2024届福建省福州市高三二模数学试卷

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（在此卷上答题无效）2023～2024学年福州市高三年级2月份质量检测数学试题（完卷时间120分钟；满分150分）友情提示：请将所有答案填写到答题卡上！请不要错位、越界答题！一、单项选择题：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.已知集合Axx1,B1,1，则AUBA.,1B.,1C.1D.1,1【答案】A【解析】集合A包含所有小于1的实数，B包含1和1两个元素，所以AUBxx1.22.已知点A2,2在抛物线C:x2py上，则C的焦点到其准线的距离为1A．B．1C．2D．42【答案】B2【解析】将点A2,2代入x2py，可得p1，故C的焦点到其准线的距离为1.3.已知e，e是两个不共线的向量，若2ee与ee是共线向量，则121212A.2B.2C.2D.2【答案】D【解析】依题意，设2ee=tee，又e，e是两个不共线的向量，121212所以t2,t，所以2.4.在△ABC中，AB2，AC4，BC27，则△ABC的面积为A.2B.23C.4D.43【答案】B222ABACBC1【解析】由余弦定理得，cosA，所以A120，2ABAC21所以SABACsinA23.△ABC2a2x5.设函数fx3在区间1,2上单调递减，则a的取值范围是A.,2B.,4C.2,D.4,【答案】Dxa2x【解析】函数y3在R上单调递增，而函数fx3在区间1,2上单调递减，1{#{QQABKYIAggggAgBAAAhCQwXICAGQkBEAAAoOxEAMsAAByRFABAA=}#} a所以y2xa在区间1,2单调递减，所以2，解得a4.故选D.26.已知正方形ABCD的四个顶点都在椭圆上，椭圆的两个焦点分别在边AD和BC上，则该椭圆的离心率为231513A．B．C．D．2222【答案】C222xyb【解析】不妨设椭圆方程为1ab0，当xc时，y，所以22aba222b2b2|AB|2c,|BC|，因为四边形ABCD为正方形，所以2c，即bac，所以aa2221551acac，所以ee10，解得e，因为e0,1，所以e.227.甲、乙、丙三个地区分别有x%，y%，z%的人患了流感，且x,y,z构成以1为公差的等差数列.已知这三个地区的人口数的比为5:3:2，现从这三个地区中任意选取一人，在此人患了流感的条件下，此人来自甲地区的概率最大，则x的可能取值为A.1.21B.1.34C.1.49D.1.51【答案】D【解析】设事件D,D,D分别为“此人来自甲、乙、丙三个地区”，事件F,F,F分别为“此123123人患了流感，且分别来自甲、乙、丙地区”，事件G为“此人患了流感”.5x3x32x410x7由题可知，PF，PF，PF，PGPFFF，12312310001000100010005x3x32x4所以PDG，PDG，PDG，因为此人患了流感来自12310x710x710x75x3x3,3甲地区的概率最大，所以解得x，故选D.5x2x4,28.已知函数fx及其导函数fx的定义域均为R，记gxfx.若gx2的图象关于点2，0对称，且g2xg2x1g12x，则下列结论一定成立的是20242024A.fxf2xB.gxgx2C.gn0D.fn0n1n1【答案】C【解析】因为gx2的图象关于点2，0对称，所以gx的图象关于原点对称，即函数gx为奇函数，则g00，又g2xg2x1g12x，所以g2xg2x1g2x1，所以gt1gtgt10，所以gtgt1gt20，所以gt1gt2，所以gtgt3，即gxgx3，202420242025所以3是g(x)的一个周期；因为gngng0g1g20，故Cn1n03正确；2{#{QQABKYIAggggAgBAAAhCQwXICAGQkBEAAAoOxEAMsAAByRFABAA=}#} 222取符合题意的函数fxcosx，则gxfxsinx，33323所以g00，又g02sing0，故2不是g(x)的一个周期,所以333gxgx2，排除B；21因为f1cos不是函数fx的最值，所以函数fx的图象不关于直线x1对称，32202420242所以fxf2x，排除A；因为fncosn10，所以排除D.n1n13二、多项选择题：本题共3小题，每小题6分，共18分。在每小题给出的四个选项中，有多项符合题目要求。全部选对的得6分，部分选对的得部分分，有选错的得0分。9.已知等差数列an的前n项和为Sn，a24，S535，则A.na的最小值为1B.nS的最小值为1nnSnanC.为递增数列D.2为递减数列nn【答案】ABC5a1a5【解析】假设a的公差为d，由S5a35，所以a7，又a4，n53322n3n1所以d3，a11，所以an3n2，Sn.2211选项A：nann3n23n，故n1时nan的最小值为1，A正确；333312331292选项B：nSnn，令fxxx，所以fxxx，可知fx在区间n222222,单调递增，所以n1时nS取得最小值1，B正确；n9Sn31Sn选项C：n，故为递增数列，C正确；n22nan23a1a2an选项D：22，因为1，21，所以2不是递减数列，D错误.nnn12n10.在长方体ABCDABCD中，AB2，AAAD1，E为AB的中点，则11111A.ABBCB.AD平面EBC111135C.点D到直线AB的距离为D.点D到平面EBC的距离为3115【答案】BC【解析】如图建立空间直角坐标系Dxyz，易知D0,0,0，A11,0,1，B1,2,0，B11,2,1，C0,2,0，E1,1,0.3{#{QQABKYIAggggAgBAAAhCQwXICAGQkBEAAAoOxEAMsAAByRFABAA=}#} 选项A，AB0,2,1，BC1,0,1，11A1BB1C0,2,11,0,110，所以A错误；选项B，显然ADBC，可得AD平面EBC，1111所以B正确；uuurAB255uuur选项C，记直线1，A1B的单位方向向量为u，则uuuur0,,，又A1D1,0,1A1B5521所以向量AD在直线AB上的投影向量为AQADuu0,,，1111552235则有D到直线AB的距离为DQADAQ，故C正确；1115选项D，设平面EB1C的法向量为mx,y,z，由B1C1,0,1，B1E0,1,1，uuuruuurDCm23可求得m1,1,1，又DC0,2,0，所以点D到平面EBC的距离d，1m3故D错误.11.通信工程中常用n元数组a,a,a,L,a表示信息，其中a0或1（i,nN*,1in）.123ni设ua,a,a,L,a,vb,b,b,L,b，du,v表示u和v中相对应的元素（a对应b，123n123niii1,2,K,n）不同的个数，则下列结论正确的是A.