综合测试2试卷
2023届广东省广州市普通高中毕业班综合测试(二)政治试卷
秘密★启用前试卷类型:B 2023年广州市普通高中毕业班综合测试(二)思想政治本试卷共8页,20小题,满分100分。
考试用时75分钟。
注意事项: 1.答卷前,考生务必用黑色字迹的钢笔或签字笔将自己的姓名、考生号、试室号和座位号填写在答题卡上。
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一、选择题:本大题共16小题,每小题3分,共48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.党的十四大提出建立社会主义市场经济体制的改革目标后,我们党对市场作用的认识主要经过了以下历程。
这说明①经济体制改革的核心问题是处理好政府和市场的关系②中国共产党在社会主义革命和建设中取得独创性理论成果③社会主义市场经济体制改革是为了使上层建筑适应经济基础的发展④中国共产党领导人民通过改革和探索,不断解放和发展社会生产力A.①②B.①④C.②③D.③④2.2020年6月,《国企改革三年行动方案(2020—2022年)》通过。
三年来,国有上市公司以混促改,完善公司治理,提高规范运作水平,深度转换机制。
推进国有企业混合所有制改革①扩大了混合所有制经济中国有成分的比重②有利于提高企业的活力和效率③有利于促进各种所有制资本优势互补④降低了企业的经营风险A.①②B.①④C.②③D.③④3.习近平总书记在党的二十大报告中指出,我们深入推进全面从严治党,坚持打铁必须自身硬,从制定和落实中央八项规定开局破题,提出和落实新时代党的建设总要求,经过不懈努力,党找到了自我革命这一跳出治乱兴衰历史周期率的第二个答案。
2022年广东省茂名市高考数学第二次综合测试试卷(解析版)
2022年广东省茂名市高考数学第二次综合测试试卷一、单选题:本题共8小题,每小题5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.已知集合A={x|﹣3≤x<5},B={x|y=},则A∩(∁R B)=()A.[﹣3,﹣)B.(﹣,5)C.[﹣3,﹣2)D.(﹣2,5)2.已知等差数列{a n}的前n项和为S n,若S3=6,S5=25,则a4=()A.6B.7C.8D.93.平面非零向量,满足||=||,|﹣|=||,则与的夹角为()A.B.C.D.4.已知f(x)=x﹣sin x,则不等式f(2m+1)+f(1﹣m)>0的解集为()A.(﹣∞,﹣2)B.(﹣2,+∞)C.(0,+∞)D.(﹣∞,0)5.由国家信息中心“一带一路”大数据中心等编写的《“一带一路”贸易合作大数据报告(2017)》发布,呈现了我国与“一带一路”沿线国家的贸易成果现状报告.由数据分析可知.在2011年到2016年这六年中,中国与“一带一路”沿线国家出口额和进口额图表如图,下列说法中正确的是中国与“一带一路”沿线国家出口额和进口额(亿美元)()A.中国与沿线国家贸易进口额的极差为1072.5亿美元B.中国与沿线国家贸易出口额的中位数不超过5782亿美元C.中国与沿线国家贸易顺差额逐年递增(贸易顺差额=贸易出口额一贸易进口额)D.中国与沿线国家前四年的贸易进口额比贸易出口额更稳定6.双碳,即碳达峰与碳中和的简称.2020年9月中国明确提出2030年实现“碳达峰”,2060年实现“碳中和”.为了实现这一目标,中国加大了电动汽车的研究与推广,到2060年,纯电动汽车在整体汽车中的渗透率有望超过70%,新型动力电池随之也迎来了蓬勃发展的机遇.Peukert于1898年提出蓄电池的容量C(单位:A•h),放电时间t(单位:h)与放电电流I(单位:A)之间关系的经验公式:C=I n•t,其中为Peukert常数.在电池容量不变的条件下,当放电电流I=10A时,放电时间t=57h,则当放电电流I=15A 时,放电时间为()A.28h B.28.5h C.29h D.29.5h7.已知0<α<,sin(﹣α)=,则的值为()A.B.C.D.8.已知双曲线的右焦点为F,左顶点为A,M为C的一条渐近线上一点,延长FM交y轴于点N,直线AM经过ON(其中O为坐标原点)的中点I,且|ON|=2|BM|,则双曲线C的离心率为()A.2B.C.D.二、多选题:本题共4小题,每小题5分,共20分.在每小题给出的选项中,有多项符合题目要求.全部选对的得5分,部分选对的得2分,有选错的得0分.(多选)9.已知复数z1=a2﹣1+ai,z2=1+(a﹣1)i(a∈R),若z1﹣2z2为实数,则下列说法中正确的有()A.B.z1z2=5+5iC.为纯虚数D.对应的点位于第三象限(多选)10.已知的展开式共有13项,则下列说法中正确的有()A.所有奇数项的二项式系数和为212B.所有项的系数和为312C.二项式系数最大的项为第6项或第7项D.有理项共5项(多选)11.已知函数f(x)=(cos x﹣|sin x|)•(cos x+sin x),下列说法正确的有()A.f(x)关于点对称B.f(x)在区间内单调递增C.若f(x1)+f(x2)=﹣2,则x1+x2=π+2kπ(k∈Z)D.f(x)的对称轴是(多选)12.棱长为4的正方体ABCD﹣A1B1C1D1中,E,F分别为棱A1D1,AA1的中点,,则下列说法中正确的有()A.三棱锥F﹣A1EG的体积为定值B.当时,平面EGC1截正方体所得截而的周长为C.直线FG与平面BCC1B1所成角的正切值的取值范围是D.当时,三棱锥A1﹣EFG的外接球的表面积为三、填空题:本题共4小题,每小题5分,共20分.13.已知正实数m,n满足m+2n=1,则的最小值为.14.正三棱锥S﹣ABC的底面边长为4,侧棱长为2,D为棱AC的中点,则异面直线SD 与AB所成角的余弦值为.15.以抛物线C:y2=4x的焦点F为圆心的圆交C于A,B两点,交C的准线于D,E两点,已知|AB|=8,则|DE|=.16.已知函数f(x)=,若存在实数t使得函数y=[f(x)]2﹣(t+2)f(x)+2t有7个不同的零点,则实数a的取值范围是.四、解答题:本题共6小题,共70分.解答应写出必要的文字说明,证明过程或演算步骤. 17.在△ABC中,角A,B,C的对边分别为a,b,c,a sin A﹣c sin C=(a﹣b)sin B,b=5,c cos A=1.(1)求C;(2)求△ABC的面积.18.冰壶是2022年2月4日至2月20日在中国举行的第24届冬季奥运会的比赛项目之一.冰壶比赛的场地如图所示,其中左端(投掷线MN的左侧)有一个发球区,运动员在发球区边沿的投掷线MN将冰壶掷出,使冰壶沿冰道滑行,冰道的右端有一圆形的营垒,以场上冰壶最终静止时距离营垒区圆心O的远近决定胜负.甲、乙两人进行投掷冰壶比赛,规定冰壶的重心落在圆O中,得3分,冰壶的重心落在圆环A中,得2分.冰壶的重心落在圆环B中,得1分,其余情况均得0分.已知甲、乙投掷冰壶的结果互不影响.甲、乙得3分的概率分别为,;甲、乙得2分的概率分别为,;甲、乙得1分的概率分别为,.(1)求甲、乙两人所得分数相同的概率;(2)设甲、乙两人所得的分数之和为X,求X的分布列和期望.19.如图所示的圆柱中,AB是圆O的直径,AA1,CC1为圆柱的母线,四边形ABCD是底面圆O的内接等腰梯形,且CD=BC=AB=AA1,E,F分别为A1D,CC1的中点.(1)证明:EF∥平面ABCD;(2)求平面AA1D与平面C1EB所成锐二面角的余弦值.20.已知数列{a n}满足,a1=2,a2=8,a n+2=4a n+1﹣3a n.(1)证明:数列{a n+1﹣a n}是等比数列;(2)若,求数列{b n}的前n项和T n.21.已知椭圆C:=1(a>2>b>0)的上顶点为A,右焦点为F,原点O到直线AF的距离为,△AOF的面积为1.(1)求椭圆C的方程;(2)过点F的直线l与C交于M,N两点过点M作ME⊥x轴于点E,过点N作NQ⊥x 轴于点Q,QM与NE交于点P,是否存在直线l使得△PMN的面积等于,若存在,求出直线l的方程;若不存在,请说明理由.22.已知函数f(x)=a cos x+x sin x+b在点处的切线方程为.(1)求函数f(x)在(﹣π,π)上的单调区间;(2)当时,是否存在实数m使得f(x)≤m(x﹣π)恒成立,若存在,求实数m的取值集合,若不存在,说明理由.(附:(π2+4)≈19.6,5π+4≈19.7).参考答案一、单选题:本题共8小题,每小题5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.已知集合A={x|﹣3≤x<5},B={x|y=},则A∩(∁R B)=()A.[﹣3,﹣)B.(﹣,5)C.[﹣3,﹣2)D.(﹣2,5)【分析】求出集合B,利用交集、补集的定义能求出A∩(∁R B).解:∵集合A={x|﹣3≤x<5},B={x|y=}={x|x≥﹣},∁R B={x|x<﹣},∴A∩(∁R B)={x|﹣3≤x<﹣}.故选:A.2.已知等差数列{a n}的前n项和为S n,若S3=6,S5=25,则a4=()A.6B.7C.8D.9【分析】利用等差数列的求和公式求解即可.解:设等差数列{a n}的公差为d,∵S3=6,S5=25,∴3a1+3d=6,5a1+10d=25,解得a1=﹣1,d=3,则a4=﹣1+3×3=8,故选:C.3.平面非零向量,满足||=||,|﹣|=||,则与的夹角为()A.B.C.D.【分析】根据题意,设与的夹角为θ,||=||=t,由数量积的运算性质可得(﹣)2=32,即2t2﹣2t2cosθ=3t2,变形可得cosθ的值,分析可得答案.解:根据题意,设与的夹角为θ,||=||=t,若|﹣|=||,则(﹣)2=32,即2t2﹣2t2cosθ=3t2,变形可得cosθ=﹣,又由0≤θ≤π,则θ=;故选:C.4.已知f(x)=x﹣sin x,则不等式f(2m+1)+f(1﹣m)>0的解集为()A.(﹣∞,﹣2)B.(﹣2,+∞)C.(0,+∞)D.(﹣∞,0)【分析】利用导数研究函数的单调性,利用定义判断函数为奇函数,把原不等式转化为关于m的一元一次不等式求解.解:f(x)=x﹣sin x,f′(x)=1﹣cos x≥0,∴f(x)=x﹣sin x在R上单调递增,又f(﹣x)=﹣x﹣sin(﹣x)=﹣x+sin x=﹣f(x),∴f(x)为R上的奇函数,则f(2m+1)+f(1﹣m)>0⇔f(2m+1)>f(m﹣1),可得2m+1>m﹣1,即m>﹣2.∴不等式f(2m+1)+f(1﹣m)>0的解集为(﹣2,+∞).故选:B.5.由国家信息中心“一带一路”大数据中心等编写的《“一带一路”贸易合作大数据报告(2017)》发布,呈现了我国与“一带一路”沿线国家的贸易成果现状报告.由数据分析可知.在2011年到2016年这六年中,中国与“一带一路”沿线国家出口额和进口额图表如图,下列说法中正确的是中国与“一带一路”沿线国家出口额和进口额(亿美元)()A.中国与沿线国家贸易进口额的极差为1072.5亿美元B.中国与沿线国家贸易出口额的中位数不超过5782亿美元C.中国与沿线国家贸易顺差额逐年递增(贸易顺差额=贸易出口额一贸易进口额)D.中国与沿线国家前四年的贸易进口额比贸易出口额更稳定【分析】根据图表中的数据,结合统计中的相关概念逐一计算判断即可得出答案.解:对于A,中国与沿海国家贸易进口额的极差这4833.6﹣3661.1=1172.5,故A错误;对于B,由已知图中的数据可得出口额的中位数为=5782.85,故B错误;对于C,2011年至2016年的贸易顺差额依次为:142.9,428.6,976.8,1536.8,2262.4,2213.7,2016年开始下降,故C错误;由图表可知中国与沿线国家前四年的贸易进口额比贸易出口额更稳定,故D正确.故选:D.6.双碳,即碳达峰与碳中和的简称.2020年9月中国明确提出2030年实现“碳达峰”,2060年实现“碳中和”.为了实现这一目标,中国加大了电动汽车的研究与推广,到2060年,纯电动汽车在整体汽车中的渗透率有望超过70%,新型动力电池随之也迎来了蓬勃发展的机遇.Peukert于1898年提出蓄电池的容量C(单位:A•h),放电时间t(单位:h)与放电电流I(单位:A)之间关系的经验公式:C=I n•t,其中为Peukert常数.在电池容量不变的条件下,当放电电流I=10A时,放电时间t=57h,则当放电电流I=15A 时,放电时间为()A.28h B.28.5h C.29h D.29.5h【分析】根据题意求出蓄电池的容量C,再把I=15A时代入,结合指数与对数的运算性质即可求出结果.解:根据题意可得C=57•10n,则当I=15A时,57•10n=15n•t,所以t=57=57=57•==28.5h,即当放电电流I=15A时,放电时间为28.5h,故选:B.7.已知0<α<,sin(﹣α)=,则的值为()A.B.C.D.【分析】由已知利用两角差的正弦公式可求得cosα﹣sinα=,两边平方,利用同角三角函数基本关系式可求得sinα•cosα=,进而可求cosα+sinα=,利用同角三角函数基本关系式化简所求即可求解.解:因为0<α<,sin(﹣α)=,可得cosα﹣sinα=,可得cosα﹣sinα=,两边平方,可得1﹣2sinα•cosα=,可得sinα•cosα=,所以cosα+sinα====,所以====.故选:C.8.已知双曲线的右焦点为F,左顶点为A,M为C的一条渐近线上一点,延长FM交y轴于点N,直线AM经过ON(其中O为坐标原点)的中点I,且|ON|=2|BM|,则双曲线C的离心率为()A.2B.C.D.【分析】由中点B,且|ON|=2|BM|得NF⊥OM,由点到直线距离公式得|FM|=b,从而得|OM|=|OA|=a,通过三角形全等证得△MNB为等边三角形,然后得,从而计算出离心率.解:记M为双曲线C:的渐近线bx﹣ay=0上的点,因为|ON|=2|BM|,且|OB|=|BN|,所以∠BOM=∠BMO,∠BMN=∠BNM.所以NF⊥OM.因为右焦点F(c,0)到渐近线bx﹣ay=0的距离,所以|OM|=|OA|=a.所以∠BMO=∠BAO,所以∠BOM=∠BAO,所以Rt△AOB全等于Rt△OMN,所以∠ABO=∠ONM,又因为∠MNB=∠NMB,∠ABO=∠NBM.所以△MNB为等边三角形,所以∠FNO=60°,所以∠MFO=30°,即,所以.故选:A.二、多选题:本题共4小题,每小题5分,共20分.在每小题给出的选项中,有多项符合题目要求.全部选对的得5分,部分选对的得2分,有选错的得0分.(多选)9.已知复数z1=a2﹣1+ai,z2=1+(a﹣1)i(a∈R),若z1﹣2z2为实数,则下列说法中正确的有()A.B.z1z2=5+5iC.为纯虚数D.对应的点位于第三象限【分析】根据已知条件,由z1﹣2z2为实数,求得a的值,依次计算,即可求解.解:∵z1﹣2z2为实数,∴a﹣2(a﹣1)=0,解得a=2,∴z1=3+2i,z2=1+i,,故A正确,z1z2=(3+2i)(1+i)=1+5i,故B错误,∵(1+i)2=2i,∴=(2i)5=32i,故C正确,∵z1=3+2i,z2=1+i,∴==,其对应的点(,)在第四象限,故D 错误.故选:AC.(多选)10.已知的展开式共有13项,则下列说法中正确的有()A.所有奇数项的二项式系数和为212B.所有项的系数和为312C.二项式系数最大的项为第6项或第7项D.有理项共5项【分析】由二项式定理结合二项式系数的求法及展开式有理项问题逐一判断即可得解.解:由的展开式共有13项,则n=12,对于选项A,由展开式二项式系数和为212,则所有奇数项的二项式系数和为211,即选项A错误;对于选项B,令x=1,得(2×)12=312,即所有项的系数和为312,即选项B正确;对于选项C,由的展开式共有13项,则二项式系数最大的项为第7项,即选项C错误;对于选项D,由(2x+)12展开式的通项公式为T r+1=212﹣r x,又0≤r≤12,则r=0、3、6、9、12时,12﹣,即展开式有理项共5项,即选项D正确,故选:BD.(多选)11.已知函数f(x)=(cos x﹣|sin x|)•(cos x+sin x),下列说法正确的有()A.f(x)关于点对称B.f(x)在区间内单调递增C.若f(x1)+f(x2)=﹣2,则x1+x2=π+2kπ(k∈Z)D.f(x)的对称轴是【分析】取特殊值判断A,化简函数表达式,作函数图象判断B,C,D.【解答】解;因为f(π)=1,f(﹣)=1,所以f(x)不关于点对称,故A错误;当x∈[2kπ,2kπ+π](k∈Z)时,f(x)=(cos x﹣sin x)(cos x+sin x)=cos2x,当x∈[2kπ+π,2kπ+2π](k∈Z)时,f(x)=(cos x+sin x)(cos x+sin x)=1+sin2x,作出f(x)的图象如图所示,由图象可知f(x)在区间内单调递增,故B正确;因为f(x1)+f(x2)=﹣2,所以f(x1)=﹣1,f(x2)=﹣1,x1=2k1π+,k1∈Z,x2=2k2π+,k2∈Z,所以x1+x2=π+2kπ(k∈Z),故C项正确;由图象可知f(x)的图象不关于x=﹣对称,故D项错误.故选:BC.(多选)12.棱长为4的正方体ABCD﹣A1B1C1D1中,E,F分别为棱A1D1,AA1的中点,,则下列说法中正确的有()A.三棱锥F﹣A1EG的体积为定值B.当时,平面EGC1截正方体所得截而的周长为C.直线FG与平面BCC1B1所成角的正切值的取值范围是D.当时,三棱锥A1﹣EFG的外接球的表面积为【分析】对于A,B1C∥平面ADD1A1,则G点到平面ADD1A1的距离为定值,从而得到三棱锥F﹣A1EG的体积为定值;对于B,延长C1G,交棱BB1于点M,则M是BB1的中点,取A1F的中点N,连接EN,MN,EC1,平面ENMC1为平面EGC1截正方体所得的截面,从而可求出截面周长;对于C,由FM⊥平面BCC1B1,则∠FGM为直线FG与平面BCC1B1所成角,由此能求出直线FG与平面BCC1B1所成角的正切值的取值范围;对于D,连接A1D,交EF于点J,则J为EF的中点,由球的性质可得球心在过点J且与GH平行的直线上,求出其半径可判断.