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第1页共6页2022届“韬智杯”高三大联考文科数学参考答案一、选择题:本题共12小题,每小题5分,共60分.题号123456789101112答案ACBABDCDABDC1.A解析:集合3,2
,1,041|xNxA,集合32|xxB,∴AB2,1,0.2.C解析:zi1)i1)(i1()i1(i2i1i2,211||22z.3.B解析:由频率分布直方图知低于60分的频率为(0.0050.01)200.3,∴
该班参加竞赛的学生人数为303.09.4.A解析:∵5.03log2log33a,0.101bee,5.05.0lnec,∴acb.5.B解析:,5,3572)(744717aaaaS,8,1377492aaaaa∴89a.
6.D解析:易知43tan,∴54cos)cos(.7.C解析:由三视图可得原几何体为右图所示的三棱锥ABCD,∴,22,13,23,3ADACABBCCDBD∴该几何体的最长棱为.22AD8.D解析:由题
知,50,90,3010∴te05.0)3090(3050,∴3105.0te,∴31ln05.0t,∴3ln05.0t,∴223ln2005.03lnt.9.A解析:∵直线AF的斜率为3,∴60PAF,又∵
抛物线的定义知|PF|=|PA|=4,∴PAF为等边三角形,∴|AF|=4,∴在AKFRt中,|KF|=2,∴2p,∴抛物线方程为24yx.10.B解析:若存在ABC△是等腰直角三角形,当A,B,C为相邻三个交点时,取得最小值,∵直角三角形斜边的中线等于斜边长的一半,∴T=22
2222422,解得2.11.D解析:如图,设圆M与12PFF△的三边12FF,1PF,2PF分别相切于点E,F,G,连接ME,MF,MG,则12MEFF,1MFPF,
2MGPF,设r为12PFF△内切圆M的半径,∴111122MPFrSPFMFPF△,221122MPFrSPFMGPF△,121212122MFFrFFMEFSF△,K第9题图第7
题图第2页共6页∵212132FMFMPFMPFSSS,∴||232||2||22121FFrPFrPFr,化简得:||32||||2121FFPFPF,由双曲线的定义可得:122PFPFa,122FFc,∴ca2322,∴离心率2
3ace.12.C解析:对任意的210xx,都有12210fxfxxx,即1122()()xfxxfx,∴2()()xxaegxxfxxxaexx在(0,)上单调递增,()20xgxaex在(0,)
上恒成立,即2xxae在(0,)上恒成立,令2()xxhxe,则2(1)()xxhxe,当(0,1)x时()0hx;当(1,)x时,()0hx,故max2()(1)hxhe,所以2ae.二、填空题:本题共4小题,每小题5分,共20
分.13.3解析:由7|2|ba得74422baba,∴21ba,设单位向量a与b的夹角为,∴21cos,∴3.14.5解析:作出可行域,如阴影部分所示:由3zxy得3yxz,平移直线
3yxz,由图象可知当直线3yxz经过点C时,直线3yxz的截距z最大,由11yxy,解得2,1C,代入得3215z,即z的最大值为5.15.32解析:补形成一个直三棱柱,设上、下底面外接圆的圆心分别为,,21OO则外接球的
球心O为12OO的中点,由正弦定理,430sin221AO∴,21AO又∵,2211SAOO,∴RAOOOOA222121,则其外接球的表面积为32224422R.16.
