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2023-2024学年高一上学期期中数学试题(Word版附答案)

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河南省实验中学2023--2024学年上期期中试卷年级:高一科目:数学(时间:120分钟,满分:150分)一、单选题(本大题共8小题,每小题5分,共40分.)1.设集合,则集合的真子集个数是()A.6B.7C.8D.152.下列命题为真命题的是()A.若൐൐,则൐B.若,则C.若൐൐,c0,则൐D.若,则3.设,,,则()A.B.C.D.4.在天文学中,天体的明暗程度可以用星等或亮度来描述.两颗星的星等与亮度满足,其中星等为的星的亮度为,.已知太阳的星等是㌳䁩,天狼星的星等是䁩䁨,则太阳与天狼星的亮度的比值为()A.䁩B.䁩C.䁩D.䁩5.已知函数൐,且,无论取何值,图象恒过定点.若点在幂函数()的图象上,则幂函数的图象大致是()A.B.C.D.6.已知൐,൐且1,若有解,则实数的取值范围是()A.,∪9,B.,]∪[9,高一数学第1页(共4页) C.9,D.[9,],7.已知函数在上是单调的函数,则实数的取值,范围是()A.,]B.,䁨]C.,],䁨]D.,,䁨]8.设函数的定义域为,为奇函数,为偶函数,当,时,,则()A.B.C.㌳为奇函数D.二、多选题(本大题共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.)9.下列说法中正确的有()A.全体奇数构成的集合可以表示为,B.“൐”是“൐”的充分不必要条件C.集合与集合,的交集是空集D.命题“൐,﹣൐”的否定是“∃,”10.给出以下四个判断,其中正确的是()A.函数的值域为[,B.若函数的定义域为[,],则函数的定义域为,]C.函数定义域,值域䁨,则满足条件的有3个D.若函数,且䁨,则实数的值为㌳11.下列命题中的真命题有()A.当൐时,的最小值是高一数学第2页(共4页) B.的最小值是䁨C.当时,的最大值是D.若,为正实数,且,则的最大值为,12.已知函数,以下结论正确的是(),A.在区间[䁨,㌳]上先增后减B.䁨C.若方程在,㌳上有㌳个不等实根i,,,䁨,,㌳,则䁨㌳㌳D.若方程恰有个实根,则,三、填空题(本大题共4小题,每小题5分,共20分.)13.已知集合,,,且,则实数的值为.14.已知是定义在上的奇函数,且当时,,当时,.15.已知,,则的取值范围为.16.已知点,9在函数൐且图象上,对于函数定义域中的任意,,有如下结论:①•;②•;③0;④上述结论中正确结论的序号是.四、解答题(本大题共6小题,17题10分,其余各题12分,共70分.)㌳䁩17.计算:(1)䁩䁩;䁨(2)•䁛.高一数学第3页(共4页) 18.已知集合,集合,.(1)当时,求:①;②;(2)若,求实数的取值范围.19.已知幂函数在,上单调递增,.(1)求实数的值;(2)当[,䁨]时,记、的值域分别为集合、,设命题:,命题:,若命题是命题的必要不充分条件,求实数的取值范围.20.定义在上的函数,满足对任意的,,都有.当൐时,,且䁨.(1)求的值;(2)判断并证明函数在R上的奇偶性;(3)解不等式.21.我国某企业为了进一步增加市场竞争力,计划在2023年利用新技术生产某款新手机.通过市场分析,生产此款手机全年需投入固定成本250万,每生产x(千部),䁨手机,需另投入可变成本万元,且,䁨,䁨由市场调研知,每部手机售价0.8万元,且全年内生产的手机当年能全部销售完.(利润销售额固定成本可变成本).(1)求2023年的利润(万元)关于年产量x(千部)的函数关系式;(2)2023年产量为多少(千部)时,企业所获利润最大?最大利润是多少?22.已知定义域为的函数是奇函数.(1)求,的值;(2)用定义证明在,上为减函数;(3)若对于任意,不等式恒成立,求的范围.高一数学第4页(共4页) 河南省实验中学2023--2024学年上期期中试卷答案(高一)一、单选题(共8小题)1-4BCCA5-8AABD二、多选题(共4小题)9.ABC10.ABC11.AC12.ABD三、填空题(共4小题)13.314.﹣ex+2x+115.[,3]16.①④四.解答题(共6小题)17.解:(1)原式䁩䁨䁩䁨䁩㌳3+27﹣21=9.··········································(5分)(2)原式=lg52+lg2•lg50+(lg2)2﹣eln8=2lg5+lg2(1+lg5)+(lg2)2﹣8=2lg5+lg2+lg2•lg5+(lg2)2﹣8=2lg5+lg2+lg2(lg5+lg2)﹣8=2(lg5+lg2)﹣8=﹣6.················································(10分)18.解:(1)当m=﹣1时,A={x|﹣3≤x≤1},集合B={x|﹣1<x≤2},所以∁UB={x|x>2或x≤﹣1},···········································(2分)所以①A∪B={x|﹣3≤x≤2};············································(4分)②A∩(∁UB)={x|﹣3≤x≤﹣1};········································(6分)(2)若A∩B=,高一数学第5页(共4页) 当A=时,2m﹣1>m+2,即m>3,······································(8分)−当A≠时,或>解得m≤﹣3或<,·············································(10分)综上,m的范围为{m|m≤﹣3或m>}.··································(12分)19.解:(1)∵为幂函数,则3m2﹣2m=1,解得m=1或−,·················································(2分)又∵幂函数在(0,+∞)上单调递增,∴−>,得m=1.