解:对于A,∵=(0≤λ≤1),∴点G为线段B1C上的一个动点,又B1C∥平面ADD1A1,则G点到平面ADD1A1的距离为定值,∴三棱锥是定值,又,故A正确;延长C1G交棱BB1于点M,则==,即M是BB1的中点,取A1F的中点N,连接EN,MN,EC1,∵EN∥D1F,D1F∥C1M,∴EN∥C1M,∴平面ENMC1为平面EGC1截面正方体所得的截面,∵EC1=,EN=,MN=,∴平面EGC1截正方体所得截而的周长为,故B错误;由上可知FM⊥平面BCC1B1,当点G在B1C上移动时,连接MG,∴∠FGM为直线FG与平面BCC1B1所成角,∵MG的最小值为,最大值为2,由tan∠FGM==,∴tan∠FGM∈,∴直线FG与平面BCC1B1所成角的正切值的取值范围是,故C正确;对于D,如图,连接A1D,交EF于点J,则J为EF的中点,A1J=JE=JF=,在A1D上取点H,使=,连接GH,则GH∥CD,∴GH⊥平面ADD1A1,则GH=4,设三棱锥A1﹣EFG的外接球的球心O,则OA1=OG=OE=OF,由OA1=OE=OF,及,得点O在过眯J且与GH平行的直线上,设OJ=h,∵=h2+()2,OG2=(4﹣h)2+(2)2,∴,解得h=,∴OA12=,∴三棱锥A1﹣EFG的外接球的表面积为4=,故D正确.故选:ACD.三、填空题:本题共4小题,每小题5分,共20分.13.已知正实数m,n满足m+2n=1,则的最小值为17.【分析】化简表达式,利用基本不等式转化求解即可.解:正实数m,n满足m+2n=1,则=+1=()(m+2n)+1=9+≥9+=17,当且仅当m=2n=时,取等号.则的最小值为17.故答案为:17.14.正三棱锥S﹣ABC的底面边长为4,侧棱长为2,D为棱AC的中点,则异面直线SD 与AB所成角的余弦值为.【分析】取BC的点E,连接SE,DE,则∠SDE(或其补角)为异面直线SD与AB所成的角,利用余弦定理计算即可.解:取BC的点E,连接SE,DE,则∠SDE(或其补角)为异面直线SD与AB所成的角,由题知,所以.故答案为:.15.以抛物线C:y2=4x的焦点F为圆心的圆交C于A,B两点,交C的准线于D,E两点,已知|AB|=8,则|DE|=.【分析】设点A在第一象限,可得A(4,4),由此可确定圆的半径,利用可求得结果.解:由抛物线方程知:,∴F(1,0),不妨设点A在第一象限,如图所示,由|AB|=8,y2=4x得:A(4,4),∴圆的半径,∴.故答案为:.16.已知函数f(x)=,若存在实数t使得函数y=[f(x)]2﹣(t+2)f(x)+2t有7个不同的零点,则实数a的取值范围是[0,+∞).【分析】利用导数研究函数f(x)的性质,得单调性和极值,并作出函数的大致图象,由y=[f(x)]2﹣(t+2)f(x)+2t=0得f(x)=2或f(x)=t,然后分类讨论,它们一个有3个根,一个有4个根,由此可得参数范围.解:当x>1时,f(x)=,f′(x)=;当1<x<e时,f′(x)<0,f(x)单调递减,当x>e时,f′(x)>0,f(x)单调递增,故x=e时,f(x)min=f(e)=1;当x≤1时,f(x)=x3﹣3x+a,f′(x)=3x2﹣3,当x<﹣1时,f′(x)>0,f(x)单调递增,当﹣1<x<1时,f′(x)<0,f(x)单调递减,所以当x=﹣1时,f(x)有极大值f(﹣1)=2+a,当x=1时,f(1)=﹣2+a,作出f(x)=的大致图象如图,由题意知[f(x)]2﹣(t+2)f(x)+2t=0,即[f(x)﹣2][f(x)﹣t]=0有7个不同的实根,当f(x)=2有三个根,f(x)=t有四个实根,此时2+a=2或﹣2+a>2,得a=0或a>4;当f(x)=2有四个根时,f(x)=t有三个实根,此时﹣2+a≤2≤2+a,得0<a≤4,所以a≥0.故答案为:[0,+∞).四、解答题:本题共6小题,共70分.解答应写出必要的文字说明,证明过程或演算步骤. 17.在△ABC中,角A,B,C的对边分别为a,b,c,a sin A﹣c sin C=(a﹣b)sin B,b=5,c cos A=1.(1)求C;(2)求△ABC的面积.【分析】(1)由已知结合正弦定理及余弦定理可求cos C,进而可求C的值.(2)由已知结合正弦定理先求a的值,然后结合三角形面积公式即可求解.解:(1)因为a sin A﹣c sin C=(a﹣b)sin B,所以a2﹣c2=ab﹣b2,由余弦定理可得cos C==,由C为三角形内角,可得C=60°.(2)因为C=60°,b=5,所以由余弦定理可得c2=a2+b2﹣2ab cos C=a2+25﹣2×,又c cos A=1,可得c•==1,可得c2=a2﹣15,所以a2+25﹣2×=a2﹣15,解得a=8,所以△ABC的面积S=ab sin C==10.18.冰壶是2022年2月4日至2月20日在中国举行的第24届冬季奥运会的比赛项目之一.冰壶比赛的场地如图所示,其中左端(投掷线MN的左侧)有一个发球区,运动员在发球区边沿的投掷线MN将冰壶掷出,使冰壶沿冰道滑行,冰道的右端有一圆形的营垒,以场上冰壶最终静止时距离营垒区圆心O的远近决定胜负.甲、乙两人进行投掷冰壶比赛,规定冰壶的重心落在圆O中,得3分,冰壶的重心落在圆环A中,得2分.冰壶的重心落在圆环B中,得1分,其余情况均得0分.已知甲、乙投掷冰壶的结果互不影响.甲、乙得3分的概率分别为,;甲、乙得2分的概率分别为,;甲、乙得1分的概率分别为,.(1)求甲、乙两人所得分数相同的概率;(2)设甲、乙两人所得的分数之和为X,求X的分布列和期望.【分析】(1)求出甲乙二人都得0分的概率,然后由两人同时得0分、1分、2分、3分计算概率并相加即可;(2)由题意X可能取值为0,1,2,3,4,5,6,分别计算出概率得分布列,由期望公式计算期望.解:(1)由题意知甲得0分的概率为,乙得0分的概率为,所以甲、乙两人所得分数相同的概率为;(2)X可能取值为0,1,2,3,4,5,6,则,,,,.,,所以,随机变量X的分布列为:X0123456P所以.19.如图所示的圆柱中,AB是圆O的直径,AA1,CC1为圆柱的母线,四边形ABCD是底面圆O的内接等腰梯形,且CD=BC=AB=AA1,E,F分别为A1D,CC1的中点.(1)证明:EF∥平面ABCD;(2)求平面AA1D与平面C1EB所成锐二面角的余弦值.【分析】(1)取AA1的中点G,连接EG,FG,AC,可证明四边形AGFC是平行四边形,从而证明平面EFG∥平面ABCD,从而得证.(2)题意知CA,CB,CC1两两垂直,以C为坐标原点,分别以CA,CB,CC1所在直线为x,y,z轴建立空间直角坐标系,利用向量法求解即可.【解答】(1)证明:取AA1的中点G,连接EG,FG,AC,因为EG∥AD,EG⊄平面ABCD,AD⊂平面ABCD,所以EG∥平面ABCD,因为AG∥CF,AG=CF,所以四边形AGFC是平行四边形,FG∥AC,又FG⊄平面ABCD,AD⊂平面ABCD,所以FG∥平面ABCD,因为FG⋂EG=G,所以平面EFG∥平面ABCD,因为EF⊂平面ABCD,所以EF∥平面ABCD.(2)解:设,由AD=CD=BC,得∠DAB=∠ABC=60°,因为AC⊥BC,所以,由题意知CA,CB,CC1两两垂直,以C为坐标原点,分别以CA,CB,CC1所在直线为x,y,z轴建立空间直角坐标系,则,,B(0,2,0),C1(0,0,4),,,所以,,设平面C1EB的一个法向量为,则,取z=1,得,连接BD,因为BD⊥AD,BD⊥AA1,AD∩AA1=A,所以BD⊥平面AA1D,所以平面AA1D的一个法向量为,所以,所以平面AA1D与平面C1EB所成锐二面角的余弦值为.20.已知数列{a n}满足,a1=2,a2=8,a n+2=4a n+1﹣3a n.(1)证明:数列{a n+1﹣a n}是等比数列;(2)若,求数列{b n}的前n项和T n.【分析】(1)由a n+2=4a n+1﹣3a n,变形为:a n+2﹣a n+1=3(a n+1﹣a n),a2﹣a1=6,即可证明结论.(2)由(1)可得:a n+1﹣a n=6×3n﹣1=2×3n,变形为:a n+1﹣3n+1=a n﹣3n,可得a n=3n ﹣1,b n==(﹣1)n[+],对n分类讨论即可得出数列{b n}的前n项和T n.解:(1)证明:a n+2=4a n+1﹣3a n,变形为:a n+2﹣a n+1=3(a n+1﹣a n),a2﹣a1=6,∴数列{a n+1﹣a n}是等比数列,首项为6,公比为3.(2)由(1)可得:a n+1﹣a n=6×3n﹣1=2×3n,变形为:a n+1﹣3n+1=a n﹣3n,a1﹣3=﹣1,∴a n﹣3n=﹣1,∴a n=3n﹣1,∴==(﹣1)n[+],∴n=2k(k∈N*),数列{b n}的前n项和T n=﹣(+)+(+)﹣(+)+…﹣[+]+[+]=﹣+.n=2k﹣1(k∈N*),数列{b n}的前n项和T n=﹣(+)+(+)﹣(+)+…+[+]﹣[+]=﹣﹣.21.已知椭圆C:=1(a>2>b>0)的上顶点为A,右焦点为F,原点O到直线AF的距离为,△AOF的面积为1.(1)求椭圆C的方程;(2)过点F的直线l与C交于M,N两点过点M作ME⊥x轴于点E,过点N作NQ⊥x 轴于点Q,QM与NE交于点P,是否存在直线l使得△PMN的面积等于,若存在,求出直线l的方程;若不存在,请说明理由.【分析】(1)根据题意列出关于a,b,c的方程,可求得答案;(2)设直线MN方程,并联立椭圆方程,得到根与系数的关系式,求得相关点的坐标,利用三角形面积之间的关系表示出,列方程求得m,可得答案.解:(1)由题意知A(0,b),F(c,0),因为△AOF的面积为1,所以.又直线AF的方程,即bx+cy﹣bc=0,因为点O到直线AF的距离为,所以,解得c=2,b=1,,所以椭圆C的方程为.(2)依题意,当直线MN斜率为0时,不符合题意;当直线斜率不为0时,设直线MN方程为x=my+2(m≠0),联立,得(m2+5)y2+4my﹣1=0,易知Δ=16m2+4(m2+5)=20(m2+1)>0.设M(x1,y1),N(x2,y2),则,,因为ME⊥x轴,NQ⊥x轴,所以E(x1,0),Q(x2,0),所以直线QM:①,直线NE:②,联立①②解得,因为ME∥NQ,ME与直线平行,所以,因为,所以,由,得m4﹣6m2+9=0,解得,故存在直线l的方程为或,使得△PMN的面积等于.22.已知函数f(x)=a cos x+x sin x+b在点处的切线方程为.(1)求函数f(x)在(﹣π,π)上的单调区间;(2)当时,是否存在实数m使得f(x)≤m(x﹣π)恒成立,若存在,求实数m的取值集合,若不存在,说明理由.(附:(π2+4)≈19.6,5π+4≈19.7).【分析】(1)根据切线方程,结合导数的几何意义,求出a,b的值,再由导数得出单调区间.(2)由题意g(x)=x sin x+cos x+1﹣m(x﹣π),只需g(x)max≤0,又注意g(π)=0,由题意π必为g(x)的最大值点,从而为g(x)的极大值点,必有g′(π)=0,再证明检验即可.解:(1)由题意知f()=+1,即+b=+1,得b=1,因为f′(x)=﹣a sin x+sin x+x cos x,所以f′()=﹣a+1=0,得a=1,所以f′(x)=x cos x,当0<x<π时,令f′(x)>0,得0<x<,令f′(x)<0,得<x<π,当﹣π<x<0时,令f′(x)>0,得﹣π<x<﹣,令f′(x)<0,得﹣<x<0,所以f(x)在(﹣π,﹣),(0,)上单调递增,在(,π),(﹣,0)上单调递减.(2)假设存在实数m,使得f(x)≤m(x﹣π)在x∈[0,]上恒成立,即x sin x+cos x+1﹣m(x﹣π)≤0在[0,]上恒成立,令g(x)=x sin x+cos x+1﹣m(x﹣π),只需g(x)max≤0,又g(π)=0,所以若x sin x+cos x+1﹣m(x﹣π)≤0在[0,]上恒成立,π必为g(x)的最大值点,从而为g(x)的极大值点,必有g′(π)=0,由g′(x)=x cos x﹣m,得g′(π)=﹣π﹣m=0,解得m=﹣π,下面证明m=﹣π符合题意,当m=﹣π时,g′(x)=x cos x+π,令h(x)=g′(x),则h′(x)=cos x﹣x sin x,①当x∈[0,]时,g′(x)>0,所以g(x)在[0,]上单调递增,当x∈[,π]时,h′(x)<0,所以h(x)=g′(x)单调递减,所以当x∈[,π)时,g′(x)>g′(π)=0,所以g(x)在[,π]上单调递增,由g(x)在[0,]和[,π]上单调递增得,g(x)在[0,π]上单调递增.②当x∈[π,]时,令F(x)=h′(x)=cos x﹣x sin x,由F′(x)=﹣2sin x﹣x cos x,得F′(x)>0,F(x)在[π,]上单调递增,因为F(π)=﹣1<0,F()=(﹣1)>0,所以由零点存在定理知存在x1∈(π,),使得F(x1)=0,当x∈[π,x1)时,F(x)<0,即h′(x)<0,h(x)单调递减,即g′(x)单调递减,当x∈(x1,]时,F(x)>0,即h′(x)>0,h(x)单调递增,即g′(x)单调递增,因为g′(π)=0,g′()=π(1﹣)>0,所以由零点的存在定理得,存在x2∈(x1,),使得g′(x2)=0,当x∈[π,x2)时,g′(x)<0,g(x)单调递减,当x∈(x2,]时,g′(x)>0,g(x)单调递增,又g(π)=0,g()=﹣(+1)++1<0,综上所述,g(x)max=0,符合题意,综上所述,m的取值集合为{﹣π}.。
期末复习综合测试卷(二)+++++2022-2023学年部编版语文九年级下册+
部编版九年级语文下册期末总复习综合性检测试卷(二)一、积累与运用。
(20分)1.下列加点字的注音完全正确的一项是()A.国殇.(shāng)伫.立(chù)诡谲.(jué)波澜.(lán)B.睥.睨(bì)稽.首(qǐ)行.头(xíng)污秽.(huì)C.踌躇.(chú)囫.囵(hú)童谣.(yáo)鞭挞.(tà)D.瘦削.(xuē)咀嚼.(jué)羡.慕(xiàn)弄.堂(nòng)2.下列词语书写完全正确的一项是()(3分)A.掬躬收揽犀利骂骂咧咧B.雷霆幌子侦辑咬牙跺脚C.怯懦忌讳拾掇摄手摄脚D.捣蛋凄惨竹匾喃喃自语3.下列各句中加点的成语运用不正确的一项是()(3分)A.母亲在家庭里极能任劳任怨....。
她性格和蔼,没有打骂过我们,也没有同任何人吵过架。
B.父亲老实厚道低三下四....累了一辈子,没人说过他有地位,父亲也从没觉得自己有地位。
C.但这是没有办法的,只得裹一条毯子,横着心躺下去。
因为实在太疲倦,一会儿就酣然..入梦了...。
D.假如一个男人跟朋友和熟人见面时彬彬有礼....,可是在家里对妻子儿女动不动就大发雷霆——那就可以肯定他不是一个有教养的人。
4.下列句子没有语病的一项是()(3分)A.是否注重用非语言沟通与孩子交流,才能让孩子培养更加优秀的语言能力。
B.那种不顾生态环境,片面强调经济效益,无疑是杀鸡取卵,到头来只能是得不偿失。
C.风油精的主要成分是由薄荷脑、樟脑桉油、丁香酚、水杨酸甲酯等配制而成的。
D.在各方重点扶持下,我们中国传统戏剧这颗璀璨明珠终会再焕光彩。
5.综合性学习材料:每年在湖北龙感湖自然保护区越冬的天鹅超过了6万只,为了让天鹅安心栖息,管理站划出300亩稻田,不予收割,为鸟留食。
但前不久发生的一件事,让管理站开始对天鹅“不近人情”:一只受伤的小天鹅在经过两周的救治后,放回水城时却趴在岸边一动不动。
2020年职业能力综合测试试卷(二)
资料一随着我国中老年人独居化趋势出现,越来越多的中老年人开始饲养宠物。
全国各地有一大批宠物食品企业通过受托加工、委托加工、合作生产等方式生产和销售各种品牌的宠物日用粮(包括宠物干粮和宠物湿粮)。
由于宠物食品行业的快速发展,且行业进入门槛不高,吸引了大量投资者关注和涌入。
目前,宠物食品企业生产和销售的宠物日用粮以口味和种类相对单一的传统宠物食品为主。
消费者认为现有的传统宠物食品同质化程度较高。
为了满足消费者对宠物食品口味和营养的需求,宠物食品市场已逐步出现口味和营养多样化的新兴宠物食品(如各种配方食品),并对传统宠物食品构成一定挑战。
A有限责任公司成立于20×3年,是一家以宠物日用粮为主要产品的宠物食品生产企业。
宠物日用粮的原材料主要由谷物、肉类和皮革构成,占宠物日用粮产品成本的较大比重。
因此,原材料价格的波动对宠物食品生产企业的经营业绩具有较大影响。
虽然原材料所属的农林牧副渔行业具有较强的周期属性,但我国农林牧副渔行业产量在保持多年稳定的基础上每年均略有增长。
并且,我国宠物食品行业原材料消耗量占上游行业生产规模的比例非常低。
A有限责任公司通过公开招标的方式选择多家原材料供应商,并与之签署长期合作协议。
随着互联网技术的大规模应用,消费者非常容易在电商平台查询到不同品牌宠物食品的市场价格及产品特点。
同时,为满足养宠人士的相互交流,市场上出现多款APP和应用小程序,既能分享宠物饲养经验,又能直接接入电商平台。
A有限责任公司顺应新的营销模式,大力拓展线上销售渠道,逐步形成“线上+线下”双渠道销售格局。
【要求】1.根据资料一,利用波特五力模型对A有限责任公司所处的国内宠物食品生产行业的五种竞争力进行分析,简要说明理由;简述应对五种竞争力的战略。
1.为满足上市要求,A有限责任公司于20×7年9月按经审计的账面净资产折股,整体变更为股份有限公司。
2.A有限责任公司原实际控制人为前任董事长刘一。
黑龙江省齐齐哈尔市第二十八中学2019——2020学年九年级下学期44月综合测试(二)英语试卷(pdf版)
二十八中综合测试(二)英语试卷第一部分语言知识运用(共计55分)I.单项选择(每小题1分,共10分)从每小题所给的A、B、C三个选项中,选出一个能填入空白处的最佳选项。
