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2023年春“荆、荆、襄、宜四地七校考试联盟”数学试题参考答案一、选择题123456789101112DDBCDAACBCDADABDACD1.D()cos4,()143fxxf=+=−+=2.D()()()()pABpApBpAB+=+−,1134691()3
241212pAB+−=+−==故()11()2()126pABpABpB===3.B14yxx=−,13xy==,切线:23(1)lyx−=−,即310xy−−=,则点()32−,到直线l的距离为233
(2)1101010312−−−==+4.C841127927m+++=,得29m=,则()842112790129273EX=+++=5.D法1:122133434337361213112376535C
CCCCC++++==法2:343744311112317653535CC−=−=−=6.A有1314133e+=,得ab由4334,得3344log,3443log,即
bc,故abc7.A1(1)221nann=+−=−,112nnb−=,122121nnnc−=−=−1123222....2212nnnccccnn+−++++=−=−−−当9n=时,1022910241110132023−−=−=当10n=时11221020481220
362023−−=−=8.C21lnln2ln222ln2axaxaxxxaxlogxaxa设()2xfxx=,则()()22ln221ln20xxxfxxx=+=+,得()fx在(0,)+
上单增则22222()()axlogxxlogxaxflogxfaxlogxaxax设2()logxgxx=,则22211lnln2()ln2xlogxxxgxxx−−==,得()gx在(0,)e上
单增,在(,)e+上单减,则21()()ln2maxlogegxgeee===,故1ln2ae9.BCDA错B对,1212133PABCDV−==C对,()21211123333333DQDPDBDPDADCDADCDP=+=++=+
+D对,221332(5)()222PACS=−=,由1311213232d=,得23d=10.ADA对,设动点(,)pxy,则2222(1)2(3)(1)xyxy+−=−+−,即22(4)(1)4xy−+−=B错,直线:(3)20lkxy−−+=过定点(3,
2)D,点D在圆C内C错,圆心(4,1)在l上,代入得1k=−.D对,max2(43)23BPBC=+=−+=11.ABDA对,22122()xfxxxx−=−=,()fx在(0,2)上单减,在(2,)+上单增B对,设()()gxfxx=−,2222(2)
()10xxxgxxx−−−+=−=,()gx在(0,)+上单减,又(1)10,(2)ln210gg==−,则()()gxfxx=−在(0,)+上有且只有一个零点.C错,22ln()xfxkxkxx+,设22ln
()xhxxx=+,则231ln2()2xhxxx−=−=224131ln1ln0xxxxxxx−−−−−=,()hx在(0,)+上单减.当x→+时,()0hx+→,则()hx无最小值,故()khx不恒成立.D对,设21(1)xttx=由121222lnlnxxxx+
=+得2211122()lnxxxxxx−=,即221122(1)lnxxxxx−=,即22122(1)2(1)2(1)ln,,lnlnttxttxxxtttt−−−====2122(1)2(1)144ln0lnln2tttxxttttt
−−−+++设21()ln(1),2thtttt−=+则2222111(1)()0.()22tthxhtttt−−−−=+=在()1,+上单减,()(1)0hth=,故124xx
+12.ACDA对,设点()()()112200AxyBxyPxy,,,,,,则有1010202022{yyxxyyxx+=+=,得10201222yyyyxx++=,22211221120121212xy
xyxxxxyxxxxxx−−===−−,又10120222xxyxx−=,得22121212012122()2()2yyxxxxxxxxx−+===−−则点1212(,)2xxPxx+,即121(,)24xx+
−,故点p在准线上1:4ly=−B错,点p在以AB为直径的圆上,则APPB⊥,即2APB=C对,设点AB,在准线l上得射影分别是11AB,,则1FPBBPB,得1FBPB⊥,即FQBQ⊥D对,由33k=,得60AFO
=,120AFy=,则1121cos1203AF==−,1211cos120FB==+得13AFFB=三、填空题:13.314.3515.0.0716.(1)1(2)8813.3245752131(1)
4(1)aaaqqaaaq++===++,22q=,得211(1)(12)9aqa+=+=,13a=14.35222222222232345633456776535123CCCCCCCCCCC++++=++++===15.0.07123668()
,(),()202020PAPAPA===,123111(),()()101520PBAPBAPBA===112233()()()()()()()PBPAPBAPAPBAPAPBA=++6161813220.07201020152020205++=++==16
.(1)1(2)88(1)22221111111111111111234132422331324++++++=+−+−+−171144=+−=,22221111111111111111234233445233445++++++=+−+−+−
111312510=+−=,所以222271111413412310+++,所以2222111112341+++=;(2)当1n=时,111112Saa=+,解得211a=,因为0na,所以111aS==,当2n时,1nn
naSS−=−,所以1112nnnnnSSSSS−−=−+−,即2211nnSS−−=,所以2nS是以1为首项,1为公差的等差数列,所以()2=nSnnN,因为0na,所以0nS,所以=nSn,当2n时,111121nnnnn+++−,即21211nnnnn+++−,所以(
)()12121nnnnn+−−−,令122023111=+++SSSS,则()()123243202420231220242+−+−++−=+−S,因为44.9844.982023.2004=,44.9944.992024.