··················································(4分)(2)由第一问得,在[1,4]上递增,所以f(x)的值域为[1,2],即集合A={x|1≤x≤2},·······················(6分)而g(x)=﹣3x+t在[1,4]上递减,所以g(x)的值域为[t﹣81,t﹣3],即B={x|t﹣81≤x≤t﹣3},··············································(8分)由命题q是命题p的必要不充分条件可得A⫋B,···························(10分)−所以,解得5≤t≤82,高一数学第6页(共4页) 即t的取值范围为[5,82].··············································(12分)20.解:(1)由f(x+y)=f(x)+f(y),令x=y=0得f(0)=f(0)+f(0),∴f(0)=0.·························································(2分)(2)f(x)是奇函数,证明:f(x)定义为R,关于原点对称·································(3分)由f(x+y)=f(x)+f(y),令y=﹣x,得f(x﹣x)=f(x)+f(﹣x),即f(x)+f(﹣x)=f(0)=0,f(﹣x)=﹣f(x),所以f(x)是奇函数.······························(6分)(3)任取x1,x2R,x1<x2,x2﹣x1>0,·······························(7分)由f(x+y)=f(x)+f(y)知f(x+y)﹣f(x)=f(y)f(x1)﹣f(x2)=f(x1﹣x2)=﹣f(x2﹣x1),···························(8分)由于x2﹣x1>0,所以f(x2﹣x1)<0,所以f(x1)﹣f(x2)=﹣f(x2﹣x1)>0,即f(x1)>f(x2),所以f(x)是减函数,·················································(9分)f(6)=f(3+3)=f(3)+f(3)=﹣8,································(10高一数学第7页(共4页) 分)所以不等式f(t﹣1)+f(t)<﹣8即f(t﹣1+t)<f(6),所以2t﹣1>6,t>,所以不等式f(t﹣1)+f(t)<﹣8的解集为(,+∞).····················(12分)21.解:(1)由题意得W(x)=800x﹣R(x)﹣250,,<<䁨∵R(x),䁨,䁨∴当0<x<40时,R(x)=10x2+200x+1000,则W(x)=800x﹣(10x2+200x+1000)﹣250=﹣10x2+600x﹣1250,·········(2分)当x≥40时,R(x)=701x8450,则W(x)=800x﹣(801x8450)﹣250=﹣x−8200,·········(4分)㌳,<<䁨综上所述,W(x);···················(6,䁨分)㌳,<<䁨(2)由(1)得W(x),,䁨则当0<x<40时,W(x)=﹣10x2+600x﹣1250=﹣10(x﹣30)2+7750,二次函数W(x)的图象开口向下,且对称轴为直线x=30,∴W(x)max=W(30)=7750,·········································(8分)高一数学第8页(共4页) 当x≥40时,W(x)=﹣x−8200,又x2⋅200,当且仅当x,即x=100时等号成立,∴W(x)=﹣x−8200≤﹣200+8200=8000,·······················(10分)∵8000>7750,∴2023年产量为100(千部)时,企业所获利润最大,最大利润是8000万元.··(12分)22.解:(1)∵f(x)为R上的奇函数,∴f(0)=0,可得b=1················································(1分)又∵f(﹣1)=﹣f(1)∴,解之得a=1··········································(2分)经检验当a=1且b=1时,f(x),满足f(﹣x)=﹣f(x)是奇函数.·······································(3分)故a=1,b=1·························································(4分)(2)由(1)得f(x)1,任取实数x1、x2,且x1<x2················································(5分)则f(x1)﹣f(x2)······················(6高一数学第9页(共4页) 分)∵x1<x2,可得<,且>∴f(x1)﹣f(x2)>0,即f(x1)>f(x2),······························(7分)∴函数f(x)在(﹣∞,+∞)上为减函数;·······························(8分)(3)根据(1)(2)知,函数f(x)是奇函数且在(﹣∞,+∞)上为减函数.∴不等式f(t2﹣2t)+f(2t2﹣k)<0恒成立,即f(t2﹣2t)<﹣f(2t2﹣k)=f(﹣2t2+k)即t2﹣2t>﹣2t2+k对任意的tR都成立.即k<3t2﹣2t对任意的tR都成立,·····································(10分)22∵3t﹣2t=3(t−)−,当t时有最小值为−························(11分)∴k<−,即k的范围是(﹣∞,−).··································(12分)高一数学第10页(共4页)

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所属: 高中 - 数学
发布时间:2023-12-21 23:55:02 页数:10
价格:¥2 大小:44.07 KB
文章作者:随遇而安

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