若u0,0,0,0,0，则存在5个5元数组v，使得du,v1B.若u1,1,1,1,1，则存在12个5元数组v，使得du,v3C.若n元数组w(0,0,L,0)，则du,wdv,wdu,v144424443n个0D.若n元数组w(1,1,L,1)，则du,wdv,wdu,v1442443n个1【答案】ACD【解析】1选项A：满足条件的数组共有C5个，故A正确；53选项B：满足条件的数组共有C10个，故B错误；5选项C：设u,v中对应项同时为0的共有m0mn个，同时为1的共有s0snm个，从而对应项一项为1与另一项为0的共有nms个，这里nms，从而du,vnms，而du,wdv,w2snmsdu,v2sdu,v，故C正确，同理D正确.三、填空题：本大题共3小题，每小题5分，共15分。12.在复平面内，复数z对应的点的坐标是2,1，则iz___________.【答案】12i【解析】依题意可知z2i，所以izi2i12i.13.底面半径为2且轴截面为正三角形的圆锥被平行于其底面的平面所截，截去一个高为34{#{QQABKYIAggggAgBAAAhCQwXICAGQkBEAAAoOxEAMsAAByRFABAA=}#} 的圆锥，所得的圆台的侧面积为.【答案】6【解析】由已知可得圆台的上底面半径r1，下底面半径r2，母线长l2，则该圆台的侧面积为rrl1226.14.在平面直角坐标系xOy中，整点P（横坐标与纵坐标均为整数）在第一象限，直线PA，22PB与eC:x2y4分别切于A，B两点，与y轴分别交于M，N两点，则使得△PMN周长为221的所有点P的坐标是_________.【答案】1,4或2,322【解析】因为直线PA，PB分别与eC:x2y4相切于A，B两点，且直线PA，PB分别与y轴交于M,N两点，所以PAPB,AMOM,BNON，所以△PMN的周长为PMMNPNPMOMONPNPMAMBNPNPAPB222PA2PCAC22PC4221，22所以PC5，设Px0,y0,x00，y00，所以x02y025，因为P为整点，所以点P的坐标为1,4或2,3.备注：只写出一个点坐标不得分.四、解答题：本大题共5小题，共77分。解答应写出文字说明、证明过程或演算步骤。15.（13分）已知函数fxsinx03，x是fx的零点.48（1）求的值；1（2）求函数yfxfx的值域.828【解析】（1）由已知可得fsin0，········································1分884解得k,kZ，············································································3分84即28k,kZ，··················································································4分又03，可得2.·············································································5分（2）由fxsin2x，可得41yfxfx8285{#{QQABKYIAggggAgBAAAhCQwXICAGQkBEAAAoOxEAMsAAByRFABAA=}#} sin2xsinx2cos2xsinx212sinxsinx2192sinx，··················································································8分48其中1sinx1，19则当sinx时取得最小值，sinx1时取得最大值2，····························12分4819故函数yfxfx的值域为,2.········································13分828816.（15分）如图，四棱锥SABCD的底面为正方形，平面SAD平面ABCD，E在SB上，且AEBC.（1）证明：SA平面ABCD；（2）若SAAB2，F为BC的中点，且EF3，求平面AEF与平面SAD夹角的余弦值.【解法一】（1）因为BCAB,BCAE,AEIABA，所以BC平面SAB，·····························································································又SA平面SAB，所以BCSA，···········································································3分又BCPAD，所以SAAD，··················································································4分又平面SAD平面ABCD，平面SADI平面ABCDAD，SA平面SAD，所以SA平面ABCD.····························································································6分（2）由（1）得BC平面SAB，又SB平面SAB，所以BCSB，······························7分因为BF1,EF3，所以BE2，·······································································8分因为SA平面ABCD，AB平面ABCD，所以SAAB，1又SAAB2，所以SB22，所以BESB，······················································9分2由（1）知SA,AD,AB两两垂直，如图，以点A为原点，分别以AB，AD，AS所在直线为x轴，y轴，z轴建立空间直角坐标系，则A0,0,0,S(0,0,2),B(2,0,0),E1,0,1,F21,0.