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高一数学试题答案第1页(共5页)三明一中2022—2023学年上学期高一第1次月考数学参考答案及评分细则一、选择题:本题共8小题,每小题5分,共40分.1.C2.B3.D4.C5.B6.C7.A8.B二、选择题:本题共4小题,每小题5分,共20分.全部选对的得5
分,有选错的得0分,部分选对的得3分.9.BCD10.AD11.CD12.ABC三、填空题:本题共4小题,每小题5分,满分20分.13.001,,14.15.216.22,33四、解答题:本大题共6小题,共70分.解答应写出文字说明、证明过程或演算步骤.17.解:
因为fx是二次函数,所以设2(0)fxaxbxca...........................2分由01f,得c=1.............................................
..........................................................3分由12fxfxx,得2211112axbxaxbxx...............
.............5分整理得220axab...........................................................................................6分所以2200a
ab,所以11ab........................................................................................8分所以21fxxx......
.............................................................................................10分18.解:(1)当2a时,集合|13Axx......................
...................................1分|13Bxx......................................................
...................................................3分所以|13ABxx......................................................................
.....................5分(2)若选择①,则ABB①,则AB........................................................................6分
因为|11Axaxa,所以A...................................................................7分又|13Bxx,所以1113aa..........................
...........................................10分解得02a...........................................................................................
...................11分所以实数a的取值范围是0,2................................................................................12分高
一数学试题答案第2页(共5页)若选择②,“xA“是“xB”的充分不必要条件,则AB.....................................6分因为|11Axaxa,所以A....................................
.................................7分又|13Bxx,所以1113aa,......................................................
............10分解得02a,...........................................................................................................11分所以实数
a的取值范围是0,2..............................................................................12分若选择③,AB,因为|11Axaxa
,|13Bxx,所以13a或11a,..............................................................................
............10分解得4a或2a,..................................................................................
.................11分所以实数a的取值范围是[,24,)..........................................................12分19.解:(1
)函数()fx在0,上单调递增........................................................1分证明:22(1)22()2111xxfxxxx.........................
................................2分任取1212,0,,xxxx且...................................................................................3分则
121221122221111xxfxfxxxxx................................................4分由120xx得,1
2120,10,10xxxx,所以12fxfx........................................5分故()fx在0,上单调递增。.................
..............................................................6分(2)因为()fx在0,上单调递增,所以(21)(1)fmfm等价于21010211mmmm
.............................................................................................................9分213m解得..........................
...................................................................................11分故实数m的取值范围2,13..............
.........................................................................12分高一数学试题答案第3页(共5页)20.解:(1)由已知得
:1510201530fxWxxxWxx....................................................2分故29030360,021080108030,261xxxfxxxx
................................................................................4分(2)若02x,则22117159030360903362fxxxxx
,.............................5分此时,对称轴为16x,故fx有最大值为2660f..............................................6分若26x,
则10801080108030111030111fxxxxx............................7分108011102301111021807501xx
,............................................................9分当且仅当10803011xx,即5x时等号成立,......................................
......................10分此时,fx有最大值为5750f,................................................................................11分综上有,fx有最大值为750
,∴当施用肥料为5千克时,该水果树的单株利润最大,最大利润是750元..................12分21.解:(1)由题意可知:方程2230axbx的两根是1,1........................
..............2分所以21103(1)11baa...................................................................
............................4分解得32ab...........................................................................................................
.........5分(2)由10f得1ba....................................................................................................
.6分xR,4fx成立,即使2210axbx恒成立,又因为1ba,代入上式可得2310axax恒成立.....................................7分高一数学试题答案第4页(共5页
)当0a时,显然上式不恒成立;................................................................................8分当0a时,要使2310axax恒成立所以20Δ340aaa
...........................................................................................10分解得91a..................................
..........................................................................11分综上可知a的取值范围是9,1....................
......................................................12分22.解:(1)5fx解集为3,12...............................................
.......................4分(2)证明:2115152488fxx.......................................................
.....................5分min8815()3558fx,所以235mm.............................................................6分解得12
m,所以2M,所以332ab..........................................................7分因为33222ababaabb.........
..................................................................8分222213024aabbabb...........................
......................................................9分所以0ab,由24abab得,23221234abaab
babababab则2ab,当且仅当ab时,等号成立..................................................................11分综上所述,02ab.............
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