的充分条件但不是必要条件B.甲是乙的必要条件但不是充分条件C.甲是乙的充要条件D.甲既不是乙的充分条件也不是乙的必要条件答案 B解析当a1=-1,q=2时,{S n}是递减数列,所以甲不是乙的充分条件;当{S n}是递增数列时,有S n+1-S n=a n+1=a1q n>0,若a1>0,则q n>0(n∈N*),即q>0;若a1<0,则q n<0(n∈N*),这样的q不存在,所以甲是乙的必要条件.故选B.17.(2020·浙江高考)已知空间中不过同一点的三条直线m,n,l,则“m,n,l在同一平面”是“m,n,l两两相交”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答案 B解析依题意m,n,l是空间中不过同一点的三条直线,当m,n,l在同一平面时,可能m∥n∥l,故不能得出m,n,l两两相交.当m,n,l两两相交时,设m∩n=A,m∩l=B,n∩l=C,根据经过两条相交直线,有且只有一个平面可知m,n确定一个平面α,而B∈m⊂α,C∈n⊂α,根据如果一条直线上的两点在一个平面内,那么这条直线在此平面内可知,直线BC即l⊂α,所以m,n,l在同一平面.综上所述,“m,n,l在同一平面”是“m,n,l两两相交”的必要不充分条件.故选B.18.(2020·北京高考)已知α,β∈R ,则“存在k ∈Z 使得α=k π+(-1)k β”是“sin α=sin β”的( )A .充分而不必要条件B .必要而不充分条件C .充分必要条件D .既不充分也不必要条件 答案 C解析 ①当存在k ∈Z 使得α=k π+(-1)k β时,若k 为偶数,则sin α=sin (k π+β)=sin β;若k 为奇数,则sin α=sin (k π-β)=sin [(k -1)π+π-β]=sin (π-β)=sin β.②当sin α=sin β时,α=β+2m π或α+β=π+2m π,m ∈Z ,即α=k π+(-1)k β(k =2m )或α=k π+(-1)k β(k =2m +1),即存在k ∈Z 使得α=k π+(-1)k β.所以,“存在k ∈Z 使得α=k π+(-1)k β”是“sin α=sin β”的充分必要条件.故选C.19.(2019·北京高考)设点A ,B ,C 不共线,则“AB →与AC →的夹角为锐角”是“|AB →+AC →|>|BC →|”的( )A .充分而不必要条件B .必要而不充分条件C .充分必要条件D .既不充分也不必要条件 答案 C解析 因为点A ,B ,C 不共线,由向量加法的三角形法则,可知BC →=AC →-AB →,所以|AB →+AC →|>|BC →|等价于|AB →+AC →|>|AC →-AB →|,不等式两边平方得AB →2+AC →2+2|AB →||AC →|cos θ>AC →2+AB →2-2|AC →||AB →|cos θ(θ为AB →与AC →的夹角),整理得4|AB →||AC →|cos θ>0,故cos θ>0,即θ为锐角.又以上推理过程可逆,所以“AB →与AC →的夹角为锐角”是“|AB →+AC →|>|BC →|”的充分必要条件.故选C.三、模拟小题20.(2022·河北衡水深州长江中学高三上开学考试)命题p :∃x ∈[0,+∞),e x <x 2-x 的否定为( )A .∃x ∈[0,+∞),e x ≥x 2-xB .∀x ∈[0,+∞),e x ≥x 2-xC .∃x ∈(-∞,0),e x ≥x 2-xD .∀x ∈(-∞,0),e x ≥x 2-x 答案 B解析 命题p :∃x ∈[0+∞),e x <x 2-x 的否定为∀x ∈[0,+∞),e x ≥x 2-x .故选B.21.(2022·福建晋江磁灶中学高三上阶段测试(一))“a <2”是“∀x >0,a ≤x +1x ”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 答案 A解析 若∀x >0,a ≤x +1x ,则a ≤⎝ ⎛⎭⎪⎫x +1x min ,因为x +1x ≥2,当且仅当x =1x 时等号成立,所以a ≤2,因为{a |a <2}{a |a ≤2},所以“a <2”是“∀x >0,a ≤x+1x ”的充分不必要条件.故选A.22.(2022·北京交通大学附属中学高三开学考试)已知数列{a n }的通项公式为a n =n +an ,则“a 2>a 1”是“数列{a n }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答案 C解析数列{a n}单调递增⇔a n+1>a n,可得n+1+an+1>n+an,化为a<n2+n,∴a<2.由a2>a1可得2+a2>1+a,∴a<2.∴“a2>a1”是“数列{a n}单调递增”的充要条件.故选C.23.(2021·山东潍坊一中模拟)已知△ABC中角A,B,C所对的边分别是a,b,c,则“a2+b2=2c2”是“△ABC为等边三角形”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答案 B解析当a=1,b=3,c=2时,满足△ABC三边关系与a2+b2=2c2,但△ABC不是等边三角形;当△ABC为等边三角形时,a2+b2=2c2成立.故“a2+b2=2c2”是“△ABC为等边三角形”的必要不充分条件.故选B.24.(多选)(2021·湖北恩施高三模拟)下列选项中,能作为x>y的充分条件的是()A.xt2>yt2B.点(x,y)是曲线x3-y3-x2=1上的点C.1x<1y<0D.点(x,y)是双曲线x2-y2=1上的点答案ABC解析由题意,对于A,由xt2>yt2,可知t2>0,可得x>y成立,所以A正确;对于B,点(x,y)是曲线x3-y3-x2=1上的点,则x3-y3=1+x2>0,可得x 3>y 3,即x >y 成立,所以B 正确;对于C ,由1x <1y <0,可得x <0,y <0,又由1y -1x =x -y xy >0,可得x -y >0,即x >y 成立,所以C 正确;对于D ,点(x ,y )是双曲线x 2-y 2=1上的点,可得x 2>y 2,不一定得到x >y 成立,所以D 不正确.故选ABC.25.