山东省烟台市2022届高三数学下学期三模试题（PDF版附答案）

\n\n2022年高考适应性练习数学参考答案一、选择题DACBCADC二、选择题9.ABC10.ACD11.BD12.AD三、填空题3π3113.14.216015.+16.41226四、解答题17.解：若选①(2bc−=)cosAaCcos，（1）由正弦定理可得，2sinBACAACcos−=sincossincos，················································1分即2sinBAcos=sin(AC+=)sinB.························································2分因为B∈(0,)π，所以sinB≠0，1所以cosA=，········································································3分2π因为A∈(0,)π，所以A=.····························································4分3113故∆ABC的面积为bcsinA=×××34=33；································6分222若选②aBbAsin=3cos，（1）由正弦定理可得，sinsinAB−=3sinBAcos0.···················································1分因为B∈(0,)π，所以sinB≠0，π则有sinAAA−3cos=2sin(−=)0，············································3分3π因为A∈(0,)π，所以A=.·················································4分3高三数学参考答案（第1页，共8页）\n113故∆ABC的面积为bcsinA=×××34=33；···································6分222若选③acosC+=3acsinAbc+，（1）由正弦定理可得，sinACcos+=3sinsinACsinBC+sin，···············································1分3sinsinAC=cossinACC+sin，·······································2分因为C∈(0,)π，所以sinC≠0，π1所以3sinAA−=cos1，可得sin(A−=),··································3分62π因为A∈(0,)π，所以A=.················································4分3113故∆ABC的面积为bcsinA=×××34=33；··································6分222222（2）在∆ABC中，由余弦定理可得，a=+−bc2bccosA1=+−×××=91623413，故a=13.·········································8分222在∆ABD中，2ADBD×∠=+−cosADBADBD16，22在∆ACD中，2ADCD×∠=+−cosADCADCD9，13又BD=CD=,∠ADC+∠ADB=π,两式相加可得，223737AD=，即AD=.···············································10分42在∆ACD中，由余弦定理可得，高三数学参考答案（第2页，共8页）\n3713222+−9AD+−CDAC447481cos∠=ADC==.························12分2ADCD⋅37134812××22········································1分18.解：（1）当n=1时，3aa11=2(−1)，则a1=−2.3Sa=2(−1）nn当n≥2时，，两式相减可得，3Sa=2(−1）nn-1-1an························································2分aann=−2−1，即=−2.an−1所以{}an是以−2为首项，−2为公比的等比数列，n故a=−(2)，···················································3分n因为b=−2，bbb,,成等比数列，1237且bd=−+2，bd=−+22，bd=−+26，2372所以(22)(2)(26)−+ddd=−+−+，解得d=3，·····························5分故bn=35−，································································6分nn（2）cn=−−(35)(2)，n123nTn=−×−+×−+×−++(2)(2)1(2)4(2)(3−×−5)(2)，························7分n234nn+1−2T=−×−+×−+×−++(2)(2)1(2)4(2)(3nn−×−8)(2)+(3−×−5)(2).n234nn+1相减得，3Tn=+−+−+−++−43[(2)(2)(2)(2)](3−−×−5)(2)，nn+1=−−−8(3n4)(2)，·············································11分n+18(3−−−n4)(2)所以T=.············································12分n319.解：（1）证明：取AC中点Q，连接PQBQ,,········································1分高三数学参考答案（第3页，共8页）\n在∆ACF中，PQ,分别为AFAC,的中点，1所以PQ//CF.·······················2分=21又BE//CF,所以PQBE//,=2=所以四边形BQPE为平行四边形，所以EPBQ//.··········································3分又因为BQ⊂面ABC，EP⊄面ABC，所以EP//平面ABC；··············································4分（2）因为CF⊥平面ABC，AB⊂面ABC，所以AB⊥CF.又AB⊥AC，ACCF=C，所以AB⊥平面ACF.·······························5分所以∠BCA就是直线BC与平面ACF所成的角，故∠=BCA45，···············6分以A为坐标原点，ABAC,所在方向分别为xy,轴正方向，建立如图所示的空间直角坐标系Axyz−，令BC=CF=2，则BE=1,AB=AC=2,··········7分可得AE=(2,0,1),AF=(0,2,2),······································8分令n=(,,)xyz为平面AEF的一个法向量，则有nAE=0，nAF=0,20xz+=即，令z=2，可得n=(−−2,22,2)，························10分220yz+=又知平面ABC的一个法向量为m=(0,0,1)，214所以cos<mn,>==，14714故平面AEF与平面ABC夹角的余弦值为.