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1/92021届高中毕业班联考(一)参考答案1.【答案】C【简析】【法一】两边取模可得:551.zz【法二】求得543,1.345iizzi2.【答案】D【简析】如图可知,MN,故N的子集有32个.3.【答案】D【简析】3袋垃圾中恰有1袋投放正确的情况有31113AC种
情形,由古典概型计算公式得三袋恰投对一袋垃圾的概率为21331113AACP,选D.4.【答案】A【简析】依题:336()201Caa,含4x项的系数516(1)6C.5.【答案】C【简析】易
知:12ab,2()14abab,1babaa,显然成立.6.【答案】B【简析】【法一】()0abc=,如图,OABC,直线OA交BC于D,30DOB,则c在a上的投影为cos3023OD
b.【法二】cos,cos,23ab=accac=bab.7.【答案】A【简析】设2211,rAFrAF,在21FAF中,由2221212122cos60FFAFAFAFAF,得
21221221212221244rrarrrrrrrrc,故222144acrr.2121AFAFAO,21122124AOAFAFAFAF,212212212122212
3441341414rrarrrrrrrra,故2214arr.222444aca,所以离心率2ace.8.【答案】B【简析】由条件可得,()cos()3gxx,作出两个函数图象,如图:yx2/9A,B,C为连续三交点
,(不妨设B在x轴下方),D为AC的中点,由对称性,则ABC是以B为顶角的等腰三角形,2ACT,由coscos3xx,整理得cos3sinxx,得3cos2x,则32CByy,所以23BBDy,要使A
BC为钝角三角形,只须4ACB即可,由3tan1BDACBDC,所以303,选.B9.【答案】BCD【简析】由表中数据可知3x,代入回归方程知142y,于是151a,B正确,将7x代入回归方程得318y,D正确,故本题答案是BCD.
10.【答案】BC【简析】若nb是等差数列,据等差数列求和公式知需1,nbbknkZ,则nb为“吉祥数列”,检验,AC可知C正确.nb是摆动数列,由并项法知:24,2nnSnSn,21242nnSS
nn,故B正确,根据等比数列求和公式知D错误.11.【答案】BCD【简析】易知准线方程为yxCpy4:,2,12.设直线11xky,代入42xy,得0142kkxx,当直线与C相切时,有0,即012kk,设
TBTA,斜率分别为21,kk,易知21,kk是上述方程两根,故121kk,故TBTA.设2211,,,yxByxA,其中4,4222211xyxy.则112124:xxxxyTA,即112yxxy,
代入点)1,1(,得02211yx,同理可得22220xy,故022:yxAB,故12ABk.3/9由21444212122212121xxxxxxxxyykAB,得1221xx,即AB中点横坐
标为1.12.【答案】ABD【简析】,()()xRfxfx,()fx是偶函数,A正确;(2)()fxfx,由函数的奇偶性与周期性,只须研究()fx在[0,2]上图象变化情况.sinsinsin2,0(),1,2xxxexfxexe
当0x时,sin()2cosxfxxe,则()fx在0,2x上单调递增,在,2上单调递减,此时()[2,2];fxe当2x时,sinsin()cosxxfxxee,则()fx在3
[,]2x上单调递增,在3,22x上单调递减,此时1()2,fxee,故当02x时,min()2fx,B正确.因()fx在,2x上单调递减,又()fx是偶函数,故()fx在,2
上单调递增,故C错误.对于D,转化为2()fxx根的个数问题.因()fx在0,2上单调递增,在,2上单调递减,在3,2上单调递增,在3,
22上单调递减.当(,)x时,()2fx,22x,2()fxx无实根.(3,)x时,max262()xefx,2()fxx无实根,3,2x时,显然x
为方程之根.sinsinsinsin(),()cos0xxxxfxeefxxee,3123322fee,单独就这段图象,3()02ff,()fx在3,2
上变化趋势为先快后慢,故()gx在3,2内有1个零点,由图象知()gx在3,32内有3个零点,又5()252fe,结合图象,知D正确.4/913.【答案】104x(答案不唯一,请阅卷教师按照题意自行把握)
.【简析】22124202xxxxx,依题有:1,(0,)2xAA.14.【答案】0【简析】()(2)1fxfx,两边同时求导可得:()(2)0fxfx,(2019)(2021)0.ff15.【答
案】24或8【简析】BPPC,AO面BPC,三棱锥APBC的外接球球心M在AO上,且4r外,2MO,4MA,6AO或2AO1126243V圆锥或112283V圆锥.1
6.【答案】31624【简析】以经过,AB的直线为x轴,线段AB的垂直平分线为y轴,建立直角坐标系,如图,则(2,0)A,(2,0)B,设(,),3PAPxyPB,2222(2)(2)3xyxy,得:12)4(0482222yx
xyx,点P的轨迹为圆(如图),其面积为12.【法一】22244PAPBxyOP,如图当P位于点D时,2OP最大,2OP最大值为31628)324(2,故PAPB最大值是31624.【法二】由极化恒等式知:2224PAPBOPOBOP,以
下同法一.四、解答题(本大题共6小题,共70分.解答应写出文字说明、证明过程或演算步骤)17.【解析】(1)成等差数列,CABcabsinsinsin22------------------------------1分当3A时,CB
