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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12fxfxx，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|13B

xx，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121221122221111xxfxfxxxxx.......................................

.........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29030360,02108010803

0,261xxxfxxxx................................................................................4分(2)若02x，则22117159030360903

362fxxxxx,.............................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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20Δ340aa

a...........................................................................................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，23221234a

baabbabababab则2ab，当且仅当ab时，等号成立..................................................................11分综上所述，02ab..............

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