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
  • 阅读 10 次
  • 下载 0 次
  • 页数 6 页
  • 大小 477.604 KB
  • 2024-10-09 上传
  • 收藏
  • 违规举报
  • © 版权认领
下载文档3.00 元 加入VIP免费下载
此文档由【管理员店铺】提供上传,收益归文档提供者,本网站只提供存储服务。若此文档侵犯了您的版权,欢迎进行违规举报版权认领
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
可在后台配置第一页与第二页中间广告代码
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
可在后台配置第二页与第三页中间广告代码
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
可在后台配置第三页与第四页中间广告代码
试读已结束,点击付费阅读剩下的3 已有10人购买 付费阅读2.40 元
/ 6
  • 收藏
  • 违规举报
  • © 版权认领
下载文档3.00 元 加入VIP免费下载
文本内容

【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+sinsinsinACACB=,·······················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12nnnaa

S+=*()n∈N,所以()1122nnnaaSn−−=≥,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=122232122

nnnTnn+×+×+×++−×+×···································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=,7A

D=,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,)2

2D−,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3015393cos,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21

60233lrr=−,··········································································2分因为6lr≥,所以02r<≤.·······························

·························································4分所以建造费用222916029232()34334yrlrmrrrmrππππ=×+

=−×+,因此22403(1),02.ymrrrππ=−+<≤···································································5分(2)由(1)得3222

406(1)406(1)(),02.1mymrrrrrmπππ−′=−−=−<≤−··········6分由于9,10,4mm>−>所以令34001rm−=−,得3401rm=−.···························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3402

1m≥−,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211

1221110014APBPyyykkxaxaxa−−=×==+−−,··············································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)2

510tyty−++=,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)m

mm−+=−,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)xfxxa′=+−,···················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()2fx

ba′−≥对任意x恒成立,即ee2xxbxax+−≤恒成立.令()ee2xxgxxax=+−,则min()bgx≤.················································7分因为()(2)e2xgxxa′=+−,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

管理员店铺
管理员店铺
管理员店铺
  • 文档 490830
  • 被下载 29
  • 被收藏 0
若发现您的权益受到侵害,请立即联系客服,我们会尽快为您处理。侵权客服QQ:12345678 电话:400-000-0000 (支持时间:9:00-17:00) 公众号
Powered by 太赞文库
×
确认删除?