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数学试题答案第1页(共8页)余姚中学2022学年第二学期期中考试高二数学参考答案一、单项选择题:12345678BDCBCCAD8.设直线l为曲线()yfx=在点()()11,xfx处的切线,()1fxx=.()1111:lnlyxxxx−=−,即111ln1yxxx=+−;设直线l为曲线()
ygx=在点()()22,xfx处的切线,()1agxax−=.()1222:aalyxaxxx−−=−,即()12221aaayxaxxax−−=+−;由题知()121121ln11aaaxxxax−
=−=−①②注意到12,0xx,故0a.由①式得()12lnln1lnxaax=−−−,代入②式得()()22ln1ln11aaaxax−−−−=−.显然1a,整理得221lnln1aaxxa+−=−.记()lnahxxx=−(0a且1a),则()111aaaxhxaxxx−−
=−=.当110axa时,()hx单调递增;当11axa时,()hx单调递减.()1max11lnaahxhaa+==−,故1ln1ln1a
aaa++−−.化简得()1ln01aaa+−,解得()10,1,ae+数学试题答案第2页(共8页)二、多项选择题:9101112ADABDBCBD12.设2FBm=,则12FBma=+,27F
Am=,172FAma=+.在12AFF△中,由余弦定理得()()()222°7272272cos120mamcmc+=+−.2222714accmam−=−.①在12BFF△中,由余弦定理得()()()222°2222cos6
0mamcmc+=+−.22222accmam−=−−.②综合①②得23ca=.故离心率32ca=,A选项错误.1212227:1AFFBFFSAFSBF==△△,B选项正确.将23ca=代入①式可得57ma=,故25FAa=,17FAa=,1197FBa=,所以12AFF△的周长为1
5a,12BFF△的周长为457a,C选项错误.设12AFF△与12BFF△的内切圆半径分别为12,rr.1212121152714527AFFBFFarSSar==△△,所以12:3:1rr=,D选项正确.三、填空题:13.3514.315.0,116.327;23516.若第n幅
图中图形的边数记为nN,则()142nnNNn−=,又13N=,故111434nnnNN−−==.注意到每次操作都是使得原来图形的每条边上长出一个小三角形,故第n幅图比第1n−幅图新增部分的面积211113499nnnnnSSNS−−−−==
,从而图形的总面积数学试题答案第3页(共8页)22334448344999559nnnnTSSSSSSS=++++=++++=−,不断地趋于883235545S==.四、解答题:17.(I)2a是14,a
a的等比中项.2214aaa=,即()()21113adaad+=+.①------------------------------------------------------------------------------1分5611aa+=.()()114511adad+++=.②--
---------------------------------------------------------------------------------------------2分又0d,联立①②可得11ad==.-----------------------------
--------------------------------------------------------------4分()11naandn=+−=.----------------------------------------
-----------------------------------------------------------------5分(II)若选①:2nnbn=.1212222nnSn=+++.-------------------------------
-----------------------------------------------------------------6分231212222nnSn+=+++.()12322222nnnSn+=−−+++-----------
-----------------------------------------------------------------------------------7分()()()1112122221212212nnnnnnnn+++−=
−=+−=−+−.-----------------------------------------------------10分若选②:()()242121nnbnn=−+.--------------------------
-----------------------------------------------------------------------6分()()111111212122121nnnn=+=+−−+−+------------------------------------
---------------------------8分2112212212121nnnnSnnnnn+=+−=+=+++.-----------------------------------------------------------------
------10分数学试题答案第4页(共8页)18.(I)设剩下的10组数据分别为()()()11221010,,,,,,uvuvuv.101211102110222535iiiiiiuvxy===−−
=.---------------------------------------------------------------------------------1分12112010.810iiux==−=,121
14322.710iivy==−=.101010.822.72451.6uv==.--------------------------------------------------------------------
--------------------------3分101222221110101194iiiiux===−−=.22101010.81166.4u==.--------------------------------------------
----------------------------------------------------------5分101102211025352451.6311941166.410iiiiiuvuvbuu==−−==−−.------------------
---------------------------------------------------------------8分22.7310.89.710avbu=−=−=−−.所以所求回归
方程为310yx=−.----------------------------------------------------------------------------------------------10分(II)当10x=时,20y
=.因为212012−=,22202−=,所以(II)中所得的线性回归方程可靠.--------------------------------------12分19.(I)2sin4cAb+=.222sinsincossin2
2CAAB+=,即sinsinsincossinCACAB+=.--------------------------------2分()sinsinsincoscossinBACACAC=+=+.---------
----------------------------------------------------------------3分sincosCC=,即tan1C=.------------------------------------------------------------
-------------------------------------4分4C=.------------------------------------------------------------------------------------------------------
-----------------------5分数学试题答案第5页(共8页)(II)因为D为AB边的中点,所以1122CDCACB=+.-------------------------------------