sin3sinsin2,即)32sin(3sinsin2BB21)6sin(B而3,66,266,320BBBB-----------------------------5分(
2)由余弦定理及cab2,222()3112cos()2842acaccaBacac,当ca时取等号.cba,,5/9结合余弦函数的单调性可知:30B.--------------------------
--------------------------------------------------10分18.【解析】(1)依题,在1122nnnaa两边同时除以12n,得:11122nnnnaa,故数列{}2nna为等差数列.-------------------
----------------------------------------------------5分(2)由上易得:1(1)2nnann,可得2nnan-----------------------------------------------
------------7分【法一裂项相消】由1122nnnaa可得:112nnnnaaa-------------------------------------------------8分则数列1{2}nna的前n项和1121()()(
)nnnnnSaaaaaa111(1)22nnaan------------------------------------------------12分【法二乘比错位】12(2)2nnnan
--------------------------------------------------------------------------------7分则数列1{2}nna的前n项和123324252(2)2nnSn
23123242(1)2(2)2nnnSnn两式相减得:2316222(2)2nnnSn----------------------------
-------------------------------10分1(1)22nnSn-------------------------------------------------------------------------------------12分19.【解析】
(1)设40个槟榔芋中,每个槟榔芋的平均质量为q,则6.19324006.022012.020044.018024.01601.014004.0q(克)------------4分所以这批槟榔芋的数量约为5176.193100000(个)------------
-------------------------------------------------------5分(2)X所有可能取值为0,1,2,3.由表中数据可知,任意挑选一个槟榔芋,质量在170,1
50的概率为101505-----------------------------------6分所以729.010903XP243.01011091213CXP027.010
110922123CXP001.010133XP------------------------10分故X的分布列为:X0123P729.0243.0027.0001.06
/9EoH所以3.0001.03027.02243.01729.00XE----------------------------------------------12分20.【解析】【法一】
(等积法)(1)即求点B到平面11ACD之距h.1111111843443333BACDABCDABCDAABDVVV,1122ADCD,1123AC,如图,5DH,111143235323BAC
DVh44555h.----------------------------------------------------------6分【法二】如图,连1111ACBDO,取1
DD中点H,在正方形11DDBB中,易证:DOHBBHODE,有BEOD①,又111111111ACBDACACBB面1BD11ACBE②,由①②可得:BE面11ACD,在BDE中,45cos5BEBDDBEBEBDBH----
----------------------------------6分【法三】如图建系,各点坐标如图,面11DAC的法向量为(,,)xyzm,1122003200yzDAxyzDCmm,取1y
,1z,3x,m可为3,1,144555BDBEmm-------------------------------------------------6分(2)如图建系,各点坐标如图,面11DAC的法向量为(,,
)xyzm,//BEm,1122003200yzDAxyzDCmm,取1y,1z,3x,m可为3,1,1,同理可求面1FCB的法向量53,1,23n,46cos,sin52003mn.----------
---------------------------------------------------------------------------12分21.【解析】(1)设公共点为P,则rPFrPF4,21,故12124PFPFFF即公共点P的轨迹为椭圆.--------
-----------------------------------------------------------------------------------------2分7/9且2,42aa,又3,12bc,故曲线134:22y