1.---Have you seen_____film Lost In Russia directed by Xu Zheng?---Yes.I think it’s______interesting one.A.a;theB.the;anC.a;an2.When we read a piece of news online,we’d better make sure it’s true beforesending it to______.If not,we may spread something bad.A.the otherB.anotherC.others3.______October1st,2019,millions of Chinese people sat by the TV,watching themilitary parade(阅兵).A.InB.OnC.At4.The5G technology can help doctors treat patients who are______kilometers away.A.hundreds ofB.hundred ofC.five hundreds5.---How do you study English for a test?---By studying with a group.I find it______to get good grades.A.thankfulB.helpfulC.careful6.---What do you usually do after getting up in the morning?---I______.ed to exerciseB.am used to readingC.am used to swim7.The government asked schools to______online courses for students from March2nd,2020.A.protectB.preventC.provide8.---Excuse me,can you tell me the way to the library?---Yes.Go down this road______you see a white building on your right.A.unlessB.untilC.since9.---The Hong Kong-Zhuhai-Macao Bridge is the world’s longest cross-seabridge.Do you know______it is?---It’s55kilometres.A.how longB.how farC.how soon10.---I am afraid that I’ll fail in the exam.It’s too difficult for me.---______You shouldn’t give up before trying.A.Don’t mention it!B.Pardon me?e on!1–5BCBAB6–10BCBACII.完形填空(每小题1分,共15分)One evening last summer,when I asked my__11__son,Ray,for help with dinner, his reply shocked me.“What’s a colander(漏勺)?”he asked.I could only blame(责备) __12__.In the family,__13__else went into the kitchen except me.But that evening, as I__14__to him that a colander is the thing with holes in it,I wondered what else I hadn’t__15__Ray for.As parents,we often focus on our children’s confidence and character.We perhaps don’t always consider that we are__16__raising someone’s future roommates, boyfriends,husbands or fathers.I wanted to know that I’d raised a boy__17__would never ask his wife“What’s for dinner?”So I__18__a plan:I would__19__Ray a private home course.I was__20__to find that he didn’t say no.For two hours,three days a week,Ray was all mine.I knew that he would rather__21__basketball with his friends than__22__to mend socks with his mother,but in fact he was learning,and more than just housekeeping.“I’m thankful for what you do__23__a mom,”he told me one day.Now,not only can he make his own__24__,but also he can make a big meal for his family.That’s what I call a man.I’m really glad that I prepared__25__great a present for his future wife.11.A.14-year-old B.14-years-old C.14years old12.A.him B.myself C.me13.A.everybody B.somebody C.nobody14.A.explained B.advised C.told15.A.planned B.prepared C.produced16.A.still B.also C.either17.A.what B.who C.which18.A.came up with B.came out C.came into19.A.provide B.advise C.offer20.A.cheerful B.nervous C.patient21.A.play B.playing C.to play22.A.ask B.learn C.check23.A.as B.for C.like24.A.decision B.dinner C.choice25.A.such B.too C.so11–15ABCAB16–20BBACA21–25ABABC阅读短文,从每小题所给的A、B、C三个选项中,选出一个能填入空白处的最佳选项。
2023届四川省宜宾市高三下学期(二诊)理科综合试卷及答案
注意事项:宜宾市普通高中2020级第二次诊断性测试理科综合能力测试I .答卷前,考生务必将自己的班级、姓名、考号填写在答题卡上。
2.回答选择题时,逃出每小题答案后,用铅笔把答题卡上对j主题目的答案标-'.}涂黑。
如需改动,用橡皮擦干净后,再选涂其它答案标号。
回答非�择题时,将答案写在答题卡上。
写在本试卷上无效。
3.考试结束后,将本试卷和答Im卡一并交囚。
可能用到的相对原子质量:H I Cl2 016 K39 Fe56 Se79一、选择题:本题共13小题,每小题6分,共78分。
在每小题给出的四个选项中,只有一项是符合题目要求的。
I.四川人爱喝茶,爱坐茶铺着川剧、I听消音、遛鸟打盹、看闲书、唠家常,迫适自在,自得其乐,这就是川入2丧事。
下列有关茶叶中元索和化合物的叙述,正确的是A.新鲜茶叶中含量最高的化学元素是CB.新鲜茶叶在炒审lj过程中主要失去的水是自由水C茶叶细胞内蛋白质中通常含有微量元素Fe或SD.晒干后的茶叶中含量最高的化合物是无机盐2.研究表明,癌细胞浴酶体中的pH低于正常细胞。
BODIPY荧光染料对pH不敏感,具良好的光学和化学稳定性。
以BODIPY为母体结构,以即匠嗦环为溶酶体定位基团,设计成溶酶体荧光探针。
该探牵十在中性或碱性条件下不显荧光,在E酸性条件下荧光强度升高。
下列说法,错误的是A.荧光探针能进入越细胞的溶酶体,是因为其对pH不敏感B.洛自每体内的酸性环境有利于其分解衰老、损伤的细胞器C若某区域的荧光强度较强,则该区域的细胞可能是癌细胞D.洛酶体执行功能的过程中,存在生物膜的流动现象3.徒步是一种健康的户外运动方式,但较长时间的徒步会使脚掌磨出“7](泡”,几天后“7)<泡”又消失了。
下列有关叙述,正确的是A.较长时间的快速徒步会使人体出汗且体温明显上升B.人体内环境中发生丙酬酸氧化分(ft!�为徒步提供能量c.7)<j包自行消失的原因是其中的液体渗入到毛细血管和毛细淋巴管中D.可以使用针成剪刀直接将水泡戳破,从而将7](泡里面的*排出4.糖皮质激素(GC)是机体内极为亟耍的一类调节分子,它对机体的发育、生长、代谢以及免疫功能等起着重要调节作用,是机体应激反应最重要的调节激素,也是l临床上使用最为广泛而有效的抗炎和免疫仰制剂。
2023-2024学年人教版七年级数学下册期末综合模拟测试2
2023-2024学年人教版七年级数学下册期末综合模拟测试2一、单选题1.下列实数是无理数的是( ) A .()01π-B .3π C .5 D .3.142.如图,一辆汽车在笔直的公路上由A 向B 行驶,M 是学校的位置,当汽车行驶到下列哪一位置时,汽车离学校最近( )A .D 点B .E 点C .F 点D .N 点3.下列说法正确的是()A .一个数的算术平方根一定是正数B .1的立方根是1±C 5=±D .2是4的平方根4.在平面直角坐标中,点A (4,-1)所在的象限是( ) A .第一象限B .第二象限C .第三象限D .第四象限5.为了了解某市参加中考的32000名学生的体重情况,抽查了其中1600名学生的体重进行统计分析.下列叙述正确的是( ) A .32000名学生是总体 B .1600名学生的体重是总体的一个样本 C .每名学生是总体的一个个体D .样本容量是1600名6.如图,直线a b ∥,一块直角三角形ABC 按如图所示放置,若150∠=︒,则2∠的度数是( )A .105︒B .110︒C .115︒D .130︒7 ) A .4和5之间B .5和6之间C .6和7之间D .7和8之间8.若31a ->,两边都除以3-,得( )A .13a <-B .13a >-C .3a <-D .3a >-9.已知点(1,3)A m -与点(2,1)B n -关于x 轴对称,则m n +的值为( ) A .1B .1-C .0D .310.在解二元一次方程组259236x y x y +=⎧⎨-=⎩①②时,用①-②消去未知数x 后,得到的方程是( )A .23y =B .215y =C .83y =D .815y =11.如果关于x 的不等式()11a x a +>+的解集为1x <,则a 的取值范围是( )A .0a <B .1a <-C .1a >D .1>-a12.如图,90C ∠=︒,将直角三角形ABC 沿着射线BC 方向平移5cm ,得三角形A B C ''',已知3cm BC =,4cm AC =,则阴影部分的面积为( )2cm .A .18B .14C .20D .2213.为了解中学生获取资讯的主要渠道,随机抽取50名中学生进行问卷调查,调查问卷设置了“A :报纸,B :电视,C :网络,D :身边的人,E :其他”五个选项(五项中必选且只能选一项),根据调查结果绘制了如图所示的条形图(D 组数据被污染).该调查的调查方式及D 组对应的频率分别为( )A .全面调查;52%B .全面调查;48%C .抽样调查;52%D .抽样调查;48%14.《九章算术》中记载这样一个问题:“以绳测井,若将绳三折测之,绳多四尺;若将绳四折测之,绳多一尺.问绳长、井深各几何?”题意是:用绳子测量水井深度,如果将绳子折成三等份,那么每等份井外余绳四尺;如果将绳子折成四等份,那么每等份井外余绳一尺.问绳长和井深各多少尺?若设绳长、井深分别为x 、y 尺,则符合题意的方程组是( )A .()()3441y x y x ⎧=+⎪⎨=+⎪⎩B .3441y x y x =+⎧⎨=+⎩C .()()3441x y x y ⎧=+⎪⎨=+⎪⎩D .3441x y x y =+⎧⎨=+⎩15.如图,在平面直角坐标系中,有若干个整点,按图中→方向排列,即()0,0→ 0,1 →()1,1→()2,2→ 2,3 →()3,3→()4,4,……,则按此规律排列下去第23个点的坐标为( )A .(13,13)B .(14,14)C .(15,15)D .(14,15)二、填空题16.图,∠1+∠2=180°,∠3=110°,则∠4=度.17.已知43x y +=,且17y -<≤则x 的取值范围是.18.在已知点A 的坐标是()2,4A -,线段AB y ∥轴,且5AB =,则B 点的坐标是. 19.已知关于x 的不等式组0521x a x -≥⎧⎨->⎩只有四个整数解,则实数a 的取值范围是.三、解答题 20.计算:1-;3π- 21.解不等式组23(1)2223x x x x +<+⎧⎪+⎨-≤⎪⎩.22.有A 、B 两种型号台灯,若购买2台A 型台灯和6台B 型台灯共需610元.若购买6台A 型台灯和2台B 型台灯共需470元. (1)求A 、B 两种型号台灯每台分别多少元?(2)采购员小红想采购A 、B 两种型号台灯共30台,且总费用不超过2200元,则最多能采购B 型台灯多少台?23.疫情期间,学校为了解学生最喜欢以下4门网课:A .数学,B .语文,C .英语,D .道德与法制中的哪一门学科,随机抽取了部分学生进行调查,并将调查结果绘制成了两幅不完整的统计图(如图1,图2),请回答下列问题:(1)这次被调查的学生共有多少人? (2)补全图2中的条形统计图;(3)图1扇形统计图中,B ,C ,D 所占的百分比各是多少?24.二元一次方程组23253x y m x y m +=+⎧⎨+=-⎩的解x ,y 的值是一个等腰三角形两边的长,且这个等腰三角形的周长为5,求腰的长.(注:等腰三角形中相等的两条边叫做等腰三角形的腰) 25.如图,已知AD BC ⊥,EF BC ⊥,垂足分别为D 、F ,23180∠+∠=︒,试说明:GDC B ∠=∠.请补充说明过程,并在括号内填上相应的理由.解:AD BC ⊥Q ,EF BC ⊥(已知)90ADB EFB ∴∠=∠=︒(), ∴EF AD ∥(), ∴2180+∠=︒().又23180∠+∠=︒Q (已知),13∠∠∴=(),∴AB P (), ∴GDC B ∠=∠().26.某商场有A 、B 两种商品,每件的进价分别为15元、35元.商场销售5件A 商品和2件B 商品,可获得利润45元;销售8件A 商品和4件B 商品,可获得利润80元. (1)求A 、B 两种商品的销售单价;(2)如果该商场计划购进A 、B 两种商品共80件,用于进货资金最多投入2 000元,但又要确保获利至少590元,请问有那几种进货方案?27.在综合与实践课上,老师让同学们以“两条平行线、AB CD 和一块含60︒角的直角三角尺EFG (90EFG ∠=︒,60EGF ∠=︒)”为主题开展数学活动.(1)如图1,若三角尺的60︒角的顶点G 放在CD 上,若221∠=∠,求1∠的度数; (2)如图2,小颖把三角尺的两个锐角的顶点E 、G 分别放在AB 和CD 上,请你探索并说明AEF ∠与FGC ∠间的数量关系;(3)如图3,小亮把三角尺的直角顶点F 放在CD 上,30︒角的顶点E 落在AB 上,请你探索并说明AEG ∠与CFG ∠间的数量关系.。
广东省广州市2024届普通高中毕业班综合测试(二)数学试卷
广东省广州市2024届普通高中毕业班综合测试(二)数学试卷学校:___________姓名:___________班级:___________考号:___________四、解答题15.治疗某种疾病有一种传统药和一种创新药,治疗效果对比试验数据如下:服用创新药的50名患者中有40名治愈;服用传统药的400名患者中有120名未治愈.(1)补全22´列联表(单位:人),并根据小概率值0.05a=的独立性检验,分析创新药的;疗效是否比传统的疗效药好(1)若000,3x y ==,求直线AB 的方程;(2)若点P 在直线3y x =+上,记PFA V 的面积为1,S PFB △的面积为2S ,求12S S ×的最小值;(3)证明:PFA BFP △∽△.19.已知函数()()21e x f x a x x -=++.(1)讨论()f x 的零点个数;(2)若()f x 存在两个极值点,记0x 为()f x 的极大值点,1x 为()f x 的零点,证明:0122x x ->.【分析】根据题意分析可知()f x为偶函数,结合偶函数可得()()f x f x++-=,进而210可知6为()f x的周期,赋值可知()21f=-,结合周期性运算求解.【详解】由题意可知:函数()f x的定义域为R,因为()()()f x f x f x-++=-,11++-=,则()()()11f x f x f x可得()()f x为偶函数,=f x f x-,所以()由()()()f x f x f x++-=+,21f x f x f x11++-=可得()()()即()()()f x f x++-=,21210++=+,整理得()()f x f x f x可得()()()()++-=++=,f x f x f x f x330则()()f x f x+=,6f x f x630+++=,可得()()所以6为()f x的周期,由()()()()f x f x f x f++-==,11,02令0x=,可得()()()f f f+==,可得()11112f=;令1x=,可得()()()+==,可得()212011f f ff=-;所以()()()()+=+=-+=.202420121f f f f故选:A.【点睛】方法点睛:函数的性质主要是函数的奇偶性、单调性和周期性以及函数图象的对称性,在解题中根据问题的条件通过变换函数的解析式或者已知的函数关系,推证函数的性质,根据函数的性质解决问题.9.BCD。