--------------------------6分22222cos4844ababCbbCD++++==.------------------------------------------------------------------------
-------8分ABC△为锐角三角形.24b.-------------------------------------------------------------------------------10分()25,10CD,即线段CD长的取值范围为()5,10.-------
-------------------------------------------------------12分20.(I)如图1,连结BD与CE交于点Q,连结PQ.由题可得//DEBC,12DEBC=,所以12DQDEBQBC==.又2APPD
=,所以12DPDQPABQ==,所以//ABPQ.----------------------------2分PQ平面PEC,AB平面PEC.AB//平面PEC.----------------------------------------
------------------------4分(II)连结点A与BE的中点O,过点O作BE的垂线与BC交于点M,易知M为BC的中点.由已知可得AEAB=,所以AOBE⊥.平面ABE⊥平面BCDE,平面ABE平面BCDEBE=,AO平面A
BE.AO⊥平面BCDE.AOOM⊥.如图所示,以点O为原点建立空间直角坐标系.()()()()()0,0,1,1,0,0,,1,2,0,2,1,0,1,0,0ABCDE−−−,所以()()()()1,0,1,1,2,1,2,1,1,0,2,0AEACADCE=−−=−−=−−=−.设()(
)2,,01APAD==−−,则点()2,,1P−−,所以()21,,1PE=−−−.--------------------------------------------------7分设平面AEC和平面PEC的法向量分别为()()111222,,,,
,mxyznxyz==.由0,0,mAEmAC==得111110,20,xzxyz−−=−+−=取()1,0,1m=−.-------------------------------
----------------------------------8分数学试题答案第6页(共8页)由0,0,nPEnCE==得()()22222110,20,xyzy−−+−=−=取(
)1,0,12n=−−.------------------------------------10分由题可知0mn=,解得23=.------------------------------
-----------------------------------------------------------------11分所以2226333APADAD===.-----------------------
---------------------------------------------------------------------12分(其他解法酌情给分)21.(I)当1a=时,()1xfxxex=−+,则(
)()11xfxxe=+−.()()2xfxex=+.()fx在(),2−−单调递减,()2,−+单调递增.---------------------------------------------------------------
-------2分注意到()00f=,①当(),0x−时,因为11x+且01xe,所以(1)1xxe+,所以()(1)10xfxxe=+−,故()fx单调递减.-----------------------------
--------------------------------------------4分②当()0,x+时,()()00fxf=,故()fx单调递增.综上,()fx的单调递减区间为(),0−,单调递增区间为()0,+.------------------
---------------------------------6分(II)法一:()lnfxax恒成立等价于ln0xxeaxaax−+−恒成立.令()()ln0xhxxeaxaaxa=−+−,则()min0hx.则()()()()11xxxxeaahxx
eaxx+−=+−−=.---------------------------------------------------------------------------7分令()xgxxea=−,则()()10xgxxe=+,故()gx在()0,
+上单调递增.()00ga=−,()()10aagaaeaae=−=−,存在()00,xa,使得()0000xgxxea=−=,此时00lnlnxxa+=.--------------------------------------
-------9分则()hx在()00,x单调递减,在()0,x+单调递增.()()()00000minln2ln0xhxhxxeaxxaaaa==−++=−.-----------------------------------------------------
--11分因为0a,所以20ae.综上,实数a的取值范围为(20,e.-----------------------------------------------------------------------
--------------------12分数学试题答案第7页(共8页)法二:()lnfxax恒成立等价于ln0xxxeaxea−+恒成立.-------------------------------------------------------8
分令()0,xxet=+,则ln0tata−+.令()lnttata=−+,则()1atattt−=−=.----------------------------------------------------------------
----------10分()t在()0,a单调递减,(),a+单调递增.故只需()()min2ln0taaaa==−,所以2ln0a−,所以20ae.-----------------------------------12分22
.(I)设点()00,Mxy,其中()220021024xybb+=,则022x−且01x.()()22222200001114bAMxyxbx=−+=−+−.-------------------------------------------
-----------------------2分2222000204bxxbx−+−,即2220002104xxxb−+−(*).若02x=,则(*)式恒成立;若022x−且01x,则248422xbxx=−++,故22b.综上,22b.------------
-----------------------------------------------------------------------------------------------------6分(II)132,,kkk或231,,kkk成等差数列.证明如下:
--------------------------------------------------------------------------7分若1b=,则22:14xCy+=.设点()()1,0Ett.①若
直线l斜率为0,则点()4,0P,不妨令点()()2,0,2,0MN−,则123,,33ttktkk=−==−,此时123,,kkk的任意排列123,,iiikkk均不成等比数列,132,,kkk或231,
,kkk成等差数列.--------------------------8分②若直线l斜率不为0,则直线():10lxmym=+,()()1122,,,MxyNxy,易知34,Pm.联立22114xmyxy=+
+=得()224230mymy++−=.数学试题答案第8页(共8页)12122223,44myyyymm+=−=−++.--------------------------------------------------------------------------------------
-10分121231233,,,1133tytytmtmkkkxxm−−−−====−−()()()211212121212121222121212231162244346223yytyytytytytytkkxxmymymyymtyytyymmmmyymmtkm−+−−−−
−+=+=+=−−−+−+++==−+−==132,,kkk或231,,kkk成等差数列.综上,132,,kkk或231,,kkk成等差数列.-------------------------------------------------------------
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