(多选)(2021·广东中山模拟)有限集合S 中元素的个数记作card(S ),设A ,B 都为有限集合,则下列命题中的真命题是( )A .A ∩B =∅的充要条件是card(A ∪B )=card(A )+card(B )B .A ⊆B 的必要条件是card(A )≤card(B )C .A ⊆/B 的充要条件是card(A )≤card(B )D .A =B 的充要条件是card(A )=card(B )答案 AB解析 A ∩B =∅,集合A 与集合B 没有公共元素,A 正确;A ⊆B ,集合A 中的元素都是集合B 中的元素,B 正确;A ⊆/B ,集合A 中至少有一个元素不是集合B 中的元素,因此A 中元素的个数有可能多于B 中元素的个数,C 错误;A =B ,集合A 中的元素与集合B 中的元素完全相同,两个集合的元素个数相同,并不意味着它们的元素相同,D 错误.故选AB.26.(2022·湖南湘潭高三模拟)用实数m (m =0或1)表示命题p 的真假,其中m =0表示命题p 为假,m =1表示命题p 为真,设命题p :∀x ∈Z ,⎪⎪⎪⎪⎪⎪x -12+⎪⎪⎪⎪⎪⎪x -23≥a (a ∈R ).(1)当a =2时,m =________;(2)当m =1时,实数a 的取值范围为________.答案 (1)0 (2)⎝ ⎛⎦⎥⎤-∞,56 解析 (1)当a =2时,不等式⎪⎪⎪⎪⎪⎪x -12+⎪⎪⎪⎪⎪⎪x -23≥2对x =1不成立,所以命题p 为假命题,故m =0.(2)因为m =1,所以命题p 为真命题,令f (x )=⎪⎪⎪⎪⎪⎪x -12+⎪⎪⎪⎪⎪⎪x -23,则f (x )=⎩⎪⎨⎪⎧76-2x ,x ≤12,16,12<x <23,2x -76,x ≥23,所以当x ≤12时,f (x )为减函数,当x ≥23时,f (x )为增函数,要使∀x ∈Z ,⎪⎪⎪⎪⎪⎪x -12+⎪⎪⎪⎪⎪⎪x -23≥a 成立,只需x =0和x =1时,⎪⎪⎪⎪⎪⎪x -12+⎪⎪⎪⎪⎪⎪x -23≥a 都成立,所以⎩⎪⎨⎪⎧a ≤76,a ≤56,得a ≤56.一、高考大题本考点在近三年高考中未涉及此题型.二、模拟大题1.(2021·山东青岛高三模拟)已知全集为R ,集合A =⎩⎨⎧⎭⎬⎫x ∈R ⎪⎪⎪x -6x +3>0,B ={x ∈R |2x 2-(a +10)x +5a ≤0}.(1)若B ⊆∁R A ,求实数a 的取值范围;(2)从下面所给的三个条件中选择一个,说明它是B ⊆∁R A 的什么条件(充分必要性).①a ∈[-7,12);②a ∈(-7,12];③a ∈(6,12].注:如果选择多个条件分别解答,按第一个解答计分.解 (1)集合A =⎩⎨⎧⎭⎬⎫x ∈R ⎪⎪⎪x -6x +3>0=(-∞,-3)∪(6,+∞), 所以∁R A =[-3,6],集合B ={x ∈R |2x 2-(a +10)x +5a ≤0}={x ∈R |(2x -a )(x -5)≤0}, 若B ⊆∁R A ,且5∈∁R A =[-3,6],只需-3≤a 2≤6,所以-6≤a ≤12.故实数a 的取值范围为[-6,12].(2)由(1)可知B ⊆∁R A 的充要条件是a ∈[-6,12].选择①,则结论是既不充分也不必要条件;选择②,则结论是必要不充分条件;选择③,则结论是充分不必要条件.2.(2021·江苏无锡惠山校级期中)已知命题p :方程x 2k +5+y 23-k =1表示焦点在x 轴上的椭圆;命题q :∀x ∈R ,x 2+kx +2k +5≥0恒成立;命题r :1-m <k <1+m (m >0).(1)若命题p 与命题r 互为充要条件,求实数m 的值;(2)若命题q 是命题r 的必要不充分条件,求正数m 的取值范围.解 若方程x 2k +5+y 23-k=1表示焦点在x 轴上的椭圆, 则k +5>3-k >0,解得-1<k <3,故p 为真命题时,-1<k <3;若∀x ∈R ,x 2+kx +2k +5≥0恒成立,则Δ=k 2-4(2k +5)≤0,解得-2≤k ≤10,故q 为真命题时,-2≤k ≤10.(1)若命题p 与命题r 互为充要条件,则(-1,3)=(1-m ,1+m ),解得m =2.(2)若命题q 是命题r 的必要不充分条件,则(1-m ,1+m )[-2,10],则⎩⎨⎧1-m ≥-2,1+m ≤10,等号不同时成立,解得m ≤3, 故正数m 的取值范围是(0,3].。
高中英语真题:2016届高考英语一轮复习Unit5Thepowerofnature导学案新人教版选修6AN EXCITING JOBI have the greatest job in the world. I travel to unusual places and work alongside people from all over the world. Sometime s working outdoors, sometimes in an office, sometimes using scientific equipment and sometimes meeting local people and tourists, I am never bored. Although my job is occasionally da ngerous, I don’t mind because danger excites me and makes me feel alive. However, the most important thing about my job is that I help protect ordinary people ____1______ one of the most powerful forces on earth — the volcano.I was appointed as a volcanologist working for the Hawaiian V olcano Observatory (HVO) twenty years ago. My job is collecti ng information for a database about Mount Kilauea, ____2___ ___(that/which) is one of the most active volcanoes in . Havin g