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高三数学参考答案（第1页，共6页）2022年高考适应性练习（一）数学参考答案一、选择题ABDCCACD二、选择题9.BCD10.AC11.ABD12.CD三、填空题13.74−14.3415.732+16.168962+四、解答题17.解：（1）由4,1,nnaS+成等比数列，得2(1)4nna

S+=,·····································1分即2214nnnaaS++=，*n∈N.①当1n=时，211(1)4aa+=，解得11a=.········································

··············2分当2n≥时，2111214nnnaaS−−−++=.②①-②得，2211224nnnnnaaaaa−−−+−=，整理得11()(2)0nnnnaaaa−−+−−=.因为{}na为正项数列，所以10nn

aa−+>，所以12nnaa−−=，·························4分即数列{}na是以1为首项、2为公差的等差数列.所以21nan=−.······························

·······················································5分（2）由（1）知，2nSn=.·················································

·····························6分所以21441111()(21)(21)22121nnnnSnbaannnn+===+−−+−+.························

······8分所以123nnTbbbb=++++11111111[(1)()()()]2335572121nnn=+×−+−+−++−−+11242nn=+−+.············································

······································10分高三数学参考答案（第2页，共6页）B1D1C1DCABA1P18.解：（1）因为cos3sinaCaCbc+=+，由正弦定理得，sincos3sinsinsinsinACACBC+=+.··

··························2分因为π()BAC=−+,所以sinsin()sincoscossinBACACAC=+=+，代入上式得，3sinsincossinsinACACC=+.·············

····························3分因为(0,π)C∈，所以sin0C≠，所以上式可化为3sincos1AA−=，···················································

······4分即π1sin()62A−=.因为(0,π)A∈，所以ππ5π(,)666A−∈−，所以ππ66A−=，所以π3A=.········································

····················································6分（2）在ABC∆中，由余弦定理得：22232232cos73aπ=+−××=，即7a=.············································8分所以22

27497cos214272acbBac+−+−===××，所以321sin14B=.················9分因为角A的平分线交BC于M，则23BMABMCAC==，所以275BM=.·······10分在ABM∆中，由正弦定理得sinsinBMAMBAMB=∠,即27142532

1AM⋅×=，所以635AM=.···············································································12分19.解：（1）证明：连结BD.

因为1AD⊥面ABCD，AD⊂面ABCD，所以1ADAD⊥.····························1分又在ADB∆中，2AB=，1AD=，60DAB∠=，由余弦定理得22221221cos603BD=+−×××=，即3BD=，由勾

股定理得90ADB∠=，即ADDB⊥.···············3分因为1ADDBD=，所以AD⊥面1ABD.··············4分又1AB⊂面1ABD，所以1ADAB⊥.···················

·····································5分高三数学参考答案（第3页，共6页）（2）由（1）知，1,,DADBDA两两垂直，以D为坐标原点，分别以向量1,,DADBDA

的方向为,,xyz轴正向，建立如图所示的空间直角坐标系Dxyz−.····························6分则(0,0,0)D，(1,0,0)A，(0,3,0)B，1(0,0,3)A，1(2,3,3)C−.所以1(0,3,3)AB

=−，1(2,3,3)DC=−，设1(01)DPDCλλ=<<，则1(2,3,3)DPDCλλλλ==−，即(2,3,3)Pλλλ−.·············

················7分所以1(2,3,33)APλλλ=−−.设111(,,)xyz=n为平面1APB的一个法向量，则1100ABAP==nn，即1111133023(33)0

yzxyzλλλ−=−++−=，令12zλ=，则12yλ=，1233xλ=−，取(233,2,2)λλλ=−n.···········8分因为DB⊥平面11AADD，所以(0,1,0)=m为平面11AADD的一个法向量.·········9分设面1APB与

面11AADD的夹角为α，则22||22cos||||3122012320λαλλλλ===⋅−+−+nmnm，01λ<<.············10分此时当12λ=，即12λ=时，cosα取得最大值，·········································1

1分因此111022DPDC==.········································································12分20.解：（1）事件A=“甲平台日销售量不低于8件”，则()PA=26241031005++=

.······································································1分设事件B=“从甲平台所有销售数据中随机抽取3天的日销售量，其

中至少有2天日销售量不低于8件”，则223333323()()()555PBCC=+·····································································

3分81125=.······················································································4分（2）设甲平台日销售收入为1Y，

则1Y所有可能的取值为6240,7270,8300,9330,10360.aaaaa−−−−−··········5分1Y的分布列为zyxPA1BACDC1D1B1高三数学参考答案（第4页，共6页）1

Y6240a−7270a−8300a−9330a−10360a−P1410026100261002410010100因此1142626()(6240)(7270)(8300)100100100EYaaa=−×+−×+−×2410(9330)(10360)100100

aa+−×+−×·······································6分7.9297a=−····························································

···············7分设乙平台日销售收入为2Y，则2Y所有可能取值为6240,7280,8320,9355,10390.aaaaa−−−−−··············8分2Y的分布列为2Y6240a−7280a−8320a−9355a−10390a−P10100251003510