1001=,21.42,所以()()12202421244.981.4288.12+−+−=,()1221324320232022+−+−+−++−S()1220231=+−,因为44.9744.972022.3009
=,44.9844.982023.2004=,,所以()()12202311244.98188.96+−+−=,所以88.1288.96S,即88=S.四、解答题17.(1)22()xxmfxe−−
=,当0m=时,()21()xxfxe−=………………………………1分由()0fx=,得1x=当1x时,()0fx,()fx在(,1)−上单增当1x时,()0fx,()fx在(1,)+上单减………………………………………
………3分故当1x=时,()fx有极大值2e,无极小值.……………………………………………………5分(2)()0fx在(3,4)上恒成立即220xxme−−在(3,4)上恒成立,22mx−……………………………………………
…7分又22(6,4)x−−−…………………………………………………………………………………8分则6m−.…………………………………………………………………………………10分(说明:结果不带等号的只扣1分)18.(1)X的可能取值为0、1、2(0)(10.
05)(10.08)0.874PX==−−=(0)0.05(10.08)(10.05)0.080.122PX==−+−=(2)0.050.080.004PX===…………………………………………………………………4分
X的分布列为X012P0.8740.1220.04()00.87410.12220.040.13EX=++=故次数X的数学期望为0.13.……………………………………………………………………6分(2)依题意,可
知A、B款手机发生屏幕意外损坏分别有24000×0.05=1200部,10000×0.08=800部………………………………………………………………………………………………8分屏幕更换总成本为1200×400+800×600=960000元碎屏险总收
入为24000a+10000×50业务收入为24000a+10000×50-960000=24000a-460000…………………………………10分则24000a+10000×50-960000≥500000,得40a,故A款手机的碎屏险费
a最低应定为40元.………………………………………………………12分19.(1)由四边形11ABBA是平行四边形,1190AAB=,得四边形11ABBA是矩形,则1ABBB⊥…………………………………………………………………
……………………………………1分由1ABBB⊥,ABBC⊥,1BBBCB=,1BB、BC面1BCCB,得AB⊥面11BCCB,又1BC面11BCCB,则1ABBC⊥……………………………………………………………3分由四边形11BCCB是菱形,得11B
CBC⊥由1ABBC⊥,11BCBC⊥,1ABBCB=,AB、1BC面1ABC,得1BC⊥面1ABC………………………………………………………………………………………………………5分(2)由(1)可知,AB⊥面11BCCB,又AB面ABC,得面ABC⊥面11B
CCB由四边形11BCCB是菱形,160BBC=,得1BCB是正三角形.取BC、AC的中点分别为O、M,连1OB,OM,则1BOBC⊥,OMOC⊥.由面ABC⊥面11BCCB,1BOBC⊥,BC=11ABCBCCB面面得1BO⊥面11BCCB………………………………………………………………
…………………6分以点O为坐标原点,1OMOCOB、、所在直线分别为xyz、、轴,建立空间直角坐标系,如下图所示则11(0,0,0),(0,1,0),(1,1,0),(0,0,3),(0,2,3)OCABC−11(0,
2,0)BC=,1(1,1,3)BA=−−,1(0,1,3)BC=−……………………………………7分面1ABC的一个法向量为1(0,1,3)BC=−………………………………………………………8分设面11ABC的一个法向量为(,,)nxyz=由11120
30nBCynBAxyz===−−=,令1z=,得3,(3,0,1)0xny===…………………………10分设二面角11BACB−−的大小为则11133coscos,224nBCnBCnBC====………………………………………
…11分213sin1cos4=−=故二面角11BACB−−的正弦值为134.…………………………………………………………12分20.(1)当2n时,113(1)nnSna−−=+,又3(2)nnSna=
+得13(2)(1)nnnanana−=+−+,即1(1)(1)(2)nnnanan−−=+……………………………2分当3n=时,3224aa=,32422aa==………………………………………………
