杭州学军中学2019学年第一学期期中考试

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浙江省杭州市学军中学(西溪校区)2019-2020学年高一上学期期中考试数学试题(解析版)

浙江省杭州市学军中学(西溪校区)2019-2020学年高一上学期期中考试数学试题(解析版)

浙江省杭州市学军中学(西溪校区)2019-2020学年高一上学期期中考试数学试题一、选择题(本大题共10小题)1.已知集合M={x|x>0},N={x|-1<x≤2},则(∁R M)∩N等于()A. B. C. D.2.下列选项中两个函数,表示同一个函数的是()A. ,xB. ,C. ,D. ,3.下列函数在其定义域上既是奇函数又是增函数的是()A. B. C. D.4.在同一直角坐标系中,函数y=,y=1og a(x+)(a>0且a≠1)的图象可能是()A. B.C. D.5.若函数f(x2+1)的定义域为[-1,1],则f(lg x)的定义域为()A. B. C. D.6.已知函数f(x)为奇函数,g(x)为偶函数,且2x+1=f(x)+g(x),则g(1)=()A. B. 2 C. D. 47.已知定义在R上的函数(m为实数)为偶函数,记a=f(log0.53),b=f(log2.53),c=f(2m),则a,b,c的大小关系为()A. B. C. D.8.已知f(x)=(x2-ax+3a)在区间(2,+∞)上是减函数,则实数a的取值范围是()A. B. C. D.9.已知a>0,设函数f(x)=(x∈[-a,a])的值域为[M,N],则M+N的值为()A. 0B. 2019C. 4037D. 403910.已知m∈R,函数f(x)=||+m在[2,5]上的最大值是5,则m的取值范围是()A. B. C. D.二、填空题(本大题共5小题,共25.0分)11.若幂函数y=f(x)的图象经过点(8,2),则f()的值是______.12.若f(1+)=,则f(3)=______.13.已知函数f(x)=x3+ln(+x).若f(a-1)+f(2a2)≤0,则实数a的取值范围是______.14.设函数f(x)=若f[f(a)]≤3,则实数a的取值范围是______.15.已知λ∈R,函数若函数f(x)恰有2个不同的零点,则λ的取值范围为______.三、解答题(本大题共6小题,共55.0分)16.若正数a,b满足log2a=log5b=lg(a+b),则的值为______ .17.化简求值:(1)-(-)0++(2)lg25+lg2+()-log29×log32.18.已知集合A={x|x2-2x-3≤0,x∈R},B={x|x2-2mx+m2-4≤0,x∈R}.(1)若A∩B={x|1≤x≤3},求实数m的值;(2)若A⊆∁R B,求实数m的取值范围.19.已知函数f(x)=log2(4x+b•2x+2),g(x)=x.(Ⅰ)当b=-3时,求函数f(x)的定义域;(Ⅱ)若对于任意x≥1,都有f(x)>g(x)成立,求实数b的取值范围.20.已知函数f(x)=log a(1-)(a>0且a≠1).(Ⅰ)判断函数f(x)的奇偶性并说明理由;(Ⅱ)当0<a<1时,判断函数f(x)在(1,+∞)上的单调性,并利用单调性的定义证明;(Ⅲ)是否存在实数a,使得当f(x)的定义域为[m,n]时,值域为[1+log a n,1+log a m]?若存在,求出实数a的取值范围;若不存在,请说明理由.21.已知函数f(x)=x2-3|x-a|.(Ⅰ)若函数y=f(x)为偶函数,求实数a的值;(Ⅱ)若a=,求函数y=f(x)的单调递减区间.(Ⅲ)当0<a≤1时,若对任意的x∈[a,+∞),不等式f(x-1)≤2f(x)恒成立,求实数a的取值范围.答案和解析1.【答案】C【解析】解:∵M={x|x>0},N={x|-1<x≤2},∴∁R M={x|x≤0},(∁R M)∩N=(-1,0].故选:C.进行补集、交集的运算即可.考查描述法、区间的定义,以及补集、交集的运算.2.【答案】B【解析】解:相同的函数必须具有相同的定义域、值域、对应关系,而函数f(x)=ln x4的定义域为非零实数集,g(x)=4ln x的定义域为正实数集合,故它们不是同一个函数;函数f(x)=x2和函数g(x)==x2,具有相同的定义域、值域、对应关系,故它们是同一个函数;函数f(x)=x-1的值域为R,而g(x)==|x-1|的值域为[0,+∞),故它们不是同一个函数;函数f(x)=x的值域为R,函数g(x)=|x|的值域为[0,+∞),故它们不是同一个函数,故选:B.由题意利用函数的三要素,判断两个函数是否为同一个函数,从而得出结论.本题主要考查函数的三要素,属于基础题.3.【答案】B【解析】【分析】本题考查函数的奇偶性与单调性的判定,关键是掌握常见函数的奇偶性与单调性,属于基础题.根据题意,依次分析选项中函数的奇偶性与单调性,综合即可得答案.【解答】解:根据题意,依次分析选项:对于A,f(x)=2x,为指数函数,不是奇函数,不符合题意;对于B,f(x)=x|x|=,既是奇函数又是增函数,符合题意;对于C,f(x)=-,在其定义域上不是增函数,不符合题意;对于D,f(x)=lg|x|,是偶函数,不符合题意;故选B.4.【答案】D【解析】【分析】本题考查了指数函数,对数函数的图象和性质,属于基础题.对a进行讨论,结合指数,对数的性质即可判断;【解答】解:由函数y=,y=1og a(x+),当a>1时,可得y=是递减函数,图象恒过(0,1)点,函数y=1og a(x+),是递增函数,图象恒过(,0);当1>a>0时,可得y=是递增函数,图象恒过(0,1)点,函数y=1og a(x+),是递减函数,图象恒过(,0);∴满足要求的图象为:D故选D.5.【答案】C【解析】解:若函数f(x2+1)的定义域为[-1,1],则1≤x2+1≤2,∴1≤lg x≤2,∴10≤x≤100,故选:C.由函数f(x2+1)的定义域为[-1,1],求出其值域,即f(lg x)的值域,从而求出其定义域.本题考查了函数的定义域,值域问题,是一道基础题.6.【答案】C【解析】解:∵函数f(x)为奇函数,g(x)为偶函数,且2x+1=f(x)+g(x),∴f(1)+g(1)=21+1=4,①f(-1)+g(-1)=2-1+1=20=1,即-f(1)+g(1)=1 ②由①+②得2g(1)=5,则g(1)=,故选:C.根据函数奇偶性的性质,建立方程组进行求解即可.本题主要考查函数值的计算,利用函数奇偶性的性质建立方程组是解决本题的关键.7.【答案】A【解析】解:根据题意,定义在R上的函数(m为实数)为偶函数,则f(-x)=f(x),即()|x-m|=()|-x-m|,分析可得m=0,则f(x)=()|x|-1=,,<,则f(x)在[0,+∞)上为减函数,又由a=f(log0.53)=f(log23),b=f(log2.53),c=f(2m)=f(0),且0<log2.53<log23,则有a<b<c;故选:A.根据题意,由偶函数的定义分析可得()|x-m|=()|-x-m|,进而可得m=0,即可得函数的解析式,分析可得f(x)在[0,+∞)上为减函数,结合对数的运算性质分析可得答案.本题考查函数的奇偶性与单调性的综合应用,注意求出m的值,确定函数的解析式,属于基础题.8.【答案】D【解析】解:令t=x2-ax+3a,则由题意可得函数t在区间(2,+∞)上是增函数,且t>0,∴ ,求得-4≤a≤4,故选:D.令t=x2-ax+3a,则由题意可得函数t在区间(2,+∞)上是增函数,且t>0,故有,由此求得a的范围.本题主要考查复合函数的单调性,二次函数的性质,体现了转化的数学思想,属于基础题.9.【答案】C【解析】解:依题意,f(x)==+2019x=2019-+2019x,f′(x)=2019+,当x∈[-a,a]时f′(x)>0,所以f(x)为[-a,a]上的增函数,所以M+N=2019--2019a+2019-+2019a=4038-=4037.故选:C.将函数f(x)分离常数后根据函数的单调性求解函数值域,即可得到M,N的值,从而得到M+N.本题考查了函数的单调性,函数的最值,考查了幂运算,主要考查分析和解决问题的能力,属于中档题.10.【答案】A【解析】解:由x∈[2,5],=1+∈[2,5],若m≤2则f(x)=的最大值为5,符合题意;当2<m≤5时,f(x)的最大值为f(2)与f(5)中较大的,由f(2)=f(5),即|5-m|+m=|2-m|+m,解得m=,显然2<m≤时,f(x)的最大值为5,m>时,f(x)的最大值不为定值.综上可得m≤时,f(x)在[2,5]上的最大值是5,故选:A.求得x∈[2,5],=1+∈[2,5],讨论m的范围,结合f(2),f(5)可得所求范围.本题考查函数的最值求法,注意运用分类讨论思想方法,考查运算能力,属于中档题.11.【答案】【解析】解:设幂函数为f(x)=xα,∵f(x)的图象经过点(8,2),∴f(8)=8α=2,即23α=2,则3α=,则α=,则f(x)=x=,则f()==,故答案为:根据幂函数的定义,利用待定系数法求出函数的解析式,然后代入求值即可.本题主要考查函数值的计算,结合幂函数的定义利用待定系数法求出是的解析式是解决本题的关键.比较基础.12.【答案】2【解析】解:∵f(1+)=,∴f(3)=f(1+)==2.故答案为:2.由f(1+)=,f(3)=f(1+),能求出结果.本题考查函数值的求法,考查函数的性质等基础知识,考查运算求解能力,是基础题.13.【答案】,【解析】解:f(x)的定义域为R,且=,∴f(x)是奇函数,且f(x)在[0,+∞)上单调递增,∴f(x)在R上单调递增,由f(a-1)+f(2a2)≤0得,f(a-1)≤f(-2a2),∴a-1≤-2a2,解得,∴实数a的取值范围是,.故答案为:,.容易判断出f(x)是R上的奇函数,且单调递增,从而根据f(a-1)+f(2a2)≤0可得出a-1≤-2a2,解出a的范围即可.本题考查了奇函数的定义及判断,增函数的定义,一元二次不等式的解法,奇函数在对称区间上的单调性,考查了计算能力,属于基础题.14.【答案】(-∞,7]【解析】解:∵函数f(x)=,先讨论f(a)的取值情况:①若f(a)≤0,则f2(a)+2f(a)≤3,解得,-3≤f(a)≤1,即-3≤f(a)≤0,②若f(a)>0,则-log2(f(a)+1)≤3,显然成立;则综上得,f(a)≥-3,再讨论a的取值情况:①若a≤0,则a2+2a≥-3,解得,a∈R,即a≤0.②若a>0,则-log2(a+1)≥-3,解得,0<a≤7,综上所述,实数a的取值范围是:(-∞,7].故答案为:(-∞,7].由已知中函数f(x)=,讨论f(a)的正负,代入求出f(a)≥-3,再讨论a的正负,求实数a的取值范围.本题考查了分段函数的应用,在已知函数值的范围时,要对自变量讨论代入函数求解,属于中档题.15.【答案】(0,2)【解析】解:根据题意,在同一个坐标系中作出函数y=x-4和y=x2-4x+2λ的图象,如图:若函数f(x)恰有2个零点,即函数f(x)图象与x轴有且仅有2个交点,可得△=16-8λ≥0,λ≤2,当λ=2时,函数f(x)恰有1个零点,所以λ<2;y=x2-4x+2λ的对称轴为x=2,(0,0)与(4,0)关于x=2对称;所以f(0)>0,可得λ>0,f(0)≤0时,函数f(x)恰有3个不同的零点,即λ的取值范围是:(0,2)故答案为:(0,2).根据题意,在同一个坐标系中作出函数y=x-4和y=x2-4x+2λ的图象,结合图象分析可得答案.本题考查分段函数的图象和函数的零点,考查数形结合思想的运用,考查发现问题解决问题的能力.16.【答案】1【解析】解:设log2a=log5b=lg(a+b)=k,∴a=2k,b=5k,a+b=10k,∴ab=10k,∴a+b=ab,则=1.故答案为:1.设log2a=log5b=lg(a+b)=k,可得a=2k,b=5k,a+b=10k,可得a+b=ab.即可得出.本题考查了对数与指数的运算性质,考查了推理能力与计算能力,属于中档题.17.【答案】解:(1)0.064-(-)0+16+0.25=.-1++.=2.5-1+8+0.5=10;(2)lg25+lg2+()-log29×log32=lg5+lg2+-2(log23×log32)=1+-2=-.【解析】本题考查了指数幂和对数的运算的性质,属于基础题.(1)根据指数幂的运算性质计算即可;(2)根据对数的运算性质计算即可.18.【答案】解:由题意:集合A={x|x2-2x-3≤0,x∈R}={x|-1≤x≤3},B={x|x2-2mx+m2-4≤0,x∈R}={x|m-2≤x≤m+2},(1)∵A∩B={x|1≤x≤3},∴ ,解得:m=3,所以:A∩B={x|1≤x≤3}时,实数m的值为3;(2)∵B={x|m-2≤x≤m+2},∴∁R B={x|m-2>x或m+2<x},∵A⊆∁R B,∴m-2>3或m+2<-1,解得:m>5或m<-3.所以:A⊆∁R B时,实数m的取值范围是:(-∞,-3)∪(5,+∞).【解析】本题考查了集合的基本运算的运用求参数的问题,属于基础题.(1)求出B,A集合,根据集合的基本运算求解实数m的值;(2)求出根据集合B,求出∁R B,在A⊆∁R B,求实数m的取值范围.19.【答案】解:(Ⅰ)当b=-3时,f(x)=log2(4x-3•2x+2),由4x-3•2x+2>0,得2x>2或2x<1,∴x>1或x<0,∴f(x)的定义域为{x|x>1或x<0};(Ⅱ)对于任意x≥1,都有f(x)>g(x)成立,即4x+b•2x+2>2x,对任意x≥1恒成立,∴b>=,对任意x≥1恒成立,∴只需b>=-2,∴b的取值范围为[-2,+∞).【解析】(Ⅰ)将b=-3代入f(x)中,由4x-3•2x+2>0,解出x的范围;(Ⅱ)根据对于任意x≥1,都有f(x)>g(x)成立,可得b>对任意x≥1恒成立,因此只需b>=-2,从而得到b的取值范围.本题考查了函数定义域的求法和不等式恒成立问题,考查了转化思想和整体思想,属中档题.20.【答案】解:(1)由1->0,可得x<-1或x>1,∴f(x)的定义域为(-∞,-1)∪(1,+∞);∵f(x)=log a(1-)=log a(),且f(-x)=log a()=log a()=-log a()=-f(x);∴f(x)在定义域上为奇函数.(2)当0<a<1时,f(x)在(1,+∞)单调递减,任取x1,x2且1<x1<x2,f(x1)-f(x2)=-=log a();由(x1-1)(x2+1)-(x1+1)(x2-1)=2(x1-x2)<0,∴0<<1,又0<a<1,∴log a()>0则f(x1)>f(x2),∴f(x)在(1,+∞)单调递减;(3)假设存在这样的实数a,使得当f(x)的定义域为[m,n]时,值域为[1+log a n,1+log a m];由0<m<n,又log a n+1<log a m+1,即log a n<log a m,∴0<a<1.由(2)知:f(x)在(1,+∞)单调递减,∴f(x)在(m,n)单调递减,∴ ,即m,n是方程log a=log a x+1的两个实根,即=ax在(1,+∞)上有两个互异实根;于是问题转化为关于x的方程ax2+(a-1)x+1=0在(1,+∞)上有两个不同的实数根,令g(x)=ax2+(a-1)x+1,则有△ >>>,解得0<a<3-2;故存在实数a∈(0,3-2),使得当f(x)的定义域为[m,n]时,值域为[1+log a n,1+log a m].【解析】(1)由1->0,可求出f(x)的定义域,利用定义法能求出f(x)在定义域上为奇函数.(2)当0<a<1时,f(x)在(1,+∞)单调递减,利用定义法能进行证明.(3)把f(x)的定义域为[m,n]时值域为[1+log a n,1+1og a m]转化为f(x)在(1,+∞)上为减函数,进一步得到=ax在(1,+∞)上有两个互异实根,令g(x)=ax2+(a-1)x+1,转化为关于a的不等式组求解.本题考查函数的定义域及奇偶性的判断,解题时要认真审题,注意函数性质的合理运用.属于中档题,21.【答案】解:(Ⅰ)函数f(x)=x2-3|x-a|,若函数y=f(x)为偶函数,则f(-x)=f(x),即(-x)2-3|-x-a|=x2-3|x-a|,∴|x+a|=|x-a|,两边平方,得x2+2ax+a2=x2-2ax+a2,∴2ax=-2ax,∴4ax=0,∴a=0,∴实数a的值为0;(Ⅱ)若,则函数y=f(x)=x2-3|x-|=,,<,画出函数f(x)的图象,如图所示;由图象知,单调减区间为(-∞,-],(,];(Ⅲ)不等式f(x-1)≤2f(x),化为(x-1)2-3|x-1-a|≤2x2-6|x-a|,即6|x-a|-3|x-1-a|≤x2+2x-1(*)对任意x∈[a,+∞)恒成立,①当0≤x≤a时,将不等式(*)可化为3a≤x2+5x+2,对0≤x≤a上恒成立,则g(x)=x2+5x+2 在(0,a]为单调递增,只需g(x)min=g(0)=2≥3a,解得0<a≤;②当a<x≤a+1时,将不等式(*)可化为9a≥-x2+7x-2,对a<x≤a+1上恒成立,由题意知h(x)=-x2+7x-2在x∈(a,a+1]上单调递增,则h(x)max=h(a+1)=-(a+1)2+7(a+1)-2≤9a,化简得a2+4a-4≥0,∴a≤-2-2或a≥-2+2;又0<a≤1,所以-2+2≤a≤1;③当x>a+1时,不等式(*)可化为3a≥-x2+x+4,则t(x)=-x2+x+4 在(a+1,+∞)为单调递减,则t(x)max=t(a+1)=-a2-a+4≤3a,解得a≤-2-2或a≥-2+2,又0<a≤1,所以-2+2≤a≤1;综上知,实数a的取值范围是(0,]∪[-2+2,1].【解析】(Ⅰ)根据偶函数的定义,化简整理,即可求得a的值;(Ⅱ)由分段函数的图象与性质,画出函数的图象,写出函数的单调区间;(Ⅲ)由题意可得,x∈[a,+∞)时,不等式恒成立,再分①当0≤x≤a时、②当x≥a+1、③当a<x<a+1时三种情况,分别求得a的取值范围,取交集即为所求.本题主要考查了分段函数的单调区间和二次函数性质的应用问题,体现了分类讨论和转化思想,属中档题.。

浙江省学军中学2019-2020年第一学期高三英语期中考试试卷 无答案

浙江省学军中学2019-2020年第一学期高三英语期中考试试卷 无答案

高三英语期中考试试卷本试卷分第I卷(选择题)和第II卷(非选择题)两个部分。

第I卷1至6页,第II卷7至8页。

满分150分,考试120分钟。

注意事项:1.答题前,考生务必将自己的姓名、准考证号填写在答题卡上。

2.在答选择题时,用2B铅笔把答题卡上对应题目的答案标号涂黑。

如需改动,用橡皮擦干净后,再选涂其他答案标号。

不能答在本试题卷上,否则无效。

第I卷第一部分: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)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 will the woman do next?A.Have a class.B. Buy some tickets.C. Change her schedule.2.What are the speakers talking about?A.The stars.B. A raindrop.C. The fireworks.3.When does the conversation probably take place?A.At breakfast.B. At lunch.C. At dinner.4.Where does the conversation probably take place?A.At a zoo.B. At a circus.C. At a camp.5.What does the man mean?A.Marcie is a busy woman.B. Marcie isn’t very polite.C. Marcie loves talking on the phone.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。

2019-2020学年浙江省杭州市学军中学西溪校区高一(上)期中数学试卷(PDF版 含答案)

2019-2020学年浙江省杭州市学军中学西溪校区高一(上)期中数学试卷(PDF版 含答案)

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浙江省杭州市学军中学2018-2019学年高一上学期期中考试数学试题 Word版含解析