collected and evaluated the information, I help other scientis ts to predict where lava from the volcano will flow next and ho w fast. Our work has saved many lives because people in the path of the lava can ____3______(warn) to leave their houses. Unfortunately, we cannot move their homes out of the way, and many houses have been covered with lava or burned ____4 ______the ground.When boiling rock erupts from a volcano and crashes back to earth, it causes less damage than you might imagine. This is because no one lives near the top of , where the rocks fall. T he lava that flows slowly like a wave down the mountain caus es far more damage because it buries everything in its path u nder the molten rock. However, the eruption itself is really exc iting to watch and I shall never forget my first sight of one. It w as in the second week after I arrived in . _______5________(w ork) hard all day, I went to bed early. I was fast asleep _____6 _____ suddenly my bed began shaking and I heard a strange sound, like a railway train _____7_____(pass) my window. Hav ing experienced quite a few earthquakes in already, I didn’t ta ke much notice. I was about to go back to sleep when sudden ly my bedroom became as bright as day. I ran out of the hous e into the back garden where I could see in the distance. Th ere had been an eruption from the side of the mountain and r ed hot lava was fountaining hundreds of metres into the air. It was an absolutely fantastic sight.The day after this eruption I was lucky enough to have a muc h closer look at it. Two other scientists and I were driven up th e mountain and dropped as close as possible to the crater that had been formed during the eruption. Having earlier collecte d special clothes from the observatory, we put them on before we went any closer. All three of us looked like spacemen. We had white protective suits that covered our whole body, helm ets, big boots and special gloves. It was not easy to walk in th ese suits, but we slowly made our way to the edge of the crat er and looked down into the red, boiling centre. The other two climbed down into the crater to collect some lava for later stud y, but this ____8______(be) my first experience, I stayed at the top and watched them.Today, I am just as enthusiastic about my job as the day I first started. Having studied volcanoes now for many years, I am still amazed at their beauty as well as their potential to cause great damage.Using LanguageTHE OFChangbaishan is in Jilin Province, Northeast China. Much of t his beautiful, mountainous area is thick forest. Changbaishan is China’s largest nature reserve and it is kept in its natural st ate for the people of and visitors from all