【文档说明】山东省烟台市2022-2023学年高三上学期期末学业水平诊断数学答案.pdf，共(6)页，477.604 KB，由管理员店铺上传

转载请保留链接：https://www.doc5u.com/view-a56b651a394fec44f96ba8db33687da8.html

以下为本文档部分文字说明：

高三数学答案（第1页，共6页）2022～2023学年度第一学期期末学业水平诊断高三数学参考答案及评分标准一、选择题DBBCACDA二、选择题9.BC10.ACD11.ACD12.ABD三、填空题13.114.3215.6716.12四、解答题17.解：（1）由正弦定理可得sincos+sins

insinACACB=，·······················1分因为πABC++=，所以sincos+sinsinsin()ACACAC=+，即sincos+sinsinsincoscoss

inACACACAC=+，····························2分整理得：sinsincossinACAC=，因为0Cπ<<，所以sin0C≠，所以tan1A=，因为0Aπ<<，所以4Aπ=.···································

·····························4分（2）在ABD∆中，由余弦定理得：2222cosBDABADABADA=+−⋅，······5分即2292(22)ABADABADABAD=+−⋅≥−⋅，·······················

·········6分整理得9(22)2ABAD+⋅≤，当且仅当ABAD=时，等号成立.所以129(21)sin2444ABDSABADABADπ+=⋅=⋅≤△，························8分因为2ADDC=，所以327(21)28ABCABDSS+

=≤△△，所以ABC△面积的最大值为27(21)8+.···············································10分18.解：（1）因为12nnnaaS+=*()n∈N，所以()1122nnn

aaSn−−=≥，两式相减得()()1122nnnnaaaan+−−=≥.·····································································1分高三数学答案（第2页，共6页）

又因为0na≠，所以()1122nnaan+−−=≥，······························································2分所以数列{}21na−和{}2na都是以2为公差的等差数列.因为11a=，所以在

12nnnaaS+=中，令1n=，得22a=，所以()2112121,nann−=+−=−()22122,nann=+−×=···········································

··3分所以nan=，·························································································

··························4分对于数列{}nb，因为112nnnbbbb+⋅==，且0nb≠，所以1*2()nnnbb+=∈N，···········6分所以数列{}nb是以2为首项，2为公比的等比数列，所以2nnb=.················

········7分（2）因为23=122232...2nnTn×+×+×++×所以()23412=122232122nnnTnn+×+×+×++−×+×···································8分两式相减得，212222nnnTn+−=+++−×·

··························································9分1122212nnn++−=−×−12(1)2nn+=−−−×······································

···················································11分所以()1122nnTn+=−×+.·······································································

······················12分19.解：（1）证明：取BC中点O，连接,OAOD，因为ABC∆是以BC为斜边的等腰直角三角形，所以OABC⊥.·························1分因为BCD∆是等边

三角形，所以ODBC⊥.···························································2分OAODO=，OA⊂平面AOD，OD⊂平面AOD，······································3分所

以BC⊥平面AOD.··························································································

·······4分因为AD⊂平面AOD，故BCAD⊥.····································································5分（2）在AOD∆中，1AO=，3OD=，7AD=，由余弦定理可得，3cos2AOD∠=−，

故150AOD∠=.··············6分如图，以,OAOB及过O点垂直于平面ABC的方向为,,xyz轴的正方向建立空间直角坐标系Oxyz−，··············7分可得33(,0,)22D−，所以33(,1,)22BD=−−

，(0,2,0)CB=，(1,1,0)AB=−，zyxODCBA高三数学答案（第3页，共6页）设111(,,)xyz=n为平面ABD的一个法向量，则11111033022xyxyz−+=−−+=，令3x=，可得(3,3,5)=

n，······················9分设222(,,)xyz=m为平面BCD的一个法向量，则22222033022yxyz=−−+=，令23x=，可得(3,0,3)=m，····················11分所以3015393co

s,313112++<>==×nm，故平面ABD与平面BCD夹角的余弦值为39331.······································12分20.解：（1）设该容器的体积为V，则2323Vrlrππ=+，又1603Vπ=，所以

2160233lrr=−，··········································································2分因为6lr≥，所以02r<≤.··

······················································································4分所以建造费用222916029232()34334yr

lrmrrrmrππππ=×+=−×+，因此22403(1),02.ymrrrππ=−+<≤······················································

·············5分（2）由（1）得3222406(1)406(1)(),02.1mymrrrrrmπππ−′=−−=−<≤−··········6分由于9,10,4mm>−>所以令34001rm−=−，得34