浙江省杭州市学军中学2018-2019学年高一上学期期中考试数学试题 Word版含解析

浙江省杭州市学军中学2018-2019学年高一上学期期中考试数学试题一、选择题(本大题共10小题,共30.0分)1.已知集合,,则()A. B. C. D.【答案】B【解析】试题分析:由题意知,故选B.【考点定位】本题考查集合的基本运算,属于容易题.2.函数f(x)=ln(1-x2)的定义域为()A. B. C. D.【答案】B【解析】【分析】由根式内部的代数式大于等于0,对数式的真数大于0联立不等式组求解.【详解】由,得0≤x<1.∴函数f(x)ln(1﹣x2)的定义域为[0,1).故选:B.【点睛】本题考查函数的定义域及其求法,是基础题.3.已知函数f(x)=,则f[f()]等于()A. B. C. D.【答案】D【解析】【分析】推导出f(),从而f[f()]=f(),由此能求出结果.【详解】∵函数f(x),∴f(),f[f()]=f().故选:D.【点睛】本题考查函数值的求法,考查函数性质等基础知识,考查运算求解能力,是基础题.4.使函数f(x)=x a的定义域为R且为奇函数的α的值可以是()A. B. C. 3 D. 以上都不对【答案】C【解析】【分析】根据题意,结合幂函数的性质依次分析选项,综合即可得答案.【详解】根据题意,依次分析选项:对于A,α=﹣1时,f(x)=x﹣1,其定义域不是R,不符合题意;对于B,α时,f(x),其定义域不是R,不符合题意;对于C,α=3时,f(x)=x3,其定义域为R且为奇函数,符合题意;对于D,错误,故选:C.【点睛】本题考查幂函数的性质,关键是掌握幂函数的性质,属于基础题.5.已知集合M,N,P为全集U的子集,且满足M⊆P⊆N,则下列结论不正确的是( )A. ∁U N⊆∁U PB. ∁N P⊆∁N MC. (∁U P)∩M=∅D. (∁U M)∩N=∅【答案】D【解析】因为P⊆N,所以∁U N⊆∁U P,故A正确;因为M⊆P,所以∁N P⊆∁N M,故B正确;因为M⊆P,所以(∁U P)∩M=∅,故C正确;因为M⊆ N,所以(∁U M)∩N∅.故D不正确.故选D.6.设函数f(x)=log a x(a>0,a≠1),若f(x1x2…x2018)=4,则f(x12)+f(x12)+…+f(x20182)的值等于()A. 4B. 8C. 16D.【答案】B【解析】【分析】由函数的解析式结合对数的运算性质即可得解.【详解】∵函数f(x)=log a x(a>0,a≠1),f(x1x2…x2018)=4,∴f(x1x2…x2018)=log a(x1x2…x2018)=4,∴f(x12)+f(x12)+…+f(x20182)=log a(x1x2…x2018)2=2log a(x1x2…x2018)=2×4=8.故选:B.【点睛】本题考查函数值的求法,考查对数性质、运算法则等基础知识,考查运算求解能力,是基础题.7.设A={x|2≤x≤4},B={x|2a≤x≤a+3},若B真包含于A,则实数a的取值范围是()A. B. C. D.【答案】C【解析】【分析】由B真包含于A,讨论B=∅与B≠∅时,求出a的取值范围.【详解】∵A={x|2≤x≤4},B={x|2a≤x≤a+3},且B真包含于A;当B=∅时,2a>a+3,解得a>3;当B≠∅时,解得a=1;此时A=B.∴a的取值范围是{a|a>3}故选:C.【点睛】本题考查了集合之间的基本运算,解题时容易忽略B=∅的情况,是易错题.8.函数f(x)=log2(-x2+ax+3)在(2,4)是单调递减的,则a的范围是()A. B. C. D.【答案】B【解析】【分析】由复合函数的单调性可知内层函数在(2,4)上为减函数,则需要其对称轴小于等于2且当函数在x=4时的函数值大于等于0,由此联立不等式组得答案.【详解】令t=﹣x2+ax+3,则原函数化为y=log2t,∵y=log2t为增函数,∴t=﹣x2+ax+3在(2,4)是单调递减,对称轴为x,∴且﹣42+4a+3≥0,解得:.∴a的范围是[,4].故选:B.【点睛】本题考查了复合函数的单调性,复合函数的单调性满足同增异减的原则,是中档题.9.对于函数f(x),若∀a,b,c∈R,f(a),f(b),f(c)为某一三角形的三边长,则称f (x)为“可构造三角形函数”.已知函数f(x)=是“可构造三角形函数”,则实数t 的取值范围是()A. B. C. D.【答案】A【解析】【分析】因对任意实数a、b、c,都存在以f(a)、f(b)、f(c)为三边长的三角形,则f(a)+f(b)>f(c)恒成立,将f(x)解析式用分离常数法变形,由均值不等式可得分母的取值范围,整个式子的取值范围由t﹣1的符号决定,故分为三类讨论,根据函数的单调性求出函数的值域,然后讨论k转化为f(a)+f(b)的最小值与f(c)的最大值的不等式,进而求出实数k 的取值范围.【详解】由题意可得f(a)+f(b)>f(c)对于∀a,b,c∈R都恒成立,由于f(x)1,①当t﹣1=0,f(x)=1,此时,f(a),f(b),f(c)都为1,构成一个等边三角形的三边长,满足条件.②当t﹣1>0,f(x)在R上是减函数,1<f(a)<1+t﹣1=t,同理1<f(b)<t,1<f(c)<t,故f(a)+f(b)>2.再由f(a)+f(b)>f(c)恒成立,可得2≥t,结合大前提t﹣1>0,解得1<t≤2.③当t﹣1<0,f(x)在R上是增函数,t<f(a)<1,同理t<f(b)<1,t<f(c)<1,由f(a)+f(b)>f(c),可得 2t≥1,解得1>t.综上可得,t≤2,故选:A.【点睛】本题主要考查了求参数的取值范围,以及构成三角形的条件和利用函数的单调性求函数的值域,同时考查了分类讨论的思想,属于难题.10.设函数,其中表示中的最小者.下列说法错误的()A. 函数为偶函数B. 若时,有C. 若时,D. 若时,【答案】D【解析】【分析】先根据定义作的图像,然后依据图像逐个检验即可.【详解】在同一坐标系中画出的图像(如图所示),故的图像为图所示.的图像关于轴对称,故为偶函数,故A正确.由图可知时,有,故B成立.从图像上看,当时,有成立,令,则,故,故C成立.取,则,,,故D不成立.综上,选D.【点睛】一般地,若(其中表示中的较小者),则的图像是由这两个函数的图像的较低部分构成的.二、填空题(本大题共7小题,共28.0分)11.若,则.【答案】10【解析】试题分析:若,则考点:对数与对数函数12.已知,则________.【答案】【解析】【分析】利用配凑法求函数的解析式.【详解】(配凑法)(1),又∈(-∞,-2]∪[2,+∞),∴.故答案为:【点睛】本题考查函数解析式的求解及常用方法,是基础题.解题时要认真审题,仔细解答.13.已知f(x)=的定义域为R,则实数a的取值范围是______.【答案】[-1,0]【解析】【分析】把f(x)的定义域为R转化为0对任意x∈R恒成立,即x2+2ax ﹣a≥0对任意x∈R恒成立,再由判别式小于等于0求解.【详解】∵f(x)的定义域为R,∴0对任意x∈R恒成立,即恒成立,即x2+2ax﹣a≥0对任意x∈R恒成立,∴△=4a2+4a≤0,则﹣1≤a≤0.故答案为:[﹣1,0].【点睛】本题考查函数的定义域及其求法,考查数学转化思想方法,是中档题.14.设max{a,b}表示a,b两数中的最大值,若f(x)=max{|x|,|x-t|}关于x=1对称,则t=______.【答案】2【解析】【分析】利用函数y=|x|的图象和函数y=|x﹣t|的图象关于直线x对称,从而得出结论.【详解】f(x)=max{|x|,|x﹣t|},由函数y=|x|的图象关于x=0对称,函数y=|x﹣t|的图象关于x=t对称,即有函数f(x)的图象关于x对称,f(x)=max{|x|,|x﹣t|}关于x=1对称,即有1,求得t=2,故答案为:2.【点睛】本题主要考查分段函数的应用,考查函数的对称性,属于基础题.15.设方程x2-mx+2=0的两根α,β,其中α∈(1,2),则实数m的取值范围是______.【答案】(2,4)【解析】【分析】由题意利用韦达定理,不等式的性质,求出实数m的取值范围.【详解】∵方程x2﹣mx+2=0的两根α,β,∴△=m2﹣8≥0,求得m≥2,或m≤﹣2①.由α•β=2,则,则,则②.由①②可得,故答案为:.【点睛】本题主要考查韦达定理,不等式的性质,属于基础题.16.已知lg2≈0.3010,则22018是______位数.【答案】608【解析】【分析】设x=22018,可得lgx=2018lg2≈607.418,即可得出.【详解】设x=22018,则lgx=2018lg2≈2018×0.3010=607.418,∴22018是608位数.故答案为:608.【点睛】本题考查了对数运算性质,考查了推理能力与计算能力,属于基础题.17.已知函数f(x)满足对任意的m,n都有f(m+n)=f(m)+f(n)-1,设g(x)=f(x)+(a>0,a≠1),g(ln2018)=-2015,则g(ln)=______.【答案】2018【解析】【分析】由已知中函数f(x)满足对任意实数m,n,都有f(m+n)=f(m)+f(n)﹣1,可得f(0)=1,进而f(x)+f(﹣x)=2,g(x)+g(﹣x)=3,结合g(ln2018)=﹣2015,由对数的运算性质计算可得所求值.【详解】∵函数f(x)满足对任意实数m,n,都有f(m+n)=f(m)+f(n)﹣1,令m=n=0,则f(0)=2f(0)﹣1,解得f(0)=1,令m=x,n=﹣x,则f(0)=f(x)+f(﹣x)﹣1,即f(x)+f(﹣x)=2,∵g(x)=f(x)(a>0,a≠0),∴g(﹣x)=f(﹣x)f(﹣x),故g(x)+g(﹣x)=f(x)+f(﹣x)+1=3,∴g(ln2018)+g(ln)=﹣2015+g(﹣ln2018)=3,即g(ln)=2018,故答案为:2018.【点睛】本题主要考查抽象函数的应用,根据条件建立方程关系是解决本题的关键,属于中档题.三、解答题(本大题共5小题,共62.0分)18.已知集合U=R,集合A={x|x2-(a-2)x-2a≥0},B={x|1≤x≤2}.(1)当a=1时,求A∩B;(2)若A∪B=A,求实数a的取值范围.【答案】(1){x|1≤x≤2};(2){a|a≤1}.【解析】【分析】(1)代入a的值,求出集合A,从而求出A∩B;(2)由A与B的并集为A,得到B为A的子集,表示出A的中不等式的解集,根据数轴确定出满足题意a的范围即可.【详解】(1)a=1时,A={x|x≥1或x≤-2},故A∩B={x|1≤x≤2};(2)∵A∪B=A,∴B⊆A,由x2-(a-2)x-2a≥0,得(x+2)(x-a)≥0,当a<-2时,如数轴表示,符合题意;同理,当-2≤a≤1,也合题意;但当a>1时,不合题意,综上可知{a|a≤1}.【点睛】本题考查了并集及其运算,熟练掌握并集的定义是解本题的关键.19.设函数f(x)=++.(1)设t=+,求t的取值范围;(2)求f(x)的最大值.【答案】(1)[,2];(2)3.【解析】【分析】(1)将t,﹣1≤x≤1,两边平方,结合二次函数的最值,即可得到所求范围;(2)由(1)可得g(t)=f(x)(t+1)2,考虑对称轴t=﹣1与区间[,2]的关系,即可得到所求最大值.【详解】(1)t=+,-1≤x≤1,可得t2=2+2,由0≤1-x2≤1,可得t2∈[2,4],由t≥0可得t的取值范围是[,2];(2)由(1)可得g(t)=f(x)=t+=(t+1)2-,由[,2]在对称轴t=-1的右边,为增区间,即有t=2,即x=0,g(t)取得最大值,且为3,即f(x)的最大值为3.【点睛】本题考查函数的最值求法,注意运用换元法和二次函数的最值求法,考查运算能力,属于中档题.20.已知函数f(x)=x+(a>0).(1)判断f(x)的奇偶性;(2)判断函数f(x)在(,+∞)上的单调性,并用定义证明.【答案】(1)见解析;(2)见解析.【解析】【分析】(1)求出函数的定义域,得到f(﹣x)=﹣f(x),判断函数的奇偶性即可;(2)根据单调性的定义证明即可.【详解】(1)f(x)的定义域是{x|x≠0},f(-x)=-x-=-(x+)=-f(x),故函数f(x)是奇函数;(2)函数在(,+∞)递增,令<m<n,则f(m)-f(n)=m+-n-=(m-n)+a•=(m-n)(1-),∵<m<n,∴m-n<0,1->0,故f(m)-f(n)<0,故f(x)在(,+∞)上递增.【点睛】本题考查了函数的奇偶性,单调性问题,考查转化思想,是一道基础题.21.已知函数f(x)=2x,g(x)=-x2+2x+b.(1)若f(x)++1≥0对任意的x∈[1,3]恒成立,求m的取值范围;(2)若x1,x2∈[1,3],对任意的x1,总存在x2,使得f(x1)=g(x2),求b的取值范围.【答案】(1)[-6,-∞);(2)见解析.【解析】【分析】(1)根据h(x)=f(x)1,结合勾函数的性质对任意的x∈[1,3]恒成立,即可求解m的取值范围;(2)根据对任意的x1,总存在x2,使得f(x1)=g(x2),可得f(x)的值域是g(x)的值域的子集,即可求解b的范围;【详解】(1)函数f(x)=2x,令h(x)=f(x)++1=;①当m=0时,可得h(x)=2x+1在x∈[1,3]恒成立;②当m<0时,可知f(x)=2x是递增函数,y=在x∈[1,3]也是递增函数,∴h(x)在x∈[1,3]是递增函数,此时h(x)min=h(1)=≥0,可得:-6≤m<0;③当m>0时,,所以函数h(x)=,满足题意.综上所述:f(x)++1≥0对任意的x∈[1,3]恒成立,可得m的取值范围是[-6,-∞);(2)由函数f(x)=2x,x∈[1,3],可得:2≤f(x)≤8;由g(x)=-x2+2x+b.其对称x=1,开口向下.∵x∈[1,3],∴g(x)在x∈[1,3]上单调递减.g(x)max=g(1)=1+b;g(x)min=g(3)=-3+b;∵对任意的x1,总存在x2,使得f(x1)=g(x2),∴f(x)的值域是g(x)的值域的子集;即,解得:无解.故x1,x2∈[1,3],对任意的x1,总存在x2,使得f(x1)=g(x2),此是b的取值范围是空集.【点睛】本题主要考查了函数恒成立问题的求解,分类讨论以及转化思想的应用,二次函数的最值以及单调性的应用,属于中档题.22.已知a∈R,f(x)=log2(1+ax).(1)求f(x2)的值域;(2)若关于x的方程f(x)-log2[(a-4)x2+(2a-5)x]=0的解集恰有一个元素,求实数a 的取值范围;(3)当a>0时,对任意的t∈(,+∞),f(x2)在[t,t+1]的最大值与最小值的差不超过4,求a的取值范围.【答案】(1)当a≥0时,值域为[0,+∞),当a<0时,值域为(-∞,0);(2)1<a≤2,或a>4;(3)(0,+∞).【解析】【分析】(1)讨论a≥0时,a<0时,由对数函数的单调性可得值域;(2)根据对数的运算法则进行化简,转化为一元二次方程,讨论a的取值范围进行求解即可;(3)根据条件得到g(x)=log2(1+ax2),a>0,函数g(x)在区间[t,t+1]上单调递增,g(t+1)﹣g(t)≤4,运用对数函数的单调性和参数分离进行求解即可.【详解】(1)f(x)=log2(1+ax),可得f(x2)=log2(1+ax2),当a≥0时,1+ax2≥1,即有log2(1+ax2)≥0;当a<0时,0<1+ax21,即有log2(1+ax2)0;即有当a≥0时,f(x)的值域为[0,+∞);当a<0时,f(x)的值域为(-∞,0];(2)由f(x)-log2[(a-4)x2+(2a-5)x]=0得log2(1+ax)=log2[(a-4)x2+(2a-5)x],即1+ax=(a-4)x2+(2a-5)x>0,①则(a-4)x2+(a-5)x-1=0,即(x+1)[(a-4)x-1]=0,②,当a=4时,方程②的解为x=-1,代入①,不成立;当a=3时,方程②的解为x=-1,代入①,不成立;当a≠4且a≠3时,方程②的解为x=-1或x=,若x=-1是方程①的解,则1-a=-a+1>0,即a<1,若x=是方程①的解,则1+=>0,即a>4或a<2,则要使方程①有且仅有一个解,则a>4或1≤a<2.综上,若方程f(x)-log2[(a-4)x2+(2a-5)x]=0的解集中恰好有一个元素,则a的取值范围是1<a≤2,或a>4;(3)当a>0时,对任意的t∈(,+∞),f(x2)=log2(1+ax2),设g(x)=log2(1+ax2),a>0,函数g(x)在区间[t,t+1]上单调递增,由题意得g(t+1)-g(t)≤4,即log2(1+at2+2at+a)-log2(1+at2)≤4,即1+at2+2at+a≤16(1+at2),即有a(15t2-2t-1)+15=a(3t-1)(5t+1)+15>0恒成立,综上可得a的范围是(0,+∞).【点睛】本题考查函数最值的求解,以及对数不等式的应用,考查对数函数的单调性,属于中档题.。

杭州学军中学2018-2019学年第一学期高二期中考试数学试题(解析版)

杭州学军中学2018-2019学年第一学期高二期中考试数学试题(解析版)

