江苏省扬州市2023-2024学年高三上学期11月期中检测 数学答案

高三数学参考答案2023.11题号123456789101112答案CACBABDCADABACDBD题号131415164答案3fxx()sin(2)20,327T17.【答案】(1)因为fx()0，fx()0，且xx的最小值为，所以，12122222则T，所以2，············································································2分T又f(0)23，所以4sin23，又||，所以，23所以fxx()4sin(2).·················································································3分35令222kxk，解得kxkkZ,,23212125所以函数fx()的单调递增区间为kkkZ,,.···································5分1221123(2)由题可知f()4sin(2)，则sin(2)，35354因为(,)，所以2(,)，12232334所以cos(2)1()2，·································································7分355所以sin2sin[(2)]sin(2)coscos(2)sin3333333143343().·····························································10分52521018.【答案】(1)证明：连接ABAB11,交于点E，连接ACAC11,交于点F，连接EF，则平面AB11C和平面A1BC交线为，因为ABCA1B1C1为直三棱柱，所以ABBA11，ACC11A为平行四边形，所以为AB1中点，F为AC1中点，所以EFBC11，··········································4分又EF平面ABC111，BC11平面ABC111，所以EF平面ABC111，即l平面ABC111．························································6分(2)直三棱柱ABCAB111C中，BABC，所以BA,,BCBB1两两垂直．以为单位正交基底，建立如图所示的空间直角坐标系，BA,,BCBB1Bxyz则C(0,1,0)，C1(0,1,1)，A1(1,0,1).····································································7分第1页（共6页）{#{QQABRYYAggCoAABAAAhCEwFCCAIQkBGCCCoOwBAIoAAAABFABCA=}#}zB1C1A1EFBACxy设Qt(0,0,)，则QC1t(0,1,1)，CA1(1,1,1)，QCCA11t3所以cos,QCCA11·····································10分23QCCA111(1)3t11解得t，所以线段BQ长为．··································································12分221a19．【答案】(1)方法一：因为f(x)是奇函数，所以f(0)＝0，即0，解得a1，····2分2221x21(21)212xxxx此时，fx()，fxfx()()，22x122(22)222xxxx111则fx()是奇函数．故a1．·····································································································4分方法二：因为f(x)是奇函数，22212xxxxaaaa所以f(xfx)()0，22222222xxxx1111即(1)(21)0ax对xR恒成立，所以a1.····································································································4分－2x＋111(2)由(1)知f(x)＝＝－＋，则f(x)在R上为减函数，··························6分2x+1＋222x＋1又f(x)是奇函数，由ffk(3sincos)(cos)02得：ff(kfk3sincos)(cos)(cos)22，所以3sincoskcos2，2即3sincoscosk在,0上有解，·············································9分4231cos21记g()3sincoscos，则g()sin2sin222622因为，则2,，63613所以sin21,，所以g(),1，62233所以k，即k．··············································································12分22第2页（共6页）{#{QQABRYYAggCoAABAAAhCEwFCCAIQkBGCCCoOwBAIoAAAABFABCA=}#}20．【答案】(1)提出假设H0：该品牌方便面中C卡片所占比例与方便面口味无关.n()150(20451075)75adbc2220.186.635，···········3分()()()()955530120418abcdacbd又P(2≥0.010)6.635，所以没有99%的把握认为“该品牌方便面中C卡片所占比例与方便面口味有关”．······4分(2)①记“小明一次购买3袋该方便面，中奖”为事件A，22124PAA()3.··············································································8分5551253②记“小明一次购买3袋该方便面，未获得C卡”为事件B.464PBA()()3，····················································································10分512564PBA()64PBA(|)125.24PA()10111256464答：①小明中奖的概率为；②小明为中奖，未获得C卡的概率为．·············12分12510121．【答案】(1)由题可知ABc，ACc3，BCc2，BABCACccc222222(2)(3)1在△ABC中，由余弦定理得cosABC，2222BABCcc又ABC(0,)，所以ABC，·································································2分31又D为BC的中点，所以BDBCc，则BDBA，2所以△ABD为等边三角形，又AD1，所以AB1，BC2，1133所以SBABCABCsin12.··········································4分△ABC2222a(2)方法一：由题可知ABc，ACc3，AD1，BDDC，2设ABC2，DAC，ADB，则ADC，ABADc1在△ABD中，由正弦定理得，即，sinsinADBABDsinsin2aDCAC3c在△ADC中，由正弦定理得，即2，sinDACsinADCsinsin()1a3所以，则a，①····························································6分sin223sincos在△ABD和△ADC中，由余弦定理得aa12()2cc212()2(3)2cosADBcosADC220，aa212122所以ac2248，②······················································································8分在△ADC中，由余弦定理得DC2AD2AC22ADACcosDAC，第3页（共6页）{#{QQABRYYAggCoAABAAAhCEwFCCAIQkBGCCCoOwBAIoAAAABFABCA=}#}113ca22a即()1(3)213cos222cc，即cos4，③························10分223cc22将ac2284代入得cos，④23c32c23(2)c22由①④得()()22，即，即9(21)(44)cccc2242，a23c8412cc22即27540ccc642，即(1)(21)(4)0ccc222，因为c0，所以c21，则ac22844，所以a2.故BC的长为2.····························································································12分Aθπ-α2θαBDC方法二：作ABC的角平分线，交AC与M.设ABC2，DAC，则ABMCBM，ABAM,sinAMBsin在△ABM和△CBM中，由正弦定理可得BCCM,sinBMCsin又AMBBMC，所以sinsin()sinAMBBMCBMC，ABAM所以.BCCMAθM2θBDCAMc3ac由题可知ABc，AC3c，BCa，所以，CM.······················7分CMaca在△ACD和△BCM中，CADCBM，ACDBCM，ACBC所以△ACD∽△BCM，所以，CDCM3ca则，即a22ac60c，即(3acac)(2)0，a3ac2ca所以ac3(舍)或ac2.··············································································9分在△ABD和△ADC中，由余弦定理得aa12()2cc212()2(3)2cosADBcosADC220，aa212122所以ac2248，························································································11分第4页（共6页）{#{QQABRYYAggCoAABAAAhCEwFCCAIQkBGCCCoOwB