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

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）