over the world to enj oy. The land varies in height from 700 metres above sea levelto over 2 000 metres and is home to a great diversity of rare plants and animals. Among the rare animals are cranes, black bears, leopards and tigers. Many people come to Changbais han to study its unique plants and animals. Others come to w alk in the mountains, to see the spectacular waterfalls or to ba the in the hot water pools. However, the attraction that arouse s the greatest appreciation in the reserve is Tianchi or the of . Tianchi is a deep lake that has formed in the crater of a dead volcano on top of the mountain. The lake is 2 194 metres abo ve sea level, and more than 200 metres deep. In winter the su rface freezes over. It takes about an hour to climb from the en d of the road to the top of the mountain. When you arrive you are rewarded not only with the sight of its clear waters, but als o by the view of the other sixteen mountain peaks that surrou nd Tianchi.There are many stories____9______(tell) about Tianchi. The m ost well-known concerns three young women from heaven. They were bathing in Tianchi when a bird flew above them and dropped a small fruit onto the dress of the youngest girl. When she pic ked up the fruit to smell it, it flew into her mouth. Having swall owed the fruit, the girl became pregnant and later gave birth t o a handsome boy. It is said that this boy, who had a great gift____10______languages and persuasion, is the father of the Manchu people.If you are lucky enough to visit the of with your loved one, do n’t forget to drop a coin into the clear blue water to guarantee your love will be as deep and lasting as the lake itself.自学加油站1. suit v. & n. a black suitChoose a computer to suit your particular needs.Blue suits you ; you should wear it more often.If we met at 10 a.m., would that suit you?She had the ability to suit her performance to the audience.2.【辨中学】suit; match; fitIf you want to go by bus, that will ______ me fine.The trousers don’t ____ him; they are too small.His white shirt ______ his black trousers well.3.. absolute adj. _____________(adv.)She said that she was right with absolute certainty.Beauty cannot be measured by any absolute standard.【思维拓展】immediate_____________ (adv.) desperate__________(adv.) extreme _________ (adv.) true_______ (adv.) gentle________(a dv.) terrible _______ (adv.) simple ________(adv.)4. 【观察思考】We lost our way in the dark.Will you be able to make your own way to the airport?The fog was so thick that I had to feel my way to my office.It was Market Day, so the farmers with their produce forced/ p ushed their way through the street.The soldiers finally fought their way out of a heavy encircleme nt. (重围)5. Although my job is occasionally dangerous, I don’t mind be cause danger excites me and makes me feel alive.虽然我的工作偶尔也有危险,但是我并不在乎,因为危险能激励我,使我感到有活力。
