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东华高级中学东华松山湖高级中学2022—2023学年第二学期高一2月考数学试卷一、单选题：本题共8小题，每小题5分，共40分．在每小题给出的四个选项中，只有一项是符合题目要求的．1.命题“Nm，21Nm+”的否定是（）A.Nm，21Nm+B.Nm，21Nm+

C.Nm，21Nm+D.Nm，21Nm+2.设Rba,，则“()02−aba”是“ba”的（）.A.充分不必要条件B.必要不充分条件C.充分必要条件D.既不充分也不必要条件3.函数()1log21+−=xxxf的零点所在的区间为（）A.1,41B.314

1，C.2131，D.1,214．若3a=，3b=r，向量a与向量b的夹角为150°，则向量a在向量b上的投影向量为（）A．32bB．32b−C．32bD．32b−5.设3log0.4a=，2log3b=，则（）A.0ab

且0ab+B.0ab且0ab+C.0ab且0ab+D.0ab且0ab+6.要得到函数()3sincosfxxx=+的图象，只需将函数()π2sin6gxx=−的图象进行如下变换得到（）A.向左平移π3个单位B.向右平移π3个单位C.向右平移π6

个单位D.向左平移π6个单位7．已知tan，tan是方程23340xx++=的两根，且ππ22−，ππ22−，则+的值为（）A．π3B．2π3−C．π3或2π3−D．π3−或2π38.若定义2023,2023−上的函数()xf满足：对任意

2023,2023,21−xx有()()()20222121−+=+xfxfxxf若()xf的最大值和最小值分别为NM,，则NM+的值为（）A.2022B.2018C.4036D.4044二、多选题：本题共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．在ABC中，D为BC中点，且EDAE2=，则（）A．CBCACE6132+=B.CBCACE3131+=C．()CBCACE+⊥D．()CBC

ACE+//10．已知函数()πsin223sincos6fxxxx=+−，则（）A．()fx的最大值为1B．直线π3x=是()fx图象的一条对称轴C．()fx在区间ππ,63−上单调递减D．()fx的图象关于点π,06对称

11.若1122ab，则下列关系式中一定成立的是（）A.33abB.abee（2.718e）C.()()sincossincosab++（是第一象限角）D.()()22ln1ln1ab++12.已知函数()+−+=2

,15820,2log22xxxxxxf，若方程()kxf=有四个不同的根4321,,,xxxx，且4321xxxx，则下列结论正确的是（）A.21−kB.22221+xxC.()84321=+xxxxD.3221+xx三、填空题：本题共4小题，每小题5分，共20分．13．已知向量a

，b满足3a=，2b=，211ab−=，则ab=______．14.请写出一个函数()fx，使它同时满足下列条件：（1）()fx的最小正周期是4；（2）()fx的最大值为2．()fx=____________．15.若()fx是定义在R上的奇函数，当0x时，()122xfxxm

=−+(m为常数)，则当0x时，()fx=_________.16.木雕是我国古建筑雕刻中很重要一种艺术形式，传统木雕精致细腻､气韵生动､极富书卷气．如图是一扇环形木雕，可视为扇形OCD截去同心扇形OAB所得部分．已知0.6mOA=，1.4mAD=，100AOB=，

则该扇环形木雕的面积为________2m．四、解答题：本题共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.（本题满分10分）的已知集合241|1,|212xAxBxaxax−==+

−(1)求集合RCA(2)若ABB=,求实数a的取值范围.18.（本题满分12分）在平面直角坐标系xOy中，O是坐标原点，角的终边OA与单位圆的交点坐标为()1,02Amm−，射线OA绕点O按逆时针方向旋转

弧度．．后交单位圆于点B，点B的纵坐标y关于的函数为()yf=.（1）求函数()yf=的解析式，并求π3f−的值；（2）若()34f=，()0,π，求4πtan3−的值.19.（本题满分12分）函数()sin2sin

fxxx=+（1）请用五点作图法画出函数()fx在0,2上的图象（先列表，再画图）（2）设()()2mFxfx=−，0,2x，当0m时，试研究函数()Fx的零点的情况.20.（本题满分12分）2020年我国面对前所未知，突如其来，来势汹汹的新冠肺炎疫情，中央出台了一系列助力复工

复产好政策．城市快递行业运输能力迅速得到恢复，市民的网络购物也越来越便利．根据大数据统计，某条快递线路运行时，发车时间间隔t（单位：分钟）满足：520t，tN，平均每趟快递车辆的载件个数()Rt(单位：个)与发车时间间隔t近似地满足()()2161810,5141618,1420ttRtt

−−=，其中tN．（1）若平均每趟快递车辆的载件个数不超过1600个，试求发车时间间隔t的值；（2）若平均每趟快递车辆每分钟的净收益5()7770()100RtStt−=+(单位：元)

，问当发车时间间隔t为多少时，平均每趟快递车辆每分钟的净收益最大？并求出最大净收益(结果取整数)．21.（本题满分12分）已知函数()21axbfxx+=+是定义域R上的奇函数，且满足()()91210

ff+=（1）判断函数()fx在区间()0,1上的单调性，并用定义证明（2）已知()12,0,xx+，且12xx，若()()12fxfx=，证明：122xx+22.（本题满分12分）若函数()xfy=对定义域内的每一个值1x，在其定

义域内都存在唯一的2x，使()()121=xfxf成立，则称函数()xfy=具有性质M．（1）判断函数()1fxx=是否具有性质M，并说明理由；（2）若函数()2144333fxxx=−+的定义域为,(,N*mnmn且2)m且具有性质M，求mn的值；（3）已知2a，函数()()2

2xfxa=−的定义域为1,2且()fx具有性质M，若存在实数1,2x，使得对任意的Rt，不等式()24fxstst++都成立，求实数s的取值范围．东华高级中学东华松山湖高级中学2022—2023学年第二学期高一2月考

数学答案一、选择题123456789101112DACDBABDBDABCBCBCD二、填空题13.2；14.x2sin2（答案不唯一）15.221xx−−+；16.9091三、解答题17.解：（1）{|13}Axx=，{

|13}RCAxxx=或····················································4分（2）由题意，若ABB=，则BA，···················

······················································5分①B=时，122aa+，解得4a；····································································

·····6分②B时，12211232aaaa++，……………………8分解得12a；…………………………………………………9分综上，a的取值范围为(1,2](4,)a+.