xE.------------------------------------------------4分(2)当直线PQ斜率不存在时,712:xPQ,代入E得712y,易知OQOP;----------
--5分当直线PQ斜率存在,设mkxyPQ:,PQ与圆O相切,22212171mrmkk---------6分将PQ方程代入E,得0124834222mkmxxk,34124,3482221221kmxxkk
mxx,221212121212121OPOQxxyyxxkxmkxmkxxkmxxm----------------------7分2222222348341241mkmkkmk341127222
kkm将171222km代入,得0OPOQ,即OQOP---------------------------------------------------------10分综上,恒有OQOP,2127APAQAPAQ
OA.--------------------------------------------12分【法二】当直线PQ斜率不存在时,712:xPQ,代入E,得712y,127A
PAQAPAQ;当直线PQ斜率存在时,设mkxyPQ:,PQ与圆O相切,rkm12,即171222km.将PQ方程代入E,得0124834222mkmxxk,34124,348222122
1kmxxkkmxx,7122171221212212122mkmxxkmkxxrOPAPkmxkkmxxm7121277122127121212,同理可得kmxAQ7121272
,故221212712127kAPAQmxxkmxx---------------------------------------------------------------------10分将21212228412,4343kmmxxxxkk
,及171222km代入,可得712AQAP.综上127APAQAPAQ.(注:本题也可联想射影定理思考)-----------------------------
-------12分8/922.【解析】(1)当1ka时,1xxye,由2(1)xxxxexexyee------------------------------------1分则函数1xxye在(,0)上单调递增,在(0,)上单调递减,当0x时,max1y--
-----------------------4分.(2)【法一】记()()()axhxfxgxekxa(0,0xa),则()()axaxkhxaekaea①当1ka时,()h
x在(0,)上单调递增,()(0)1hxha,由10a,可得01a---------5分②当1ka时,()hx在1(0,ln)kaa上单调递减,在1(ln)kaa,上单调递增,min1()(ln)khxhaalnkkkaaaa,由m
in()0hx,可得21ln0kaak,令2()1lnkaaak---------------------7分1.若1k时,对于①,1ka,此时a不唯一(舍去)2.若1k时,对于①,11a,即1a对于②
,01a时,由221()1ln=1+lnaaaaa,所以当01a时,2112()2aaaaa,当212a时,()a为减函数,()(1)0a,此时a不唯一(舍去)---------------------------------8分3.若1k
时,对于①,a无解,对于②,转化为2()1ln0kaaak在(0,)k内仅有唯一解的问题.2122()akaaakak,()a在(0,)2k上单调递增,在(,2k)k上单调递减,1(,)2kak,()10kk
;21(0,)2kae,211()ln0keek,要使a唯一,只须()02k,即11202k,解得2ek,此时22kea符合题意------------------------------
---------------------11分综上:存在2ek,有唯一的22kea符合题意--------------------------------------------------------------12分【法二】原问题等价于()10axkxaxe
恒成立,2()axakxakxe,(0)1a--------------5分1°.当00,()0,()(0,)kxxx时,在上单调递增,要使()0x恒成立,只须10a即可,得01a,
此时a不唯一(舍去)---------------------------------------------------------------------------------------6分2
°.当0k时,220,(),()kakaxxxxakak可得单调递增;单调递减9/9若01a时,221(0)0,=10,()0akkakxakae满足,此时a不唯一(舍去)
若1a时,221(0)0,=10,akkakakae由零点存在性定理得:200()00,kaxxak,且,当00,()0xxx时,与题意矛盾(舍去)--------------------
------------------------------------------------------9分3°.当20ka时,()(0,),(0)10xa在上单调递增只须,此时a不唯一.当222,,()0,()kakakaxxxx
akak单调递增;时单调递减,要使()0x恒成立,且a唯一,只须221=10akkakakae,所以有21lnakka,令222()lnln1,()akaHaakHakak,故()Ha在(0,)2ka上单调递增,在(,)2ka
上单调递减,则须使02kH,解得,22eeka符合题意,综上:,22eeka----------------12分