·························································10分uuuruuur所以AE1,0,1，AF2,1,0，···········································································显然平面SAD的一个法向量n1,0,0，·································································1uuurn2AE,设平面AEF的法向量为n2(x,y,z)，则uuurnAF,2uuurn2AExz0,即uuur取x1，则n21,2,1，·····················································nAF2xy0,26{#{QQABKYIAggggAgBAAAhCQwXICAGQkBEAAAoOxEAMsAAByRFABAA=}#} nn1612所以cosn,n，·····································································14分12nn66126设平面AEF与平面SAD的夹角为，则coscosn,n，1266所以平面AEF与平面SAD夹角的余弦值为.························································15分6【解法二】（1）因为BCAB,BCAE,AEABA，所以BC平面SAB，·································2分又SA平面SAB，所以BCSA，···········································································3分因为平面ABCD平面SAD，平面ABCD平面SADAD，ABAD，AB平面ABCD，所以AB平面SAD，又SA平面SAD，所以ABSA，···················5分又BCABB，BC平面ABCD，AB平面ABCD，所以SA平面ABCD.····························································································6分（2）由（1）知SA,AD,AB两两垂直，如图，以点A为原点，分别以AB，AD，AS所在直线为x轴，y轴，z轴建立空间直角坐标系，则S0,0,2,B2,0,0,F2,1,0.··················7分设SESB01，则AEASSE0,0,22,0,22,0,22，所以EFAFAE2,1,02,0,2222,1,22，······································9分2213由EF3，得221223，解得，或（舍去），22所以E1,0,1，····································································································10分uuuruuur所以AE1,0,1，AF2,1,0，···········································································显然平面SAD的一个法向量n1,0,0，·································································1uuurn2AE,设平面AEF的法向量为n2(x,y,z)，则uuurnAF,2uuurn2AExz0,即uuur取x1，则n21,2,1，·····················································nAF2xy0,2nn1612则cosn1,n2，········································································14分nn66126设平面AEF与平面SAD的夹角为，则coscosn,n，1266所以平面AEF与平面SAD夹角的余弦值为.··························································15分617.（15分）人的性格可以大体分为“外向型”和“内向型”两种，树人中学为了了解这两种性格特征与人的性别是否存在关联，采用简单随机抽样的方法抽取90名学生，得到如下数据：7{#{QQABKYIAggggAgBAAAhCQwXICAGQkBEAAAoOxEAMsAAByRFABAA=}#} 外向型内向型男性4515女性2010（1）以上述统计结果的频率估计概率，从该校男生中随机抽取2人、女生中随机抽取1人担任志愿者.设这三人中性格外向型的人数为X，求X的数学期望．（2）对表格中的数据，依据0.1的独立性检验，可以得出独立性检验的结论是这两种性格特征与人的性别没有关联.如果将表格中的所有数据都扩大为原来10倍，在相同的检验标准下，再用独立性检验推断这两种性格特征与人的性别之间的关联性，得到的结论是否一致？请说明理由.nadbc20.10.050.012附：参考公式：.abcdacbdx2.7063.8416.635【解法一】3（1）由统计结果可知，外向型男生在所有男生中占比为，外向型女生在所有女生中占比423为，故从该校男生中随机抽取一人为外向型男生的概率是，从该校女生中随机抽取一342人为外向型女生的概率是.····················································································2分3则X的所有可能取值为0，1，2，3.·········································································3分221111311121则PX0，PX1C，243484434362231131221323PX2C2，PX3，·······························7分43443484381121313所以EX0123.·······························································8分4864886（2）零假设为H：这两种性格特征与人的性别无关联.················································9分0由所获得的所有数据都扩大为原来10倍，可知22900450100150200906.9232.706x0.1···········································13分60030065025013依据0.1的独立性检验，可以推断这两种性格特征与人的性别有关联，与原来的结论不一致，原因是每个数据扩大为原来的10倍，相当于样本量变大为原来的10倍，导致推断结论发生了变化.