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山东省烟台市2021-2022学年高二上学期期末考试数学试题答案

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2021~2022学年度第一学期期末学业水平诊断高二数学参考答案及评分标准一、选择题CDBABBAC二、选择题9.ACD10.BD11.ACD12.BCD三、填空题17213.14.915.3416.yx=2,27四、解答题54×17.解:由已知aad=+=4,Sa=+5dad=+=51030.················2分215112ad+=41联立方程组,解得a1=2,d=2.··········································4分ad+=261所以an=2.·····················································································5分nnn(22)+S==nn(+1),··································································6分n222由题意Snnn=++≥8λ,即λ≤nn−7.·········································7分n27令fxx()=−7x,其图象为开口向上的抛物线,对称轴为x=,···············8分2所以当n=3或4时,fn()取得最小值−12,·········································9分故λ≤−12.······································································10分222cab+b2218.解:(1)因为离心率e===+=15,所以ba=4.·······3分2aaa412又因为点M(2,23)−在双曲线C上,所以−=1.························4分22ab22联系上述方程,解得a=1,b=4,即a=1,b=2.·····························6分22y(2)设所求双曲线的方程为x−=≠λλ(0),······································8分4高二数学答案(第1页,共6页)\n20由双曲线经过点P(3,25),得3−=λ,即λ=−2.························10分42222yyx所以双曲线的方程为x−=−2,其标准方程为−=1.··················12分48219.解:(1)由已知得,Sna+=−1,nn+1当n≥2时,Sna+−=−11.·····················································1分nn−1两式相减得,aa=21+.······························································2分nn+1a+1n+1于是aa+=12(+1),即=2,············································4分nn+1a+1na+12又a=3,a+=14,a+=≠120,所以=2满足上式,··················5分221a+11an+1+1∗所以=2对∀∈nN都成立,故数列{a+1}是等比数列.··················6分na+1nn(2)由(1)得a=21−,···························································7分nnbn=+−21,··························································8分n23nTnn=+++++++++−(2222)[123(1)]·····························9分2n+1nn−=22−+.·················································12分2p20.解:(1)由抛物线定义|PF|1=+=2,解得p=2,···························2分22所以抛物线C的方程为yx=4.·············································3分2(2)因为点Pt(1,)在C:yx=4上,且t>0,所以t=2,即P(1,2).············4分由题意可知m≠0,设Axy(,),Bxy(,),1122高二数学答案(第2页,共6页)\n2yx=42联立,得y−4my−=40,x=my+1所以yym+=4,yy=−4,·························································5分1212yy−−224411于是k===,则直线PA的方程为yx−=2(−1),PA2xy11−+12y1y1+2−14y2y11令y=0,则x=−;令x=0,则y=,2y+21y2y11所以C点的坐标为(−,0),D点的坐标为(0,).······························7分2y+21y2y22同理M点的坐标为(−,0),N点的坐标为(0,).···························8分2y+22||yy−yy−1212所以||CM=,|DN|4|=|.·····························9分2(yy++2)(2)1211||yy−−yy1212S=|CM||⋅=DN|××4||四边形CDMN222(yy++2)(2)1222(yy−)(y+−y)4yy121212==|(y++2)(y2)||yy+++2(yy)4|1212122216mm++16222===+≥2||m4,···········································11分|8|mm||||m当且仅当m=±1时等号成立,此时四边形CDMN的面积最小值为4,直线l的方程为yx=−1或yx=−+1.································································12分aaa++=1412321.解:(1)由得,a2=4.······································1分2(a+=+1)aa213442又a=,aq=4,所以++=44q14,即2520qq−+=,13qq高二数学答案(第3页,共6页)\n1解得q=2或q=(舍去).·······························································3分2n∗所以a=2(n∈N),·································································4分n当n=1时,bS==1,11111221当n≥2时,bSS=−=+−−+−=(nn)[(n1)(n1)]n,············5分nnn−12222经检验,n=1时,b=1适合上式,1∗故bn=(n∈N).·······································································6分nn,n为奇数an(2)由(1)可知,c=n22nn−−4(2),n为偶数ann当n为奇数时,c=,nn222nn(−2)当n为偶数时,c=−,····················································7分nnn−22213521n−由题意,有Tccc=++++c=++++,①奇13521n−3521n−222211352321nn−−T=+++++,②奇35721nn−+2142222211−312222211nn−−44n21①−②得,T=+++++−=+−4奇2222357221nn−+22121221n+1−45421565nn−+=−−=−,························9分nnn63424664×××106n+5所以T=−.····················································10分奇n−19184×22222222204264(2)nn(2−2)Tccc=++++=−+−+−c()()()+−偶2462n2042642nn2−222222222高二数学答案(第4页,共6页)\n2222(2)n04nn=−==.························································11分20nnn−1224422106n+5n1018nn−−65故T=−+=+.································12分2nnn−−11n−19184××4918422.解:(1)由已知得b=1,··················································1分2b2离心率e1=−=,·································································2分2a222所以ab=22=,·····································································3分2x2故椭圆E的方程为+=y1.··········································4分2(2)当直线l的斜率存在时,设l:ykx=(+1),Axy(,),Bxy(,),11222x2+=y12222联立方程组2得,(12+kx)+4kxk+−=220,ykx=(+1)224k22k−所以xx+=−,xx=.··········································5分12212212+k12+k222||1|AB=+−kxx|=+1k(x+−x)4xx121212222224kk88−288k+=+−1(k)−=+×1k22212++kk1212+k222(1+k)=.···························································7分212+k22212x1+2|AF|=++=++−=(x1)y(x1)1x,111112222212x2+2|BF|=++=++−(x1)y(x1)1x=.·····························8分1222222高二数学答案(第5页,共6页)\nxx++22xx+++2(xx)4121212所以|AF||BF|=×=1122222228kk−−+4212++kk22121+k==.··································9分2212+k所以|AB|22|=AF||BF|.·······················································10分11当直线l的斜率不存在时,l:x=−1,2x2+=y122联立方程组2,得A(1,−),B(1,−−).=−22x1221||2AB=,|AF||BF|=×=,········································11分11222所以|AB|22|=AF||BF|.11综上,存在实数t=22使得|AB|=tAF|||BF|恒成立.··························12分11高二数学答案(第6页,共6页)

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所属: 高中 - 数学
发布时间:2022-08-10 13:00:05 页数:6
价格:¥5 大小:299.12 KB
文章作者:fenxiang

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