1.直线10x y ++=的倾斜角是( ) A.34π B.23π C.4π D.4π- 【答案】A【解析】∵直线方程为10x y ++=,∴化成斜截式得1y x =--,直线的斜率为1k =-,设直线的倾斜角为α,则tan 1α=-,∵(0,)απ∈,∴34πα=,即直线10x y ++=的倾斜角是34πα=. 2.如果直线210ax y ++=与直线20x y +-=互相垂直,则实数a =( ) A.1 B.2- C.23- D.13- 【答案】B【解析】直线210ax y ++=的斜率12ak =-,直线20x y +-=的斜率为21k =-,因为两直线垂直,所以121k k ⋅=-,即()(1)12a -⋅-=-,解得2a =-.3.设x ,y 满足约束条件2330233030x y x y y +-≤⎧⎪-+≥⎨⎪+≥⎩,则2z x y =+的最小值是( )A.1B.9C.15-D.9- 【答案】C【解析】由题意约束条件作出可行域如图所示,当目标函数2z x y =+过(6,3)--点时取得最小值,最小值为2(6)315z =⨯--=-.4.圆222210x y x y +--+=上的点到直线2x y -=的距离的最大值是( )A.22+11+2【解析】圆222210x y x y +--+=即22(1)(1)1x y -+-=,表示以点(1,1)C 为圆心,以1为半径的圆,由于圆心到直线2x y -=的距离d ==,故圆上的点到直线的距离的最大值是1r d +=5.已知(4,0)A ,(0,4)B ,从点(2,0)P 射出的光线经直线AB 反射后再射到直线OB 上,最后经直线OB 反射后又回到P 点,则光线所经过的路程是( )A.6D. 【答案】D 【解析】如图:作点P 关于AB 的对称点1P ,作点1P 关于OB 的对称点2P ,设两次反射的入射点分别为C ,D ,则1P ,C ,D 三点共线,2P ,D ,P 三点共线,则光线所经过的路程即为2P P .40:4404AB l y x x -=+=-+-,:0OB l x =,设111(,)P x y ,222(,)P x y ,则1PP AB ⊥,1PP 中点在AB 上,所以11110(1)1202422y x y x -⎧⋅-=-⎪-⎪⎨++⎪=-+⎪⎩,解得1124y x =⎧⎨=⎩.同理,关于y 轴对称,所以24x =-,22y =,所以2P P ==6.在长方体1111ABCD A BC D -中,2AB BC ==,11AA =,则1AC 与平面1111A B C D所成角的正弦值为( )A.3 B.23C.4D.13【答案】D【解析】连接1AC ,在长方体1111ABCD A BC D -中,1A A ⊥平面1111A B C D ,则11AC A ∠为1AC 与平面1111A B C D 所成角.在11AC A ∆中,11111sin 3AA AC A AC ∠===.7.如图,长方体1111ABCD A BC D -中,12AA AB ==,1AD =,E 、F 、G 分别是1DD 、AB 、1CC 的中点,则异面直线1A E 与GF 所成角的余弦值是( )A.5B.2C.5D.0 【答案】D【解析】如图,连接EG ,1B F ,CF ,因为E ,G 均为中点,所以11//A B EG ,11A B EG =,故四边形11A EGB 为平行四边形,11//A E B G ,所以求1A E 与GF 的夹角即为求1B G 与GF 的夹角,即1B GF ∠.因为112BF AB ==,根据勾股定理,CF =1112CG CC ==,所以根据勾股定理,FG ==,1B G ==,1B F ==22211B F B G FG =+满足勾股定理,所以1B G FG ⊥,所以1cos 0B GF ∠=.8.已知集合{(,)|(1)(1)}A x y x x y y r =-+-≤,集合222{(,)|}B x y x y r =+≤,若A B ⊆,则实数r 可以取的一个值是( )12 D.12+ 【答案】A【解析】(1)(1)x x y y r -+-=可化为22111()()222x y r -+-=+,由题意,集合A 表示的圆面所对应的圆内含或内切于集合B 表示的圆面所对应的圆,则r ≤1r ≥9.已知圆22:(2)(3)4M x y -+-=,过x 轴上的点0(,0)P x 存在圆M 的割线PAB ,使得PA AB =,则0x 的取值范围是( )A.[-B.[-C.[2-+D.[2-+ 【答案】C【解析】max 4AB =,故max6PM=6≤,解得022x -≤+.10.在棱长为1的正方体1111ABCD A BC D -中,E 为线段1B C 的中点,F 是棱11C D 上的动点,若点P 为线段1BD 上的动点,则PE PF +的最小值为( )A.6 B.12+2 D.2【答案】A【解析】注意到虽然点P 、F 都是动点,但它们都在面11BC D 上,将该平面提取出来,如图,取11Rt BC D ∆,作点E 关于直线1BD 的对称点E ',作11E G C D '⊥,点G 在直线11C D 上.则PE PF PE PF E G ''+=+≥,连接EE '(与1BD 垂直),作E H BC '⊥,点H 在直线1BC 上,则cos 2sin sin 3EH EE E EH EB B B ''=⋅∠=⋅⋅∠⋅∠=,1E G C E EH '=+=+=.二、填空题11.直线310x y -+=关于直线0x y +=对称的直线方程是 . 【答案】310x y -+= 【解析】将x y =-,y x =-代入直线310x y -==得310x y -+=.12.如图是一个正三棱柱的三视图,若三棱柱的体积是a = .【答案】【解析】由题意知三棱柱的底面是一个正三角形,一条边上的高是a ,得到三棱柱的底面边,∴底面面积是212a ⨯=,三棱柱的高是2,∴三棱柱的体积是223a ⨯=a =13.已知(,)P x y 满足0102x x y ≤≤⎧⎨≤+≤⎩,则点(,)Q x y y +构成的图形的面积为 .【答案】2【解析】设点(,)Q u v ,则x y u +=,y v =,则点(,)Q u v 满足0102u v u ≤+≤⎧⎨≤≤⎩,在uOv 平面内画出点(,)Q u v 所构成的平面区域如图,可知它是一个平行四边形,边长为1,高为2,故其面积为212⨯=.14.有且只有一对实数(,)x y 同时满足:20x y m +-=与223(0)x y y +=≥,则实数m 的取值范围是 .【答案】[{15}- 【解析】根据题意画出图形,如图所示:当直线2y x m =-+与半圆相切时,圆心到直线的距离等于圆的半径,即35=,解得15m =或15m =- (舍去), 当直线过(3,0)-时,把此点代入直线方程求得23m =-,当直线过时,把此点代入直线方程求得m =观察图象可知满足题意的实数m 的取值范围是[{15}-.15.异面直线a ,b 成60︒角,直线a c ⊥,则直线b ,c 所成角的范围是 . 【答案】[,]62ππ 【解析】由题意可作出大致图象,如下:直线a c ⊥,则把c 放在α上,只需要a α⊥即可,a ,b 成60︒角,那么可将b 平移到b '与a 相交,相当于是圆锥的母线,所以b 与c 所成角即为b '与面α上任意一条直线所成角,所以最小值为30︒,最大值为90︒,即取值范围是[,]62ππ. 16.在平面直角坐标系xOy 中,A 为直线:2l y x =上在第一象限内的点,(5,0)B ,以AB 为直径的圆C (C 为圆心)与直线l 交于另一点D .若0AB CD ⋅=,则点A 的坐标为 . 【答案】(3,6)A【解析】根据题干,可作出大致图象,如下:∵AB 为直径,且0AB CD ⋅=,则AB CD ⊥,且ABD ∆为等腰直角三角形,又(5,0)B ,可求出BD ==,OD ==,从而AO OD AD =+==(,2)(0)A x x x >,则222(2)x x +=,解得3x =,所以(3,6)A .17.在平面直角坐标系xOy 中,点(3,0)A ,直线:24l y x =+,设圆C 的半径为1,圆心C 在直线l 上,若圆C 上存在点M ,使2MA MO =,则圆心C 的横坐标a 的取值范围 是 .【答案】98419[2][,]555+---【解析】设(,)M x y ,由2MA MO =22(1)4x y ++=,故点M 的轨迹是圆心为(1,0)-,半径为2的圆;设圆C 的圆心(,24)C a a +,则两圆外切时22(1)(24)9a a +++=,即251880a a ++=,解得95a -=; 两圆内切时22(1)(24)1a a +++=,即2518160a a ++=,解得2a =-或85-,所以得a 的取值范围是98419[2][,]555+---. 三、解答题18.已知圆22:(1)5C x y +-=,直线:120l mx y m -+-= (1)求证:不论m 取何实数,直线l 与圆C 总有两个不同的交点;(2)设直线l 与圆C 交于点A ,B,当AB =时,求直线l 的方程. 【答案】(1)略;(2)10x y --=,30x y +-=【解析】直线(2)10m x y --+=经过定点(2,1),222(11)5+-<,∴定点在圆内,故无论m 取实数,直线l 与圆C 总有两个不同的交点. (1)由圆心(0,1)到直线120mx y m -+-=的距离d ==,而圆的弦长AB ===,解得1m =±,故所求的直线方程为10x y --=和30x y +-=.19.已知菱形ABCD 的边长为2,120ABC ∠=︒,四边形BDEF 是矩形,且BF ⊥平面ABCD,BF =(1)求证://CF 平面ADE ;(2)设EF 中点为G ,求证AG ⊥平面CEF .【答案】(1)略;(2)略【解析】(1)证明://BC AD ,//BF DE ⇒平面//BCF 平面//ADE CF ⇒平面ADE . (2)证明:因为EF 中点为G ,则由AF AE AG EF =⇒⊥,且计算可得:AG CG ==又AC =222AG CG AC AG CG +=⇒⊥,又EF CG G =,所以AG ⊥平面CEF.20.已知:以点2(,)(,0)C t t R t t∈≠为圆心的圆与x 轴交于点O ,A ,与y 轴交于点O ,B ,其中O 为原点.(1)求证:OAB ∆的面积为定值;(2)设直线24y x =-+与圆C 交于点M ,N ,若OM ON =,求圆C 的方程. 【答案】(1)略;(2)22(2)(1)5x y -+-= 【解析】(1)∵圆C 过原点O ,∴2224OC t t =+. 设圆C 的方程是222224()()x t y t t t -+-=+,0x =,得10y =,24y t=;令0y =,得10x =,22x t = ∴1142422OAB S OA OB t t∆=⨯=⨯⨯=,即:OAB ∆的面积为定值. (2)∵OM ON =,CM CN =,∴OC 垂直平分线段MN . ∵2MN k =-,∴12OC k =,∴直线OC 的方程是12y x =.∴212t t =,解得2t =或2t =-,当2t =时,圆心C 的坐标为(2,1),OC =此时C 到直线24y x =-+的距离d =<C 与直线24y x =-+相交于两点当2t =-时,圆心C 的坐标为(2,1)--,OC = 此时C 到直线24y x =-+的距离d =>圆C 与直线24y x =-+不相交,∴2t =-不符合题意舍去.∴圆C 的方程为22(2)(1)5x y -+-=.21.如图,在三棱柱111ABC A B C -中,1CC ⊥平面ABC ,D ,E ,F ,G 分别为1AA ,AC ,11AC ,1BB 的中点,且AB BC ==,AC =1AA =(1)证明:AC FG ⊥;(2)证明:直线FG 与平面BCD 相交;(3)求直线BD 与平面1BEC 所成角的正弦值.【答案】(1)略;(2)略;(3)14【解析】(1)在三棱柱111ABC A B C -中,∵1CC ⊥平面ABC ,∴四边形11A ACC 为矩形. 又E ,F 分别为AC ,11AC 的中点,∴AC EF ⊥∵AB BC =.∴AC BE ⊥,∴AC ⊥平面BEF .又G 是1BB 中点,1//BB EF ,∴G 在平面BEF 内,∴AC FG ⊥.(2)设EF CD M =,则4FM =2BG =,所以,四边形BGFM 是梯形,所以,直线FG 与直线MB 相交,又MB ⊂平面BCD ,可得FG 与平面BCD 相交.(3)过D 作1DO C E ⊥于点O ,连BO ,易证BE ⊥平面11ACC A ,∴DO BE ⊥,∴DO ⊥平面1BEC ,从而DBO ∠就是直线BD 与平面1BEC 所成角的平面角,计算得,2BD =,sin 414DO DO DBO BD =⇒∠==.。

杭州学军中学2019学年第一学期高三地理期中考试试题卷

杭州学军中学2019学年第一学期高三地理期中考试试题卷

杭州学军中学2019学年第一学期期中考试高三地理试卷命题人:赵平 审题人:陈欣考生须知:1.全卷分为试题卷和答题卷两部分。

满分100分。

考试时间90分钟。

2.所有试题的答案均应做在答题卷相应的空格中,做在试题卷上无效。

3.请用钢笔或圆珠笔将姓名、考号、座位号分别填写在答题卷的相应位置上。

第Ⅰ卷一、选择题(本大题共20题,每小题3分,共60分。

每小题列出的四个备选项中只有一个是符合题目要求的,不选、多选、错选均不得分)十九世纪,德国对易北河、多瑙河以及莱茵河等主要河道进行了大规模的裁弯取直。

近年来,德国开始重塑成有凹岸和凸岸、浅滩、沙洲的自然河湾。

完成1~2题。

1. 十九世纪,德国进行河流整治的主要目的是促进A. 航运发展B. 旅游业C. 水产养殖业D. 水质改善 2. 近年来,德国重塑自然河湾的影响是A. 减小河道摆动B. 减缓河流流速C. 增强河流泄洪能力D. 增大河流侵蚀能力光伏农业是太阳能与现代农业相结合的大型并网发电项目,光伏大棚采用棚顶发电,为棚内提供自动化浇水、光照、通风、供暖等需要的电能,再在棚内种植花卉、水果等农作物。

完成3~4题。

3.影响光伏大棚棚顶与地面夹角南北差异的主要因素是A. 降水B.纬度C. 地形D.风力4.光伏农业的主要优势是①优化农业结构 ②提高土地利用率 ③提高能源利用率 ④削弱进入棚内的日照A. ①②③B. ②③④C. ①②④D. ①③④ 下表为我国某省级行政区1978-2016年城镇人口比重和产业比重变化数据。

完成5~6题。

年份 城镇人口比重(%) 第一产业比重(%) 第三产业比重(%)1978 11.3 50.7 21.6 1994 16.6 46.0 36.9 2010 22.67 13.5 54.2 2016 29.56 9.1 53.5 第1、2题图第3、4题图5.该省级行政区可能是A .西藏B .上海C .北京D .江苏6.该省第一产业所占比重大幅缩小,主要是由于A .人口数量减少,市场需求量缩小B .耕地面积缩小,种粮积极性降低C .人口向城市集中,农业劳动力减少D .自然条件独特,旅游业快速发展干旱土是指发育在干旱水分条件下的土壤。

2019-2020学年浙江省杭州市学军中学(西溪校区)2019级高一上学期期中考试数学试卷及解析

2019-2020学年浙江省杭州市学军中学(西溪校区)2019级高一上学期期中考试数学试卷及解析

2019-2020学年学军中学(西溪校区)2019级高一上学期期中考试数学试卷★祝考试顺利★一、选择题(本大题共10小题)1.已知集合M={x|x>0},N={x|-1<x≤2},则(∁R M)∩N等于()A. B. C. D.2.下列选项中两个函数,表示同一个函数的是()A. ,xB. ,C. ,D. ,3.下列函数在其定义域上既是奇函数又是增函数的是()A. B. C. D.4.在同一直角坐标系中,函数y=,y=1og a(x+)(a>0且a≠1)的图象可能是()A. B.C. D.5.若函数f(x2+1)的定义域为[-1,1],则f(lg x)的定义域为()A. B. C. D.6.已知函数f(x)为奇函数,g(x)为偶函数,且2x+1=f(x)+g(x),则g(1)=()A. B. 2 C. D. 47.已知定义在R上的函数(m为实数)为偶函数,记a=f(log0.53),b=f3),c=f(2m),则a,b,c的大小关系为()(log2.5A. B. C. D.8.已知f(x)=(x2-ax+3a)在区间(2,+∞)上是减函数,则实数a的取值范围是()A. B. C. D.9.已知a>0,设函数f(x)=(x∈[-a,a])的值域为[M,N],则M+N的值为()A. 0B. 2019C. 4037D. 403910.已知m∈R,函数f(x)=||+m在[2,5]上的最大值是5,则m的取值范围是()A. B. C. D.二、填空题(本大题共5小题,共25.0分)11.若幂函数y=f(x)的图象经过点(8,2),则f()的值是______.12.若f(1+)=,则f(3)=______.13.已知函数f(x)=x3+ln(+x).若f(a-1)+f(2a2)≤0,则实数a的取值范围是______.14.设函数f(x)=若f[f(a)]≤3,则实数a的取值范围是______.15.已知λ∈R,函数若函数f(x)恰有2个不同的零点,则λ的取值范围为______.三、解答题(本大题共6小题,共55.0分)16.若正数a,b满足log2a=log5b=lg(a+b),则的值为______ .17.化简求值:(1)-(-)0++(2)lg25+lg2+()-log29×log32.18.已知集合A={x|x2-2x-3≤0,x∈R},B={x|x2-2mx+m2-4≤0,x∈R}.(1)若A∩B={x|1≤x≤3},求实数m的值;(2)若A⊆∁R B,求实数m的取值范围.19.已知函数f(x)=log2(4x+b•2x+2),g(x)=x.(Ⅰ)当b=-3时,求函数f(x)的定义域;(Ⅱ)若对于任意x≥1,都有f(x)>g(x)成立,求实数b的取值范围.20.已知函数f(x)=log a(1-)(a>0且a≠1).(Ⅰ)判断函数f(x)的奇偶性并说明理由;(Ⅱ)当0<a<1时,判断函数f(x)在(1,+∞)上的单调性,并利用单调性的定义证明;(Ⅲ)是否存在实数a,使得当f(x)的定义域为[m,n]时,值域为[1+log a n,m]?若存在,求出实数a的取值范围;若不存在,请说明理由.1+loga21.已知函数f(x)=x2-3|x-a|.(Ⅰ)若函数y=f(x)为偶函数,求实数a的值;(Ⅱ)若a=,求函数y=f(x)的单调递减区间.(Ⅲ)当0<a≤1时,若对任意的x∈[a,+∞),不等式f(x-1)≤2f (x)恒成立,求实数a的取值范围.2019-2020学年学军中学(西溪校区)2019级高一上学期期中考试数学参考答案1.【答案】C【解析】解:∵M={x|x>0},N={x|-1<x≤2},∴∁R M={x|x≤0},(∁RM)∩N=(-1,0].故选:C.进行补集、交集的运算即可.考查描述法、区间的定义,以及补集、交集的运算.2.【答案】B【解析】解:相同的函数必须具有相同的定义域、值域、对应关系,而函数f(x)=ln x4的定义域为非零实数集,g(x)=4ln x的定义域为正实数集合,故它们不是同一个函数;函数f(x)=x2和函数g(x)==x2,具有相同的定义域、值域、对应关系,故它们是同一个函数;函数f(x)=x-1的值域为R,而g(x)==|x-1|的值域为[0,+∞),故它们不是同一个函数;函数f(x)=x的值域为R,函数g(x)=|x|的值域为[0,+∞),故它们不是同一个函数,故选:B.由题意利用函数的三要素,判断两个函数是否为同一个函数,从而得出结论.本题主要考查函数的三要素,属于基础题.3.【答案】B【解析】【分析】本题考查函数的奇偶性与单调性的判定,关键是掌握常见函数的奇偶性与单调性,属于基础题.根据题意,依次分析选项中函数的奇偶性与单调性,综合即可得答案.【解答】解:根据题意,依次分析选项:对于A,f(x)=2x,为指数函数,不是奇函数,不符合题意;对于B,f(x)=x|x|=,既是奇函数又是增函数,符合题意;对于C,f(x)=-,在其定义域上不是增函数,不符合题意;对于D,f(x)=lg|x|,是偶函数,不符合题意;故选B.4.【答案】D【解析】【分析】本题考查了指数函数,对数函数的图象和性质,属于基础题.对a进行讨论,结合指数,对数的性质即可判断;【解答】解:由函数y=,y=1og a(x+),当a>1时,可得y=是递减函数,图象恒过(0,1)点,函数y=1og a(x+),是递增函数,图象恒过(,0);当1>a>0时,可得y=是递增函数,图象恒过(0,1)点,函数y=1og a(x+),是递减函数,图象恒过(,0);∴满足要求的图象为:D故选D.5.【答案】C【解析】解:若函数f(x2+1)的定义域为[-1,1],则1≤x2+1≤2,∴1≤lg x≤2,∴10≤x≤100,故选:C.由函数f(x2+1)的定义域为[-1,1],求出其值域,即f(lg x)的值域,从而求出其定义域.6.【答案】C【解析】解:∵函数f(x)为奇函数,g(x)为偶函数,且2x+1=f(x)+g∴f(1)+g(1)=21+1=4,①f(-1)+g(-1)=2-1+1=20=1,即-f(1)+g(1)=1②由①+②得2g(1)=5,则g(1)=,故选:C.根据函数奇偶性的性质,建立方程组进行求解即可.7.【答案】A【解析】解:根据题意,定义在R上的函数(m为实数)为偶函数,则f(-x)=f(x),即()|x-m|=()|-x-m|,分析可得m=0,则f(x)=()|x|-1=,则f(x)在[0,+∞)上为减函数,又由a=f(log0.53)=f(log23),b=f(log2.53),c=f(2m)=f(0),且0<log2.53<log23,则有a<b<c;故选:A.根据题意,由偶函数的定义分析可得()|x-m|=()|-x-m|,进而可得m=0,即可得函数的解析式,分析可得f(x)在[0,+∞)上为减函数,结合对数的运算性质分析可得答案.8.【答案】D【解析】解:令t=x2-ax+3a,则由题意可得函数t在区间(2,+∞)上是增函数,且t>0,∴,求得-4≤a≤4,故选:D.令t=x2-ax+3a,则由题意可得函数t在区间(2,+∞)上是增函数,且t>0,故有,由此求得a的范围.9.【答案】C【解析】解:依题意,f(x)==+2019x=2019-+2019x,当x∈[-a,a]时f′(x)>0,所以f(x)为[-a,a]上的增函数,所以M+N=2019--2019a+2019-+2019a=4038-=4037.故选:C.将函数f(x)分离常数后根据函数的单调性求解函数值域,即可得到M,N 的值,从而得到M+N.10.【答案】A【解析】解:由x∈[2,5],=1+∈[2,5],若m≤2则f(x)=的最大值为5,符合题意;当2<m≤5时,f(x)的最大值为f(2)与f(5)中较大的,由f(2)=f(5),即|5-m|+m=|2-m|+m,解得m=,显然2<m≤时,f(x)的最大值为5,m>时,f(x)的最大值不为定值.综上可得m≤时,f(x)在[2,5]上的最大值是5,故选:A.求得x∈[2,5],=1+∈[2,5],讨论m的范围,结合f(2),f(5)可得所求范围.11.【答案】【解析】解:设幂函数为f(x)=xα,∵f(x)的图象经过点(8,2),∴f(8)=8α=2,即23α=2,则3α=,则α=,则f(x)=x=,则f()==,故答案为:根据幂函数的定义,利用待定系数法求出函数的解析式,然后代入求值即可.12.【答案】2【解析】解:∵f(1+)=,∴f(3)=f(1+)==2.故答案为:2.由f(1+)=,f(3)=f(1+),能求出结果.13.【答案】【解析】解:f(x)的定义域为R,且=,∴f(x)是奇函数,且f(x)在[0,+∞)上单调递增,∴f(x)在R上单调递增,由f(a-1)+f(2a2)≤0得,f(a-1)≤f(-2a2),∴a-1≤-2a2,解得,∴实数a的取值范围是.故答案为:.容易判断出f(x)是R上的奇函数,且单调递增,从而根据f(a-1)+f(2a2)≤0可得出a-1≤-2a2,解出a的范围即可.14.【答案】(-∞,7]【解析】解:∵函数f(x)=,先讨论f(a)的取值情况:①若f(a)≤0,则f2(a)+2f(a)≤3,解得,-3≤f(a)≤1,即-3≤f(a)≤0,②若f(a)>0,则-log2(f(a)+1)≤3,显然成立;则综上得,f(a)≥-3,再讨论a的取值情况:①若a≤0,则a2+2a≥-3,解得,a∈R,即a≤0.②若a>0,则-log2(a+1)≥-3,解得,0<a≤7,综上所述,实数a的取值范围是:(-∞,7].故答案为:(-∞,7].a的正负,求实数a的取值范围.15.【答案】(0,2)【解析】解:根据题意,在同一个坐标系中作出函数y=x-4和y=x2-4x+2λ的图象,如图:若函数f(x)恰有2个零点,即函数f(x)图象与x轴有且仅有2个交点,可得△=16-8λ≥0,λ≤2,当λ=2时,函数f(x)恰有1个零点,所以λ<2;y=x2-4x+2λ的对称轴为x=2,(0,0)与(4,0)关于x=2对称;所以f(0)>0,可得λ>0,f(0)≤0时,函数f(x)恰有3个不同的零点,即λ的取值范围是:(0,2)故答案为:(0,2).根据题意,在同一个坐标系中作出函数y=x-4和y=x2-4x+2λ的图象,结合图象分析可得答案.16.【答案】1【解析】解:设log2a=log5b=lg(a+b)=k,∴a=2k,b=5k,a+b=10k,∴ab=10k,∴a+b=ab,则=1.故答案为:1.设log2a=log5b=lg(a+b)=k,可得a=2k,b=5k,a+b=10k,可得a+b=ab.即可得出.17.【答案】解:(1)0.064-(-)0+16+0.25 =-1++=2.5-1+8+0.5=10;(2)lg25+lg2+()-log29×log32=lg5+lg2+-2(log23×lo g32)=1+-2=-.【解析】本题考查了指数幂和对数的运算的性质,属于基础题.(1)根据指数幂的运算性质计算即可;(2)根据对数的运算性质计算即可.18.【答案】解:由题意:集合A={x|x2-2x-3≤0,x∈R}={x|-1≤x≤3},B={x|x2-2mx+m2-4≤0,x∈R}={x|m-2≤x≤m+2},(1)∵A∩B={x|1≤x≤3},∴,解得:m=3,所以:A∩B={x|1≤x≤3}时,实数m的值为3;(2)∵B={x|m-2≤x≤m+2},B={x|m-2>x或m+2<x},∴∁R∵A⊆∁R B,∴m-2>3或m+2<-1,解得:m>5或m<-3.所以:A⊆∁R B时,实数m的取值范围是:(-∞,-3)∪(5,+∞).【解析】本题考查了集合的基本运算的运用求参数的问题,属于基础题.(1)求出B,A集合,根据集合的基本运算求解实数m的值;(2)求出根据集合B,求出∁R B,在A⊆∁R B,求实数m的取值范围.19.【答案】解:(Ⅰ)当b=-3时,f(x)=log2(4x-3•2x+2),由4x-3•2x+2>0,得2x>2或2x<1,∴x>1或x<0,∴f(x)的定义域为{x|x>1或x<0};(Ⅱ)对于任意x≥1,都有f(x)>g(x)成立,即4x+b•2x+2>2x,对任意x≥1恒成立,∴b>=,对任意x≥1恒成立,∴只需b>=-2,∴b的取值范围为[-2,+∞).【解析】(Ⅰ)将b=-3代入f(x)中,由4x-3•2x+2>0,解出x的范围;(Ⅱ)根据对于任意x≥1,都有f(x)>g(x)成立,可得b>对任意x≥1恒成立,因此只需b>=-2,从而得到b的取值范围.20.【答案】解:(1)由1->0,可得x<-1或x>1,∴f(x)的定义域为(-∞,-1)∪(1,+∞);∵f(x)=log a(1-)=log a(),且f(-x)=log a()=log a()=-log a()=-f(x);∴f(x)在定义域上为奇函数.(2)当0<a<1时,f(x)在(1,+∞)单调递减,任取x1,x2且1<x1<x2,f(x1)-f(x2)=-=log a();由(x1-1)(x2+1)-(x1+1)(x2-1)=2(x1-x2)<0,∴0<<1,又0<a<1,∴log a()>0则f(x1)>f(x2),∴f(x)在(1,+∞)单调递减;(3)假设存在这样的实数a,使得当f(x)的定义域为[m,n]时,值域为[1+loga n,1+logam];由0<m<n,又log a n+1<log a m+1,即log a n<log a m,∴0<a<1.由(2)知:f(x)在(1,+∞)单调递减,∴f(x)在(m,n)单调递减,∴,即m,n是方程log a=log a x+1的两个实根,即=ax在(1,+∞)上有两个互异实根;于是问题转化为关于x的方程ax2+(a-1)x+1=0在(1,+∞)上有两个不同的实数根,令g(x)=ax2+(a-1)x+1,则有,解得0<a<3-2;故存在实数a∈(0,3-2),使得当f(x)的定义域为[m,n]时,值域为[1+log a n,1+logam].【解析】(1)由1->0,可求出f(x)的定义域,利用定义法能求出f(x)在定义域上为奇函数.(2)当0<a<1时,f(x)在(1,+∞)单调递减,利用定义法能进行证明.(3)把f(x)的定义域为[m,n]时值域为[1+log a n,1+1og a m]转化为f(x)在(1,+∞)上为减函数,进一步得到=ax在(1,+∞)上有两个互异实根,令g(x)=ax2+(a-1)x+1,转化为关于a的不等式组求解.21.【答案】解:(Ⅰ)函数f(x)=x2-3|x-a|,若函数y=f(x)为偶函数,则f(-x)=f(x),即(-x)2-3|-x-a|=x2-3|x-a|,∴|x+a|=|x-a|,两边平方,得x2+2ax+a2=x2-2ax+a2,∴2ax=-2ax,∴4ax=0,∴a=0,∴实数a的值为0;(Ⅱ)若,则函数y=f(x)=x2-3|x-|=,画出函数f(x)的图象,如图所示;由图象知,单调减区间为(-∞,-],(,];(Ⅲ)不等式f(x-1)≤2f(x),化为(x-1)2-3|x-1-a|≤2x2-6|x-a|,即6|x-a|-3|x-1-a|≤x2+2x-1(*)对任意x∈[a,+∞)恒成立,①当0≤x≤a时,将不等式(*)可化为3a≤x2+5x+2,对0≤x≤a上恒成立,则g(x)=x2+5x+2 在(0,a]为单调递增,只需g(x)min=g(0)=2≥3a,解得0<a≤;②当a<x≤a+1时,将不等式(*)可化为9a≥-x2+7x-2,对a<x≤a+1上恒成立,由题意知h(x)=-x2+7x-2在x∈(a,a+1]上单调递增,则h(x)max=h(a+1)=-(a+1)2+7(a+1)-2≤9a,化简得a2+4a-4≥0,∴a≤-2-2或a≥-2+2;又0<a≤1,所以-2+2≤a≤1;③当x>a+1时,不等式(*)可化为3a≥-x2+x+4,则t(x)=-x2+x+4 在(a+1,+∞)为单调递减,则t(x)max=t(a+1)=-a2-a+4≤3a,解得a≤-2-2或a≥-2+2,又0<a≤1,所以-2+2≤a≤1;综上知,实数a的取值范围是(0,]∪[-2+2,1].【解析】(Ⅰ)根据偶函数的定义,化简整理,即可求得a的值;(Ⅱ)由分段函数的图象与性质,画出函数的图象,写出函数的单调区间;(Ⅲ)由题意可得,x∈[a,+∞)时,不等式恒成立,再分①当0≤x≤a时、②当x≥a+1、③当a<x<a+1时三种情况,分别求得a的取值范围,取交集即为所求.。

