武汉市2024届九所重点中学第一次联考数学试题参考答案

武汉市2024届九所重点中学第一次联考数学参考答案与评分标准选择题：题号123456789101112答案ADCDBCCABCDACABDAD填空题：xy22413．（答案不唯一）+=114．233n33444215．或16．；n+422793解答题：17．（10分）解：（1）设数列{}an的公差为d，于是aandndn=+−=−−2(2)(2)1，因为SS95=5，所以925aa53=，所以9(31)25(1)dd−=−，解得d=−8，则ann=−158；··························································································3分22naa()1+nSnnnaa()1+nnaa1+n2n（2）a1=7，Sn=，考虑，即，即，2ankn+2ankn+ann+k由于d0，则n3时，aan20，且aann+=−12280，an44n−结合上述不等式得knn−2，整理得kn，································7分aa1+n411n−44n−任取整数nk，则nnk，原不等式成立，411n−2Snn于是对于任意正数k，均存在n(,3)nn+使得成立．························10分ankn+18．（12分）解：2（1）fx()的导函数fxxx'()366=−−，令fx'()0=得x1=−13，x2=+13，列表如下：x(,13)−−13−(1−+3,13)13+(13,)++fx'()+0−0+↑极大值↓极小值↑32所以有一个极大值为fx(1)=(1−3)−3(1−3)−6(1−3)+9=1+63，32一个极小值为fx(2)=(1+3)−3(1+3)−6(1+3)+9=1−63．············4分232（2）切线l的方程为y−f(x0)=f'(x0)(x−x0)，整理得y=(3x0−6x0−6)x−2x0+3x0+9，32232设gxfxyx()()=−=−3(36)23x−x0−xxx0+0−x0，且易知gx(0)=0，{#{QQABaQSQogAoABAAAAgCAwnQCEOQkAECCIoOhBAEoAAAQQFABAA=}#}232则gxxxxmxn()()()=−++0，从而gxxmxxnmxxnx()()()=+−+−−000，mx−=−3mx=−3于是0，解得0，322−=−nxxx00023nx=x−+2300222于是g()()[(3)23]()xxxxxxxxxxxx=−+−−+=−+−000000(23)，由题意可知，方程gx()0=有另一解x110x()x，于是xx10=−32，所以|||33xxx010−=−=|6，解得x0=3或x0=−1，·············································10分时，l的斜率kfx=='()30，时，l的斜率；综上，切线l的斜率为3．···········································································12分19．（12分）解：cb（1）在ABC中，由正弦定理得==2b，则2bsinC=c，sinsinCB又2bcoAsCD=c，且bc0，所以sincosCACD=，考虑到sin0C，则ACD是锐角，若C为钝角，则CA−CD=，即=DCB，2255此时有C，又A0，则CAB=−−，则C(,)；·························4分2626若C为锐角，则CACD+=，由题意知DCB0，则CACD，2所以CACDC+=2，又ACD0，所以C，所以C(,)，22425综上所述，C(,)(,)．···································································6分4226（2）由题知BDc，若C为钝角，则DBC中，BDCDc==22，不合题意，故C为锐角，也即，································································7分ACCDADc222+−ACD中，由余弦定理得cos==ACD，22ACCDb解得ADb22=，因为AD0，所以AD=b，···················································9分结合正弦定理知=ACDADC，于是AACD=−2，又，则ACC=−−=2()2，235又ABC++=，将上式代入，得A+=，解得A=．···························12分26920．（12分）解：16yy−（）设关于的回归方程为ˆˆˆ，时，，61，1zyz=+abyi=2,3,4,5,6yy==i4.7z==1.625i=25n(z−−z)(yy)ii11.33则i=1，ˆ，b=n=0.34a=z−by=0.02233.7()zzi−i=1{#{QQABaQSQogAoABAAAAgCAwnQCEOQkAECCIoOhBAEoAAAQQFABAA=}#}于是z关于y的回归方程为zyˆ=+0.020.34．····················································6分（2）选择②，理由如下：由（1）问知z和y有线性关系，故用zˆ估计z，知yyyiii−=+−10.020.34，则0.66yyii=+0.02−1，于是0.66(yyii+0.06)=0.06+−1，1即{0.06y+}是首项为y+=0.061.06，公比为q=的等比数列，i10.661−i(ln−0.66)xi从而yi=−=−1.06(0.66)0.060.7e0.06，满足②式的形式，故②适宜作为y与x的回归方程类型．