(苏科版)初中数学八年级上册 第2章综合测试试卷02及答案
第2章综合测试一、单选题1.图书馆的标志是浓缩了图书馆文化的符号,下列图书馆标志中,不是轴对称的是( ).A B C D2.如图,ABC △中,70A Ð=°,点E 、F 在AB 、AC 上,沿EF 向内折叠AEF △,得DEF △,则图中12Ð+Ð的和等于( )A .70°B .90°C .120°D .140°3.点P 在AOB Ð的平分线上,点P 到OA 边的距离等于6,点Q 是OB 边上的任意一点,则下列选项正确的是()A .6PQ >B .6PQ ≥C .6PQ <D .6PQ ≤4.如图,在四边形ABCD 中,50C Ð=°,90B D Ð=Ð=°,E ,F 分别是BC ,DC 上的点,当AEF △的周长最小时,EAF Ð的度数为( )A .50°B .60°C .70°D .80°5.如图,射线OC 是AOB Ð的角平分线,D 是射线OC 上一点,DP OA ^于点P ,4DP =,若点Q 是射线OB 上一点,3OQ =,则ODQ △的面积是( )A .3B .4C .5D .66.如图,在ABC △中,BA BC =,120ABC Ð=°,AB 的垂直平分线交AC 于点M ,交AB 于点E ,BC的垂直平分线交AC 于点N ,交BC 于点F ,连接BM ,BN ,若24AC =,则BMN △的周长是( )A .36B .24C .18D .167.已知30AOB Ð=°,点P 在AOB Ð的内部,点1P 和点P 关于OA 对称,点2P 和点P 关于OB 对称,则1P 、O 、2P 三点构成的三角形是()A .直角三角形B .钝角三角形C .等腰直角三角形D .等边三角形8.如图,在ABC △中AB AC =,4BC =,面积是20,AC 的垂直平分线EF 分别交AC ,AB 边于E ,F 点,若点D 为BC 边的中点,点M 为线段上一动点,则CDM △周长的最小值为( ).A .6B .8C .10D .129.如图在ABC △中,BO ,CO 分别平分ABC Ð,ACB Ð,交于O ,CE 为外角ACD Ð的平分线,BO 的延长线交CE 于点E ,记1BAC Ð=Ð,2BEC Ð=Ð,则以下结论①122Ð=Ð,②32BOC Ð=Ð,③901BOC Ð=°+Ð,④902BOC Ð=°+Ð正确的是( )A .①②③B .①③④C .①④D .①②④10.两组邻边分别相等的四边形叫做“筝形”,如图,四边形ABCD 是一个筝形,其中AD CD =,AB CB =,在探究筝形的性质时,得到如下结论:①ABD CBD ≌△△;②AC BD ^;③12ABCD AC BD =×四边形的面积,其中正确的结论有( )A .①②B .①③C .②③D .①③②二、填空题11.已知等腰三角形的其中两边长为6 cm 和8 cm ,则这个三角形的周长为________cm .12.等腰三角形的顶角是50°,则它一腰上的高与底边的夹角为________.13.若等腰三角形一腰上的高与腰长之比为1:2,则该等腰三角形顶角的度数为________.14.如图,Rt ABC △中,90ACB Ð=°,50A Ð=°,将其折叠,使点A 落在边CB 上'A 处,折痕为CD ,则'A DB Ð=________度.15.如图,ABC △的三边AB 、BC 、CA 长分别是40、60、80,其三条角平分线将ABC △分为三个三角形,则BCO ::ABO CAO S S S △△△等于________.16.如图,已知钝角三角形ABC 的面积为20,最长边10AB =,BD 平分ABC Ð,点M 、N 分别是BD 、BC 上的动点,则CM MN +的最小值为________.17.如图,等边ABC △中,D ,E 分别是AB 、BC 边上的一点,且,则DPC Ð=________°.18.如图,已知:=30MON а,点1A 、2A 、3A 在射线ON 上,点1B 、2B 、3B …在射线OM 上,112A B A △、223A B A △、433A B A △…均为等边三角形,若1OA a =,则667A B A △的边长为________.三、综合题19.作图题(保留作图痕迹,不写画法).(1)请在坐标系中,画出ABC △关于y 轴对称的''C'A B △.(2)如图(2),A 与B 是两个居住社区,OC 与OD 是两条交汇的公路,欲建立一个超市M ,使它到A 、B 两个社区的距离相等,且到两条公路OC 、OD 的距离也相等.请利用尺规作图,确定超市M 的位置.20.如图,已知点D ,E 分别是ABC △的边BA 和BC 延长线上的点,作DAC Ð的平分线AF ,若AF BC ∥.(1)求证:ABC △是等腰三角形;(2)作ACE Ð的平分线交AF 于点G ,若40B Ð=°,求ACG Ð的度数.21.如图,已知ABC △为等边三角形,点D 、E 分别在BC 、AC 边上,且AE CD =,AD 与BE 相交于点F .(1)求证:BE AD =;(2)求BFD Ð的度数.22.如图,在ABC △中,AB 边的垂直平分线1l 交BC 于点D ,AC 边的垂直平分线2l 交BC 于点E ,1l 与2l 相交于点O ,联结OB 、OC ,若ADE △的周长为6 cm ,OBC △的周长为16 cm .(1)求线段BC 的长;(2)联结OA ,求线段OA 的长;(3)若120BAC Ð=°,求DAE Ð的度数.23.如图,已知:E 是AOB Ð的平分线上一点,EC OB ^,ED OA ^,C 、D 是垂足,连接CD ,且交OE 于点F .(1)求证:OE 是CD 的垂直平分线;(2)若60AOB Ð=°,请你探究OE ,EF 之间有什么数量关系?并证明你的结论.24.如图,ABC △中,AD 平分BAC Ð,DG BC ^且平分BC ,DE AB ^于E ,DF AC ^于F .(1)求证:BE CF =;(2)如果8AB =,6AC =,求AE 、BE 的长.25.如图,在ABC △中,AB c =,AC b =.AD 是ABC △的角平分线,DE A ^于E ,DF AC ^于F ,EF 与AD 相交于O ,已知ADC △的面积为1.(1)证明:DE DF =;(2)试探究线段EF 和AD 是否垂直?并说明理由;(3)若BDE △的面积是CDF △的面积2倍.试求四边形AEDF 的面积.第2章综合测试答案解析一、1.【答案】D【解析】解:A .为轴对称图形;B .为轴对称图形;C .为轴对称图形;D .不是轴对称图形。
黑龙江省哈尔滨市香坊区2020届九年级下学期综合测试(二)英语试卷(WORD版含答案)
2020年香坊区初中毕业学年综合测试(二)英语试题第I卷一、单项选择(本题共20分,毎小题1分)选择最佳答案.1. Which pair of the words with the underlined letters has the same sound?A. down allowB. check schoolC. fear early( )2. Which of the following words has the same sound as the underlined letters of the word “lost”?A. loveB. post C topic3. Which word of the following doesn't have the same stress as the others?A. ExactB. SubjectC. Reflect4. You can never imagine what_____ excellent surprises you'll achieve based on your hardwork. Come on, boys and girls!A./B. anC. a5,Mary and Joan look exactly the same. Their friends often mistake them_____ each other.A. ofB. forC. with6. —How do you like your___________ National Day Holiday ?—Very much! But I regretted _____too much time on games.A.7 days, takingB. 7-day, spendingC. 7 days', to spend7. —What do you think of your school, Linda?—It's a good place for as to____________ ourselves for the future.A. predictB. provideC. prepare8. —I heard you took your son to Inner Mongolia (内蒙古)last month.—Yes, My son was_____ to see____ little sheep on the grassland.A. exciting, suchB. excited, soC. excited, such9. 一When would you please help me with the exam skills?—_________day is OK from this Sunday to next Saturday.A Either B. All C. Any10. —I wonder if our teacher Miss Green __________us to the zoo tomorrow?—Don’t worry. I'm sure fine, she __________us there.A. takes, will takeB. will take, will takeC. will take, takes11. The number of volunteers in our school, including some teachers, _______in the past few years.A. was increasedB. has increasedC. have increased12. E-safety needs more attention these days. Don't give any response (答复)to any e-mail_______your personal information^ no matter how official it is.A. requiringB. to requireC. required13. Mary took up singing to deal with her shyness. As she got better she _________in public.How excited she was!A. dared singingB. dared to singC. dares sing14. —What happened to France^ most famous landmarks (地标)in Paris?一The Notre-Dame Cathedral (巴黎圣母院)caught fire last year. It’s said that many people saw the heavy smoke ________in the sky at that time.A. rakeB. risingC. to rise15. —When I study in France as an exchange student, the host family _________make me feel athome, Pm very thankful to them.—What good luck!A. get used toB. get in the way ofC. go out of their way to16. —I wonder_______________________.—I'm truly sorry. I missed the early bus this morning and had to wait for another ten minutes.A. how come you were late for school againB. why did you come late to school againC. bow you are feeling if you are late for school again17. —Is this film ______many children watched during the holiday?―Right It tells us some stories about a boy _______is disabled but overcame a lot ofdifficulties.A. that, whoB. the one, thatC. which, who18. Thanksgiving is a special day in North America. This festival is always on the fourth Thursdayin November in America. But it falls on the second Monday in October in __________.A. RussiaB. CanadaC. Japan19. Our parents always have high expectation on our study, but we have just failed in a veryimportant exam. What kind of attitude should we have?①Talk to our parents to make them understand and support us.②We feel so sorry to our parents that we feel the life is hopeless.③Never mind. We can do better next time if we have a never-give up attitude.④Always complain about those unnecessary reasons.⑤Learn from the mistakes and keep working hard.A. ①③⑤B. ①②③C. ①④⑤20. Today is Friday, March 6th' There are some concessionary(减价的)activities at WandaPlayground. If Li Jun, a boy, goes to the Wanda Playground with his parents the day afterA. 140 yuanB. 150 yuanC. 200 yuan二、完形填空(本題共10分,毎小题1分)Have you ever started your career planning yet? No matter in 21 year you are studying, you should start getting yourself prepared for your future career as early as you can. There are four stages (阶段)when getting _ 22 ready for your future career. Let's have a look First, you should have s better understanding of your abilities and 23 interests. You also need to know whether your classmates, friends, family members and others agree 24disagree with what you think aboutyourself.Make your own choice. Remember that 25 you should be given a sense of autonomy (自主权)to express your own opinion.You should find out what job 26 your abilities and character best You should know about the requirements(要求)of the career that are interesting to you. Then you can decide if you are fit for the job. You probably need help and care 27 career advisors(顾问). They will provide you with information about the real job, The Internet is another important resource (资源)to turn to.Decide what you wish to achieve in the short, medium and long term. Get to know what skills you should have to achieve your goals and then 28 develop those skills. You will probably change your goals from time to time as you get a better understanding from stages One and Two.It is a good idea to get some experience to get ready for your careen 29 an internship(实习)for example, is a very good way to discover the job first-hand. It brings you one big step 30 to you career goal than before.