杭州学军中学2019学年上学期期中考试

杭州学军中学2019学年上学期期中考试

杭州学军中学2018学年上学期期中考试高二年级语文试卷一、语言基础(共19分,选择题每题3分)1.下列加点字的读音全都正确的一项是(3分)A.累.积(lěi)青睐.(lái)脚踵.(zhònɡ)殒.身不恤(yǔn)B.懊丧.(sànɡ)命中.(zhònɡ)精湛.(zhàn)外强中干.(ɡān)C.浩淼.(miǎo)藏.蓝(zànɡ)装盛.(shènɡ)强.颜欢笑(qiánɡ)D.倜.傥(tì)更.相(ɡènɡ)踌躇.(chú)流泉淙.淙(zōnɡ)2.下列句子中没有错别字的一项是(3分)A.我目睹中国女子的办事,是始于去年的,虽然是少数,但看那干炼坚决,百折不回的气慨,曾经屡次为之感叹。

B.我寻找爱情,最后是因为在爱情的结合中,我看到圣徒和诗人们所想象的天堂的神密的缩影。

C.这些作品讴歌真善美,抨击假丑恶,有鲜明的立场、健康的格调和浓烈的情感,他们震撼过你的心灵,令你难以忘怀。

D.想象着数十万年前的一只昆虫停歇于树枝之上,其貌不洋,不经意间从头顶落下一滴汁液,便永恒地凝固而成为琥珀。

3.下列加点的词语使用不正确的一项是(3分)A.这三首诗歌有异曲同工....之妙,同样的语言平白如话,情感却直率奔放,而又毫无矫揉作态之感,这样决绝的誓言,任你心似铁也要化为绕指柔吧。

B.这两年矿难不断发生的原因可能是多方面的,而有些领导对安全生产规程和规章制度的不屑一顾....,肯定是酿成安全生产事故的一个重要原因。

C.中新8月24日报道,国际奥委会主席罗格在致辞中宣布第29届奥林匹克运动会闭幕,并称本届奥运会是一届真正的无与伦比....的奥运会。

D.党的十七届四中全会强调必须居安思危....,增强忧患意识,加强防腐倡廉的工作,实际上也反映出党和国家领导人也认识到了腐败问题的严重性。

4.下列没有语病的一句是(3分)A.我们只有对一个问题的两方面的事实和论点加以充分地比较和叙述,才能得到良好的结果。

2019-2020学年浙江省杭州市西湖区学军中学高三第一学期期中考试试卷

2019-2020学年浙江省杭州市西湖区学军中学高三第一学期期中考试试卷

2019-2020学年浙江省杭州市西湖区学军中学高三第一学期期中考试试卷第一部分听力略第二部分阅读理解(共两节,计35分)第一节:(共10小题,每小题2.5分,计25分)阅读下列短文,从每题所给的四个选项(A、B、C和D)中选出最佳选项。

AA man who broke his neck outdoors in freezing conditions survived lying in snow for nearly 20 hours thanks to his dog, who kept him warm through the night and barked for help.The Michigan man, named only as Bob, was alone when he left his farmhouse on New Year's Eve to collect firewood. Anticipating a journey of only several meters. Bob was wearing just long johns, a shirt and slippers when he went outside, despite the temperature being around -4 degrees centigrade. However, he slipped and broke his neck."I was yelling for help but my nearest neighbor is about a quarter of a mile away and it was 10:30p.m., but my Kelsey came, "said Bob.Kelsey is Bob's five-year-old Golden Retriever. She kept Bob warm by lying on top of him, and kept him awake by licking his hands and face.Bob said, "She kept barking for help but never left my side. She kept me warm and awake. I knew I had to persevere (get) through this and that it was my choice to stay alive.""By morning my voice was gone and I couldn't yell for help, but Kelsey didn't stop barking.""She was letting out this screaming howl that got my neighbor's attention. He found me at 6:30 pm on New Year's Day."Bob's neighbor eventually discovered him after hearing Kelsey's howls and called the emergency services. When Bob arrived at the hospital, his body temperature was below 21 degrees centigrade. Normal body temperature is37 degrees centigrade and hypothermia(低体温症) occurs when the body drops below 35 degrees centigrade."I was surprised to find out that I didn't have any frost bite," said Bob, "I am sure it was because of Kelsey's determination to keep me warm and safe."And to the surprise of doctors, Bob made a quick initial recovery from his neck injury."I think his dog really kept him alive and helped him, and he was very fortunate," said Chaim Colen, MD, Neurosurgeon at McLaren."I am so thankful for my two heroes," Bob said. "Kelsey kept me warm, alert, and never stopped barking for help Dr.Colen saved my life. They are truly heroes and I will be forever grateful."1. What happened to Bob on New Year's Eve?A. He fell on some firewood.B. He slipped inside his house.C. He got lost while walking his dog.D. He was injured and couldn't move in snow.2. Which of the following did Kelsey not do to save Bob?A. She kept barking for help.B. She found the emergency services.C. She lay on top of him to keep him warm.D. She licked his hands and face to keep him alert.3. The best title for this passage isA. A New Year Eve's AccidentB. An Old Dog Is Still UsefulC. A Loyal Dog Saves Her Owner's Life.D. The Love between a Dog and Her Master【答案】:DBC【解答】:1. 细节理解题;根据第二段“However, he slipped and broke his neck. ”可知选D。

2019-2020学年杭州市学军中学高三英语上学期期中考试试卷及参考答案

2019-2020学年杭州市学军中学高三英语上学期期中考试试卷及参考答案

2019-2020学年杭州市学军中学高三英语上学期期中考试试卷及参考答案第一部分阅读(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,从每题所给的A、B、C、D四个选项中选出最佳选项AThe Origins of Famous BrandsOur lives are full of brand names and trademarked products that we use every day. Although many brand names are simple acronyms(首字母缩略词) or versions of their founders names, some of the companies we trust every day actually have fascinating and surprising back stories.StarbucksIt seems fitting that the most famous coffee brand in the world would take its name from one of the world’s greatest works of literature. The inspiration for the name of the coffeehouse came from Herman Melville’sMoby Dick. The founders’ original idea was to name the company after the Captain Ahab’s ship, but they eventually decided that Pequot wasn’t a great name for coffee, so they chose Ahab’s first mate, Starbucks, as the name instead.GoogleGoogle was originally called Backrub, for it searched for links in every corner of the Web. In 1997, when the founders of the company were searching for a new name showing a huge amount of data for their rapidly improving search technology, a friend suggested the word “googol”. When a friend tried to register the new domain (域) name, he misspelled “googol” as “google”.NikeOriginally founded as a distributor for Japanese running shoes, the company was originally named BRS, or Blue Ribbon Sports. In 1971, BRS introduced its own soccer shoe, a model called Nike, which is alsothe name for the Greek goddess of victory. In 1978, the company officially renamed itself as Nike, Inc.The right name is essential to a company’s success, and a great origin story is just as important as a great product. An attractive origin story is one more thing that keeps customers guessing, wondering, and buying its products.1. What is the name of the Captain Ahab’s ship?A. Moby Dick.B. Starbucks.C. Pequot.D. Herman Melville.2. Why did the founders of the Google want to change its name?A. They mistook their name.B. They wanted new customers.C. The company’s original name was too long.D. The company’s search technology was improving rapidly.3. Where does the importance of the origin story of one company lie in?A. It can change the company’s image.B. It can add myth to the company.C. It explains the development of the company to customers.D. It makes customers imagine and purchase its goods.BI started out in life with few advantages. I didn't graduate from high school. I worked at menial (不体面的) jobs. I had limited education, limited skills and a limited future.And then I began asking, "Why are some people more successful than others?" This question changed my life.Over the years, I have read thousands of books and articles on the subjects of success and achievement(成就). It seems that the reasons have been discussed and written about for more than two thousand years, in every possible way. One quality that most philosophers, teachers and experts agree on is the importance of self-discipline (自律). As Al Tomsik summarized it years ago, "Success is tons of discipline."Some years ago, I attended a conference in Washington. It was the lunch break and I was eating at a nearby food fair. The area was crowded and I sat down at the last open table by myself, even though it was a table for four.A few minutes later, an older gentleman and a younger woman who was his assistant came along carrying trays of food, obviously looking for a place to sit. With plenty of room at my table, I immediately invited the older gentleman to join me. He was hesitant (犹豫), but I insisted. Finally, thanking me as he sat down, we began to chat over lunch.It turned out that his name was Kop Kopmeyer. As it happened, I immediately knew who he was. He was a legend in the field of success and achievement. Kop Kopmeyer had written four large books, each of which contained 250 success principles that he had obtained from more than fifty years of research and study. I had read all four books from cover to cover, more than once.After we had chatted for a while, I asked him the question that many people in this situation would ask, "Of all the one thousand success principles that you have discovered, which do you think is the most important?”He smiled at me, as if he had been asked this question many times, and replied, without hesitating, "The most important success principle of all was stated by Thomas Huxley many years ago. He said, 'Do what you should do, when you should do it, whether you feel like it or not.'"He went on to say, "There are 999 other success principles that I have found in my reading and experience, but without self-discipline, none of them work."4. Why did the writer ask the question in Paragraph 2 ?A. Because he wasn't satisfied with himself.B. Because he was a person of self-discipline.C. Because he dislike those successful people.D. Because he wanted to share his idea on success.5. What made the writer invite the older gentleman to join him ?A. His great kindness.B. The gentleman's fame.C. His eagerness for success.D. The gentleman's habit.6. What are the four large books about ?A. Personal changesB. The secret of successC. Sayings of wisdomD. The gentleman's manners.7. What's the best title for the text ?A. The Magic of ReadingB. An Unexpected ConversationC. A Question that Changed MyLifeD. The Power of Self-disciplineCIn Australia, plenty of wild things can bite or sting(刺伤) you. Strangely enough, one of them is a tree. Now scientists have figured out what makes the tree’s sting so bad.The rainforests of eastern Australia are home to a stinging tree known as Dendrocnide. Many people callit the gympie-gympie tree—a name given to the tree by native Australians. It’s covered with sharp, needle-like hairs that carry poison. If you touch a gympie-gympie tree, you won’t forget it anytime soon. The pain can stay with you for hours, days or weeks. In some cases, it’s been reported to stay for months.Scientists have long looked for the source of this powerful sting. Now researchers at the University of Queensland have discovered what makes this stinging plant so painful. After carefully studying different kinds of gympie-gympie trees, the scientists were able to separate out different chemicals that the trees produce. This allowed them to identify a group of chemicals that they believed was responsible for the pain.The researchers created artificial versions of these chemicals, which they call “gympietides”. Sure enough, when the scientists injected mice with gympietides, the mice licked(舔) at the places where they’d been injected, indicating that they hurt in those places. When the scientists studied the way gympietides were built, they found that they formed a knot-like shape. The shape makes the chemicals very stable, which helps explain why the pain stays so long.The knot-like shape of the gympietides was similar to the shape of poisons produced by poisonous spiders and cone snails. The scientists were surprised to see three very different kinds of life all using similar poisons. Spiders and cone snails carry poisons because they catch food by stinging other creatures. It’s not clear how stinging helps the gympie-gympie tree.Though the tree’s sting may stop some animals from eating it, it doesn’t stop all animals. Beetles and pademelons (small s of the kangaroo) are able to eat the plant without trouble.8. Why is a touch on the stinging tree unforgettable?A. Because it has so unusual an appearance.B. Because it is extremely rare in existence.C. Because touching it creates a quite strange feeling.D. Because the pain caused by it doesn’t go away quickly.9. What do scientists fail to find out about the stinging tree?A. How it produces poisons.B. What poisons it produces.C. How it benefits from the sting.D. The consequences of its sting.10. What does the text imply about the stinging tree?A. It produces the same poisons as spiders.B. Poisonous as it is, it also has natural enemies.C. Animals are wise enough to stay away from it.D. Only one chemical in it causes pain to the toucher.11. What’s the best title for the text?A. Scientists Discover Stinging Tree's SecretB. Caution: Stinging Tree Can Bite and Poison YouC. Scientists Discover a Strange Species in AustraliaD. Effective Ways to Avoid Being Hurt by Stinging TreeDI started out in life with few advantages. I didn't graduate from high school. I worked at menial (不体面的) jobs. I had limited education, limited skills and a limited future.And then I began asking, "Why are some people more successful than others?" This question changed my life.Over the years, I have read thousands of books and articles on the subjects of success and achievement(成就). It seems that the reasons have been discussed and written about for more than two thousand years, in every possible way. One quality that most philosophers, teachers and experts agree on is the importance of self-discipline (自律). As Al Tomsik summarized it years ago, "Success is tons of discipline."Some years ago, I attended a conference in Washington. It was the lunch break and I was eating at a nearby food fair. The area was crowded and I sat down at the last open table by myself, even though it was a table for four.A few minutes later, an older gentleman and a younger woman who was his assistant came along carrying trays of food, obviously looking for a place to sit. With plenty of room at my table, I immediately invited the older gentleman to join me. He was hesitant (犹豫), but I insisted. Finally, thanking me as he sat down, we began to chat over lunch.It turned out that his name was Kop Kopmeyer. As it happened, I immediately knew who he was. He was a legend in the field of success and achievement. Kop Kopmeyer had written four large books, each of which contained 250 success principles that he had obtained from more than fifty years of research and study. I had read all four books from cover to cover, more than once.After we had chatted for a while, I asked him the question that many people in this situation would ask, "Of all the one thousand success principles that you have discovered, which do you think is the most important?”He smiled at me, as if he had been asked this question many times, and replied, without hesitating, "The most important success principle of all was stated by Thomas Huxley many years ago. He said, 'Do what you should do, when you should do it, whether you feel like it or not.'"He went on to say, "There are 999 other success principles that I have found in my reading and experience, but without self-discipline, none of them work."12. Why did the writer ask the question in Paragraph 2 ?A. Because he wasn't satisfied with himself.B. Because he was a person of self-discipline.C. Because he dislike those successful people.D. Because he wanted to share his idea on success.13. What made the writer invite the older gentleman to join him ?A. His great kindness.B. The gentleman's fame.C. His eagerness for success.D. The gentleman's habit.14. What are the four large books about ?A. Personal changesB. The secret of successC. Sayings of wisdomD. The gentleman's manners.15. What's the best title for the text ?A. The Magic of ReadingB. An Unexpected ConversationC. A Question that Changed MyLifeD. The Power of Self-discipline第二节(共5小题;每小题2分,满分10分)阅读下面短文,从短文后的选项中选出可以填入空白处的最佳选项。

浙江省杭州市西湖区学军中学2019-2020学年高一上学期期中物理试卷-含答案解析

浙江省杭州市西湖区学军中学2019-2020学年高一上学期期中物理试卷-含答案解析

精品文档-可编辑浙江省杭州市西湖区学军中学2021-2020学年高一上学期期中物理试卷一、单选题(本大题共13小题,共39.0分)1.下列哪些现象中的物体可视为质点()A.研究“神十一”飞船绕地球飞行的高度B.地面上放置一个木箱,在上面的棱角处用水平力推,研究它是滑动还是翻转的问题C.研究汽车在行驶过程中汽车中齿轮的转动问题D.研究“神十一”与空间站的对接问题2.下列说法正确的是()A.地球对物体的万有引力就是物体所受重力B.将物体向上搬运要比向下搬运更费劲,是因为物体向上运动时所受的重力比较大C.形状固定的物体其重心相对于自身位置不会因为物体的运动而发生变化D.静止在水平地面上的物体所受的重力即是物体对地面的正压力3.关于摩擦力,下列说法正确的是()A.运动的物体只可能受到滑动摩擦力B.滑动摩擦力的方向不可能和运动方向一致C.若接触面是曲面,滑动摩擦力与接触面上的压力不垂直D.物体间的接触面有压力不一定有摩擦力,有摩擦力则一定有压力4.如表是四种交通工具的速度改变情况,下列说法正确的是()①②的速度变化最慢C.③的速度变化最快D.④的末速度最大,但加速度最小5.如图所示,水桶保持静止,两位同学对水桶拉力分别为F 1、F 2,则F 1、F 2的合力方向为()A.竖直向上B.竖直向下C.沿F 1方向D.沿F 2方向6.某列火车在一段长30k m的笔直铁轨上行驶,行驶的平均速度是60km/ℎ,下列说法正确的是()A.这列火车如果行驶60k m的路程一定需要1hB.这列火车一定以60km/ℎ的速度在这段铁轨上行驶C.这列火车通过这段铁轨的时间是0.5ℎD.60km/ℎ是火车在这一路段中的最高速度7.如图所示,一质量为m的物块恰好静止在倾角为θ、质量为M的斜劈上.物块与斜劈和斜劈与地面间的动摩擦因数均为μ.现对物块施加一个水平向右的恒力F,如果物块和斜劈都仍处于静止状态(最大静摩擦力等于滑动摩擦力),则()A.物块受到的合外力增大B.地面对斜劈的摩擦力可能减小C.水平恒力F不可能大于mgsinθ+μmgcosθcosθ−μsinθD.水平恒力F可能大于μ(m+M)g8.自由落体运动的物体,先后经过空中M、N两点时的速度分别是v1和v2,则下列说法正确的是()A.M N的间距为v22−v12g B.经过M N的平均速度为v2−v12C.经过M N所需时间为v2−v12g D.经过M N中点时速度为√v12+v2229.《龟兔赛跑》是大家耳熟能详的寓言故事,它告诉我们勤能补拙的道理。