02010010100因此2102535()(6240)(7280)(8320)100100100EYaaa=−×+−×+−×2010(9355)(10390)100100aa+−×+−×···································

····9分7.95316a=−········································································10分于是21()()0.0519EYEYa−=−.令0

.05190a−≥，得380a≥.所以当300380a≤<时，选择甲平台；当380a=时，甲、乙平台均可；当380500a<≤时，选择乙平台.12分21.解：（1）因为双曲线的实轴长为2，所以1a=.又因为点(7,

1)−在C上，所以22171b−=，所以216b=.所以双曲线C的方程为2261xy−=.····························································2分因为抛物线的准线为1y=−，所以抛物线E的方程为24

xy=.·························4分（2）设00(,)Pxy，直线PA方程为010()yykxx−=−，将其与抛物线方程24xy=联立，得2101044()0xkxykx−−−=，因为直线PA与抛物线相切，所以210101616()0kykx∆=+−=，整

理得211000kkxy−+=，①············5分点A的横坐标12Axk=，211(2,)Akk.同理，直线PB方程为020()yykxx−=−，有222000kkxy−+=，②高三数学参考

答案（第5页，共6页）点B的横坐标22Bxk=，222(2,)Bkk.由①②知，12,kk是方程2000kkxy−+=的两个根，所以120kkx+=，120kky=，··························

··········································6分由211(2,)Akk，222(2,)Bkk，得22121212222ABkkkkkkk−+==−.直线AB方程为12122kkyxkk+=−，即002xyxy=−.···

·································7分()2222201201212000||12244444xABkkxkkkkxxy=+−=+⋅+−=+⋅−···8分点00(,)Pxy到直线AB的距离2002044xydx−=+，所以()3200

11||422PABSABdxy∆=⋅=−.····················································9分因为点00(,)Pxy在双曲线上，所以22006

1xy=+，所以()320016412APBSyy∆=−+，0y∈R.················································10分当013y=时，APBS∆取得最小值318，故PAB∆面积的取值范围为3[,)18+∞.····12分22.解

：（1）2()()2e2xgxfxax′==−，所以2()4e2xgxa′=−.·························2分当0a≤时，()0gx′<，所以()gx在(,)−∞+∞上单减，又当x→−∞时，()g

x→+∞，当x→+∞时，()gx→−∞，由零点存在定理，此时()gx存在唯一零点；···············3分当0a>时，令2()4e20xgxa′=−>，得ln22ax>−，故当ln2(,)2ax∈−+

∞，()0gx′>，()gx单增.同理，ln2(,)2ax∈−∞−时，()gx单减，所以min()gx=ln2()1ln22aga−=+.································

···················································4分又当x→−∞，()gx→+∞，当x→+∞，()gx→+∞，所以当ln2()02ag−<，即102ea<<时，由零点存在定理()gx存在两个零点；当12e

a=时，()gx存在一个零点；当12ea>时，()gx在(,)−∞+∞无零点.·····························································5分高三数学参

考答案（第6页，共6页）综上，当0a≤或12ea=时，()gx存在1个零点；当102ea<<时，()gx存在2个零点；当12ea>时，()fx无零点.·······························································

·····6分（2）由（1）知，()fx的极值点为12,xx，即方程2()2e20xgxax=−=的两根，······7分且102ea<<，所以121e0xax=>，222e0xax=>.两式相除并取对数得22112()ln0xxxx−=>.由1212e

e2xxxxλ+≥−，得22112121e2()()lne2xxxxxxxxλ−+≥−，·················8分所以2112212e2()e2e2lnxxxxxxλ+−−−≤.···················································

···········9分令211xtx=>，令2e12()e2e2()(1)lnttthtt+−−−=>,则()htλ≤恒成立.2222e2e()ln()e2e2e2()2lnttttthtt+−+−−−−′=，令22e2e()()ln()e2e2e2ttttt

ϕ=+−+−−−−，则e12()2lne2e2tttttϕ′=−+−−−，因为2e1()2ln1e2tttϕ′′=+−−在(1,)+∞单增，且(1)0ϕ′′<，(e)0ϕ′′>，所以0(1,e)t∃∈，使得0()0tϕ′′

=，且当0(1,)tt∈时，()0tϕ′′<，()tϕ′单减；0(,)tt∈+∞时，()0tϕ′′>，()tϕ′单增.又(1)0ϕ′=，(e)0ϕ′>，由零点存在定理，存在1(1,e)t∈，当1(1,)tt∈时，(t)0ϕ′<，()tϕ单减，当1(,)tt∈+∞时，(t)0ϕ′>，

()tϕ单增.又(1)(e)0ϕϕ==，所以当(1,e)x∈时，()0tϕ<，()ht单减，当(e,+)x∈∞时，()0tϕ>，()ht单增.···········································································

······················11分所以当et=时，2min2(e1)()(e)e2hth−==−，λ的取值范围为22(e1)e2λ−≤−.······12分获得更多资源请扫码加入享学资源网微信公众号www.xiangxue100.com