·····································································································15分【解法二】3（1）由统计结果可知，外向型男生在所有男生中占比为，外向型女生在所有女生中占比423为，故从该校男生中随机抽取一人为外向型男生的概率是，从该校女生中随机抽取一342人为外向型女生的概率是.····················································································2分38{#{QQABKYIAggggAgBAAAhCQwXICAGQkBEAAAoOxEAMsAAByRFABAA=}#} 从该校男生中随机抽取2人，抽到性格外向型的人数记为Y；从该校女生中随机抽取1人，132抽到性格外向型的人数记为Y，则Y:B2,,Y:B1,，·······································4分212433322所以EY2，EY1，··································································6分1242333213所以EXEYYEYEY.························································8分1212236（2）略，同解法一.18.（17分）22y已知双曲线W:x1，A3,0，动直线l:xmy30与x轴交于点B，且与W8交于C，D两点，tCD是BC，BD的等比中项，tR．（1）若C，D两点位于y轴的同侧，求t取最小值时△ACD的周长；（2）若t1，且C，D两点位于y轴的异侧，证明：△ACD为等腰三角形．【解法一】22y（1）因为动直线l:xmy30与x轴交于点B3,0，因为W:x1的右焦点为3,0，8所以点B为W的右焦点.设BCp，BDq，因为C,D两点位于y轴的同侧，所以CDpq，2因为tCD是BC，BD的等比中项，所以tCDBCBD，·······························2分2pqpq1t1所以pq2pq24，当且仅当pq时取等号，所以tmin，···················4分221当t时pq，所以AC2p2qAD，所以lx轴，··································5分2x3,由解得y8，228xy8,所以BCBD8，所以CD16，·································································6分由双曲线的定义得AC10，所以ACADCD10101636，即△ACD的周长为36.····················································································8分（2）设Cx1,y1，Dx2,y2，xmy30,22由22得8m1y48my640，8xy8,因为直线l与W交于C，D两点，28m10,48m64所以2且y1y22，y1y22，····························10分256m10,8m18m19{#{QQABKYIAggggAgBAAAhCQwXICAGQkBEAAAoOxEAMsAAByRFABAA=}#} 22222由t1，可得BCBDCD，故1my11my21my1y2，22又C,D两点位于y轴的异侧，所以y1y20，所以y1y2y1y2，即5y1y2y1y2，26448m25所以5，解得m，····················································13分228m18m146422564所以y1y2，所以CDBCBD1my1y2116，949所以CD4，·····························································································15分BCAC2，4BDAC2，不妨设点C在第二象限，根据双曲线定义，得，即ADBD2，ADBD2，解得ACAD，所以△ACD是等腰三角形.····················································17分【解法二】（1）设Cx1,y1,Dx2,y2xmy30,22由22得8m1y48my640，8xy8,因为直线l与W交于C，D两点，28m10,48m64所以2且y1y22，y1y22，·····································2分256m10,8m18m16421由C,D两点位于y轴的同侧，可得y1y220，解得m，8m1822又tCD是BC，BD的等比中项，故可得BCBDtCD，22222故1my11my2t1my1y2，6422y1y28m21118m即t222，············································5分yy48m644m1124228m18m12212118m19又0m，故t8，2284m14m121111可得0t,即t且t0，所以tmin，··················································6分42221x3,当t即m0时，所以lx轴，由22解得y8，28xy8,所以BCBD8，所以CD16，22又AB6，所以AC8610，所以ACADCD10101636，即△ACD的周长为36.····················································································8分10{#{QQABKYIAggggAgBAAAhCQwXICAGQkBEAAAoOxEAMsAAByRFABAA=}#} 6421（2）因为C,D两点位于y轴的异侧，故y1y220，所以m，8m18222118m18m1且由（1）知t221，4m14m155解得m或m，·····················································································12分225yy4512当m时，设CD的中点E的坐标为xE,yE，y，E22355451145xy33，所以点E的坐标为，，························13分EE33223354551又CD的垂直平分线的斜率为，所以CD的垂直平分线方程为yx，2323535即yx，······························································································15分22535又点A3,0在直线yx上，所以ACAD，即△ACD为等腰三角形．