……………3分又10a,得0na,111nnanan−+=−当2n时,121211143(1)1123212nnnaannnnnaaaannn−−+−+===−−−当1n=时,11a=符合上式综上,得(1)2nnna+=……………………
…………………………………………………………6分(2)12112()(1)1nannnn==−++…………………………………………………………………8分1211111111112()2()2()2(1)122311nnTaaannn=+++=−+−++−=−++……………10分由1n,得1
1012n+,12(1)[1,2)1n−+,即12nT.………………………………12分21.(1)依题意,得2222212222caabcab==+=,解得222abc===则椭圆22:142xy
C+=.………………………………………………………………………………4分(2)设直线:ABykxt=+,点11(,)Axy,22(,)Bxy由221420xykxyt+=−+=,消y,得222(42)84(2)0kxktxt+++−=得
12221228424(2)42ktxxktxxk−+=+−=+,且()22442420kt=+−…………………………………………6分由2PAPBkk+=,得1212222yyxx+=−−,即1212222kxtkxtxx+++=
−−,即1212(22)(24)()480kxxktxxt−+−+++−−=则2224(2)8(22)(24)4804242tktkkttkk−−−+−++−−=++即224420ktktkt−−−−−=,2(
2)(2)0ktkt+++=即(2)(21)0ktkt+++=,2tk=−或21tk=−−………………………………………………9分当2tk=−时,直线:2(2)ABykxkkx=−=−过定点(2,0)P,不合题
意,故舍去.当21tk=−−时,直线:21(2)1ABykxkkx=−−=−−过定点(2,1)−……………………11分又ABMN⊥,故直线AB与MN的交点在以(6,0)和(2,1)−所连线段为直径的定圆上.…12分22.(1)当1,0ab==时,()ln1fxxx=−+,11()1xfxxx−
=−=…………………………1分得当01x时,()0fx,()fx在(0,1)上单增当1x时,()0fx,()fx在(1,)+上单减则当1x=时,()fx有最大值(1)0f=………………
……………………………………………4分(2)当,1aeb=时,()()ln1xfxaxxe=−−,2()xxaaexfxexxx−=−=①2()xgxaex=−,2()(2)0xgxexx=−+,()gx在(0,)+上
单减……………………6分由ae,得ln1a,2ln1a,则21ln0a−又(1)0gae=−,()22ln(ln)ln1ln0agaaeaaa=−=−(说明:也可以()10aagaaeaae=−=−())由零点存在性定
理可知,存在唯一(1,ln)a使()0g=,即20ae−=………………7分得当(0,)时,()0gx,()0fx,()fx在(0,)上单增当()x+时,()0gx,()0fx,()fx在()+上单减则()fx在x=处取得极
大值,即()fx存在唯一的极值点.…………………………………8分②由①可知,20ae−=,即2ae=由,且(1,ln)a,得1由()0f=,得ln(1)0ae−−=,ln1ae=−两式相除,得2ln1e−=−……………………………………
…………………………………9分由(1)可知,()0fx,即ln10xx−+,则ln10−+,ln11−则22ln1e−=−,2ln−,()()()21221ln−++……………1
0分法一设11()2hln=−−,则()2221111()1022h−−=−+=得()h在(1,)+上单减,则()(1)0hh=,得12ln−,则()()()()1212ln21
21−++−+………………11分又()()()()2221112131110−−+−−=−+−+=−+得()()212131−+−故()()()
22131−+−成立.证毕………………………………………………………12分法二(同上)………………………………………………………………………………………………10分设21()2321ln−=−
+,则()()23222222231()322121++−−−=−=−=++()()()()()2222121[3]1321332202121−++−++−=−++得()在(1,)+上单减,则()(1)0
=,得212321ln−+,则()()22ln2131+−故()()()22131−+−成立.证毕………………………………………………………12分获得更多资源请扫码加入享学
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