根据短文内容选择最佳答案.21. A. which B. that C. what22. A. something B. everything C. nothing23. A. person B. person's C personal24. A. and B. or C but25. A. in that case B. in fact C. in short2&A. polishes B. processes C. matches27. A. about B. for C. from28. A. care B. careful C. carefully29. A. Taking B. Taken C. Tike30.A. close B. closer C. closest三、阅读理解(本题共20分,每小题1分)31. According to the passage we can probably find the message________,A. in a school newspaperB. in a local magazineC. on the Internet32. Cyber Boy organized the discussion in order to___________.A. work out a travel planB. find much pleasureC. suggest a group tour33. Yoshi suggested Cyber Boy should____________.A. take a simple tour to three countriesB. choose England for future travelC. pick Germany for this summer34. Esperanto was considering Cyber Boy’s difficulty with ___________A. customB. transportationC. money35. Yoshi was ____________Esperanto’s idea when he said “That sounds OK.”A. agreeing withB. arguing againstC. following(B)This is a story of a man who works in a big factory. I have seen him for years but Fve never paid any attention to him. He was a little bit weird. He always wore an old red hat and carried a rubbish bag. He usually spent his break time and his lunchtime walking around in that old big factory and collecting used tins (铝罐) .One day, I was fixing one of the broken machines in the factory when this “tin man” came with his bag. As usual, he picked up the tins, which were all around the place. My manager was standing there watching me.When I finished my job, I heard my manager ask the “tin man” about what he was going to do with those tins he had collected. I never even thought about this kind of question, because I always guessed that “tin man” would take those tins to the recycling center.Unexpectedly, the “tin man” answered, “I will give these tins to my neighbor. He has been ill for many years and cannot work. ” I was so shocked to hear that, so I asked him:“You mea n you collect all those tins just to help your neighbor?” ‘‘I know this does not help much ,” he said “But I give everything to him, because he cannot work. He needs help. ”It was the most beautiful moment in my life. It has made me feel humble every day since then.根据短文内容判断正、误.(正确的涂A错误的涂B)36. The writer has paid much attention to the tin man for years.37. The tin man always spent his all day collecting used tins.38. The tin man*s neighbor didn't haw ability to work because of his illness.39. Those tins were taken to the recycling center by the tin man.40. The passage mainly tells us that we should help others in our own ways.(C)When you think of Scotland, what comes to your mind? Maybe the beautiful green scenery (风景) or the national dish of haggis(羊杂碎). But what makes a Scot stand out straight away is the national dress of Scotland: the kilt.The kilt is a special dress. Most Scottish men wear it. The dress dates back 500 years. TheScottish highlands were wet at that time. It was difficult for soldiers to move around without getting stuck in the mud (陷在泥中).The kilt was made to give them an advantage over their enemy (敌人).Kilts are very flexible , making soldiers easier to move in. What’s more, you can use the kilt as a blanket for those cold nights.In modem times, the kilt has become a symbol of Scottish spirit, culture and patriotism (爱国心). People wear kilts for special events, such as weddings, ceremonies (庆典)and the traditional Scottish sporting event known as the Highland games.But that’s n ot all. The colors of a kilt show your family history. That's because each design stands for a different Scottish family. So a Scotsman is not only wearing the national dress of his country, but also the spirit of his family. There’s a famous saying in Scotland, “A man in a kilt is a man and a half.”根据短文内容选择最佳答案.41. The underlined word "flexible" in the passage means "________" in Chinese.A.保暖的B.优雅的C.灵活的42. The kilt is a special dress because_________.A. it gave the soldier many advantages over their enemy.B. it has many colors and looks wonderfulC. it stands for Scottish pride, culture and patriotism43. "A man in a kilt is a man and a half" means_________.A. a kilt can make ft man look strongerB. a man in a kilt looks a real manC. a man in a kilt looks taller than the others44. Which of the following is Not True according to the passage?A. The kilt was first made more than five hundred years ago^B. Scotsmen wear the kilts for special events whether in the past or in modem times.C. A Scotsman wears the kilts not only far his country but for his family.45. What is the second paragraph mainly about?A. The origin (起源)of the kiltB. How to wear a kiltC. Kilts in wartime(D)Cartoon has a long history. It became popular in the 20th century along with the film and newspaper industries. This important art form has been around for many thousands of years.Human ancestors drew on caves and rocks, 46 They produced pictures of animals, hunting scenes, people dancing, and so on. 47 In later centuries, other societies, such as the Mayans(玛雅人)and the Egyptians(埃及人)carved strange cartoons into solid rock In fact, the languages of these two peoples were represented(代表)by figures of animals and people carved into their tombs(坟墓)and temples.Through the 17th, 18th and 19th centuries, 48 , being used to illustrate(说明)stories in books, magazines and newspapers.Today, cartoons are everywhere. In addition to books and newspapers, 49 You'll have a bard time spending a single day without seeing a cartoon somewhere, whether it is on a TV commercial(电视广告)a magazine, or even an advertisement in your mailbox, ___50 Because of this, a cartoonist can get a job in almost any field!根据短文内容,将下面方框内的句子还原到文中空白处,使短文内容完整,通顺(每个选项只能用一次);A. This is good news for cartoonists.B. cartoon form became an important part of the printed worldC. cartoons can be found on billboards(布告板), posters, television and moviesD. Using paint and charcoal(碳), cave artists drew what was important to them.E. Cave painting was s way for early people to nuke w record of their daily life.第II卷四.交际应用(本题共10分,毎空[分)(A)从A-G选项中选出能填入空白处的最佳选项补全对话。
法医鉴定综合测试(共13份)
综合测试试卷一一、选择题1.尸全解剖检查的目的是:A 确定死亡原因B 推定至伤物体C 分析死亡方式D 进行个人识别E 推断死亡时间2.尸斑扩散期一般出现在A 死亡后立即出现B 常开始于死后第二天C 一般在死后2-4小时D 一般在死后12-24小时E 无法确定3.头部外伤,软脑膜血管破裂,在蛛网膜于软脑膜之间有血液积聚称:A 帽状腱膜下出血B 硬膜外血肿C 外伤性脑出血D 蛛网膜下腔出血E 硬脑膜下出血4.以下哪一项不是锐器的特征:A 创角尖锐B 创缘整齐C 有组织间桥D 无表皮剥落E 创口深,出血多5.生前溺水的尸体的特征是:A 可有白钯泡沫团在口鼻空B 仅上呼吸道有少量溺液、异物C 有水性肺气肿,肺表面有肋骨压痕D 胃内有少量溺液,一般不多E 可握有水草或泥沙6.电损伤主要依据:A 电击现场B 电流斑C 案情调查D 法医学解剖E 实验室检查7.可导致迟速死亡的是:A 氧化物中毒B 冠心病C CO中毒D 有机磷中毒E 土的宁中毒8.锐器造作伤损伤的特征有:A 多数呈大小一致B 创伤集中,密度大,间距小C 排列整齐,方向一致D 程度深,大都危及生命E有试刀痕9.扼死一定是指:A 自杀B 他杀C 意外D 以上都不是10.早期尸体显现时间是指:A 12小时B 24小时C 36小时D 48小时二、名词解释1.尸僵2.组织间桥3.挫伤三、问答题1.检查水中所发现的尸体,主要解决哪些问题?2.有机磷中毒尸体检查有和特征?3.勒死与缢死的鉴别要点是什么参考答案:一、选择题1.ABCDE2.D3.D4.C5.ACE6.ABCDE7.CD8.ABCE9.B 10.B二、名词解释1.尸僵:是一种早期死后变化。
人死后,各个肌群发生僵硬并关节固定的现象称为尸僵。
2.组织间桥:挫裂创时,由于结缔组给纤维、神经纤维和血管皆具有韧性,遭受钝器打击时,常有一部分纤维或血管末发生断裂,横贯两创壁之间,这种末完全断裂的血管和结缔组织,称为组织间桥。
2022年职业能力综合测试真题及答案--试卷二
2022年职业能力综合测试(试卷二)真题、答案及依据说明:本试卷共50分。
资料一近年来我国经济结构发生了明显的改变,消费成为拉动国民经济增长的主要动力之随着消费观念的变化,消费需求更加多样化,超市大卖场“大而全”的经营模式不再唱主角,“小而美”的模式逐渐受到消费者的青睐,便利店、生鲜店等社区门店开始兴起.便利店成为近几年零售业中发展速度较快的业态,特别是品牌连锁便利店,其增速位居零售业前列。
2019年底,全国范围内便利店总数共有10万多个,但尚未出现真正意义上的全国性连锁品牌。
2019年底,国家出台加快发展品牌连锁便利店的相关政策,提出优化便利店营商环境,推动便利店品牌化、连锁化、智能化发展,织密便民消费网格,更好地发挥便利店服务民生和促进消费的重要作用,并提出到2022年全国品牌连锁便利店总量达到30万家的发展目标。
2020年开始,由于新冠疫情爆发,餐饮、旅游、零售等线下消费受到较大冲击。
品牌便利店由于其贴近写字楼和居民区的特点,受到的冲击较小,多数品牌便利店进一步扩张门店数量,规模快速拓展;少数品牌便利店受到疫情负面影响,出现关店现象。