浙江省杭州市学军中学2018_2019学年高一数学上学期期中试题(含解析)

浙江省杭州市学军中学2018_2019学年高一数学上学期期中试题(含解析)

浙江省杭州市学军中学2018-2019学年高一上学期期中考试数学试题一、选择题(本大题共10小题,共30.0分)1.已知集合,,则()A. B. C. D.【答案】B【解析】试题分析:由题意知,故选B.【考点定位】本题考查集合的基本运算,属于容易题.2.函数f(x)=ln(1-x2)的定义域为( )A. B. C. D.【答案】B【解析】【分析】由根式内部的代数式大于等于0,对数式的真数大于0联立不等式组求解.【详解】由,得0≤x<1.∴函数f(x)ln(1﹣x2)的定义域为[0,1).故选:B.【点睛】本题考查函数的定义域及其求法,是基础题.3.已知函数f(x)=,则f[f()]等于( )A. B. C. D.【答案】D【解析】【分析】推导出f(),从而f[f()]=f(),由此能求出结果.【详解】∵函数f(x),∴f(),f[f()]=f().故选:D.【点睛】本题考查函数值的求法,考查函数性质等基础知识,考查运算求解能力,是基础题.4.使函数f(x)=x a的定义域为R且为奇函数的α的值可以是( )A. B. C. 3 D. 以上都不对【答案】C【解析】【分析】根据题意,结合幂函数的性质依次分析选项,综合即可得答案.【详解】根据题意,依次分析选项:对于A,α=﹣1时,f(x)=x﹣1,其定义域不是R,不符合题意;对于B,α时,f(x),其定义域不是R,不符合题意;对于C,α=3时,f(x)=x3,其定义域为R且为奇函数,符合题意;对于D,错误,故选:C.【点睛】本题考查幂函数的性质,关键是掌握幂函数的性质,属于基础题.5.已知集合M,N,P为全集U的子集,且满足M⊆P⊆N,则下列结论不正确的是( )A. ∁U N⊆∁U PB. ∁N P⊆∁N MC. (∁U P)∩M=∅D. (∁U M)∩N=∅【答案】D【解析】因为P⊆N,所以∁U N⊆∁U P,故A正确;因为M⊆P,所以∁N P⊆∁N M,故B正确;因为M⊆P,所以(∁U P)∩M=∅,故C正确;因为M⊆ N,所以(∁U M)∩N∅.故D不正确.故选D.6.设函数f(x)=log a x(a>0,a≠1),若f(x1x2…x2018)=4,则f(x12)+f(x12)+…+f(x20182)的值等于( )A. 4B. 8C. 16D.【答案】B【解析】【分析】由函数的解析式结合对数的运算性质即可得解.【详解】∵函数f(x)=log a x(a>0,a≠1),f(x1x2…x2018)=4,∴f(x1x2…x2018)=log a(x1x2…x2018)=4,∴f(x12)+f(x12)+…+f(x20182)=log a(x1x2…x2018)2=2log a(x1x2…x2018)=2×4=8.故选:B.【点睛】本题考查函数值的求法,考查对数性质、运算法则等基础知识,考查运算求解能力,是基础题.7.设A={x|2≤x≤4},B={x|2a≤x≤a+3},若B真包含于A,则实数a的取值范围是( )A. B. C. D.【答案】C【解析】【分析】由B真包含于A,讨论B=∅与B≠∅时,求出a的取值范围.【详解】∵A={x|2≤x≤4},B={x|2a≤x≤a+3},且B真包含于A;当B=∅时,2a>a+3,解得a>3;当B≠∅时,解得a=1;此时A=B.∴a的取值范围是{a|a>3}故选:C.【点睛】本题考查了集合之间的基本运算,解题时容易忽略B=∅的情况,是易错题.8.函数f(x)=log2(-x2+ax+3)在(2,4)是单调递减的,则a的范围是( )A. B. C. D.【答案】B【解析】【分析】由复合函数的单调性可知内层函数在(2,4)上为减函数,则需要其对称轴小于等于2且当函数在x=4时的函数值大于等于0,由此联立不等式组得答案.【详解】令t=﹣x2+ax+3,则原函数化为y=log2t,∵y=log2t为增函数,∴t=﹣x2+ax+3在(2,4)是单调递减,对称轴为x,∴且﹣42+4a+3≥0,解得:.∴a的范围是[,4].故选:B.【点睛】本题考查了复合函数的单调性,复合函数的单调性满足同增异减的原则,是中档题.9.对于函数f(x),若∀a,b,c∈R,f(a),f(b),f(c)为某一三角形的三边长,则称f(x)为“可构造三角形函数”.已知函数f(x)=是“可构造三角形函数”,则实数t 的取值范围是( )A. B. C. D.【答案】A【解析】【分析】因对任意实数a、b、c,都存在以f(a)、f(b)、f(c)为三边长的三角形,则f(a)+f(b)>f(c)恒成立,将f(x)解析式用分离常数法变形,由均值不等式可得分母的取值范围,整个式子的取值范围由t﹣1的符号决定,故分为三类讨论,根据函数的单调性求出函数的值域,然后讨论k转化为f(a)+f(b)的最小值与f(c)的最大值的不等式,进而求出实数k的取值范围.【详解】由题意可得f(a)+f(b)>f(c)对于∀a,b,c∈R都恒成立,由于f(x)1,①当t﹣1=0,f(x)=1,此时,f(a),f(b),f(c)都为1,构成一个等边三角形的三边长,满足条件.②当t﹣1>0,f(x)在R上是减函数,1<f(a)<1+t﹣1=t,同理1<f(b)<t,1<f(c)<t,故f(a)+f(b)>2.再由f(a)+f(b)>f(c)恒成立,可得2≥t,结合大前提t﹣1>0,解得1<t≤2.③当t﹣1<0,f(x)在R上是增函数,t<f(a)<1,同理t<f(b)<1,t<f(c)<1,由f(a)+f(b)>f(c),可得 2t≥1,解得1>t.综上可得,t≤2,故选:A.【点睛】本题主要考查了求参数的取值范围,以及构成三角形的条件和利用函数的单调性求函数的值域,同时考查了分类讨论的思想,属于难题.10.设函数,其中表示中的最小者.下列说法错误的()A. 函数为偶函数B. 若时,有C. 若时,D. 若时,【答案】D【解析】【分析】先根据定义作的图像,然后依据图像逐个检验即可.【详解】在同一坐标系中画出的图像(如图所示),故的图像为图所示.的图像关于轴对称,故为偶函数,故A正确.由图可知时,有,故B成立.从图像上看,当时,有成立,令,则,故,故C成立.取,则,,,故D不成立.综上,选D.【点睛】一般地,若(其中表示中的较小者),则的图像是由这两个函数的图像的较低部分构成的.二、填空题(本大题共7小题,共28.0分)11.若,则.【答案】10【解析】试题分析:若,则考点:对数与对数函数12.已知,则________.【答案】【解析】【分析】利用配凑法求函数的解析式.【详解】(配凑法) (1),又∈(-∞,-2]∪[2,+∞),∴.故答案为:【点睛】本题考查函数解析式的求解及常用方法,是基础题.解题时要认真审题,仔细解答.13.已知f(x)=的定义域为R,则实数a的取值范围是______.【答案】[-1,0]【解析】【分析】把f(x)的定义域为R转化为0对任意x∈R恒成立,即x2+2ax﹣a≥0对任意x∈R恒成立,再由判别式小于等于0求解.【详解】∵f(x)的定义域为R,∴0对任意x∈R恒成立,即恒成立,即x2+2ax﹣a≥0对任意x∈R恒成立,∴△=4a2+4a≤0,则﹣1≤a≤0.故答案为:[﹣1,0].【点睛】本题考查函数的定义域及其求法,考查数学转化思想方法,是中档题.14.设max{a,b}表示a,b两数中的最大值,若f(x)=max{|x|,|x-t|}关于x=1对称,则t=______.【答案】2【解析】【分析】利用函数y=|x|的图象和函数y=|x﹣t|的图象关于直线x对称,从而得出结论.【详解】f(x)=max{|x|,|x﹣t|},由函数y=|x|的图象关于x=0对称,函数y=|x﹣t|的图象关于x=t对称,即有函数f(x)的图象关于x对称,f(x)=max{|x|,|x﹣t|}关于x=1对称,即有1,求得t=2,故答案为:2.【点睛】本题主要考查分段函数的应用,考查函数的对称性,属于基础题.15.设方程x2-mx+2=0的两根α,β,其中α∈(1,2),则实数m的取值范围是______.【答案】(2,4)【解析】【分析】由题意利用韦达定理,不等式的性质,求出实数m的取值范围.【详解】∵方程x2﹣mx+2=0的两根α,β,∴△=m2﹣8≥0,求得m≥2,或m≤﹣2①.由α•β=2,则,则,则②.由①②可得,故答案为:.【点睛】本题主要考查韦达定理,不等式的性质,属于基础题.16.已知lg2≈0.3010,则22018是______位数.【答案】608【解析】【分析】设x=22018,可得lgx=2018lg2≈607.418,即可得出.【详解】设x=22018,则lgx=2018lg2≈2018×0.3010=607.418,∴22018是608位数.故答案为:608.【点睛】本题考查了对数运算性质,考查了推理能力与计算能力,属于基础题.17.已知函数f(x)满足对任意的m,n都有f(m+n)=f(m)+f(n)-1,设g(x)=f(x)+(a>0,a≠1),g(ln2018)=-2015,则g(ln)=______.【答案】2018【解析】【分析】由已知中函数f(x)满足对任意实数m,n,都有f(m+n)=f(m)+f(n)﹣1,可得f(0)=1,进而f(x)+f(﹣x)=2,g(x)+g(﹣x)=3,结合g(ln2018)=﹣2015,由对数的运算性质计算可得所求值.【详解】∵函数f(x)满足对任意实数m,n,都有f(m+n)=f(m)+f(n)﹣1,令m=n=0,则f(0)=2f(0)﹣1,解得f(0)=1,令m=x,n=﹣x,则f(0)=f(x)+f(﹣x)﹣1,即f(x)+f(﹣x)=2,∵g(x)=f(x)(a>0,a≠0),∴g(﹣x)=f(﹣x)f(﹣x),故g(x)+g(﹣x)=f(x)+f(﹣x)+1=3,∴g(ln2018)+g(ln)=﹣2015+g(﹣ln2018)=3,即g(ln)=2018,故答案为:2018.【点睛】本题主要考查抽象函数的应用,根据条件建立方程关系是解决本题的关键,属于中档题.三、解答题(本大题共5小题,共62.0分)18.已知集合U=R,集合A={x|x2-(a-2)x-2a≥0},B={x|1≤x≤2}.(1)当a=1时,求A∩B;(2)若A∪B=A,求实数a的取值范围.【答案】(1){x|1≤x≤2};(2){a|a≤1}.【解析】【分析】(1)代入a的值,求出集合A,从而求出A∩B;(2)由A与B的并集为A,得到B为A的子集,表示出A的中不等式的解集,根据数轴确定出满足题意a的范围即可.【详解】(1)a=1时,A={x|x≥1或x≤-2},故A∩B={x|1≤x≤2};(2)∵A∪B=A,∴B⊆A,由x2-(a-2)x-2a≥0,得(x+2)(x-a)≥0,当a<-2时,如数轴表示,符合题意;同理,当-2≤a≤1,也合题意;但当a>1时,不合题意,综上可知{a|a≤1}.【点睛】本题考查了并集及其运算,熟练掌握并集的定义是解本题的关键.19.设函数f(x)=++.(1)设t=+,求t的取值范围;(2)求f(x)的最大值.【答案】(1)[,2];(2)3.【解析】【分析】(1)将t,﹣1≤x≤1,两边平方,结合二次函数的最值,即可得到所求范围;(2)由(1)可得g(t)=f(x)(t+1)2,考虑对称轴t=﹣1与区间[,2]的关系,即可得到所求最大值.【详解】(1)t=+,-1≤x≤1,可得t2=2+2,由0≤1-x2≤1,可得t2∈[2,4],由t≥0可得t的取值范围是[,2];(2)由(1)可得g(t)=f(x)=t+=(t+1)2-,由[,2]在对称轴t=-1的右边,为增区间,即有t=2,即x=0,g(t)取得最大值,且为3,即f(x)的最大值为3.【点睛】本题考查函数的最值求法,注意运用换元法和二次函数的最值求法,考查运算能力,属于中档题.20.已知函数f(x)=x+(a>0).(1)判断f(x)的奇偶性;(2)判断函数f(x)在(,+∞)上的单调性,并用定义证明.【答案】(1)见解析;(2)见解析.【解析】【分析】(1)求出函数的定义域,得到f(﹣x)=﹣f(x),判断函数的奇偶性即可;(2)根据单调性的定义证明即可.【详解】(1)f(x)的定义域是{x|x≠0},f(-x)=-x-=-(x+)=-f(x),故函数f(x)是奇函数;(2)函数在(,+∞)递增,令<m<n,则f(m)-f(n)=m+-n-=(m-n)+a•=(m-n)(1-),∵<m<n,∴m-n<0,1->0,故f(m)-f(n)<0,故f(x)在(,+∞)上递增.【点睛】本题考查了函数的奇偶性,单调性问题,考查转化思想,是一道基础题.21.已知函数f(x)=2x,g(x)=-x2+2x+b.(1)若f(x)++1≥0对任意的x∈[1,3]恒成立,求m的取值范围;(2)若x1,x2∈[1,3],对任意的x1,总存在x2,使得f(x1)=g(x2),求b的取值范围.【答案】(1)[-6,-∞);(2)见解析.【解析】【分析】(1)根据h(x)=f(x)1,结合勾函数的性质对任意的x∈[1,3]恒成立,即可求解m的取值范围;(2)根据对任意的x1,总存在x2,使得f(x1)=g(x2),可得f(x)的值域是g(x)的值域的子集,即可求解b的范围;【详解】(1)函数f(x)=2x,令h(x)=f(x)++1=;①当m=0时,可得h(x)=2x+1在x∈[1,3]恒成立;②当m<0时,可知f(x)=2x是递增函数,y=在x∈[1,3]也是递增函数,∴h(x)在x∈[1,3]是递增函数,此时h(x)min=h(1)=≥0,可得:-6≤m<0;③当m>0时,,所以函数h(x)=,满足题意.综上所述:f(x)++1≥0对任意的x∈[1,3]恒成立,可得m的取值范围是[-6,-∞);(2)由函数f(x)=2x,x∈[1,3],可得:2≤f(x)≤8;由g(x)=-x2+2x+b.其对称x=1,开口向下.∵x∈[1,3],∴g(x)在x∈[1,3]上单调递减.g(x)max=g(1)=1+b;g(x)min=g(3)=-3+b;∵对任意的x1,总存在x2,使得f(x1)=g(x2),∴f(x)的值域是g(x)的值域的子集;即,解得:无解.故x1,x2∈[1,3],对任意的x1,总存在x2,使得f(x1)=g(x2),此是b的取值范围是空集.【点睛】本题主要考查了函数恒成立问题的求解,分类讨论以及转化思想的应用,二次函数的最值以及单调性的应用,属于中档题.22.已知a∈R,f(x)=log2(1+ax).(1)求f(x2)的值域;(2)若关于x的方程f(x)-log2[(a-4)x2+(2a-5)x]=0的解集恰有一个元素,求实数a 的取值范围;(3)当a>0时,对任意的t∈(,+∞),f(x2)在[t,t+1]的最大值与最小值的差不超过4,求a的取值范围.【答案】(1)当a≥0时,值域为[0,+∞),当a<0时,值域为(-∞,0);(2)1<a≤2,或a>4;(3)(0,+∞).【解析】【分析】(1)讨论a≥0时,a<0时,由对数函数的单调性可得值域;(2)根据对数的运算法则进行化简,转化为一元二次方程,讨论a的取值范围进行求解即可;(3)根据条件得到g(x)=log2(1+ax2),a>0,函数g(x)在区间[t,t+1]上单调递增,g(t+1)﹣g(t)≤4,运用对数函数的单调性和参数分离进行求解即可.【详解】(1)f(x)=log2(1+ax),可得f(x2)=log2(1+ax2),当a≥0时,1+ax2≥1,即有log2(1+ax2)≥0;当a<0时,0<1+ax21,即有log2(1+ax2)0;即有当a≥0时,f(x)的值域为[0,+∞);当a<0时,f(x)的值域为(-∞,0];(2)由f(x)-log2[(a-4)x2+(2a-5)x]=0得log2(1+ax)=log2[(a-4)x2+(2a-5)x],即1+ax=(a-4)x2+(2a-5)x>0,①则(a-4)x2+(a-5)x-1=0,即(x+1)[(a-4)x-1]=0,②,当a=4时,方程②的解为x=-1,代入①,不成立;当a=3时,方程②的解为x=-1,代入①,不成立;当a≠4且a≠3时,方程②的解为x=-1或x=,若x=-1是方程①的解,则1-a=-a+1>0,即a<1,若x=是方程①的解,则1+=>0,即a>4或a<2,则要使方程①有且仅有一个解,则a>4或1≤a<2.综上,若方程f(x)-log2[(a-4)x2+(2a-5)x]=0的解集中恰好有一个元素,则a的取值范围是1<a≤2,或a>4;(3)当a>0时,对任意的t∈(,+∞),f(x2)=log2(1+ax2),设g(x)=log2(1+ax2),a>0,函数g(x)在区间[t,t+1]上单调递增,由题意得g(t+1)-g(t)≤4,即log2(1+at2+2at+a)-log2(1+at2)≤4,即1+at2+2at+a≤16(1+at2),即有a(15t2-2t-1)+15=a(3t-1)(5t+1)+15>0恒成立,综上可得a的范围是(0,+∞).【点睛】本题考查函数最值的求解,以及对数不等式的应用,考查对数函数的单调性,属于中档题.。

浙江省杭州市学军中学2018_2019学年高一数学上学期期中试题(含解析)

浙江省杭州市学军中学2018_2019学年高一数学上学期期中试题(含解析)