···························································12分21．（12分）（1）证明：记ABC和A1B11C的中心分别为点OO,1，当AA11OO时，记点A1B1,,1C所在位置分别为点A000B,,C，易知ABCA−BC000为正三棱柱，且点23ABCABC,,,,,均在以O为圆心的圆上，O的半径为AO=；11100011013因为=AOBAOB010111，所以=AOABOB011011，同理可得=A011011011=OABOBCOC，记=AOA011(0)，则在中，AABBCC010101==，由上知OO1⊥，于是AABBCCOO0001====4，且AAAA001⊥，BBBB001⊥，CCCC001⊥，则AABBCC111==，··························3分下面只需证明AABBBBCCCCAA111111==，因为AA1100=++=+=+BBAAA1001000()()16A10BBB1010BAA1BBAABBAABB，同理有BBCCB110101=+BC16C，CCAAC110101=+CA16A，而AABBBBCCBBAACCBBACAC0101−0101=01()()01−01=0100−11=+++B0100001111BA()BBCABBC在中，由于BCAB0011=，所以AB10和BC10所对弧长相等，即=A110100BBBBC，所以=A0B10101BBBC，则B0BB10=AB0101BB1C，B0BB10=CB0101BB1A，也即AABBBBCC0101−0101=0，则AABBBBCC01=010101，同理有BBCCCCAA01=010101，AABBBBCCCCAA所以，所以11==1111，|AA1||BB1||BB1||CC1||CC1||AA1|即AA1,BB1,CC1两两夹角相等．······································································6分{#{QQABaQSQogAoABAAAAgCAwnQCEOQkAECCIoOhBAEoAAAQQFABAA=}#}（2）解：记直线CA11和CA00的交点为P，连接PA，PC，设A1和C1到平面ACC00A的距离分别为d1，d2，作C100GA⊥C于G，A100HA⊥C于H，由（1）问知AA0⊥，CC0⊥，则CC01⊥CG，AA01⊥AH，于是AH1⊥平面ACC00A，CG1⊥平面，1从而AHd=，CGd=，又APC的面积SACAA==4，1112202若P在线段AC上（如上图），则0，因为==CAOCAO，0030011116所以==APAAOA01011，于是d11C=Psin，dA21P=sin，1148则VVVSdSd=+=+=+=sin()sinCPPA，ACC1111AACPAACCP331233112若P在线段的延长线上（如下图），则，因为，3所以+=PAOA110，所以P=−，于是dCPCP111=−=sin()sin，，1148则VVVSdSdPAC=−=−=−=Psin()sin，ACC1AACPAACC111P33331211综上，当sin1=时，四面体ACCA的体积取最大值，此时=，·····················9分11226242则AOAO⊥，则AAA==O2，AABBCCAAA===+=A22，0111010131110013由（1）问知，AA110=+=+++BBA1010A161611B10BA()11OO()1ABOOB23222444=++−++=−=16()[2coscos()2cos()]16，33333344AABB11设AA和BB的夹角为，则cos===113．······························12分1156||||AABB1114322．（12分）解：5232259（1）将P(,)代入双曲线C得：−=1，2222ab22b双曲线的渐近线为yx=，即bxay=0；取l:0bx+=ay，l:0bx−=ay，a34|5ba+3||5ba−3|则点P到l3的距离d1=，点P到l4的距离d2=，2(ab22+)2(ab22+)因为a0且b0，所以|5b+3a||5b−3a|，所以dd12，于是dd12=4，即|5b+3a|=4|5b−3a|，259又−=10，所以53ba，所以534(53baba+=−)，解得ba=，22ab222592598xy22所以−=−==1，则a2=8，则C:1−=．························4分2a22b22a22a2a288{#{QQABaQSQogAoABAAAAgCAwnQCEOQkAECCIoOhBAEoAAAQQFABAA=}#}（2）i．设l11yk:xm=+，lykxnmn21:()=+，Qx(,y)11，3252k将点P代入直线l得：=+1m，即23mk5=−，1221222联立直线和双曲线的方程，消去y得：(1)280kxkmxm11−+++=，2km1则xxPQ+=−2，（式①）·······································································5分k1−1nnn因为点A在第一象限，故联立直线yx=与l2解得x=−，则A(,)−−，k1−1kk11−−11