疫情的反复,也对品牌便利店门店的正常经营、员工稳定、物流通畅、成本控制及消费意愿等产生了较多不利影响。
同时,线上零售业务吸引了众多零售巨头的加入,线上社区团购的兴起加剧了竞争态势,品牌便利店的线下门店销售受到了一定程度的影响。
悦来悦股份有限公司(以下简称“悦来悦公司”)是一家成立于2017年的民营企业主要从事城市零售便利店业务。
悦来悦公司创始人兼董事长李悦峰曾在消费零售领域特别是大型商超领域闯荡多年,积累了丰富的零售行业经验,具有敏锐的市场视角。
悦来悦公司依靠其核心技术团队,运用大数据技术选取贴近写字楼和居民区的门店,建立完善的供应链,采用直营模式和自动化管理方式,运营24小时的便利店。
同时悦来悦公司自行开发手机应用程序(APP)及微信小程序,打造全新的社区便民服务新模式,开启智能社区生活圈。
(北师大版)高中数学必修第一册 第二章综合测试试卷02及答案
第二章综合测试一、单选题(每小题5分,共40分),1.函数()f x = )A .[]12-,B .(]12-,C .[)2+¥,D .[)1+¥,2.设函数()221121x x f x x x x ì-ï=í+-ïî,≤,,>,则()12f f öæ÷çç÷èø的值为( )A .1-B .34C .1516D .43.已知()32f x x x =+,则()()f a f a +-=( )A .0B .1-C .1D .24.幂函数223a a y x --=是偶函数,且在()0+¥,上单调递减,则整数a 的值是( )A .0或1B .1或2C .1D .25.函数()34f x ax bx =++(a b ,不为零),且()510f =,则()5f -等于( )A .10-B .2-C .6-D .146.已知函数22113f x x x x öæ+=++ç÷èø,则()3f =( )A .8B .9C .10D .117.如果函数()2f x x bx c =++对于任意实数t 都有()()22f t f t +=-,那么( )A .()()()214f f f <<B .()()()124f f f <<C .()()()421f f f <<D .()()()241f f f <<8.定义在R 上的偶函数()f x 满足对任意的[)()12120x x x x Î+¥¹,,,有()()21210f x f x x x --,且()20f =,则不等式()0xf x <的解集是( )A .()22-,B .()()202-+¥U ,,C .()()8202--U ,,D .()()22-¥-+¥U ,,二、多选题(每小题5分,共20分,全部选对得5分,选对但不全的得3分,有选错的得0分)9.定义运算()()a ab a b b a b ìï=íïî≥□<,设函数()12x f x -=□,则下列命题正确的有( )A .()f x 的值域为[)1+¥,B .()f x 的值域为(]01,C .不等式()()12f x f x +<成立的范围是()0-¥,D .不等式()()12f x f x +<成立的范围是()0+¥,10.关于函数()f x =的结论正确的是( )A .定义域、值域分别是[]13-,,[)0+¥,B .单调增区间是(]1-¥,C .定义域、值域分别是[]13-,,[]02,D .单调增区间是[]11-,11.函数()f x 是定义在R 上的奇函数,下列命题中是正确命题的是( )A .()00f =B .若()f x 在[)0+¥,上有最小值1-,则()f x 在(]0-¥,上有最大值1C .若()f x 在[)1+¥,上为增函数,则()f x 在(]1-¥-,上为减函数D .若0x >时,()22f x x x =-,则0x <时,()22f x x x =--12.关于函数()f x )A .函数是偶函数B .函数在()1-¥-,)上递减C .函数在()01,上递增D .函数在()33-,上的最大值为1三、填空题(每小题5分,共20分)13.已知函数()()f x g x ,分别由表给出,则()()2g f =________.x 123()f x 131()g x 32114.已知()f x 为R 上的减函数,则满足()11f f x öæç÷èø>的实数x 的取值范围为________.15.已知函数()f x 是奇函数,当()0x Î-¥,时,()2f x x mx =+,若()23f =-,则m 的值为________.16.符号[]x 表示不超过x 的最大整数,如[][]3.143 1.62=-=-,,定义函数:()[]f x x x =-,则下列说法正确的是________.①()0.80.2f -=;②当12x ≤<时,()1f x x -;③函数()f x 的定义域为R ,值域为[)01,;④函数()f x 是增函数,奇函数.四、解答题(共70分)17.(10分)已知一次函数()f x 是R 上的增函数,()()()g x f x x m =+,且()()165f f x x =+.(1)求()f x 的解析式.(2)若()g x 在()1+¥,上单调递增,求实数m 的取值范围.18.(12分)已知()()212021021 2.f x x f x x x x x +-ìï=+íï-î,<<,,≤<,,≥(1)若()4f a =,且0a >,求实数a 的值.(2)求32f öæ-ç÷èø的值.19.(12分)已知奇函数()q f x px r x =++(p q r ,,为常数),且满足()()5171224f f ==,.(1)求函数()f x 的解析式.(2)试判断函数()f x 在区间102æùçúèû,上的单调性,并用函数单调性的定义进行证明.(3)当102x æùÎçúèû,时,()2f x m -≥恒成立,求实数m 的取值范围.20.(12分)大气中的温度随着高度的上升而降低,根据实测的结果,上升到12km 为止,温度的降低大体上与升高的距离成正比,在12km 以上温度一定,保持在55-℃.(1)当地球表面大气的温度是a ℃时,在km x 的上空为y ℃,求a x y 、、间的函数关系式.(2)问当地表的温度是29℃时,3km 上空的温度是多少?21.(12分)已知函数()f x 是定义在[]11-,上的奇函数,且()11f =,对任意[]110a b a b Î-+¹,,,时有()()0f a f b a b++成立.(1)解不等式()1122f x f x öæ+-ç÷èø<.(2)若()221f x m am -+≤对任意[]11a Î-,恒成立,求实数m 的取值范围.22.(12分)已知函数()[](]2312324.x x f x x x ì-Î-ï=í-Îïî,,,,,(1)画出()f x 的图象.(2)写出()f x 的单调区间,并指出单调性(不要求证明).(3)若函数()y a f x =-有两个不同的零点,求实数a 的取值范围.第二章综合测试答案解析一、1.【答案】B【解析】选B .由10420x x +ìí-î>,≥,得12x -<≤.2.【答案】C【解析】选C .因为()222224f =+-=,所以()211115124416f f f öæööææ==-=÷çç÷ç÷ç÷èèøøèø.3.【答案】A【解析】选A .()32f x x x =+是R 上的奇函数,故()()f a f a -=-,所以()()0f a f a +-=.4.【答案】C【解析】选C .因为幂函数223aa y x --=是偶函数,且在()0+¥,上单调递减,所以2223023a a a z a a ì--ïÎíï--î<,,是偶数.解得1a =.5.【答案】B【解析】选B .因为()51255410f a b =++=,所以12556a b +=,所以()()51255412554642f a b a b -=--+=-++=-+=-.6.【答案】C【解析】选C .因为22211131f x x x x x x ööææ+=++=++ç÷ç÷èèøø,所以()21f x x =+(2x -≤或2x ≥),所以()233110f =+=.7.【答案】A【解析】选A .由()()22f t f t +=-,可知抛物线的对称轴是直线2x =,再由二次函数的单调性,可得()()()214f f f <<.8.【答案】B【解析】选B .因为()()21210f x f x x x --<对任意的[)()12120x x x x Î+¥¹,,恒成立,所以()f x 在[)0+¥,上单调递减,又()20f =,所以当2x >时,()0f x <;当02x ≤<时,()0f x >,又()f x 是偶函数,所以当2x -<时,()0f x <;当20x -<<时,()0f x >,所以()0xf x <的解集为()()202-+¥U ,,.二、9.【答案】AC【解析】选AC .根据题意知()10210xx f x x ìöæïç÷=íèøïî,≤,,>,()f x 的图象为所以()f x 的值域为[)1+¥,,A 对;因为()()12f x f x +<,所以1210x x x +ìí+î>≤,或2010x x ìí+î<>,所以11x x ìí-î<≤,或01x x ìí-î<>,所以1x -≤或10x -<<,所以0x <,C 对.10.【答案】CD【解析】选CD .由2230x x -++≥可得,2230x x --≤,解可得,13x -≤≤,即函数的定义域为[]13-,,由二次函数的性质可知,()[]22231404y x x x =-++=--+Î,,所以函数的值域为[]02,,结合二次函数的性质可知,函数在[]11-,上单调递增,在[]13,上单调递减.11.【答案】ABD【解析】选ABD .()f x 为R 上的奇函数,则()00f =,A 正确;其图象关于原点对称,且在对称区间上具有相同的单调性,最值相反且互为相反数,所以B 正确,C 不正确;对于D ,0x <时,()()()22022x f x x x x x --=---=+>,,又()()f x f x -=-,所以()22f x x x =--,即D 正确.12.【答案】ABD【解析】选ABD .函数满足()()f x f x -=,是偶函数;作出函数图象,可知在()1-¥-,,()01,上递减,()10-,,()1+¥,上递增,当()33x Î-,时,()()max 01f x f ==.三、13.【答案】1【解析】由题表可得()()2331f g ==,,故()()21g f =.14.【答案】()()01-¥+¥U ,,【解析】因为()f x 在R 上是减函数,所以11x,解得1x >或0x <.15.【答案】12【解析】因为()f x 是奇函数,所以()()223f f -=-=,所以()2223m --=,解得12m =.16.【答案】①②③【解析】()[]f x x x =-,则()()0.80.810.2f -=---=,①正确,当12x ≤<时,()[]1f x x x x =-=-,②正确,函数()f x 的定义域为R ,值域为[)01,,③正确,当01x ≤<时,()[]f x x x x =-=;当12x ≤<时,()1f x x =-,当0.5x =时,()0.50.5f =;当 1.5x =时,()1.50.5f =,则()()0.5 1.5f f =,即有()f x 不为增函数,由()()1.50.5 1.50.5f f -==,,可得()()1.5 1.5f f -=,即有()f x 不为奇函数,④错误.四、17.【答案】(1)由题意设()()0f x ax b a =+>.从而()()()2165f f x a ax b b a x ab b x =++=++=+,所以21655a ab ì=í+=î,,解得41a b =ìí=î,或453a b =-ìïí=-ïî,(不合题意,舍去).所以()f x 的解析式为()41f x x =+.(2)()()()()()()()414241g x f x x m x x m x m x m g x =+=++=+++,图象的对称轴为直线418m x +=-.若()g x 在()1+¥,上单调递增,则4118m +-≤,解得94m -≥,所以实数m 的取值范围为94öé-+¥÷êëø.18.【答案】(1)若02a <<,则()214f a a =+=,解得32a =,满足02a <<;若2a ≥,则()214f a a =-=,解得a =或a =,所以32a =或a =.(2)由题意,3311222f f f öööæææ-=-+=-ç÷ç÷ç÷èèèøøø1111212222f f ööææ=-+==´+=ç÷ç÷èèøø.19.【答案】(1)因为()f x 为奇函数,所以()()f x f x -=-,所以0r =.又()()5121724f f ì=ïïíï=ïî,即52172.24p q q p ì+=ïïíï+=ïî解得212p q =ìïí=ïî,,所以()122f x x x =+.(2)()122f x x x =+在区间102æùçúèû,上单调递减.证明如下:设任意的两个实数12x x ,,且满足12102x x <<≤,则()()()12121211222f x f x x x x x -=-+-()()()()21211212121214222x x x x x x x x x x x x ---=-+=.因为12102x x <<≤,所以2112121001404x x x x x x -->,<<,>,所以()()120f x f x ->,所以()122f x x x =+在区间102æùçúèû,上单调递减.(3)由(2)知()122f x x x =+在区间102æùçúèû,上的最小值是122f öæ=ç÷èø.要使当102x æùÎçúèû,时,()2f x m -≥恒成立,只需当102x æùÎçúèû,时,()min 2f x m -≥,即22m -≥,解得0m ≥即实数m 的取值范围为[)0+¥,.20.【答案】(1)由题意知,可设()0120y a kx x k -=≤≤,<,即y a kx =+.依题意,当12x =时,55y =-,所以5512a k -=+,解得5512a k +=-.所以当012x ≤≤时,()()5501212x y a a x =-+≤≤.又当12x >时,55y =-.所以所求的函数关系式为()55012125512.x a a x y x ì-+ï=íï-î,≤≤,,>(2)当293a x ==,时,()3295529812y =-+=,即3km 上空的温度为8℃.21.【答案】(1)任取[]121211x x x x Î-,,,<,()()()()()()()()1212121212f x f x f x f x f x f x x x x x +--=+-=-+-g 由已知得()()()12120f x f x x x +-+->,所以()()120f x f x -<,所以()f x 在[]11-,上单调递增,原不等式等价于112211121121x x x x ì+-ïïï-+íï--ïïî<,≤≤≤,所以106x ≤<,原不等式的解集为106öé÷êëø,.(2)由(1)知()()11f x f =≤,即2211m am -+≥,即220m am -≥,对[]11a Î-,恒成立.设()22g a ma m =-+,若0m =,显然成立;若0m ¹,则()()1010g g -ìïíïî≥≥,即2m -≤或2m ≥,故2m -≤或2m ≥或0m =.22.【答案】(1)由分段函数的画法可得()f x 的图象.(2)单调区间:[]10-,,[]02,,[]24,,()f x 在[]10-,,[]24,上递增,在[]02,上递减.(3)函数()y a f x =-有两个不同的零点,即为()f x a =有两个实根,由图象可得,当11a -<≤或23a ≤<时,()y f x =与y a =有两个交点,则a 的范围是(][)1123-U ,,.。
2024届广东省广州市普通高中毕业班综合测试(二)英语试卷
绝密★启用前2024年广州市普通高中毕业班综合测试(二)英语注意事项:微信公众号IAI English1.答卷前,考生务必将自己的姓名、准考证号填写在答题卡上。
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微信公众号IAI English第二部分阅读理解(共两节,满分50分)第一节(共15小题;每小题2.5分,满分37.5分)阅读下列短文,从每题所给的A、B、C和D四个选项中,选出最佳答案。