········································································10分18.解：（1）因为1sin2=−，且0m，所以7π6=，···································

·····················2分由此得()7πsin6f=+···························································································4分ππ7

π5π1sinsin33662f−=−+==.········································································5分（2）由()34f=知7

ππ3sinsin664+=−+=，即π3sin64+=−···················7分由于()0,π，得ππ7π,666+，与此同时πsin06+，所以πcos06

+由平方关系解得：π13cos64+=−，·········································································9分ππsi

ncos4π3936tantanππ333cossin36−−−+−=−===−−+·····························12分19、（1）(3sin,0,()

sin,,2xxfxx=−··················································································2分按五个关键点列表：x02322sinx010-10(

)sin2sinfxxx=+03010描点并将它们用光滑的曲线连接起来如图1：········································································7分（2）因为()()2mFxfx=−，所以()Fx的零点个数等价

于()yfx=与2my=图象交点的个数，··········································8分设2mt=，0m，则1t·······································································

······················9分当20log3m，即13t时，()Fx有2个零点；当2log3m=，即3t=时，()Fx有1个零点；当2log3m，即3t时，()Fx有0个零点.·············································

··············12分20、解：（1）当1420t时，16181600，不满足题意，舍去．······································1分当514t时，21618(10)1600t−−，

即220820tt−+．···········································3分解得1032t+（舍）或1032t−．·······························································

··········4分∵514t且tN，∴5t=．·····················································································5分所以发车时间间隔为5分钟．··················

········································································6分（2）由题意可得()1805200,514320100,1420tttSttt−++=

+．····················································8分当514t时，180()25200140Sttt−+=（元），·········

············································9分当且仅当1805tt=，即6t=时，等号成立，········································

·······························10分当2014t时，()tS单调递减，14t=时，()12310014320+tS（元）···························

11分所以发车时间间隔为6分钟时，净收益最大为140（元）．····················································12分21.解：（1）由()fx为奇函数，(0

)0f=可得0b=；···························································1分又9(1)(2)10ff+=，得1a=；··························

····························································2分所以2()1xfxx=+.2()1xfxx=+在(0,1)上单调递增，理由如下：····················

··············································3分12,(0,1)xx，且12xx，则1221121222221212()(1)()()11(1)(1)xxxxxxfxfxxxxx−−−=−=++++···················

4分因为1201xx，所以210xx−，1210xx−，2110x+，2210x+所以12()()0fxfx−，12()()fxfx，()fx在(0,1)上单调递增·····························

·········6分（2）证法一：由题意，12()()fxfx=，则有2112122212()(1)()()0(1)(1)xxxxfxfxxx−−−==++·················8分因为120xx，所以1210xx−=，即121xx=，········

·······················································10分所以121222xxxx+=，得证.···································

················································12分证法二：由（1）知，()fx在(0,1)上单调递增，同理可证()fx在(1,)+上单调递减.因为12,(0,)xx+，12()()fxfx=，所以1(0,1)x，2(1,)x+，

所以12(1,2)x−·······························································8分要证122xx+，即证212xx−，即证21()(2)fxfx−，即证11()(2)fxfx−，··

····························································9分代入解析式得11221121(2)1xxxx−+−+，即证221111[(2)1](2)(1)x

xxx−+−+化简整理得321113310xxx−+−，即证31(1)0x−，························································10分因为1(0,1)x，31(1)0x−显然成

立，··········································································11分所以原不等式得证，所以122xx+.················································

·····························12分22、解：（1）对于函数()1fxx=的定义域()(),00,−+U内任意的1x，取211xx=，则()()121fxfx=，···················································

································1分结合()1fxx=的图象可知对()(),00,−+U内任意的1x，211xx=是唯一存在的，·················2

分所以函数()1fxx=具有性质M.（2）因为()221441(2)3333fxxxx=−+=−，且m>2，所以()fx在,mn上是增函数，········3分又函数()xf具有性质M，所以()()1=nfmf，即()()1229122=−−nm，·

···························4分因为2nm，所以()()223mn−−=且220nm−−，又*,Nmn，所以2123mn−=−=，解得35mn==，所以15mn=．

·································································5分（3）因为1,2x，所以22,4x，且2xy=在定义域上单调递增，又因为2a，()2yxa=−在2,4上单调递增，所

以()()22xfxa=−在上[1,2]单调递增，········································································6分又因为()fx具有性质

M，从而()()121ff=，即()()241aa−−=，所以2670aa−+=，解得32a=−或32a=+（舍去），·····························································

···············7分因为存在实数1,2x，使得对任意的Rt，不等式()24fxstst++都成立，所以2max()4fxstst++，·······························································

····························8分因为()()22xfxa=−在上1,2单调递增，所以()222(12)4fstst=+++即21220stst++−对任意的Rt恒成立．··············································

···················9分所以()20Δ41220sss=−−或0s=，········································································11分解得4820s−或0s=

，综上可得实数s的取值范围是482,0−………………12分