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［辽宁大联考］辽宁省2024届高三年级9月联考数学试卷答案

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(Text6)W:Whathaveyoubeendoing?M:I'vebeenworkingreallyhard.SometimesIgoroundtoLucinda'splaceandwestudyto-gether.W:Lucinda?Ihaven'theardaboutherbefore.Whoisshe?M:YouknowLucinda.I'msureI'vetoldyouabouther.I'veknownherforages.Weoftenhelpeachotherwithwork.Sometimeswegotothepuborcookamealtogether.Todaywe'vebeentestingeachotheroneconomicsandmarketing.That'stomorrow'sexam.She'sgoneouttogetaChinesetakeaway.Oh,Lucinda'sjustcomebackwiththefood.I'llringagainbeforeIcomehome.LovetoDad.(Text7)W:DoyouknowanythingaboutDrMiller'sclasses?M:Yes.Areyouinhisclass?W:No.I'mtakinghisclassnextterm.M:Heisveryseriousandpatient.Heisreallyclearwhatyouneedtolearntogetagoodgrade.Butheisverystrictabouthomework.Areyouwillingtostudyhard?W:Yes,Iguessso.M:WhatIlikemostabouthimisthatheisateacherwhoalwaysunderstandsothers.Doyouenjoythatinateacher?W:Yes,Ihadsuchateacherbefore.M:DidyouknowthathecametoEnglandlastyearafter10yearsofteachinginAmerica?W:No,Ididn't,butthatcouldbeagoodthing.M:Well,takealookateverythingandworkoutwhatisbestforyou.Goodluck!(Text8)W:SometimesIjustcan'tfigureoutwhysomepeoplealwaysbuyfamousbrandssuchasLVorGucci.They'renotverygoodvalueformoney.Whatdoyouthinkaboutthat,Tom?M:Maybetheythinkthatfamousbrandscanmakethemfeelgood.Youknow,producersoffamousbrandsspendalotofmoneyontheiradvertising.Theywantpeopletoknowthattheyareofferingthebest.W:Icanunderstandtheinfluenceofadvertisementsonpeople.Somepeopledonotmakethatmuchmoneybutspendhalfoftheirincomesonfamousbrands.M:Oh,Isee.Somepeopleprobablyneedtowearbrand-nameclothingtomakethemselvesuniqueandfashionable.Theydon'twanttobeanobodyinthecrowd.W:Well,that'soneofthereasonswhytheyspendsomuch.Theywanttokeepupwithchan-gingfashions,sotheybuythemregardlessoftheprice.M:AsfarasIamconcerned,Iwillbuythingsofhighqualityatreasonableprices.Therearealsomanycheapthingswithgoodquality.Ineverwanttobeavictimofadvertisements.W:Icouldn'tagreemore.·118:【23·G3DY(新教材老高考)·英语·参考答案一WYB一必考一QG】

分析（1）对二次方程分类讨论：当在（0，1]上有两相等实根，和在（0，1）上有且仅有一个实根和恰有一根为x=1，根据不同情况分别求m的范围．
（2）对x分类，去绝对值，利用二次函数求出区间内的最大值．

解答解：方程f（x）=g（x）在（0，1]上有且仅有一个实根，
∴方程x²-（m-1）x+2m=0在（0，1]上有且仅有一个实根，
当方程x²-（m-1）x+2m=0在（0，1]上有两相等实根，
∴△=（m-1）²-8m=0，
0＜$\frac{m-1}{2}$≤1，得出m无解；
当方程x²-（m-1）x+2m=0有两相等实根，且在（0，1]上有且仅有一个实根，
当在（0，1）上有且仅有一个实根，
∴f（0）f（1）＜0，
∴2m（m+2）＜0，
∴-2＜m＜0，
当f（1）=0时，m=-2，x²+3x-4=0，
∴x₁=1，x₂=4符合题意，
∴m的取值范围是[-2，0）；
（2）当x∈[0，1]时，f（x）=x（1-x）+m=-（x-$\frac{1}{2}$）²+m+$\frac{1}{4}$
当x=$\frac{1}{2}$时，f（x）_max=m+$\frac{1}{4}$，
当x∈（1，m]时，f（x）=（x-$\frac{1}{2}$）²+m-$\frac{1}{4}$，
函数在（1，m]时递增，
∴f（x）_max=f（m）=m²，
由m²＞m+$\frac{1}{4}$得m≥$\frac{1+\sqrt{2}}{2}$，
当m≥$\frac{1+\sqrt{2}}{2}$时，f（x）_max=f（m）=m²，
当1＜m＜$\frac{1+\sqrt{2}}{2}$时，f（x）_max=m+$\frac{1}{4}$．

点评考查了二次函数区间内根的个数的判定和绝对值的分类讨论问题．难点是对参数的分类方法和二次函数性质的运用．