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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．

15．216．22,33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12

fxfxx，得2211112axbxaxbxx............................5分整理得220axab......................................

.....................................................6分所以2200aab，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ABB①，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AB.....................................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12122

1122221111xxfxfxxxxx................................................4分由120xx得，12120,10,10xxxx，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108010801080

30111030111fxxxxx............................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)11

baa...............................................................................................4分解得32ab

....................................................................................................................5

分（2）由10f得1ba...........................................................................................

..........6分xR，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12m，所以2M，所以332ab.............................

.............................7分因为33222ababaabb...........................................................

................8分222213024aabbabb.............................................................................

....9分所以0ab，由24abab得，23221234abaabbabababab则2ab，当且仅当ab时，等号成立...........................................

.......................11分综上所述，02ab.......................................................................

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