高三数学参考答案2023.11题号123456789101112答案CACBABDCADABACDBD题号131415164答案3fxx()sin(2)20,327T17.【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)23,所以4sin23,又||,所以,23所以fxx()4sin(2).·················································································3分35令222kxk,解得kxkkZ,,23212125所以函数fx()的单调递增区间为kkkZ,,.···································5分1221123(2)由题可知f()4sin(2),则sin(2),35354因为(,),所以2(,),12232334所以cos(2)1()2,·································································7分355所以sin2sin[(2)]sin(2)coscos(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ABCA1B1C1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ABCAB111C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,,BCBB1Bxyz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),QCCA11t3所以cos,QCCA11·····································10分23QCCA111(1)3t11解得t,所以线段BQ长为.··································································12分221a19.【答案】(1)方法一:因为f(x)是奇函数,所以f(0)=0,即0,解得a1,····2分2221x21(21)212xxxx此时,fx(),fxfx()(),22x122(22)222xxxx111则fx()是奇函数.故a1.·····································································································4分方法二:因为f(x)是奇函数,22212xxxxaaaa所以f(xfx)()0,22222222xxxx1111即(1)(21)0ax对xR恒成立,所以a1.····································································································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)02得:ff(kfk3sincos)(cos)(cos)22,所以3sincoskcos2,2即3sincoscosk在,0上有解,·············································9分4231cos21记g()3sincoscos,则g()sin2sin222622因为,则2,,63613所以sin21,,所以g(),1,62233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)75adbc2220.186.635,···········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()10111256464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,ACc3,BCc2,BABCACccc222222(2)(3)1在△ABC中,由余弦定理得cosABC,2222BABCcc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AD1,所以AB1,BC2,1133所以SBABCABCsin12.··········································4分△ABC2222a(2)方法一:由题可知ABc,ACc3,AD1,BDDC,2设ABC2,DAC,ADB,则ADC,ABADc1在△ABD中,由正弦定理得,即,sinsinADBABDsinsin2aDCAC3c在△ADC中,由正弦定理得,即2,sinDACsinADCsinsin()1a3所以,则a,①····························································6分sin223sincos在△ABD和△ADC中,由余弦定理得aa12()2cc212()2(3)2cosADBcosADC220,aa212122所以ac2248,②······················································································8分在△ADC中,由余弦定理得DC2AD2AC22ADACcosDAC,第3页(共6页){#{QQABRYYAggCoAABAAAhCEwFCCAIQkBGCCCoOwBAIoAAAABFABCA=}#}113ca22a即()1(3)213cos222cc,即cos4,③························10分223cc22将ac2284代入得cos,④23c32c23(2)c22由①④得()()22,即,即9(21)(44)cccc2242,a23c8412cc22即27540ccc642,即(1)(21)(4)0ccc222,因为c0,所以c21,则ac22844,所以a2.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,sinAMBsin在△ABM和△CBM中,由正弦定理可得BCCM,sinBMCsin又AMBBMC,所以sinsin()sinAMBBMCBMC,ABAM所以.BCCMAθM2θBDCAMc3ac由题可知ABc,AC3c,BCa,所以,CM.······················7分CMaca在△ACD和△BCM中,CADCBM,ACDBCM,ACBC所以△ACD∽△BCM,所以,CDCM3ca则,即a22ac60c,即(3acac)(2)0,a3ac2ca所以ac3(舍)或ac2.··············································································9分在△ABD和△ADC中,由余弦定理得aa12()2cc212()2(3)2cosADBcosADC220,aa212122所以ac2248,························································································11分第4页(共6页){#{QQABRYYAggCoAABAAAhCEwFCCAIQkBGCCCoOwB
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
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