01rm=−.···························7分高三数学答案（第4页，共6页）若34021m<−，即6m>，当340(0,)1rm∈−时，0y′<，()yr为减函数，当340(,2)1rm∈−时，0y′>，()yr为

增函数，此时3401rm=−为函数()yr的极小值点，也是最小值点.··················································································

····································9分若34021m≥−，即964m<≤，当(0,2]r∈时，0y′<，()yr为减函数，此时2r=是()yr的最小值点.·········································

··································································11分综上所述，当964m<≤时，建造费用最小时2r=；当6m>时，建造费用最小时340.1rm=−··········

··············································································································12分21.解：（1）设(,0)Aa−，(

,0)Ba，11(,)Pxy，则2111221110014APBPyyykkxaxaxa−−=×==+−−，··············································1分又因为点11(,)Pxy在双

曲线上，所以2211221xyab−=.·····································2分于是2222221112144abyxxba=−=−，对任意10x≠恒成立，所以2214ba=，即224ab

=.···································································3分又因为5c=，222cab=+，可得24a=，21b=，所以双曲线C的方程为2214xy

−=.······················································5分（2）设直线l的方程为：5xty=+，3344(,),(,)MxyNxy，由题意可知2t≠±，高三数

学答案（第5页，共6页）联立22145xyxty−==+，消x可得，22(4)2510tyty−++=，则有342254tyyt−+=−，34214yyt=−，································

···················6分假设存在定点(,0)Dm，则3434()()DMDNxmxmyy=−−+3434(5)(5)tymtymyy=+−+−+·························

·····························7分223434(1)(5)()(5)tyymtyym=++−++−22222125(5)(5)44tmtmtt+−=−+−−−2222(4)(48519)4mtmmt−−−+=

−····························································8分令22485194(4)mmm−+=−，解得758m=，····················

······················10分此时224511446464DMDNm=−=−=−，························································

11分所以存在定点75(,0)8D，使得DMDN为定值1164−.···································12分22.解：（1）2()e2xfxxaxax=−−，则()(1)(e2)xfx

xa′=+−，···················1分当0a>时，方程e20xa−=的根为ln(2)xa=.当ln(2)1a>−，即12ea>时，当(,1)x∈−∞−和(ln(2),)xa∈+∞时，()0fx′>，

()fx单调递增，当(1,ln(2))xa∈−时，()0fx′<，()fx单调递减.·············2分当ln(2)1a<−，即102ea<<，当(,ln(2))xa∈−∞和(1,)x∈−+∞时，()0fx′>，()fx单调递增，当(ln(2)

,1)xa∈−时，()0fx′<，()fx单调递减.·············4分当ln(2)1a=−，即12ea=时，0y′≥恒成立，函数在R上单调递增，··············5分高三数学答案（第6页，共6页）综上所述，当102ea<<时，()fx在(,ln(2))a

−∞，(1,)−+∞上单调递增，在(ln(2),1)a−上单调递减；当12ea=时，()fx在R上单调递增，当12ea>时，()fx在(,1)−∞−，(ln(2),)a+∞上单调递增，在(1,ln(2))a−上单调递减.········6分

（2）存在实数a使得()2fxba′−≥对任意x恒成立，即ee2xxbxax+−≤恒成立.令()ee2xxgxxax=+−，则min()bgx≤.················································7分因为()(2)e

2xgxxa′=+−，当2x−≤时，()0gx′<恒成立；当2x>−时，()(3)e0xgxx′′=+>，函数()gx′在(2,)−+∞上单调递增，且(2)20ga′−=−<，2(2)(22)e20agaaa′=+−>

，所以，存在0(2,2)xa∈−，使得0()0gx′=，且()gx在0(2,)x−上单调递减，在0(,)x+∞上单调递增，所以0min000()()(1)e2xgxgxxax==+−.···············9

分于是，原命题可转化为存在a使得000(1)e2xbxax+−≤在(2,)−+∞上成立，又因为000()(2)e20xgxxa′=+−=，所以002(2)exax=+.所以存在0(2,)x∈−+∞，使得0002200000(1)e(2)ee(

1)xxxbxxxxx+−+=−−+≤成立.····················································10分令2()e(1)xhxxx=−−+，(2,)x∈−+∞，则2()e(3

)xhxxx′=−−，所以当(2,0)x∈−时，()0hx′>，()hx单调递增，当(0,)x∈+∞时，()0hx′<，()hx单调递减，所以max()(0)1hxh==，所以1b≤.··············································

·············12分