225当m时，同理可证，△ACD为等腰三角形．2综上所述，△ACD为等腰三角形．··········································································17分19.（17分）2已知函数fxxlnxx1.（1）讨论fx的单调性；x12（2）求证：fxe1；2xx（3）若p0，q0且pq1，求证：fpfq4.【解法一】（1）fx的定义域为0,，··············································································1分fxlnx2x1，·····························································································2分112x记txfx，tx2，xx11当x0,时，tx0，tx单调递增；当x,时，tx0，tx单调递减.····3分221所以txtln20，即fx0，·····························································4分max2所以fx在区间0,上单调递减.·········································································5分2（2）先证fxx1，记gxfxx1，则gxxlnxxxxlnxx1，1记mxlnxx1，则mx1，所以x0,1时，mx0，mx递增；xx1,时，mx0，mx递减.所以mxm10，所以mx0，又x0，所以gx0，故fxx1.max11{#{QQABKYIAggggAgBAAAhCQwXICAGQkBEAAAoOxEAMsAAByRFABAA=}#} ··························································································································8分x12x12x12再证e1x1，即证ex0，记hxex，222xxxxxx2x1x则hxex11ex1，xxx记pxex1，则px1e0，所以px在x0,递增，x12所以pxp00，所以hx0，即e1x1，2xxx12所以fxe1.····················································································11分2xx（3）由（2）知mxlnxx1的最大值为0.因为p0，q0且pq1，则p，q之中至少有一个大于1，·····································12分11不妨设p1，则q0，由（1）可知fx为减函数，所以fqf，pp1所以fpfqfpf，··········································································14分p221211111因为fpfplnpp1ln1plnpp4pppppp11111plnpp4，记splnpp，则spmp110，ppppp11因为p1，所以p，所以fpf4，所以fpfq4.······················17分pp【解法二】（1）略，同解法一.·······························································································5分xx（2）构造函数hxex1x0，hxe1，x当x0时，hx0，hx单调递增，hxh00，所以ex1，························6分1构造函数φxlnxx1，φx1，x当x0,1时，φx0，φx单调递增；当x1,时，φx0，φx单调递减.·········所以φxmaxφ10，即φx0，即lnxx1成立.·············································7分22所以fxxlnxx1xx1x1x1，······················································8分x121212所以e1x11x，····················································9分222xxxxxx212121则只需证明xx1，即10，而10显然成立，·····················10分22xxxxxx12所以fxe1.····················································································11分2xx2（3）先证fxx1，记gxfxx1，则gxxlnxxxxlnxx1，1记mxlnxx1，则mx1，所以x0,1时，mx0，mx递增；x12{#{QQABKYIAggggAgBAAAhCQwXICAGQkBEAAAoOxEAMsAAByRFABAA=}#} x1,时，mx0，mx递减.······································································13分所以mxm10，所以mx0，又x0，所以gx0，故fxx1.max························································································································14分所以fpp1，fqq1，因为p0，q0且pq1，所以fpfqpq2，··············································································15分所以pq2pq212，所以pq2，则fpfq224.························································································································17分13{#{QQABKYIAggggAgBAAAhCQwXICAGQkBEAAAoOxEAMsAAByRFABAA=}#}