121n解析:∵当2n时,an=2221nnSS,∴21221nnnnSSSS,第14题图第11题图第3页共6页整理可得112nnnnSSSS,∴1112nnSS,∴数列{1nS}是以1为首项,2为公差的等差数列,∴121nnS,∴121nSn.三、解答题:共
70分,解答应写出文字说明、证明过程或演算步骤.17.解:(1),ab的取值情况有(23,25),(23,30),(23,26),(23,16),(25,30),(25,26),(25,16),(30,26),(30,16),(2
6,16),基本事件的总数为10.·······················································································3分事件A包括的基
本事件有(25,30),(25,26),(30,26)共3个,·························5分所以事件A的概率为3()10PA.································································
········6分(2)由列表可知5521111,24,615,1351,iiiiixyxxy···································8分设回归方程为^^^ybxa,5^152215313
.1105iiiiixyxyxxb,·······································10分^^10.1aybx,···············································
······································11分故所求方程为^3.110.1yx···················································
························12分18.解:(1)在ABC中,因为,0sin2coscosBbAcCa由正弦定理得sincossincos2sinsin0ACCABB,····································1分所以sin
2sinsin0ACBB,即sin12sin0BB,······························3分又因为sin0B,所以1sin2B,········································
·····························4分因为B是三角形的内角,所以6B或56.·····················································
····6分(2)由(1)知6B,·····················································································8分所以ABC为等腰三角形,且23C
,在ABC中,设2ACBCx,···············9分在ADC中,由余弦定理得222222cos773ADACDCACDCx,解得1x,所以2BCAC,所以3si
n21CBCACSABC,所以三角形的面积为3.··············································································12分第4页共6页19.解:(1)证明:在PD上取点M使得13PMP
D,连结,FMMA,(如图)因为11,33PFPCAEAB,所以//,//FMCDAECD,······································1分且FMAE,所以四边形
AEFM是平行四边形,·················································2分所以//FEAM,·······················································
·······································3分又FE平面PAD,AM平面PAD,·····················································
······4分所以直线//EF平面PAD················································································5分(2)取CD中点O,连
结PO,因为侧面PCD与底面ABCD垂直,CD为两个面的交线,所以PO平面ABCD,·································································
···················7分又因为1,3PFPC,所以F到面ABCD的距离等于24333PO··························9分0128344sin120233BCES·······································
························10分所以1183432333339BCEVSh···························································12分20.解:(1)由题意,设椭圆C:22221
(0)xyabab,焦距为2c,则3c,椭圆的另一个焦点为23,0F,·····················································1分由椭圆定义得12712422aPFPF,2a,221bac,···········
···3分所以C的方程2214xy.·············································································4分(2)由已知得
0,1D,由2214ykxmxy得222148440kxkmxm,·········5分22=410km,,设11,Axy,22,Bxy,则122814kmxxk,2122441
4mxxk,·························6分121222214myykxxmk,2212122414mkyykxmkxmk,·········7分由ADBD得
1212110DADBxxyy,整理得22523014mmk,········9分所以,25230mm,解得1m或35m,···············
································10分①当1m时,直线l经过点D,舍去;·····························································11分②当35m时,显然有,直线l经过定点30,
5.····································12分21.解:(1)当3a时,3ln1fxxx,第19题图第5页共6页32'111xfxxx,······································
·········································1分当1,2x时,'0fx,fx是减函数,··················································2分2,x,'0fx,fx
是增函数,·······················································3分所以,fx的减区间为1,2,增区间为2,.··············································4分(2)当1a
时,ln1fxxx,2kxfx,即2ln10kxxx.······························································5分设
2ln1gxkxxx,0x,则只需0gx在0,恒成立即可.···········6分易知00g,22(21)[2(1'2111]112)kxkgxkxxxxxxkxk
,································7分①当0k时,'0gx,此时gx在0,上单调递减,所以00gxg,与题设矛盾;······································
···························8分②当102k时,由'0gx得1102xk,当10,12k时,'0gx,当11,2xk
时,'0gx,此时gx在10,12k上单调递减,所以,当10,12xk时,00gxg,与题设矛盾;····························10分③当12k时,'0gx,故
gx在0,上单调递增,所以00gxg恒成立.····································································
···············································11分综上,12k.················································································
················12分22.解:(1)(1)(1)22(2)2210122(2)xtxyyt,····················2分22cos22coscos22sinsin2cos2sin444
222222cos2sin22(1)(1)2xyxyxy,·············4分第6页共6页∴直线l的方程为2210xy,圆C的
方程为:22(1)(1)2xy.···········5分(2)圆C的圆心坐标为(1,1),半径为2,圆心到直线l的距离为2222(1)(1)1223(22)(1)d,······································6分∴2222210|
|2(2)()33AB·····································································7分∵点P到直线AB距离的最大值为2
252233rd,··································9分∴max1210521052339S.·····································
·································10分23.解:(1)3321()2211221332xxfxxxxxxx,>,,<.····················
···················2分∵()3fx,∴3332xx>或13122xx或33312xx<,·······················
·············3分解得2x或0x,∴不等式的解集为02|xxx或.··································································5分(2)由
(1)知,函数()fx在1(,)2上单调递减,在1[,)2上单调递增,所以min13()()22fxf,则1322abcm,···············································6分由柯西不等式,有222222211()[
()9411]()22abcabc,·····················8分∴2221abc,当且仅当cba2,即32,31cba时取等号,···················9分∴222abc的最小值为1.··················
·························································10分