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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()121121ln11aaaxxxa
x−=−=−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221lnln1
aaxxa+−=−.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1ln1ln1aaaa++−−.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:9101112ADABDBCBD1
2.设2FBm=,则12FBma=+,27FAm=,172FAma=+.在12AFF△中,由余弦定理得()()()222°7272272cos120mamcmc+=+−.2222714accmam−=−.①在12BFF△中,由余弦定理得()()
()222°2222cos60mamcmc+=+−.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15a,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,aa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==−=,12114322.710iivy==−=.1010
10.822.72451.6uv==.----------------------------------------------------------------------------------------------3分1
01222221110101194iiiiux===−−=.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+=.222sinsincossin22CAAB+=,即sinsinsincossinCACAB
+=.--------------------------------2分()sinsinsincoscossinBACACAC=+=+.-------------------------------------------------------------------------3分sin
cosCC=,即tan1C=.-------------------------------------------------------------------------------------------------4分
4C=.-----------------------------------------------------------------------------------------------------------------------------5分数学试题答案第5页(共8页)(I
I)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ABE.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()()()()11xxxxeaahxxeaxx+−=+−−=.------------------------------------------------------------------
---------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.()()()00000minln2ln0xhxhxxeaxxaaa
a==−++=−.-------------------------------------------------------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()()1,0Et
t.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−−−−====−−()()()211212121212121222121212231162244346223yytyytytytytytkkxxmymym
yymtyytyymmmmyymmtkm−+−−−−−+=+=+=−−−+−+++==−+−==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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