···································12分7c22220.解：（1）由题意，=，可得ac=2，·······································1分a222222又abc=+，可得ab=2，·················································2分高三数学参考答案（第4页，共8页）\n22xy61所以椭圆方程为+=1，将(6,1)代入方程得：+=1，22222bb2bb22解得b=4，所以a=8，·······················································3分22xy故椭圆C的方程为+=1；····················································4分84（2）由（1）可得A(0,2)，将(6,1)代入抛物线可得，62=p，p=3，············5分32所以抛物线方程为xy=6，所以F(0,)，··········································6分23则直线MN的方程为y=kx+，设MxyNxy(,),(,)，1122222xy+=18422联立直线和椭圆方程，，消y得，(24+kx)+12kx−=70，3y=kx+2−12k−7所以xx+=，xx=，········································8分12212224+k24+ky−2y−2因为12k=，k=，AMANxx12112k1(kx−)(kx−)kxx−++(xx)yy−−22122212212412kk⋅=×==····10分AMANxxxxxx12121222kk−+12124k1=−×k−×=−，2−747141所以直线AMAN,的斜率之积为定值−.········································12分14高三数学参考答案（第5页，共8页）\na21．解：（1）易知fx()的定义域为(0,+∞)，fx′()1=−，····························1分xa①当a<0时，fx′()1=−>0，所以函数fx()单调递增，x1111又fa(e)aa=−elneaa=−<e10，f(1)=>10，所以函数fx()有唯一零点，··············································2分②当a=0时，fxx()=>0恒成立，所以函数fx()无零点，·······················3分axa−③当0e<<a时，令fx′()1=−==0，得xa=，xx当0<<xa时，fx′()0<，函数fx()单调递减，当xa>时，fx′()0>，函数fx()单调递增.则fx()==faaaaa()−=−>ln(1ln)a0min所以函数fx()无零点.···································································4分综上所述，当0e≤<a时，函数fx()无零点；当a<0时，函数fx()有一个零点.······································5分ax（2）当x∈(1,+∞)时，fx()≥−axlnxxe恒成立,xa即xe+≥xaxlnxa+lnx对x∈(1,+∞)恒成立，xaxln亦即xxaxe+≥(ln)e+axln对x∈(1,+∞)恒成立，······························6分x设函数gxx()=e+x，则gx()≥gax(ln)对x∈(1,+∞)恒成立，x则gx′()=++(x1)e1，xx设ϕ()xx=++(1)e1，所以ϕ′()xx=(+2)e，·······································7分令ϕ′()0x=得，x=−2，高三数学参考答案（第6页，共8页）\n当x∈−∞−(,2)时，ϕ′()0x<，函数gx′()在(−∞−,2)单调递减，当x∈−+∞(2,)时，ϕ′()0x>，函数gx′()在(2,−+∞)单调递增，1所以gxg′′()≥−=−>(2)10，故gx()在R上单调递增，2e又gx()≥gax(ln)，所以xax≥ln在x∈(1,+∞)恒成立，························9分x即a≤在x∈(1,+∞)恒成立，lnxx令hx()=，x∈(1,+∞)，lnxlnx−1则hx′()==0，得x=e，················································10分2(ln)x当x∈(1,e)时，hx′()0<，函数hx()在(1,e)单调递减，当x∈∞(e,+)时，hx′()0>，函数hx()在(e,+)∞单调递增，所以hx()=h(e)=e，故a≤e.min即a的取值范围是(−∞,e].··························································12分22.解：（1）设第一轮测试产生消杀效果的载片个数为X，则XBp(8,)，·········1分77808PX(=7)=C(1−pp)，PX(=8)=C(1−pp)，····························3分88所以经过一轮检测该消毒液即合格的概率PPX==(7)+=PX(8)=7780878C(1−+−=−+ppC)(1pp)8(1pp)p；··································4分8878（2）由（1）可知，P(ξ=8)=−+8(1pp)p，设第二轮检测次数为k，则k=2,3,,8，88−−kkk可得P(ξ=+=8kC)(1−pp)，·········································6分8高三数学参考答案（第7页，共8页）\n8所以Eξξ==8(8P)(++=∑8)(8)kPξ+k····································7分k=28788−−kkk8=−+++8(8(1pp)p)∑(8kC)8(1−pp)k=288788888−−−−kkkkkk=−++8(8(1ppp))8∑∑C88(1−pp)+kC(1−pp)kk=22=888−−kkk8kkk81−7=8∑∑C8(1−+−−pp)kC88(1pp)C(1−pp)kk=00=88kkk87−=−++8(1pp)∑kC8(1−p)p−−8(1pp)，····························9分k=08kkk8−令YB(8,1−p)，则EY=∑kC8(1−=p)p8(1−p)，k=0877则Eξ=8(1−++−−−pp)8(1p)8(1pp)=+−−−88(1p)8(1pp)，77一方面88(1+−−−p)8(1pp)=+−−88(1p)(1p)，··························10分7因为1−>−>pp0,10，所以Eξ>8，···································11分77另一方面88(1+−−−p)8(1pp)=−−−168p8(1pp)<−168p，综上可得，8<<−Epξ168.·················································12分高三数学参考答案（第8页，共8页）