ATrain the important skills modern editors use to evaluate and enhance writing for clarity,precision and accuracy.In this course,students will learn how an editor approaches a submitted piece,going beyond sentence-level error and looking at the big picture around accuracy,style and organization.Understanding the different challenges in an editor's job,students will get a behind-the-scenes look at this sometimes busy and often exciting career. Meanwhile,students will grow their own technical editing skills and return home a more competent editor.Using Gen Z Era as their case study,students will meet and study under the people who decide what topics are relevant and valuable to the audience and who determine the overall editorial strategy,ensuring that the content meets the standards and tone of the publication.微信公众号IAI EnglishCourse Highlights●Visit the media city and attend lectures by award-winning guest speakers.●Assess articles as well as question and coach the authors to get the best piece possible.●Connect with professionals who manage the development and publication of accurate and worthy content.●Edit one piece into a well-written and fact-checked article in the style of Gen Z Era.Price●Residential Program(Students live on campus):$6,600●Day Program(Students commute to class every day):$5,500(Graduating seniors can have a$400discount if applying before May2,2024.)Term Date:July9-July21,2024Application Deadline:Friday,May31,2024Contact Admission:******************21.Which is the probable name of the course?A.Career Development in Media.B.Editorial Decision-making.C.Fundamentals of Editing.D.Introduction to Publication.22.What will students do in the course?A.Interview award-winning guests.B.Help authors improve their articles.C.Connect with professional publishers.D.Edit one article for Gen Z Era.23.A graduating senior applying for a Day Program on April30,2024should pay_______.A.$5,100B.$5,500C.$6,200D.$6,600BCourage is a huge theme in my life,a quality I constantly seek,appreciate,and analyze.The root of“courage”is“cor,”the Latin word for heart.Originally,courage meant“to speak one’s mind by telling all one's heart.”While courage is often associated with heroism nowadays,I believe true courage lies in being open and honest about who we are and how we feel.I recently witnessed an example of true courage.During a mountain-climbing trip with my15-year-old daughter and some college students,I noticed her struggling to keep up with the group.Despite my suggestions to rest,she persisted until she couldn't breathe properly.Panicked,I called out to the front for help,but there was no response,and we had no cellphone signal. Fortunately,two students just came back to check out on us.They offered assistance and calmed us down.As we continued at a slower pace,they shared their own experiences,from starting out as beginners like my daughter to becoming consistently among the first to reach the peak.“You know,”one of them said,looking at my daughter,“I was just like you when I started.But with practice and proper pace,you'll get there too.”“Yeah,don't let your lack of experience stop you,”the other added.“It's okay to admit when you're struggling or not feeling alright.In fact,it's important to speak up and ask for help when you need it.That's how we improve and grow.”微信公众号IAI EnglishReaching the mountain top was a huge relief for both my daughter and me.However,the two students addressed the celebrating group directly,emphasizing the importance of staying together in tough environments. Their words led the group to apologize to us for overlooking our struggle.I was totally amazed at their bravery,and my daughter learned that it's okay to be the least experienced in a group.Courage,I've come to realize,has a ripple effect.Each time we choose courage,we inspire those around us to be a little brave r and make the world a little better.24.Why does the author mention the original meaning of courage?A.To argue for the true essence of courage.B.To question the common belief of courage.C.To show the changing meaning of courage.D.To compare different interpretations of courage.25.What did the two students suggest the daughter do?A.Challenge her own limits.B.Seek help whenever possible.C.Keep to a suitable pace.D.Stick with experienced climbers.26.Which action in the mountain-climbing story is an example of true courage?A.The mother asked the girl to rest.B.The girl tried hard not to fall behind.C.The group celebrated the reach of the top.D.The two students pointed out the group's fault.27.What does the author intend to tell us?A.Kindness connects us all.B.Being a beginner takes courage.C.With courage,everyone can be perfect.D.We don't have to be a hero to be brave.C“It's not unusual for guests to feel emotional when they discover the story behind our food,”says Patrick Navis.“Not to mention when they taste it.One even cried with happiness.”The setting for these tearful scenes? Navis's restaurant in a Dutch city.Here,the owner and his team create experimental food using herbs,roots, flowers and nuts—some common,others less so.微信公众号IAI EnglishMost of these ingredients(食材)come from the Ketelbroek Food Forest nearby.To the untrained eye,it's like an ordinary wood.But there's one key difference:everything in it is edible.It was set up in2009by Dutch botanist and environmentalist Noah Eck as an experiment in slow farming,to see what would happen if the right combination of food plants were left to grow together like a natural forest,without chemicals.“It's the first‘food forest’of its kind in Europe and we’re one of the few restaurants around the world cooperating in this way,”says Navis.“We have over400different species of edible plants we plan our menus around, including some we previously knew little about."He harvests the ingredients and,with his fellow chefs,works them into beautifully presented tasting menus,served in a dining room hidden in the backstreets of the city,“To us,fine dining is not about the fame of a restaurant,its location,expensive decoration,fancy cooking and wine list,”says Navis.“It's about adding value through creativity and using ingredients nobody knows of,which are grown with great attention.”However,he adds,luxury cooking can be about enhancing everyday ingredients, too.“When looking at cooking in this way,who can argue that caviar(鱼子酱),for example,is more valuable than a carrot grown with specialist knowledge?”微信公众号IAI EnglishExperimentation is extremely important to Navis.In the next five years,he hopes to open an outdoor restaurant.But for now,the most important thing is to continue focusing on how plants are being grown and the perennial system used in the Food Forest,reducing the need for replanting each season.28.What can we learn about Navis's restaurant?A.It is well received by its guests.B.It serves food with moving stories.C.It offers experimental food for free.D.It is known for its rare food sources.29.How is Ketelbrock Food Forest different from ordinary woods?A.It is a natural forest.B.Diverse plants coexist in it.C.Plants there take longer to grow.D.It provides safe food ingredients.30.What is the key element of fine dining according to Navis?A.Convenient locations.B.Expensive ingredients.C.Innovative menus.D.Fancy cooking techniques.31.What does“the perennial system”in the last paragraph probably refer to?A.The sustainable farming practice.B.Farming with proper use of chemicals.C.Natural farming without human intervention.D.An experimental farm for an outdoor restaurant.DMy father started learning French at57,drawn by the potential benefits of bilingualism in delaying dementia (失智症).Now,20years later,he's on his third teacher.Many people like my father have attempted to pick up a new language.But can this really boost brain health?According to experts,regularly using a new language brings cognitive(认知的)benefits.If you're trying to recall the right words in another language,your brain is forced to inhibit your mother tongue.This process,called cognitive inhibition,helps improve your brain function.Repeating this process makes your brain more resistant to diseases like dementia.The more you challenge your brain,the better it functions,even if your brain health starts to decline.However,evidence for the benefits of learning a second language in your60s is weaker.Research by Dr.Leo Antoniou found that older Italians who took English lessons for four months didn't see any difference in their cognition scores,but people who didn't saw their scores decline.Prof.Diana Smith's2023studies found similarresults.