浙江省杭州市学军中学2018-2019学年高一上学期期中考试数学试题一、选择题(本大题共10小题,共30.0分)1.已知集合,,则()A. B. C. D.【答案】B【解析】试题分析:由题意知,故选B.【考点定位】本题考查集合的基本运算,属于容易题.2.函数f(x)=ln(1-x2)的定义域为()A. B. C. D.【答案】B【解析】【分析】由根式内部的代数式大于等于0,对数式的真数大于0联立不等式组求解.【详解】由,得0≤x<1.∴函数f(x)ln(1﹣x2)的定义域为[0,1).故选:B.【点睛】本题考查函数的定义域及其求法,是基础题.3.已知函数f(x)=,则f[f()]等于()A. B. C. D.【答案】D【解析】【分析】推导出f(),从而f[f()]=f(),由此能求出结果.【详解】∵函数f(x),∴f(),f[f()]=f().故选:D.【点睛】本题考查函数值的求法,考查函数性质等基础知识,考查运算求解能力,是基础题.4.使函数f(x)=x a的定义域为R且为奇函数的α的值可以是()A. B. C. 3 D. 以上都不对【答案】C【解析】【分析】根据题意,结合幂函数的性质依次分析选项,综合即可得答案.【详解】根据题意,依次分析选项:对于A,α=﹣1时,f(x)=x﹣1,其定义域不是R,不符合题意;对于B,α时,f(x),其定义域不是R,不符合题意;对于C,α=3时,f(x)=x3,其定义域为R且为奇函数,符合题意;对于D,错误,故选:C.【点睛】本题考查幂函数的性质,关键是掌握幂函数的性质,属于基础题.5.已知集合M,N,P为全集U的子集,且满足M⊆P⊆N,则下列结论不正确的是( )A. ∁U N⊆∁U PB. ∁N P⊆∁N MC. (∁U P)∩M=∅D. (∁U M)∩N=∅【答案】D【解析】因为P⊆N,所以∁U N⊆∁U P,故A正确;因为M⊆P,所以∁N P⊆∁N M,故B正确;因为M⊆P,所以(∁U P)∩M=∅,故C正确;因为M⊆ N,所以(∁U M)∩N∅.故D不正确.故选D.6.设函数f(x)=log a x(a>0,a≠1),若f(x1x2…x2018)=4,则f(x12)+f(x12)+…+f(x20182)的值等于()A. 4B. 8C. 16D.【答案】B【解析】【分析】由函数的解析式结合对数的运算性质即可得解.【详解】∵函数f(x)=log a x(a>0,a≠1),f(x1x2…x2018)=4,∴f(x1x2…x2018)=log a(x1x2…x2018)=4,∴f(x12)+f(x12)+…+f(x20182)=log a(x1x2…x2018)2=2log a(x1x2…x2018)=2×4=8.故选:B.【点睛】本题考查函数值的求法,考查对数性质、运算法则等基础知识,考查运算求解能力,是基础题.7.设A={x|2≤x≤4},B={x|2a≤x≤a+3},若B真包含于A,则实数a的取值范围是()A. B. C. D.【答案】C【解析】【分析】由B真包含于A,讨论B=∅与B≠∅时,求出a的取值范围.【详解】∵A={x|2≤x≤4},B={x|2a≤x≤a+3},且B真包含于A;当B=∅时,2a>a+3,解得a>3;当B≠∅时,解得a=1;此时A=B.∴a的取值范围是{a|a>3}故选:C.【点睛】本题考查了集合之间的基本运算,解题时容易忽略B=∅的情况,是易错题.8.函数f(x)=log2(-x2+ax+3)在(2,4)是单调递减的,则a的范围是()A. B. C. D.【答案】B【解析】【分析】由复合函数的单调性可知内层函数在(2,4)上为减函数,则需要其对称轴小于等于2且当函数在x=4时的函数值大于等于0,由此联立不等式组得答案.【详解】令t=﹣x2+ax+3,则原函数化为y=log2t,∵y=log2t为增函数,∴t=﹣x2+ax+3在(2,4)是单调递减,对称轴为x,∴且﹣42+4a+3≥0,解得:.∴a的范围是[,4].故选:B.【点睛】本题考查了复合函数的单调性,复合函数的单调性满足同增异减的原则,是中档题.9.对于函数f(x),若∀a,b,c∈R,f(a),f(b),f(c)为某一三角形的三边长,则称f (x)为“可构造三角形函数”.已知函数f(x)=是“可构造三角形函数”,则实数t 的取值范围是()A. B. C. D.【答案】A【解析】【分析】因对任意实数a、b、c,都存在以f(a)、f(b)、f(c)为三边长的三角形,则f(a)+f(b)>f(c)恒成立,将f(x)解析式用分离常数法变形,由均值不等式可得分母的取值范围,整个式子的取值范围由t﹣1的符号决定,故分为三类讨论,根据函数的单调性求出函数的值域,然后讨论k转化为f(a)+f(b)的最小值与f(c)的最大值的不等式,进而求出实数k 的取值范围.【详解】由题意可得f(a)+f(b)>f(c)对于∀a,b,c∈R都恒成立,由于f(x)1,①当t﹣1=0,f(x)=1,此时,f(a),f(b),f(c)都为1,构成一个等边三角形的三边长,满足条件.②当t﹣1>0,f(x)在R上是减函数,1<f(a)<1+t﹣1=t,同理1<f(b)<t,1<f(c)<t,故f(a)+f(b)>2.再由f(a)+f(b)>f(c)恒成立,可得2≥t,结合大前提t﹣1>0,解得1<t≤2.③当t﹣1<0,f(x)在R上是增函数,t<f(a)<1,同理t<f(b)<1,t<f(c)<1,由f(a)+f(b)>f(c),可得 2t≥1,解得1>t.综上可得,t≤2,故选:A.【点睛】本题主要考查了求参数的取值范围,以及构成三角形的条件和利用函数的单调性求函数的值域,同时考查了分类讨论的思想,属于难题.10.设函数,其中表示中的最小者.下列说法错误的()A. 函数为偶函数B. 若时,有C. 若时,D. 若时,【答案】D【解析】【分析】先根据定义作的图像,然后依据图像逐个检验即可.【详解】在同一坐标系中画出的图像(如图所示),故的图像为图所示.的图像关于轴对称,故为偶函数,故A正确.由图可知时,有,故B成立.从图像上看,当时,有成立,令,则,故,故C成立.取,则,,,故D不成立.综上,选D.【点睛】一般地,若(其中表示中的较小者),则的图像是由这两个函数的图像的较低部分构成的.二、填空题(本大题共7小题,共28.0分)11.若,则.【答案】10【解析】试题分析:若,则考点:对数与对数函数12.已知,则________.【答案】【解析】【分析】利用配凑法求函数的解析式.【详解】(配凑法)(1),又∈(-∞,-2]∪[2,+∞),∴.故答案为:【点睛】本题考查函数解析式的求解及常用方法,是基础题.解题时要认真审题,仔细解答.13.已知f(x)=的定义域为R,则实数a的取值范围是______.【答案】[-1,0]【解析】【分析】把f(x)的定义域为R转化为0对任意x∈R恒成立,即x2+2ax ﹣a≥0对任意x∈R恒成立,再由判别式小于等于0求解.【详解】∵f(x)的定义域为R,∴0对任意x∈R恒成立,即恒成立,即x2+2ax﹣a≥0对任意x∈R恒成立,∴△=4a2+4a≤0,则﹣1≤a≤0.故答案为:[﹣1,0].【点睛】本题考查函数的定义域及其求法,考查数学转化思想方法,是中档题.14.设max{a,b}表示a,b两数中的最大值,若f(x)=max{|x|,|x-t|}关于x=1对称,则t=______.【答案】2【解析】【分析】利用函数y=|x|的图象和函数y=|x﹣t|的图象关于直线x对称,从而得出结论.【详解】f(x)=max{|x|,|x﹣t|},由函数y=|x|的图象关于x=0对称,函数y=|x﹣t|的图象关于x=t对称,即有函数f(x)的图象关于x对称,f(x)=max{|x|,|x﹣t|}关于x=1对称,即有1,求得t=2,故答案为:2.【点睛】本题主要考查分段函数的应用,考查函数的对称性,属于基础题.15.设方程x2-mx+2=0的两根α,β,其中α∈(1,2),则实数m的取值范围是______.【答案】(2,4)【解析】【分析】由题意利用韦达定理,不等式的性质,求出实数m的取值范围.【详解】∵方程x2﹣mx+2=0的两根α,β,∴△=m2﹣8≥0,求得m≥2,或m≤﹣2①.由α•β=2,则,则,则②.由①②可得,故答案为:.【点睛】本题主要考查韦达定理,不等式的性质,属于基础题.16.已知lg2≈0.3010,则22018是______位数.【答案】608【解析】【分析】设x=22018,可得lgx=2018lg2≈607.418,即可得出.【详解】设x=22018,则lgx=2018lg2≈2018×0.3010=607.418,∴22018是608位数.故答案为:608.【点睛】本题考查了对数运算性质,考查了推理能力与计算能力,属于基础题.17.已知函数f(x)满足对任意的m,n都有f(m+n)=f(m)+f(n)-1,设g(x)=f(x)+(a>0,a≠1),g(ln2018)=-2015,则g(ln)=______.【答案】2018【解析】【分析】由已知中函数f(x)满足对任意实数m,n,都有f(m+n)=f(m)+f(n)﹣1,可得f(0)=1,进而f(x)+f(﹣x)=2,g(x)+g(﹣x)=3,结合g(ln2018)=﹣2015,由对数的运算性质计算可得所求值.【详解】∵函数f(x)满足对任意实数m,n,都有f(m+n)=f(m)+f(n)﹣1,令m=n=0,则f(0)=2f(0)﹣1,解得f(0)=1,令m=x,n=﹣x,则f(0)=f(x)+f(﹣x)﹣1,即f(x)+f(﹣x)=2,∵g(x)=f(x)(a>0,a≠0),∴g(﹣x)=f(﹣x)f(﹣x),故g(x)+g(﹣x)=f(x)+f(﹣x)+1=3,∴g(ln2018)+g(ln)=﹣2015+g(﹣ln2018)=3,即g(ln)=2018,故答案为:2018.【点睛】本题主要考查抽象函数的应用,根据条件建立方程关系是解决本题的关键,属于中档题.三、解答题(本大题共5小题,共62.0分)18.已知集合U=R,集合A={x|x2-(a-2)x-2a≥0},B={x|1≤x≤2}.(1)当a=1时,求A∩B;(2)若A∪B=A,求实数a的取值范围.【答案】(1){x|1≤x≤2};(2){a|a≤1}.【解析】【分析】(1)代入a的值,求出集合A,从而求出A∩B;(2)由A与B的并集为A,得到B为A的子集,表示出A的中不等式的解集,根据数轴确定出满足题意a的范围即可.【详解】(1)a=1时,A={x|x≥1或x≤-2},故A∩B={x|1≤x≤2};(2)∵A∪B=A,∴B⊆A,由x2-(a-2)x-2a≥0,得(x+2)(x-a)≥0,当a<-2时,如数轴表示,符合题意;同理,当-2≤a≤1,也合题意;但当a>1时,不合题意,综上可知{a|a≤1}.【点睛】本题考查了并集及其运算,熟练掌握并集的定义是解本题的关键.19.设函数f(x)=++.(1)设t=+,求t的取值范围;(2)求f(x)的最大值.【答案】(1)[,2];(2)3.【解析】【分析】(1)将t,﹣1≤x≤1,两边平方,结合二次函数的最值,即可得到所求范围;(2)由(1)可得g(t)=f(x)(t+1)2,考虑对称轴t=﹣1与区间[,2]的关系,即可得到所求最大值.【详解】(1)t=+,-1≤x≤1,可得t2=2+2,由0≤1-x2≤1,可得t2∈[2,4],由t≥0可得t的取值范围是[,2];(2)由(1)可得g(t)=f(x)=t+=(t+1)2-,由[,2]在对称轴t=-1的右边,为增区间,即有t=2,即x=0,g(t)取得最大值,且为3,即f(x)的最大值为3.【点睛】本题考查函数的最值求法,注意运用换元法和二次函数的最值求法,考查运算能力,属于中档题.20.已知函数f(x)=x+(a>0).(1)判断f(x)的奇偶性;(2)判断函数f(x)在(,+∞)上的单调性,并用定义证明.【答案】(1)见解析;(2)见解析.【解析】【分析】(1)求出函数的定义域,得到f(﹣x)=﹣f(x),判断函数的奇偶性即可;(2)根据单调性的定义证明即可.【详解】(1)f(x)的定义域是{x|x≠0},f(-x)=-x-=-(x+)=-f(x),故函数f(x)是奇函数;(2)函数在(,+∞)递增,令<m<n,则f(m)-f(n)=m+-n-=(m-n)+a•=(m-n)(1-),∵<m<n,∴m-n<0,1->0,故f(m)-f(n)<0,故f(x)在(,+∞)上递增.【点睛】本题考查了函数的奇偶性,单调性问题,考查转化思想,是一道基础题.21.已知函数f(x)=2x,g(x)=-x2+2x+b.(1)若f(x)++1≥0对任意的x∈[1,3]恒成立,求m的取值范围;(2)若x1,x2∈[1,3],对任意的x1,总存在x2,使得f(x1)=g(x2),求b的取值范围.【答案】(1)[-6,-∞);(2)见解析.【解析】【分析】(1)根据h(x)=f(x)1,结合勾函数的性质对任意的x∈[1,3]恒成立,即可求解m的取值范围;(2)根据对任意的x1,总存在x2,使得f(x1)=g(x2),可得f(x)的值域是g(x)的值域的子集,即可求解b的范围;【详解】(1)函数f(x)=2x,令h(x)=f(x)++1=;①当m=0时,可得h(x)=2x+1在x∈[1,3]恒成立;②当m<0时,可知f(x)=2x是递增函数,y=在x∈[1,3]也是递增函数,∴h(x)在x∈[1,3]是递增函数,此时h(x)min=h(1)=≥0,可得:-6≤m<0;③当m>0时,,所以函数h(x)=,满足题意.综上所述:f(x)++1≥0对任意的x∈[1,3]恒成立,可得m的取值范围是[-6,-∞);(2)由函数f(x)=2x,x∈[1,3],可得:2≤f(x)≤8;由g(x)=-x2+2x+b.其对称x=1,开口向下.∵x∈[1,3],∴g(x)在x∈[1,3]上单调递减.g(x)max=g(1)=1+b;g(x)min=g(3)=-3+b;∵对任意的x1,总存在x2,使得f(x1)=g(x2),∴f(x)的值域是g(x)的值域的子集;即,解得:无解.故x1,x2∈[1,3],对任意的x1,总存在x2,使得f(x1)=g(x2),此是b的取值范围是空集.【点睛】本题主要考查了函数恒成立问题的求解,分类讨论以及转化思想的应用,二次函数的最值以及单调性的应用,属于中档题.22.已知a∈R,f(x)=log2(1+ax).(1)求f(x2)的值域;(2)若关于x的方程f(x)-log2[(a-4)x2+(2a-5)x]=0的解集恰有一个元素,求实数a 的取值范围;(3)当a>0时,对任意的t∈(,+∞),f(x2)在[t,t+1]的最大值与最小值的差不超过4,求a的取值范围.【答案】(1)当a≥0时,值域为[0,+∞),当a<0时,值域为(-∞,0);(2)1<a≤2,或a>4;(3)(0,+∞).【解析】【分析】(1)讨论a≥0时,a<0时,由对数函数的单调性可得值域;(2)根据对数的运算法则进行化简,转化为一元二次方程,讨论a的取值范围进行求解即可;(3)根据条件得到g(x)=log2(1+ax2),a>0,函数g(x)在区间[t,t+1]上单调递增,g(t+1)﹣g(t)≤4,运用对数函数的单调性和参数分离进行求解即可.【详解】(1)f(x)=log2(1+ax),可得f(x2)=log2(1+ax2),当a≥0时,1+ax2≥1,即有log2(1+ax2)≥0;当a<0时,0<1+ax21,即有log2(1+ax2)0;即有当a≥0时,f(x)的值域为[0,+∞);当a<0时,f(x)的值域为(-∞,0];(2)由f(x)-log2[(a-4)x2+(2a-5)x]=0得log2(1+ax)=log2[(a-4)x2+(2a-5)x],即1+ax=(a-4)x2+(2a-5)x>0,①则(a-4)x2+(a-5)x-1=0,即(x+1)[(a-4)x-1]=0,②,当a=4时,方程②的解为x=-1,代入①,不成立;当a=3时,方程②的解为x=-1,代入①,不成立;当a≠4且a≠3时,方程②的解为x=-1或x=,若x=-1是方程①的解,则1-a=-a+1>0,即a<1,若x=是方程①的解,则1+=>0,即a>4或a<2,则要使方程①有且仅有一个解,则a>4或1≤a<2.综上,若方程f(x)-log2[(a-4)x2+(2a-5)x]=0的解集中恰好有一个元素,则a的取值范围是1<a≤2,或a>4;(3)当a>0时,对任意的t∈(,+∞),f(x2)=log2(1+ax2),设g(x)=log2(1+ax2),a>0,函数g(x)在区间[t,t+1]上单调递增,由题意得g(t+1)-g(t)≤4,即log2(1+at2+2at+a)-log2(1+at2)≤4,即1+at2+2at+a≤16(1+at2),即有a(15t2-2t-1)+15=a(3t-1)(5t+1)+15>0恒成立,综上可得a的范围是(0,+∞).【点睛】本题考查函数最值的求解,以及对数不等式的应用,考查对数函数的单调性,属于中档题.。

2019-2020学年杭州市学军中学高三英语上学期期中试题及答案

2019-2020学年杭州市学军中学高三英语上学期期中试题及答案

2019-2020学年杭州市学军中学高三英语上学期期中试题及答案第一部分阅读(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,从每题所给的A、B、C、D四个选项中选出最佳选项AI once taught in a small private school. Each morning at nine o’clock all the students, ranging in age from three to seven years old, gathered in the Great Room for a warm-up in preparation for the day.One morning the headmistress made an announcement to all the children gathered,“Today we begin a great experiment of the mind.” She held up two ivy(常春藤) plants, each potted in an identical container. She continued, “Do they look the same?”All the children nodded. So did I, for, in this way, I was alsoa child.“We will give the plants the same amount of light, the same amount of water, but not the same amount of attention,” She said. “Together we are going to see what will happen when we put one plant in the kitchen away from our attention and the other plant right here in this room. Each day for the next month, we shall sing to our plant in the Great Room and tell it how much we love it, and how beautiful it is. We will use our good minds to think good thoughts about it.”Four weeks later my eyes were as wide and disbelieving as the children’s. The kitchen plant was leggy and sick-looking, and it hadn’t grown at all. But the Great Room plant, which had been sung to and surrounded by positive thoughts and words, had increased threefold in size with dark leaves that were filled with energy.In order to prove the experiment, the kitchen ivy was brought to the Great Room to join the other ivy. Within three weeks, the second plant had caught up with the first ivy. Within four weeks, they could not be distinguished, one from the other.I took this lesson to heart and made it my own.1. Why did the headmistress do the experiment?A. She wanted to teach me a lesson.B. She expected the students to learn to grow plants.C. She meant to prove the impact of good minds on growth.D. She intended to show students how to save a sick-looking plant.2. What happened to the ivy in the kitchen at last?A. It stopped growing and died.B. It was leggy and sick with dark leaves.C. It looked almost the same as the other one.D. It grew better than the one in the Great Room.3. What can be a suitable title for the passage?A. Life Means GrowthB. Things Grow with LoveC. Equality Makes a DifferenceD. Positive Thoughts Really CountBWolves have a certain undeserved reputation: fierce, dangerous, good forhunting down deer and farmers’ livestock. However, wolves have a softer, more social side, one that has been embraced by a heart-warming new initiative.In a bid to save some of Europe’s last wolves, scientists have explored the willingness of these supposedly fierce creatures to help others of their kind. Female wolves, the scientists have discovered, make excellent foster parents to wolf cubs that are not their own. The study, published in Zoo Biology, suggests that captive-bred wolfcubs(幼兽)could be placed with wild wolf families, boosting the wild population.The gray wolf was once the world’s most widely distributed mammal, but it became extinct as a result of widespread habitat destruction and the deliberate killing of wolves suspectedof preying on livestock. Fear and hatred of the wolf have since become culturally rooted, fuelled by myths, fables and stories.In Scandinavia, the gray wolf is endangered, the remaining population found by just five animals. As a result, European wolves are severely inbred and have little geneticvariability(变异性), making them vulnerable to threats, such as outbreaks of disease that they can’t adapt to quickly. So Inger Scharis and Mats Amundin of Linkoping University, in Sweden, started Europe’s first gray wolf-fostering program. They worked with wolves keptat seven zoos across Scandinavia. Eight wolf cubs between four and six days old were removed from their natural parents and placed with other wolf packs in other zoos. The foster mothers accepted the new cubs placed in their midst.The welfare of the foster cubs and the wolves’ natural behavior were monitored using a system of surveillance cameras. The foster cubs had a similar growth rate as their step siblings in the recipient litter, as well as their biological siblings in the source litter. The foster cubs had a better overall survival rate, with 73% surviving until 33 weeks, than their biological siblings left behind, of which 63% survived. That rate of survival is similar tothat seen in wild wolf cubs. Scientists believe that wolves can recognize their young, but this study suggests they can only do so once cubs are somewhere between three to seven weeks of age.If captive-bred cubs can be placed with wild-living families, which already have cubs of a similar age, not only will they have a good chance of survival, but they could help dramatically increase the diversity of the wild population, say the researchers. Just like the wild wolves they would join, these foster cubs would need protection from hunting. Their arrival could help preserve the future of one of nature’s most iconic and polarizing animals.4. What’s the theme of the passage?A. Giving wolf cubs a new lifeB. Foster wolf parents and foster cubsC. The fate of wild wolvesD. Changing diversity of wild wolves5. Which of the following flow chart best demonstrates the relationship between the wolves?A. B.C. D.6. Which of the following statements is true?A. Female wolves are willing to raise wolf cubs of 3 to 7 weeks old.B. Foster cubs are accepted by foster parents and are well bred.C. Man’s hostile attitude towards wolves roots in myths, fables and stories.D. Foster cubs and their biological siblings have similar growth rate and survival rate.7. What’s the purpose of the research?A. To help wolves survive various threatsB. To improve wolves’ habitat and stop deliberate killingC. To save endangered wolves by increasing their populationD. To raise man’s awareness of protecting wolvesCIt was New Year time, but I wasn’tlooking forward to it. That winter, my mother and my stepfather moved our family toSouthern California. My brother and I were leaving our ruralAlabamabehind. This would be our first New Year away fromAlabama. My mother took toCalifornialike a swan to a royal lake. My athletic little brother, Paul, was keyed up at a climate that allowed him to go to the beach whenever he wanted.I, however, was a fat child with heavy southern pronunciation. My first day in the new class, I introduced myself in a low voice. The moment I opened my mouth to speak, the whole class burst into laughter, “He talks funny.” It was so frustrating that I went to place a call to Granny Smith after school, who was my biggest support, But I didn’t get through.On Sunday evening, the phone rang. It was Granny. She often took advantage of the discounted long-distance rates on Sundays. She said she’d shipped a New Year package. Sure enough, it arrived. Surprised at the box, large enough to hold a small refrigerator, we eagerly tore it open. The smell of Granny’s house filled the room: a combination of fried meat, sausages, furniture polish and decorations. Her house was tiny and always filled withtackyholiday decorations and homemade food before New Year. But in my childhood eyes, it was precious and fantastic.There were countless tins and containers. We open hem to discover piles of holiday treats. She even included our traditional candy bats. The box was as bottomless as a magical box. There, beneath all these, was familiar holiday.Every New Year that we spent inCalifornia, the postal service would call and say our package was arrived. Over the years, many treasures arrived in the box. For me, it’s always been the best part of the holiday.8. How did the author’s brother feel when they were moving toCalifornia?A. Indifferent.B. Joyful.C. Appreciative.D. Disappointed.9. Why did the author’s classmates laugh at him?A. He spoke in a low voice.B. He made a humorous talk.C. He looked overweight.D. He had a strong accent.10. What does the underlined word “tacky” probably mean?A. Suitable.B. Expensive.C. Cheap.D. Attractive.11. Which of the following can be the best title of the text?A. Granny’s Care PackageB. An UnforgettableHolidayC. Our Move toCaliforniaD. A Telephone Call from GrannyDPeople saved a 20-foot orca (虎鲸) that was stuck between rocks on an Alaskan shore by continuously pouring water over it and protecting it from birds who circled above the defenseless whale.The whale was ultimately saved after a six-hour, labor-intensive life-saving operation. Someone spotted the large whale on the Prince of Wales Island near the coast of British Columbia on the morning of July 29th. The Coast Guard was called around 9 a.m. local time. Chance Strickland, the captain of a private yacht in Alaska, and his crew anchored and began life-saving action that were videoed by Aroon Melane and posted on the Internet.Strickland could hear the orca calling out to killer whales swimming in the area. People on other boats stopped with water and buckets to pour water over the animal. “There were tears coming out of its eyes,” Mr. Strickland told the local newspaper. “It was pretty sad.”The group of people formed a chain that passed buckets of seawater back and forth and poured the water on the orca, which seemed to liven it up. It made a noise and raised its tail when it got water.The National Oceanic and Atmospheric Administration (NOAA) was called in, which can be seen on the video using a machine to spray amist of seawater on the orca, which doubled as a way to keep the whale cool and scare the large group of birds that were hoping to feast on the beast.Melane said in her video that the orca was stranded (搁浅) for about six hours until the tide came in andswept it back into the ocean. The group efforts of Strickland’s crew and the NOAA saved the 13-year-old killer whale.12. Why did birds circle above the orca?A. They were eager to eat it.B. They wished to protect it.C. They were attracted by the people.D. They wanted to find a place to rest.13. What did Strickland do immediately after finding the whale?A. Posted pictures online.B. Called friends for help.C. Took action to save it.D. Videoed the trapped animal.14. Why did the whale make a noise and raise its tail?A. To express its eager for water.B. To extend its thanks to people.C. To call out to its fellow whales.D. To show its power and sadness.15. Which can be the best title for the text?A. Killer Whale Got SavedB. The Orca Inspired KindnessC. Combined Efforts WantedD. Animals and Humans United第二节(共5小题;每小题2分,满分10分)阅读下面短文,从短文后的选项中选出可以填入空白处的最佳选项。

