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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|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，|1

3Bxx，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29030360,021080108030,261xxxfxxxx

................................................................................4分(2)若02x，则22117159030360903362fxx

xxx,.............................5分此时，对称轴为16x，故fx有最大值为2660f..............................................6分若26x，则10

801080108030111030111fxxxxx............................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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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