吉林省"通化优质高中联盟”2023~2024学年度高一上学期期中考试(24-103A)数学.考卷答案

吉林省"通化优质高中联盟”2023~2024学年度高一上学期期中考试(24-103A)数学.考卷答案试卷答案,我们目前收集并整理关于吉林省"通化优质高中联盟”2023~2024学年度高一上学期期中考试(24-103A)数学.考卷答案得系列试题及其答案,更多试题答案请关注微信公众号:考不凡/直接访问www.kaobufan.com(考不凡)

试题答案

吉林省"通化优质高中联盟”2023~2024学年度高一上学期期中考试(24-103A)数学.考卷答案试卷答案

以下是该试卷的部分内容或者是答案亦或者啥也没有,更多试题答案请关注微信公众号:考不凡/直接访问www.kaobufan.com(考不凡)

上期参考答案(上接A2版)A2-A4版1-5BAABB6-10CBBCA26.WhydidBryanShawexaminethepicturesofNoah?ousedoesthewomanprefer?11-15BACAC16-20BACCAA.TorecallNoah'sbabydays.21-25CDBBBagarden.26-30ACCDAB.Toclassifythesephotos.wofloors.31-35BCBDC36-40GEBD/C.Tofindacureforretinoblastomomsdoesthehousehavetheds?41-45CCDBD46-50CBAABB.13.51-55ABDCA56-60ABDDCuldthewomanpayt61.hasworkedtheman62.whichNearlyallparentschildrenwhobeliey63.toarriveB.10,000yuan.abilities,buthow64.although/thoughchildrentosee1答第17至20题

competent?Will65.prepared66.tohelptoimproveowaboutthehospital?Research00patients.67.bigger68.wastouchedchildrenforbeitsrulesat4pm.69.collecting70.motivationmorenegativeatmostineachroomchildrenforthepeakersayaboutthevisiting短文改错whenchildrenAsawell-knownpainterinthetheytendtohoursareinthemorninganditselfandmorQingDynastyZhengBanqiaothemselvesanopleareallowedtovisitoneenjoyedgreatpopularityintheChinawhoarepraisametime.whenfacedvtaywiththepatientsfortwoenjoysbeing“smartme.allthetime.Heisfamousaspaintingbeingsmartenoakfast?justgiveupB.At7:00am.forChildrenisedatientsprobablydoafterthisflowers,treesorstones,especiallyconcludethateywhentheyfaandwalk.bepersistenbamboos.Thispaintinghappenstobepraisedfoometea.intelligenccprogram.oneofhismasterpieceonbamboo.Itmasterpiecesutesifyoudonottakeoffatmeasures138centimeterslongerandeforAllAgeslongntseeingTouristheperfecthingthatSanAntoniohasto38centimeterswide.trickBesidesfororlowsgueststocompletelytheirvacationtosuittheir

分析①直接取x1=x2=0,利用f(x1+x2)≥f(x1)+f(x2)可得:f(0)≤0,再结合已知条件f(0)≥0即可求得f(0)=0;
②按照“友谊函数”的定义进行验证;
③由0≤x1<x2≤1,则0<x2-x1<1,故有f(x2)=f(x2-x1+x1)≥f(x2-x1)+f(x1)≥f(x1),即得结论成立.

解答解:①∵f(x)为“友谊函数”,
则取x1=x2=0,得f(0)≥f(0)+f(0),即f(0)≤0,
又由f(0)≥0,得f(0)=0,故①正确;
②g(x)=x在[0,1]上满足:(1)g(x)≥0;(2)g(1)=1,
若x1≥0,x2≥0,且x1+x2≤1,
则有g(x1+x2)-[g(x1)+g(x2)]=(x1+x2)-(x1+x2)=0,即g(x1+x2)≥g(x1)+g(x2),
满足(3).故g(x)=x满足条件(1)﹑(2)﹑(3),
∴g(x)=x为友谊函数.故②正确;
③∵0≤x1<x2≤1,∴0<x2-x1<1,
∴f(x2)=f(x2-x1+x1)≥f(x2-x1)+f(x1)≥f(x1),
故有f(x1)≤f(x2).故③正确.
故答案为:①②③.

点评本题考查命题的真假判断与应用,是在新定义下对抽象函数进行考查,在做关于新定义的题目时,一定要先研究定义,在理解定义的基础上再做题,是中档题.