杭州学军中学(西溪校区)2019学年第一学期期中考试高二数学(数学答案)(1)(1)(1)

杭州学军中学(西溪校区)2019学年第一学期期中考试高二数学(数学答案)(1)(1)(1)

杭州学军中学2019学年第一学期期中考试高二数学答案一、选择题:本大题共10小题,每小题4分,共40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。

10.解析:),3()2,(+∞⋃--∞=A ,令43)(+-=ax x x f ,由题意,0169>-=∆a ,又0>a ,所以34>a ,设]21693,21693[22-+--=a a a a B ,又2169302--<a a =2169382<-+a a 。

所以要使B A ⋂中恰好有两个整数解,则只能是4和5,所以应满足⎪⎩⎪⎨⎧>≤≤0)6(,0)5(,0)4(f f f ⇒⎪⎩⎪⎨⎧>+-≤+-≤+-041836,041525,041216a a a 解得9201529<≤a 。

二、填空题:本大题共7小题,多空题每题6分,单空题每题4分,共36分。

11.4π,a 22。

12.410;13.32,26-。

14.1,)5,2(-;15.223; 16.平行,423;17.5342t ≤≤。

17.解析:原不等式⇔()2110,41log 04x a t x t ⎧--≥⎪⎨-≥⎪⎩或()2110,41log 04x at x t ⎧--≤⎪⎨-≤⎪⎩。

即11,214xt t x ⎧≥+⎪⎪⎨⎪≤-⎪⎩(1)或11,214x t t x ⎧≤+⎪⎪⎨⎪≥-⎪⎩(2)。

注意到当1=x 时,(2)成立,此时2343≤≤t 。

当2,≥∈x Z x 时,(1)成立,在(1)中,41211-≤≤+x t x ,又4521)(--=x x x g 为单调递增函数,所以要使11,214x t t x ⎧≥+⎪⎪⎨⎪≤-⎪⎩(1)对2,≥∈x Z x 成立,只需2=x 时成立。

又2=x 时,4745≤≤t 。

所以要使不等式对任意的正整数x 恒成立,所以t 的取值范围是2345≤≤t 。

三、解答题(本大题共5小题,共74分。

杭州学军中学2019学年第一学期期中考试高三数学试卷(6页)

杭州学军中学2019学年第一学期期中考试高三数学试卷(6页)

杭州学军中学2019学年第一学期期中考试高三数学试卷一、选择题(本大题共10小题,每小题4分,共40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.设全集U =R ,集合2{|1},{|1}M x x P x x =>=>则下列关系中正确的是( )A.P M =B.M P M =C.M P M =D.()U C M P =∅ 2.设纯虚数z 满足 11i ai z-=+(其中i 为虚数单位),则实数a 等于( )A.1B.-1C.2D.-23.若,x y 满足约束条件03020x x y x y ≥⎧⎪+-≥⎨⎪-≤⎩,则2z x y =+的取值范围是( )A.[]6,0B.[]0,4C.[)6,+∞D.[)4,+∞4.已知,a b R ∈,下列四个条件中,使a b >成立的充分不必要的条件是( )A.1a b >-B.1a b >+C.a b >D.22a b > 5.函数2ln x x y x=的图象大致是( )A B CD6.已知函数1()0x D x x ⎧=⎨⎩为有理数为无理数,则( )A.(())1D D x =,0是()D x 的一个周期B.(())1D D x =,1是()D x 的一个周期C.(())0D D x =,1是()D x 的一个周期D.(())0D D x =,()D x 最小正周期不存在7.若关于x 的不等式222213x t x t t t +-+++-<无解,则实数t 的取值范围是( )A.1,15⎡⎤-⎢⎥⎣⎦B.(],0-∞C.(],1-∞D.(],5-∞ 8.若O 是ABC ∆垂心,6A π∠=且sin cos sin cos 2sin sin B C AB C BAC m B C AO +=,则m =( )A.12B.2 C.3 D.69.已知二次函数2()(2)f x ax bx ba =+≤,定义{}1()max ()11f x f t t x =-≤≤≤,{}2()min ()11f x f t t x =-≤≤≤,其中{}m a x ,a b 表示,a b中的较大者,{}min ,a b表示b a ,中的较小者,下列命题正确的是( ) A.若11(1)(1)f f -=,则(1)>(1)f f - B.若22(1)(1)f f -=,则(1)(1)f f ->C.若21(1)(1)f f =-,则11(1)(1)f f -< D.若21(1)(1)f f =-,则22(1)(1)f f ->10.已知数列{}n a 满足2111,312n n n a a a a +=-=++,若12n n b a =+,设数列{}n b 的前项和为n S ,则使得2019S k -最小的整数k 的值为( )A.0B.1C.2D.3二、填空题(本大题共7小题,多空题每题6分,单空题每题4分,共36分.)11. ()512x -展开式中3x 的系数为 ,所有项的系数和为 .12.等比数列{}n a 中,12a a ==,则2201382019a a a a +=+ ,1234a a a a = .13.在ABC ∆中,角,,A B C 所对的边分别为,,a b c ,已知sin cos c A C =,则C = ,若c =ABC ∆的面积为2,则a b += . 14.已知函数222,0()2(1),0x x x f x f x x -⎧+-≥=⎨+<⎩,则3()2f -= ,若函数()()g x f x k =-有无穷多个零点,则k 的取值范围是 .15.已知,x y R ∈且221x y xy ++=,则x y xy ++的最小值为 . 16.已知平面向量,,a b c 满足,,015a b c a c b c ⋅==-=-=,则a b -的最大值为 .17.当[]1,4x ∈时,不等式322044ax bx a x ≤++≤恒成立,则7a b +的取值范围是 .三、解答题(本大题共5小题,共74分.写出文字说明、证明过程或演算步骤.)18.(本题满分14分)已知函数()2sin cos()3f x x x π=+ (Ⅰ)求函数()f x 的单调递减区间;(Ⅱ)求函数()f x 在区间0,2π⎡⎤⎢⎥⎣⎦上的最大值及最小值.19.(本题满分15分)已知在ABC ∆中,1AB =,2AC =.(Ⅰ)若BAC ∠的平分线与边BC 交于点D ,求()2AD AB AC ⋅-uuu r uu u r uuu r;(Ⅱ)若点E 为BC 的中点,求2211AE BC+uu u r uu u r 的最小值.20.(本题满分15分)已知正项等差数列{}n a 满足:233312n n S a a a =+++,其中n S 是数列{}n a 的前n 项和. (Ⅰ)求数列{}n a 的通项公式; (Ⅱ)令()()()1412121n n n n nb a a -=--+,证明:122221n n b b b n ++++≤+.21.(本题满分15分)设函数(),x f x e ax a a R =-+∈,其图象与x 轴交于12(,0),(,0)A x B x两点,且12.x x <(Ⅰ)求a 的取值范围;(Ⅱ)证明:0f '<.22.(本题满分15分)已知函数2()ln 2,.f x x ax bx a R =---∈ (Ⅰ)当2b =时,试讨论()f x 的单调性;(Ⅱ)若对任意的3(,)b e∈-∞-,方程()0f x =恒有2个不等的实根,求a 的取值范围.。

浙江省杭州市学军中学2018-2019学年高一上学期期中考试数学试题含答案解析

浙江省杭州市学军中学2018-2019学年高一上学期期中考试数学试题含答案解析

x 1浙江省杭州市学军中学 2018-2019 学年高一上学期期中考试数学试题一、选择题(本大题共 10 小题,共 30.0 分)1.已知集合 , ,则( )A.B.C.D.【答案】B【解析】由题意知,故选 B.2.函数 f (x )= A.【答案】B【解析】由∴函数 f (x )ln (1-x 2)的定义域为( )B. C. D.,得 0≤<1.ln (1﹣x 2)的定义域为[0,1).故选:B .3.已知函数 f (x )=,则 f [f ( )]等于( )A.B. C. D.【答案】D【解析】∵函数 f (x ) ,∴f ( ) ,f [f ( )]=f ( ).故选:D .4.使函数 f (x )=x a 的定义域为 R 且为奇函数的 α 的值可以是()A.B. C. 3 D. 以上都不对【答案】C【解析】根据题意,依次分析选项:对于 A ,α=﹣1 时,f (x )=x ﹣,其定义域不是 R ,不符合题意;对于 B ,α 时,f (x ) ,其定义域不是 R ,不符合题意;f f f f f对于 C ,α=3 时,f (x )=x 3,其定义域为 R 且为奇函数,符合题意;对于 D ,错误,故选:C .5.已知集合 M ,N ,P 为全集 U 的子集,且满足 M ⊆P ⊆N ,则下列结论不正确的是()A. ∁U N ⊆∁U PC. (∁U P )∩M =∅ B. ∁N P ⊆∁N MD. (∁U M )∩N =∅【答案】D【解析】因为 P ⊆N ,所以∁U N ⊆∁U P ,故 A 正确; 因为 M ⊆P ,所以∁N P ⊆∁N M ,故 B 正确; 因为 M ⊆P ,所以(∁U P )∩M =∅,故 C 正确;因为 M ⊆ N ,所以(∁U M )∩N ∅.故 D 不正确.故选 D.6.设函数 (x )=log a x (a >0,a≠1),若 (x 1x 2…x 2018)=4,则 (x 12)+ (x 12)+…+ (x 20182)的值等于( )A. 4B. 8C. 16D.【答案】B【解析】∵函数 f (x )=log a x (a >0,a ≠1),f (x 1x 2…x 2018)=4,∴f (x 1x 2…x 2018)=log a (x 1x 2…x 2018)=4,∴f (x 12)+f (x 12)+…+f (x 20182)=log a (x 1x 2…x 2018)2=2log a (x 1x 2…x 2018)=2×4=8. 故选:B .7.设 A ={x |2≤x ≤4},B ={x |2a≤x ≤a+3},若 B 真包含于 A ,则实数 a 的取值范围是()A.B. C. D.【答案】C【解析】∵A ={x |2≤x ≤4},B ={x |2a ≤x ≤a +3},且 B 真包含于 A ;当 B =∅时,2a >a +3,解得 a >3;当 B ≠∅时,解得 a =1;此时 A=B.∴a 的取值范围是{a |a >3}.故选:C .8.函数 f (x )=log 2(-x 2+ax +3)在(2,4)是单调递减的,则 a 的范围是()f f f fA.B. C. D.【答案】B【解析】令 t =﹣x 2+ax +3,则原函数化为 y =log 2t ,∵y =log 2t 为增函数,∴t =﹣x 2+ax +3 在(2,4)是单调递减,对称轴为 x,∴且﹣42+4a +3≥0,解得: .∴a 的范围是[ ,4].故选:B .9.对于函数 f (x ),若∀a ,b ,c ∈R ,f (a ),f (b ),f (c )为某一三角形的三边长,则称f (x )为“可构造三角形函数”.已知函数 f (x )=值范围是()A.B.C.是“可构造三角形函数”,则实数 t 的取D.【答案】A【解析】由题意可得 f (a )+f (b )>f (c )对于∀a ,b ,c ∈R 都恒成立,由于 f (x )1 ,①当 t ﹣1=0,(x )=1,此时 (a ),(b ),(c )都为 1,构成一个等边三角形的三边长, 满足条件.②当 t ﹣1>0,f (x )在 R 上是减函数,1<f (a )<1+t ﹣1=t ,同理 1<f (b )<t ,1<f (c )<t ,故 f (a )+f (b )>2.再由 f (a )+f (b )>f (c )恒成立,可得 2≥t ,结合大前提 t ﹣1>0,解得 1<t ≤2.③当 t ﹣1<0,f (x )在 R 上是增函数,t <f (a )<1,同理 t <f (b )<1,t <f (c )<1,由 f (a )+f (b )>f (c ),可得 2t ≥1,解得 1>t .综上可得, t ≤2,故选:A .10.设函数,其中表示中的最小者.下列说法错误的()A.函数C.若为偶函数时,B.若D.若时,有时,【答案】D【解析】在同一坐标系中画出的图像(如图所示),故的图像如图所示.的图像关于轴对称,故为偶函数,故A正确.由图可知从图像上看,当时,有时,有,故B成立.成立,令,则,故,故C成立.取,则,,,故D不成立.综上,选D.二、填空题(本大题共7小题,共28.0分)11.若,则.【答案】10【解析】若,则b b f ) |x12.已知【答案】【解析】(配凑法)∴故答案为:13.已知 f (x )=【答案】[-1,0]【解析】∵f (x ) ,则 ________.,又 ∈(-∞,-2]∪[2,+∞),.的定义域为 R ,则实数 a 的取值范围是______.的定义域为 R ,∴即0 对任意 x ∈R 恒成立,恒成立,即 x 2+2ax ﹣a ≥0 对任意 x ∈R 恒成立,∴ =△4a 2+4a ≤0,则﹣1≤a ≤0.故答案为:[﹣1,0].14.设 ma x {a ,}表示 a , 两数中的最大值,若(x =ma x {|x |, -t|}关于 x =1 对称,则 t =______.【答案】2【解析】f (x )=max{|x |,|x ﹣t |},由函数 y =|x |的图象关于 x =0 对称,函数 y =|x ﹣t |的图象关于 x =t 对称,即有函数 f (x )的图象关于 x 对称,f (x )=max{|x |,|x ﹣t |}关于 x =1 对称,即有1,求得 t =2,故答案为:2.15.设方程 x 2-mx +2=0 的两根 α,β,其中 α∈(1,2),则实数 m 的取值范围是______.【答案】(2,4)【解析】∵方程 x 2﹣mx +2=0 的两根 α,β,∴=△m 2﹣8≥0,求得m≥2,或m≤﹣2①.由α•β=2,则由①②可得,故答案为:,则..,则②.16.已知lg2≈0.3010,则22018是______位数.【答案】608【解析】设x=22018,则lgx=2018lg2≈2018×0.3010=607.418,∴22018是608位数.故答案为:608.17.已知函数f(x)满足对任意的m,n都有f(m+n)=f(m)+f(n)-1,设g(x)=f(x)+(a>0,a≠1),g(ln2018)=-2015,则g(ln)=______.【答案】2018【解析】∵函数f(x)满足对任意实数m,n,都有f(m+n)=f(m)+f(n)﹣1,令m=n=0,则f(0)=2f(0)﹣1,解得f(0)=1,令m=x,n=﹣x,则f(0)=f(x)+f(﹣x)﹣1,即f(x)+f(﹣x)=2,∵g(x)=f(x)(a>0,a≠0),∴g(﹣x)=f(﹣x)f(﹣x),故g(x)+g(﹣x)=f(x)+f(﹣x)+1=3,∴g(ln2018)+g(ln)=﹣2015+g(﹣ln2018)=3,即g(ln)=2018,故答案为:2018.三、解答题(本大题共5小题,共62.0分)18.已知集合U=R,集合A={x|x2-(a-2)x-2a≥0},B={x|1≤x≤2}.(1)当a=1时,求A∩B;(2)若A∪B=A,求实数a的取值范围.解:(1)a=1时,A={x|x≥1或x≤-2},故A∩B={x|1≤x≤2};(2)∵A∪B=A,∴B A,由x2-(a-2)x-2a≥0,得(x+2)(x-a)≥0,当a<-2时,如数轴表示,符合题意;同理,当-2≤a≤1,也合题意;但当a>1时,不合题意,综上可知{a|a≤1}.19.设函数f(x)=(1)设t=+++.,求t的取值范围;(2)求f(x)的最大值.解:(1)t=+,-1≤x≤1,可得t2=2+2,由0≤1-x2≤1,可得t2∈[2,4],由t≥0可得t的取值范围是[,2];(2)由(1)可得g(t)=f(x)=t+=(t+1)2-,由[,2]在对称轴t=-1的右边,为增区间,即有t=2,即x=0,g(t)取得最大值,且为3,即f(x)的最大值为3.20.已知函数f(x)=x+(a>0).(1)判断f(x)的奇偶性;(2)判断函数f(x)在(,+∞)上的单调性,并用定义证明.(1)解:f(x)的定义域是{x|x≠0},f(-x)=-x-=-(x+)=-f(x),故函数f(x)是奇函数;(2)证明:函数在(,+∞)递增,令 <m <n ,则 f (m )-f (n )=m + -n - =(m -n )+a •=(m -n )(1- ),∵ <m <n ,∴m -n <0,1->0,故 f (m )-f (n )<0,故 f (x )在( ,+∞)上递增.21.已知函数 f (x )=2x ,g (x )=-x 2+2x +b .(1)若 f (x )+ +1≥0 对任意的 x ∈[1,3]恒成立,求 m 的取值范围;(2)若 x 1,x 2∈[1,3],对任意的 x 1,总存在 x 2,使得 f (x 1)=g (x 2),求 b 的取值范围.解:(1)函数 f (x )=2x ,令 h (x )=f (x )++1= ;①当 m =0 时,可得 h (x )=2x +1 在 x ∈[1,3]恒成立;②当 m <0 时,可知 f (x )=2x 是递增函数,y = 在 x ∈[1,3]也是递增函数,∴h (x )在 x ∈[1,3]是递增函数,此时 h (x )min =h (1)=≥0,可得:-6≤m <0;③当 m >0 时,,所以函数 h (x )= ,满足题意.综上所述:f (x )++1≥0 对任意的 x ∈[1,3]恒成立,可得 m 的取值范围是[-6,-∞);(2)由函数 f (x )=2x ,x ∈[1,3],可得:2≤f (x )≤8;由 g (x )=-x 2+2x +b .其对称 x =1,开口向下.∵x ∈[1,3],∴g (x )在 x ∈[1,3]上单调递减.g (x )ma x =g (1)=1+b ;g (x )min =g (3)=-3+b ;∵对任意的 x 1,总存在 x 2,使得 f (x 1)=g (x 2), ∴f (x )的值域是 g (x )的值域的子集;即,解得:无解.f t故 x 1,x 2∈[1,3],对任意的 x 1,总存在 x 2,使得 f (x 1)=g (x 2),此是 b 的取值范围是空集.22.已知 a ∈R ,f (x )=log 2(1+ax ). (1)求 f (x 2)的值域;(2)若关于 x 的方程 f (x )-log 2[(a -4)x 2+(2a -5)x ]=0 的解集恰有一个元素,求实数 a的取值范围;(3)当 a >0 时,对任意的 t ∈( ,+∞),(x 2)在[t , +1]的最大值与最小值的差不超过 4,求 a 的取值范围.解:(1)f (x )=log 2(1+ax ),可得 f (x 2)=log 2(1+ax 2),当 a ≥0 时,1+ax 2≥1,即有 log 2(1+ax 2)≥0;当 a <0 时,0<1+ax 2 1,即有 log 2(1+ax 2) 0; 即有当 a ≥0 时,f (x )的值域为[0,+∞);当 a <0 时,f (x )的值域为(-∞,0];(2)由 f (x )-log 2[(a -4)x 2+(2a -5)x ]=0,得 log 2(1+ax )=log 2[(a -4)x 2+(2a -5)x ], 即 1+ax =(a -4)x 2+(2a -5)x >0,①则(a -4)x 2+(a -5)x -1=0,即(x +1)[(a -4)x -1]=0,②,当 a =4 时,方程②的解为 x =-1,代入①,不成立;当 a =3 时,方程②的解为 x =-1,代入①,不成立;当 a ≠4 且 a ≠3 时,方程②的解为 x =-1 或 x =,若 x =-1 是方程①的解,则 1-a =-a +1>0,即 a <1,若 x =是方程①的解,则 1+ = >0,即 a >4 或 a <2,则要使方程①有且仅有一个解,则 a >4 或 1≤a <2.综上,若方程 f (x )-log 2[(a -4)x 2+(2a -5)x ]=0 的解集中恰好有一个元素,则 a 的取值范围是 1<a ≤2,或 a >4;(3)当a>0时,对任意的t∈(,+∞),f(x2)=log2(1+ax2),设g(x)=log2(1+ax2),a>0,函数g(x)在区间[t,t+1]上单调递增,由题意得g(t+1)-g(t)≤4,即log2(1+at2+2at+a)-log2(1+at2)≤4,即1+at2+2at+a≤16(1+at2),即有a(15t2-2t-1)+15=a(3t-1)(5t+1)+15>0恒成立,综上可得a的范围是(0,+∞).。