高考模拟·随堂集训
【高考题组——明考向】
光电效应规律
考向一
〈高考常见设题点5年4考,以选择题、计算题为主〉
1. [2014·广东高考](多选)在光电效应实验中,用频率为ν的光照射光电管阴极,发生了光电效应,下列说法正确的是()
A. 增大入射光的强度,光电流增大
B. 减小入射光的强度,光电效应现象消失
C. 改用频率小于ν的光照射,一定不发生光电效应
D. 改用频率大于ν的光照射,光电子的最大初动能变大
解析:增大入射光强度,单位时间内逸出的光电子数目增多,光电流增大,A项正确;光电效应的发生与入射光的强度无关,B项错误;入射光频率小于ν时,若仍大于金属的截止频率,仍能发生光电效应,C项错误;增大入射光的频率时,由爱因斯坦光电效应方程E k=hν-W0可知,光电子的最大初动能变大,D项正确。
答案:AD
2. [2014·江苏高考]已知钙和钾的截止频率分别为7.73×1014Hz 和5.44×1014 Hz,在某种单色光的照射下两种金属均发生光电效应,比较它们表面逸出的具有最大初动能的光电子,钙逸出的光电子具有较大的()
A. 波长
B. 频率
C. 能量
D. 动量
解析:由爱因斯坦光电效应方程E k=hν-W0,金属钙的逸出功
大,则逸出的光电子的最大初动能小,即能量小,频率低,波长长,动量小,选项A正确。
答案:A
3. [2013·北京高考]以往我们认识的光电效应是单光子光电效应,即一个电子在极短时间内只能吸收到一个光子而从金属表面逸出。
强激光的出现丰富了人们对于光电效应的认识,用强激光照射金属,由于其光子密度极大,一个电子在极短时间内吸收多个光子成为可能,从而形成多光子光电效应,这已被实验证实。
光电效应实验装置示意如图。
用频率为ν的普通光源照射阴极K,没有发生光电效应。
换用同样频率ν的强激光照射阴极K,则发生了光电效应;此时,若加上反向电压U,即将阴极K接电源正极,阳极A接电源负极,在KA之间就形成了使光电子减速的电场。
逐渐增大U,光电流会逐渐减小;当光电流恰好减小到零时,所加反向电压U可能是下列的(其中W为逸出功,h为普朗克常量,e为电子电量)()
A. U=hν
e-
W
e B. U=
2hν
e-
W
e
C. U=2hν-W
D. U=5hν
2e-
W
e
解析:同频率的光照射K极,普通光不能使其发生光电效应,而强激光能使其发生光电效应,说明一个电子吸收了多个光子。
设吸收的光子个数为n,光电子逸出的最大初动能为E k,由光电效应方程知:E k=nhν-W(n≥2)①;光电子逸出后克服减速电场做功,由动
能定理知E k=eU②,联立上述两式得U=nhν
e -W
e
,当n=2时,即
为B选项,其他选项均不可能。
答案:B
考向二
波粒二象性与新科技结合的考查
〈高考灵活设题点5年2考,以选择题为主〉
4. [2012·北京高考]“约瑟夫森结”由超导体和绝缘体制成。
若在结两端加恒定电压U,则它会辐射频率为ν的电磁波,且ν与U成正比,即ν=kU。
已知比例系数k仅与元电荷e的2倍和普朗克常量h有关。
你可能不了解此现象的机理,但仍可运用物理学中常用的方法,在下列选项中,推理判断比例系数k的值可能为()
A. h
2e B.
2e
h
C. 2he
D.
1 2he
解析:根据物理单位知识,表达式及变形式两侧单位是一致的,由本题中涉及的物理量:ν、U、e、h及与其有联系的能量表达式E
=hν①,E=Ue②,由①②得h的单位与Ue
ν
的单位相同,即h单位可
用V·C·s 表示,题中ν=kU ,即k =νU ③,单位可用1V·s
表示,选项B 中2e h 单位等效于C V·C·s =1V·s
,故选项B 正确,A 、C 、D 错误。
答案:B
【模拟题组——提考能】
1. [2014·广东深圳一模](多选)关于光电效应,下列说法正确的是
( )
A. 爱因斯坦用光子说成功解释了光电效应
B. 入射光的频率低于极限频率就不能发生光电效应
C. 光电子的最大初动能与入射光的强度成正比
D. 光电子的最大初动能与入射光频率成正比
解析:爱因斯坦提出了光子说并成功地解释了光电效应现象,A 正确;当入射光的频率小于极限频率,不会发生光电效应,B 正确;根据光电效应方程知E k =hν-W 0,光电子的最大初动能与入射光的频率成一次函数关系,不是正比关系,D 错误;根据光电效应方程知光电子的最大初动能与入射光的强度无关,C 错误。
答案:AB
2. [2015·广州模拟](多选)如图是某金属在光的照射下,光电子最
大初动能E k与入射光频率ν的关系图象,由图象可知()
A. 该金属的逸出功等于E
B. 该金属的逸出功等于hν0
C. 入射光的频率为2ν0时,产生的光电子的最大初动能为2E
D. 入射光的频率为ν0
2时,产生的光电子的最大初动能为
E
2
解析:由光电效应方程hν=W+E k得E k=hν-W。
若ν=0,则W=-E k=-E,A对。
由逸出功的意义知W=hν0,B对。
由光电效应方程得,当ν加倍或减半时,E k不会加倍或减半,C、D均错。
答案:AB
3. [2014·河北石家庄一模]如图所示,N为金属板,M为金属网,它们分别与电池的两极相连,各电池的电动势和极性如图所示。
已知金属板的逸出功为
4.8 eV。
现分别用不同能量的光子照射金属板(各光子的能量已在图上标出),那么各图中没有光电子到达金属网的是________(填正确答案标号)。
能够到达金属网的光电子的最大动能是________eV。
解析:因为金属板的逸出功为4.8 eV,所以能发生光电效应的是B、C、D,B所加的电压为正向电压,则电子一定能到达金属网,到达金属网时最大动能为1.5 eV;C光电子的最大初动能为1.0 eV,根
据动能定理知电子不能到达金属网;D光电子的最大初动能为2.0 eV,根据动能定理光电子能够到达金属网。
故没有光电子达到金属网的是
A、C。
D项中逸出的光电子最大初动能为E k=E光-W逸=6.8 eV-
4.8 eV=2.0 eV,到达金属网时最大动能为0.5 eV。
答案:AC 1.5。