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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，∴

AB2,1,0．2．C解析：zi1)i1)(i1()i1(i2i1i2，211||22z．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303.0

9．4．A解析：∵5.03log2log33a，0.101bee，5.05.0lnec，∴acb．5．B解析：,5,3572)(744717aaaaS,8,1377492aaaaa∴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，∴,22,13,23,3ADACABBCCDBD∴该几何体的最长棱为.22AD8．D解析：由题知,50,90,3010∴te05.0)309

0(3050，∴3105.0te，∴31ln05.0t，∴3ln05.0t，∴223ln2005.03lnt.9．A解析：∵直线AF的斜率为3，∴60PAF，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，∴2p，∴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=222222422

,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23ace.12．C解析：对任意的210xx，都有12210fxfxxx，即1122()()xfxxfx，∴2()()xxaegxxf

xxxaexx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74422baba，∴21ba，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3215

z，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,430sin221AO∴,21AO又∵,2211SAOO，∴RAOOOOA

222121，则其外接球的表面积为32224422R.16．121n解析：∵当2n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），（26,16），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，0sin2coscosBbAcCa由正弦定理得sincossincos2sinsin0ACCABB，··································

··1分所以sin2sinsin0ACBB，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2BCAC，所以3sin21CBCACSABC，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13PMPD，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0128344sin1

20233BCES·······························································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，21224414mxxk，·························6分121222214myykxxmk，2212122414mky

ykxmkxmk，·········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，3ln1f

xxx，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2252233rd，··································9分∴max12105210523

39S.······································································10分23.解：（1）3321()2211221332xxfxxxxxxx

，＞，，＜．·······································2分∵()3fx，∴3332xx＞或13122xx或3

3312xx＜，····································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cba2，即

32,31cba时取等号，···················9分∴222abc的最小值为1．························································

···················10分