微信公众号IAI EnglishResearchers offered a few potential explanations for their disappointing results.One is that the participants were highly motivated volunteers,probably of high cognitive level for their age,making it hard to see any improvements.“When choosing participants,we have to be careful,are they really representative of the population?”said Dr.Judith Ware.Another is that the language interventions were perhaps too short.These studies have used language lessons that“were very different in their length and frequency,”said ura Grossman.To Dr.Antoniou,the limited findings are not entirely surprising.No one would say that learning a new language for six months would be the same as having used two languages for your entire life.But he does think that language lessons can provide cognitive benefits by being cognitively stimulating.微信公众号IAI English Perhaps more important,Prof.Grossman said,learning another language offers other potential advantages, like traveling or connecting with new communities.My father,for example,has remained pen friends with his first teacher and traveled to France numerous times.And at76,he's as sharp as ever.32.What happens in the process of“cognitive inhibition”?A.Memory improves.B.Native language is held back.C.Dementia is cured.D.Brain health worsens.33.Which is a possible explanation for the disappointing research results?A.The intervention of the first language.B.The great length of the language course.C.The poor choice of research participants.D.The age difference of the research subjects.34.Whose opinion does the author support by mentioning his father's experience?A.Leo Antoniou's.B.Diana Smith's.C.Judith Ware's.ura Grossman's.35.Which of the following is a suitable title for the text?A.Is it never too late to learn a new language?B.Can learning a new language delay dementia?C.Why does my father start learning a new language?D.How does learning a new language benefit aging brain?第二节(共5小题;每小题2.5分,满分12.5分)根据短文内容,从短文后的选项中选出能填入空白处的最佳选项。
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长沙理工大学考试试卷(A)………………………………………………………………………………………………………………试卷编号拟题教研室(或教师)签名叶吉祥教研室主任签名………………………………………………………………………………………………………………课程名称(含档次)面向对象与可视化编程课程代号专业城南学院通信工程层次(本、专)本考试方式(开、闭卷)闭注意:卷面上所有程序代码中大写字母都认为是相应的小写字母(类名和字符串除外)1一、判断题(每题1分,共20分,对用Y表示,错用X表示,答案写在答题纸上)1、函数模板定义中的类属参数个数只能有一个()2、派生类如果没有全部重新定义其从抽象基类继承的纯虚函数,那么他也还是一个抽象类()3、namespace ABC{int a;int b;} ,上面语句定义了一个名为ABC的名字空间()4、只有类的非静态成员函数才能被定义为虚函数()5、一旦将一个类的成员函数定义为虚函数,该函数在该类派生类的继承体系中就永远是虚函数了。
()6、x++的运算符号(后加)重载函数申明方式(X是类型):X operator++(X &a,int);( )7、标准C++中类的静态成员函数不能访问非静态数据成员,只能访问静态数据成员()8、下面程序在编译时没有语法错( )#include<iostream>using namespace std;class B{private:int i;protected:int j;public:int k; };void main(){B b;b.i =1;b.j=2;b.k=3; }9、基类构造函数必须在其派生类构造函数的初始化列表中调用()10、X a;X *is=&a;(X是一个合法的类)上面的两条语句将会调用X的构造函数两次()11、一个派生类只能继承一个基类()12、一个类的常量成员函数可以修改该类的所有数据成员()13、因为构造函数和析构函数都不能被继承和重写,所以他们都不能定义为虚函数()14、Template <class T> T min(T x,T y){...},其中代码class T定义了一个类T ( )15、基类对象不可以赋值给派生类对象,但派生类对象也可赋值给基类对象()16、类的Protected成员可以在派生类中直接使用()17、const int a;这条语句如果出现在函数的数据定义中有语法错误()18、char &ir;这条语句如果定义在main()中会因为没有绑定变量而有语法错( )19、函数df声明如下:Int &df(int a);df函数的用法:df(5)=67;有语法错误 ( )20、类的对象成员初始化赋值必须在构造函数初始化列表中进行()二、选择题(每空2分,共20分)1、#include <iostream>using namespace std;int &f(int &a,int b=20){ a=a*b;return a;}void main(){int j=10; int &m=f(j);int *p=&m;m=20;*p=f(j,5);cout<<j<<m<<*p; }上面程序的运行结果是:()A)100100100 B)102050 C) 505050 D)1020202、假定一个二维数组的定义语句为“int a[3][4]={{3,4},{2,8,6}};”,则元素a[2][1]的值( )A) 0 B) 4 C) 8 D) 63、下面是类X的派生定义过程,main()中用X定义对象后,构造函数的调用顺序正确的是()class E{private:int a;public:E(int a1){a=a1;}}; class F{Private:int b; public:F(int b1){b=b1}};class X:public E,virtual public F{ public:X(int a,int b):E(b),F(a){}};A)E(b),F(a),X(a,b) B)X(a,b),F(a),E(b) C) F(a),E(b),X(a,b) D)X(a,b),E(b),F(a)4、定一个类X,希望含有一个整数常量私有数据成员,下面定义正确的是 ( )A)class X{private:const int a=90;public:X(int a1){a=a1;}};B)class X{private:const int a;public:X(int a1):a(a1){}};C)class X{private:const int a;public:X(){a=90;}};D)class X{const a=90;};5、下面程序的运行结果是( A )#define A 10+10#define B A-A#include<iostream>int main(){ const int C=10+10;const int D=C-C;std::cout<<"B="<<B<<" D="<<D<<std::endl;return 0;}A)B=20 D=20 B)B=0 D=20 C)B=20 D=0 D)B=0 D=06、下面程序的运行结果是( )#include<iostream> using namespace std; class A{const int i;int &j;public: A(int a=0,int b=0):i(a),j(b){cout<<"i="<<i<<"\t"<<"j="<<j<<"\t"<<endl; }};void main(){int m=6;A x(4,m);}A)i=4 j=6 B)程序有错C)i=4 j=0 D)i=0 j=67、下面程序的运行结果是()#include<iostream>using namespace std;class A{int a;public:A(int i){a=i;cout<<"constructing A:"<<a<<endl; }};class B{int b;public: B(int i){b=i;cout<<"constructing B:"<<b<<endl;}};class C{A a1;B b1,b2;public:C(int i1,int i2,int i3):b1(i1),a1(i2),b2(i3){} };void main(){C x(1,2,3);}A) constructing A:2 B)constructing B:1constructing B:3 constructing A:2constructing B:1 constructing A:3C) constructing A:2 D)constructing B:1constructing B:1 constructing B:3constructing B:3 constructing A:2 8、下面程序的运行结果是 ( )#include<iostream>using namespace std;class A{private:int x,y;public:A(int i,int j){x=i;y=j;}int getx(){return x;}int gety(){return y;}friend class B; //声明类B是类A的友元类};class B{ private:int z;public:int add(A a){return a.x+a.y+z;} int mul(A a){return a.x*a.y*z;} B(int i=0){z=i;}};void main(){A a(2,3);B b(4);cout<<b.add(a);cout<<b.mul(a)<<endl;}A)249 B)942 C)2 3 4 D)924 9、下面程序的运行结果是()#include<iostream>using namespace std;class B {public:B(){ cout<<"constructing B"<<endl; } ~B(){ cout<<"destructing B"<<endl; }}; class D:public B {public: D(){ cout<<"constructing D"<<endl; } ~D(){ cout<<"destructing D"<<endl; }}; void main(){B *p;p=new D;//注意析构函数是否被调用delete p;}长沙理工大学考试试卷(A)………………………………………………………………………………………………………………试卷编号拟题教研室(或教师)签名叶吉祥教研室主任签名………………………………………………………………………………………………………………课程名称(含档次)面向对象与可视化编程课程代号专业城南学院通信工程层次(本、专)本考试方式(开、闭卷)闭注意:卷面上所有程序代码中大写字母都认为是相应的小写字母(类名和字符串除外)4A)constructing B B)constructing B C)destructing B D)constructing Bconstructing D destructing D destructing D constructing Ddestructing D destructing B constructing B destructing Bdestructing B destructing D10、下面程序的运行结果是()#include<iostream>using namespace std;class B{public:virtual void f(){ cout << "B::f"<<endl; }; };class D : public B{public:void f(){ cout << "D::f"<<endl; }; };void main(){D d;B *pB = &d, &rB=d;pB->f();rB.f();}A)B::f B)D::f C)B::f D)程序运行出错 D::f D::f B::f三、阅读程序写出程序运行结果(每题5分,共25分)1、#include<iostream>using namespace std;class Sample{int n;public:Sample(int i){n=i;}friend int add(Sample &s1,Sample &s2); }; int add(Sample &s1,Sample &s2) {return s1.n+s2.n;}void main(){Sample s1(20),s2(20);cout<<add(s1,s2)<<endl;}2、#include<iostream> using namespace std; class A{public:A();A(int i,int j);void print(); private:int a,b;};A::A(){ a=b=10;}A::A(int i,int j){ a=i,b=j;}void A::print(){cout<<"a="<<a<<",b="<<b<<endl;} void main() {A m,n(15,18);m.print();n.print();}3、#include<iostream>using namespace std;class Figure{ //抽象类protected:double x,y;public:void set(double i,double j){ x=i; y=j; } virtual void area()=0;//纯虚函数};class Triangle:public Figure{public:void area(){cout<<"三角形面积:"<<x*y*0.5<<endl;} };//重写基类纯虚函数void main(){Figure *pF;Triangle t;t.set(10,20);pF=&t;pF->area();Figure &rF=t;rF.set(20,20);rF.area();}4、#include<iostream> using namespace std; class T{public:T(){}~T(){cout<<"T";} };void main(){T a[2],*p[2];}5、#include<iostream> using namespace std; class point{public:static int n;public:point() { n++;}~point() {n--;} };int point::n=0;void main(){ point A,B,*p,*ptr=new point[3]; p=ptr;;delete[] p;cout<<point::n;}四、程序编写题(选择一题,共35分,另外一题为附加题20分)1、建立一个CPoint类,该类有两个私有成员变量x,y,表示点的坐标。