2019-2020学年浙江省杭州市学军中学西溪校区高一(上)期中数学试卷 (含答案解析)

2019-2020学年浙江省杭州市学军中学西溪校区高一(上)期中数学试卷 (含答案解析)

2019-2020学年浙江省杭州市学军中学西溪校区高一(上)期中数学试卷一、选择题(本大题共10小题,共40.0分)1. 己知M ={x|−1≤x ≤2},N ={x|x ≤3},则(∁R M)∩N =( )A. [2,3]B. (2,3]C. (−∞,−1]∪[2,3]D. (−∞,−1)∪(2,3] 2. 下列四组中,f(x)与g(x)表示同一个函数的是( ) A. f(x)=√x 44,g(x)=(√x 4)4 B. f(x)=x ,g(x)=√x 33C. f(x)=1,g(x)={1,x >0−1,x <0D. f(x)=x2−4x+2,g(x)=x −2 3. 下列函数中,在其定义域内既为奇函数又为增函数的是( ) A. f (x )=−1xB. f (x )=x 3C. f (x )=|x |D. f (x )=3x +3−x24. 已知实数a ≥0且a ≠1,则在同一直角坐标系中,函数f(x)=a −x (x >0),g(x)=log a (−x)的图象可能是( )A. B.C. D.5. 函数f(x)=lg(−x +4)的定义域为( )A. (−∞,4]B. (−∞,4)C. (0,4)D. (0,4]6. 已知函数f(x)={−2x 2+x,x ≥0x 2−g(x),x <0是奇函数,则g(−2)的值为( ) A. 0 B. −1C. −2D. −3 7. 定义在R 上的函数f(x)=(13)|x−m|−2为偶函数,a =f(log 212),b =f((12)13),c =f(m),则( ) A. c <a <bB. a <c <bC. a <b <cD. b <a <c 8. 函数的单调减区间为( )A. (0,1]B. (0,2)C. (1,2)D. [0,2] 9. 若函数f(x)=1+2x+12x +1+sinx 在区间[−k,k](k >0)上的值域为[m,n],则m +n 的值是( )A. 0B. 1C. 2D. 4 10. 若函数f(x)=3−|x|−m 的最大值为2,则实数m 的值为( )A. −1B. −2C. −3D. −4 二、填空题(本大题共5小题,共25.0分)11. 已知幂函数f(x)的图象经过点(2,4),则f(5)= ______ .12. 已知f(1x −x)=x 2+1x 2,则f(2)=______.13. 已知函数f(x)=x 3+x ,且f(3a −2)+f(a −1)< 0,则实数a 的取值范围是__________.14. 设则________. 15. 已知函数f(x)={1−e x ,x ⩽0x 2−2x,x >0,若函数y =f(x)−m 有两个不同的零点,则m 的取值范围___. 三、解答题(本大题共6小题,共55.0分)16. 已知2x =7y =196,则1x +1y = ______ .17. (Ⅰ)(0.064)−13−(−78)0+[(−2)3] −43+(16)−0.75; (Ⅱ)log 3√27+lg25+lg4+7 log 72+(−9.8)0.18. 设集合A ={x|2−5≤2−x ≤4},B ={x|x 2+2mx −3m 2<0,m >0}.(1)若m =2,求A ∩B ;(2)若B ⊆A ,求实数m 的取值范围.19.已知函数f(x)=4−log3x,g(3x)=2x.(1)求函数g(x)的表达式;(2)设函数ℎ(x)=f(x)+g(x)的定义域为(1,9),求函数y=ℎ(x)·ℎ(x2)的值域;,27],不等式f(x3)·f(x2)>kg(x)恒成立,求实数k的取值范围.(3)若对任意的x∈[1920.已知函数f(x)=log a(x+1)−log a(4−2x),(a>0,且a≠1).(1)求函数f(x)的定义域;(2)求使函数f(x)>0时实数x的取值范围.21.已知y=f(x)是定义在R上的偶函数,当x≥0时,f(x)=x2−2x(1)求f(1),f(−2)的值;(2)求f(x)的解析式,(3)画出y=f(x)简图;写出y=f(x)的单调递增区间(只需写出结果,不要解答过程)-------- 答案与解析 --------1.答案:D解析:解:M ={x|−1≤x ≤2},N ={x|x ≤3},∴∁R M ={x|x <−1或x >2},∴(∁R M)∩N ={x|x <−1或2<x ≤3}=(−∞,−1)∪(2,3].故选:D .根据补集与交集的定义计算即可.本题考查了补集与交集的定义和应用问题,是基础题.2.答案:B解析:选项A 、C 、D 中两个函数的定义域不相同.3.答案:B解析:【分析】本题考查函数的奇偶性与单调性的判断,属于基础题.根据题意,依次分析选项中函数的奇偶性与单调性,综合即可得答案.【解答】对于A ,是奇函数,但在定义域内不是增函数,因为当x <0时,f(x)>0;当x >0时,f(x)<0,所以不符合;对于B ,f (x )=x 3是奇函数也是增函数,符合;对于C ,f (x )=|x |是偶函数,不符合题意;对于D ,f (x )=3x +3−x 2是偶函数,不符合题意.故选B . 4.答案:D解析:【分析】本题主要考查了对数函数和指数函数的图象和性质,属于基础题.根据对数函数和指数函数的图象和性质,即可判断出问题.【解答】解:若0<a <1, ∴y =a −x =(1a )x ,故为增函数,且过定点(0,1),∴y =log a (−x)的定义域为(−∞,0),故为增函数,且过定点(−1,0),以上图象都不成立.若a >1,y =a −x =(1a )x 为减函数,且过定点(0,1),y =log a (−x)的定义域为(−∞,0),故为减函数,且过定点(−1,0),所以D 成立.故选D . 5.答案:B解析:解:由题意得:−x +4>0,解得:x <4,故函数的定义域是(−∞,4),故选:B .根据对数函数的性质求出函数的定义域即可.本题考查了求函数的定义域问题,考查对数函数的性质,是一道基础题.6.答案:C解析:解:∵f(x)是奇函数,∴f(−2)=(−2)2−g(−2)=−f(2),即4−g(−2)=−(−8+2)=6,即g(−2)=4−6=−2,故选:C .根据函数f(x)是奇函数,进行转化求解即可.本题主要考查函数值的计算,根据函数奇偶性的性质进行转化求解是解决本题的关键.7.答案:C解析:【分析】本题考查函数的奇偶性和单调性的应用,对数与对数运算,属于基础题.利用偶函数的定义得m =0,即f(x)=(13)|x|−2,再利用对数运算得结论.【解答】解:因为函数f(x)=(13)|x−m|−2是R 上的偶函数,所以(13)|x−m|−2=(13)|−x−m|−2 对x ∈R 恒成立,即(13)|x−m|=(13)|x+m| 对x ∈R 恒成立,所以m =0,即f(x)=(13)|x|−2.故函数f(x)=(13)|x|−2在(0,+∞)上为减函数,又log 212=−1,0<(12)13<(12)0=1,m =0 所以f (log 212)<f ((12)13)<f (0), 即a <b <c .故选C . 8.答案:A解析:【分析】本题考查了求复合函数的单调区间,按照“同增异减”的方法求单调区间,属于容易题.【解答】解:令t =2x −x 2>0,求得0<x <2,可得函数的定义域为{x|0<x <2},且y =log 13t , 本题即求函数t 在定义域内的增区间,再利用二次函数的性质可得函数t 在定义域内的增区间为(0,1].故选A .9.答案:D解析:【分析】本题考查了函数的值域,考查了函数的对称性,根据已知条件求出f(−x),由f(x)+f(−x)的值可以求出函数的对称中心,故答案可得.【解答】解:∵f (x )=1+2(2x +1)−22x +1+sinx =3−22x +1+sinx , f (−x )=3−22−x +1+sin (−x )=3−2×2x 1+2x −sinx ,∴f (x )+f (−x )=4,∴函数f(x) 以 (0,2)为对称中心,∴其最大值与最小值的和m +n =4 .故选D .10.答案:A解析:解:f(x)=3−|x|−m,当x≥0时,函数f(x)为减函数,当x<0时,函数f(x)为增函数,故f(x)max=f(0)=1−m=2,解得m=−1,故选:A.根据当x≥0时,函数f(x)为减函数,当x<0时,函数f(x)为增函数,即可求出f(x)max=f(0),解得即可.本题考查了函数的单调性和最值关系,属于基础题.11.答案:25解析:解:设幂函数f(x)=xα,它的图象经过点(2,4),∴2α=4,即α=2,∴f(x)=x2;∴f(5)=52=25.故答案为:25.设出幂函数f(x)的解析式,根据图象过点(2,4),求出解析式,计算f(5)的值.本题考查了幂函数的图象与性质的应用问题,也考查了待定系数法求函数解析式的问题,是基础题.12.答案:6解析:【分析】本题考查函数值的求法,考查函数性质等基础知识,考查运算求解能力,是基础题.推导出f(1x−x)=x2+1x2=(1x−x)2+2,由此能求出f(2).【解答】解:∵f(1x −x)=x2+1x2=(1x−x)2+2,∴f(2)=22+2=6.故答案为6.13.答案:(−∞,34)解析:【分析】本题考查函数的奇偶性和单调性,根据函数的奇偶性得到f(3a−2)<f(1−a),再由单调性可知3a−2<1−a,即可求出答案,属于中档题.【解答】解:f(−x)=−x3−x=−f(x),且y1=x3与y2=x在R上都为增函数,函数f(x)=x3+x是R上的递增的奇函数,f(3a−2)+f(a−1)< 0即f(3a−2)<f(1−a),也即3a−2<1−a,解得a<34.故答案为(−∞,34).14.答案:解析:【分析】本题考查了分段函数,属于基础题.利用分段函数计算函数值即可得出结论.【解答】解:因为所以,因此.故答案为.15.答案:(−1,1)解析:【分析】本题考查了函数的零点的判断及分段函数的应用,考查数形结合思想,属于中档题.画出函数y=f(x)与y=m的图象,由图象可得m的取值范围.【解答】解:函数f(x)={1−e x,x≤0x2−2x,x>0,画出函数y=f(x)与y=m的图象,如图所示,∵函数y=f(x)−m有2不同的零点,∴函数y=f(x)与y=m的图象有2交点,由图象可得m的取值范围为(−1,1).故答案为(−1,1).16.答案:12解析:【分析】本题考查了指数式与对数式的互化、对数换底公式,考查了推理能力与计算能力,属于中档题.由2x=7y=196转化为对数形式,再利用换底公式换成同底的进行计算求值.【解答】解:∵2x=7y=196,,,则1x+1y=lg2+lg7lg196=lg14lg142=12,故答案为:12.17.答案:解:(Ⅰ)原式=(0.43)−13−1+(−2)−4+(24)−34=52−1+116+18=2716,(Ⅱ)原式=32+2+3=132.解析:本题考查了对数的运算性质和指数幂的运算性质,属于基础题.(Ⅰ)根据指数幂的运算性质计算即可.(Ⅱ)根据对数的运算性质计算即可.18.答案:解:(1)集合A ={x|2−5≤2−x ≤4}={x|2−5≤2−x ≤22}={x|−2≤x ≤5}当m =2时,B ={x|x 2+2mx −3m 2<0}={x|−6<x <2},那么:A ∩B ={x|−2≤x <2}.(2)B ={x|x 2+2mx −3m 2<0}由x 2+2mx −3m 2<0可得:(x +3m)(x −m)<0∵m >0∴−3m <x <m故得集合B ={x|−3m <x <m}要使B ⊆A 成立,只需{−3m ≥−2m ≤2,解得:m ≤23. 所以:0<m ≤23综上可得m 的取值范围是(0,23].解析:本题主要考查集合的基本运算,属于基础题.(1)化简集合A ,当m =2时,求解集合B ,根据集合的基本运算即可求A ∩B ;(2)根据A ⊇B ,建立条件关系即可求实数m 的取值范围.19.答案:解:(1)令3x =t >0,则x =log 3t ,从而g(t)=2log 3t ,即g(x)=2log 3x,(x >0).(2)函数ℎ(x)=f(x)+g(x)=4+log 3x 的定义域为(1,9),故函数y =ℎ(x)·ℎ(x 2)的定义域为(1,3).又y =ℎ(x)·ℎ(x 2)=(4+log 3x)·(4+2log 3x),令n =log 3x ∈(0,1),则y =(4+n)·(4+2n)=2(n +3)2−2在(0,1)上单调增,所以函数y =ℎ(x)·ℎ(x 2)的值域为(16,30).(3)由f(x 3)·f(x 2)>kg(x)得,(4−3log 3x)·(4−2log 3x)>2klog 3x ,令m =log 3x ,由x ∈[19,27]得m ∈[−2,3],原问题转化为:(4−3m)(4−2m)>2km ,即km <3m 2−10m +8对任意m ∈[−2,3]恒成立. 当m =0时,km <3m 2−10m +8恒成立,故k ∈R .当m ∈(0,3]时,k <3m 2−10m+8m =3m +8m −10,所以k <(3m +8m −10)min ,因为3m +8m −10≥2√3m ·8m −10=4√6−10,当且仅当3m =8m 即m =2√63∈(0,3]时取等号, 所以k <4√6−10.当m∈[−2,0)时,k>3m2−10m+8m =3m+8m−10,所以k>(3m+8m−10)max,因为m<0,3m+8m −10=−(−3m+8−m)−10≤−2√3m·8m−10=−4√6−10,当且仅当−3m=8−m 即m=−2√63∈[−2,0)时取等号,所以k>−4√6−10.综上所述,实数k的取值范围是(−4√6−10,4√6−10).解析:本题考查函数的定义域、值域,考查求函数的解析式,考查不等式恒成立问题,属于较难题.(1)令3x=t>0,用换元法求解即可;(2)令n=log3x,由题可知0<n<1,用换元法得y=2(n+3)2−2求解即可;(3)令m=log3x∈[−2,3],换元后分离参数,分类讨论,利用基本不等式求解.20.答案:解:(1)要使函数f(x)=log a(x+1)−log a(4−2x),(a>0,且a≠1)的解析式有意义自变量x须满足:{x+1>04−2x>0解得−1<x<2故函数f(x)的定义域为(−1,2)(2)∵f(x)=log a(x+1)−log a(4−2x)=log a x+14−2x(−1<x<2)当0<a<1时,若f(x)>0,则0<x+14−2x<1,解得−1<x<1,即此时实数x的取值范围为(−1,1)当a>1时,若f(x)>0,则x+14−2x>1,解得1<x<2,即此时实数x的取值范围为(1,2)解析:(1)根据使函数f(x)=log a(x+1)−log a(4−2x),(a>0,且a≠1)的解析式有意义的原则,结合对数型函数真数大于0,构造不等式组,可求出函数f(x)的定义域;(2)分当0<a<1时,和当a>1时,两种情况,结合对数函数的单调性及(1)中函数的定义域,分别求出x的取值范围本题考查的知识点是对数函数图象与性质的综合应用,熟练掌握对数函数的图象和性质是解答的关键.21.答案:解:(1)由题意可得f(1)=1−2=−1,∵f(x)为R上的偶函数,∴f(−2)=f(2)=22−4=0;(2)∵x<0,∴−x>0,∴f(−x)=(−x)2−2(−x)=x2+2x,∵f(−x)=f(x),∴f(x)=x2+2x,∴f(x)={x 2+2x,x<0x2−2x,x≥0;(3)由(2)作出函数图象如图所示:由函数图象可得f(x)的单调递增区间为[−1,0],[1,+∞).解析:本题考查函数的奇偶性、单调性及函数解析式求解,以及函数图象的做法,属于中档题目.(1)利用函数的奇偶性,直接代入即可求解相应的函数值;(2)利用函数的奇偶性求出函数解析式;(3)利用二次函数的图象和性质,作出函数图象,由函数图象得出函数的单调递增区间即可.。

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杭州学军中学2019学年第一学期期中考试
高三数学试卷
命题人:王 馥 审题人:纪向胜
一、选择题(本大题共10小题,每小题4分,共40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.设全集U =R ,集合2{|1},{|1}M x x P x x =>=>则下列关系中正确的是( ) A.P M = B.M P M = C.M
P M = D.()U C M P =∅
2.设纯虚数z 满足 11i
ai z
-=+(其中i 为虚数单位)
,则实数a 等于( ) A.1
B.-1
C.2
D.-2
3.若,x y 满足约束条件03020x x y x y ≥⎧⎪
+-≥⎨⎪-≤⎩
,则2z x y =+的取值范围是( )
A.[]6,0
B.[]0,4
C.[)6,+∞
D.[)4,+∞
4.已知,a b R ∈,下列四个条件中,使a b >成立的充分不必要的条件是( ) A.1a b >- B.1a b >+ C.a b > D.22a
b
> 5.函数2ln x x y x
=
的图象大致是( )
A B C D 6.已知函数1()0
x D x x ⎧=⎨
⎩为有理数为无理数
,则( )
A.(())1D D x =,0是()D x 的一个周期
B.(())1D D x =,1是()D x 的一个周期
C.(())0D D x =,1是()D x 的一个周期
D.(())0D D x =,()D x 最小正周期不存在
7.若关于x 的不等式222213x t x t t t +-+++-<无解,则实数t 的取值范围是( ) A.1,15⎡⎤-⎢⎥⎣⎦
B.(],0-∞
C.(],1-∞
D.(],5-∞ 8.若O 是ABC ∆垂心,6
A π
∠=且sin cos sin cos 2sin sin B C AB C BAC m B C AO +=,
则m =( ) A.
1
2
B.2
C.3
D.6
9.已知二次函数2
()(2)f x ax bx b a =+≤,定义{}1()max ()11f x f t t x =-≤≤≤,
{}2()min ()11f x f t t x =-≤≤≤,其中{}max ,a b 表示,a b 中的较大者,{}min ,a b
表示b a ,中的较小者,下列命题正确的是( )
A.若11(1)(1)f f -=,则(1)>(1)f f -
B.若22(1)(1)f f -=,则(1)(1)f f ->
C.若21(1)(1)f f =-,则11(1)(1)f f -<
D.若21(1)(1)f f =-,则22(1)(1)f f -> 10.已知数列{}n a 满足2111
,312
n n n a a a a +=-=++,若1
2
n n b a =+,设数列{}n b 的前项和为n S ,则使得2019S k -最小的整数k 的值为( )
A.0
B.1
C.2
D.3
二、填空题(本大题共7小题,多空题每题6分,单空题每题4分,共36分.) 11. ()5
12x -展开式中3
x 的系数为 ,所有项的系数和为 .
12.等比数列{}n a
中,12a a =22013
82019
a a a a +=+ ,1234a a a a = .
13.在ABC ∆中,角,,A B C 所对的边分别为,,a b c ,已
知s i n c o s c A a C =,则
C =
,若c =,ABC ∆
,则a b += . 14.已知函数222,0()2(1),0
x x x f x f x x -⎧+-≥=⎨+<⎩,则3
()2f -= ,若函数()()g x f x k =-有
无穷多个零点,则k 的取值范围是 .
15.已知,x y R ∈且221x y xy ++=,则x y xy ++的最小值为 .
16.已知平面向量,,a b c 满足,,015a b c a c b c ⋅==-=-=,则a b -的最大值 为 .
17.当[]1,4x ∈时,不等式322044ax bx a x ≤++≤恒成立,则7a b +的取值范围是 .
三、解答题(本大题共5小题,共74分.写出文字说明、证明过程或演算步骤.)
18.(本题满分14分)已知函数()2sin cos()3
f x x x π
=++
(Ⅰ)求函数()f x 的单调递减区间; (Ⅱ)求函数()f x 在区间0,2π⎡⎤⎢⎥⎣⎦
上的最大值及最小值.
19.(本题满分15分)已知在ABC ∆中,1AB =,2AC =.
(Ⅰ)若BAC ∠的平分线与边BC 交于点D ,求()
2AD AB AC ⋅-uuu r uu u r uuu r

(Ⅱ)若点E 为BC 的中点,求2211
AE BC
+uu u r uu u r 的最小值.
20.(本题满分15分)已知正项等差数列{}n a 满足:233
312n n S a a a =+++,其中n S 是数
列{}n a 的前n 项和.
(Ⅰ)求数列{}n a 的通项公式; (Ⅱ)令()()()
1
412121n n n n n
b a a -=--+,证明:1222
21
n n b b b n +++
+≤
+.
21.(本题满分15分)设函数(),x
f x e ax a a R =-+∈,其图象与x 轴交于12(,0),(,0)A x B x
两点,且12.x x <
(Ⅰ)求a 的取值范围;(Ⅱ)证明:0f '<.
22.(本题满分15分)已知函数2
()ln 2,.f x x ax bx a R =---∈ (Ⅰ)当2b =时,试讨论()f x 的单调性;
(Ⅱ)若对任意的3(,)b e
∈-∞-,方程()0f x =恒有2个不等的实根,求a 的取值范围.
杭高2019年11月期中考卷高二数学卷答案 BDDDC BCCBA 11.
4
π
,02=--y x
12. )
,(3,31 13. 2,)122,221(---
14. 5104,π
15. 0
16. 1- 17. )2,1( 18. 11 19. 12·
-=n n n a 20. (1)略;(2)
28
7
5 21. (1)1=y 或0143=-+y x ;(2)
324;(3)2
2
min =ST ,0=t 22. (1)2=AB (2)0 (3)45+±=x y。

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