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第1页共5页2022—2023学年度(下)联合体高二期中检测数学参考答案及评分标准一、单选题（本大题共8小题，每小题5分，共40分）1．D【解析】因为𝑋∼𝑁(3，𝜎2)，所以𝑃(𝑋<2)=𝑃(𝑋>4)=15，所以𝑃(𝑋<4)=1−𝑃(𝑋>4)=45．故选：D．

2．B【解析】在等比数列{𝑎𝑛}中，由a2＝8，根据等比中项可得𝑎1𝑎3＝𝑎22＝64，所以log2𝑎1+log2𝑎3＝log2𝑎22=log264=6．故选：B．3．A【解析】由题意知随机变量𝜉服从超几何分布．因为𝜉表示这6本书中理科书籍的本数，所以C86C40C1

26+C85C41C126=𝑃(𝜉=0)+𝑃(𝜉=1)=𝑃(𝜉≤1)．故选：A．4．A【解析】设公差为𝑑，则𝑎2+𝑎3=2𝑎1+3𝑑=8，解得𝑑=4，所以数列{𝑎𝑛}的通项公式是𝑎𝑛=−2+4(𝑛−1)=4𝑛−6．故选：A．5．C【解析

】由𝑃(𝐴)=14，得𝑃(𝐴)=1−𝑃(𝐴)=34．因为𝑃(𝐵𝐴)=𝑃(𝐴)𝑃(𝐵|𝐴)=𝑃(𝐵)𝑃(𝐴|𝐵)，所以𝑃(𝐴|𝐵)=𝑃(𝐴)𝑃(𝐵|𝐴)𝑃(𝐵)=34×131

2=12．故选：C．6．A【解析】设𝑎𝑛=𝑎1𝑞𝑛−1．由题意得{𝑎3=16，𝑎2+𝑎4=2×20，即{𝑎1𝑞2=16，𝑎1𝑞+𝑎1𝑞3=40，解得{𝑎1=4，𝑞=2，（𝑞=12舍去），所以该数列的前n项和为4×(1−2𝑛)1−2=2𝑛+

2−4．故选：A．7．D【解析】由题意，得𝑎+𝑏+𝑐+13=1，所以𝑎+𝑏+𝑐=23①．因为𝐸(𝑋)=(−1)×𝑎+0×𝑏+1×𝑐+2×13=34，所以−𝑎+𝑐=112②．由𝑃(𝑋≥1)=𝑐+13=712，得𝑐=14，依次代入②、①，解得�

�=16，𝑏=14，所以𝐷(𝑋)=(−1−34)2×16+(0−34)2×14+(1−34)2×14+(2−34)2×13=1916．故选：D．8．C【解析】不妨设𝑂𝐷1=𝐷𝐶1=𝐶𝐵1=𝐵𝐴1=1，则𝐷𝐷1=0.11

4，𝐶𝐶1=𝑘1，𝐵𝐵1=𝑘2，𝐴𝐴1=𝑘3．由题意，知𝐷𝐷1+𝐶𝐶1+𝐵𝐵1+𝐴𝐴1𝑂𝐷1+𝐷𝐶1+𝐶𝐵1+𝐵𝐴1=0.414，即0.114+𝑘1+𝑘2+𝑘34=0.41

4．设该数列的公差为𝑑．因为𝑘1=0.114+𝑑，𝑘2=0.114+2𝑑，𝑘3=0.114+3𝑑，所以0.114×4+6𝑑4=0.414，解得𝑑=0.2．故选：C．二、多选题（本大题共4小题，每小题5分，共20分）9．ABD【解析】由题意，得{𝑎1+𝑑=0，7𝑎1

+7×62𝑑=𝑎1+3𝑑+12，解得{𝑎1=−1，𝑑=1，故A正确；所以𝑎𝑛=−1+(𝑛−1)×1=𝑛−2，故B正确；所以𝑎4+𝑎10=2𝑎1+12𝑑=−2+12=10，故C错误；所以𝑆𝑛=(−1)×𝑛+𝑛(𝑛

−1)2×1=𝑛2−3𝑛2=12(𝑛−32)2−98．因为𝑛∈𝐍∗，所以当𝑛=1或𝑛=2时，𝑆𝑛取得最小值，故D正确．故选：ABD．第2页共5页10．BC【解析】因为𝐷(𝑋)=1225，所以𝐷(5𝑋+2)=25𝐷(𝑋)=12，故A错误；由𝑦̂=0.3−0.7�

�，得样本点(2，−3)的残差为−3−(0.3−0.7×2)=−1.9，故B正确；由𝑃(𝐵)=0.3，得𝑃(𝐵)=0.7，所以𝑃(𝐵|𝐴)=𝑃(𝐵)，即事件A与B独立，故C正确；根据𝜒2=3.712<3.841，故没有95%的把握认为X与Y有关，故D错误．故选：BC．1

1．ABD【解析】由𝑎1＝1，𝑎𝑛𝑎𝑛+1=2𝑛，得𝑎1𝑎2=2，则𝑎2=2，所以𝑎2𝑎3=4，所以𝑎3=2，故A正确；由𝑎𝑛𝑎𝑛+1=2𝑛，得𝑎𝑛+1𝑎𝑛+2=2𝑛+1，所以𝑎𝑛+1𝑎𝑛+2𝑎𝑛𝑎𝑛+

1=2𝑛+12𝑛=2，即𝑎𝑛+2𝑎𝑛=2，所以数列{𝑎𝑛}的奇数项和偶数项，均是以2为公比的等比数列，𝑎2𝑛=2×2𝑛−1=2𝑛，𝑎2𝑛−1=1×2𝑛−1=2𝑛−1，故B正确；𝑎2𝑛−1+𝑎2𝑛=2𝑛−1+2𝑛=3×2𝑛−1，�

�2𝑛−𝑎2𝑛−1=2𝑛−2𝑛−1=2𝑛−1，故C错误，D正确．故选：ABD．12．AD【解析】设事件𝐴=“向右下落”，则𝐴=“向左下落”，且𝑃(𝐴)=𝑃(𝐴)=12．因为小球最后落入格子的号码�

�等于事件𝐴发生的次数，而小球下落的过程中共碰撞小木钉10次，所以𝑋∼𝐵(10，12)，所以𝑃(𝑋=1)=C10112×(12)9=5512，𝑃(𝑋=9)=C109(12)9×12=5512，

故A正确，B错误；所以𝐷(𝑋)=10×12×(1−12)=52，故C错误，D正确．故选：AD．三、填空题（本大题共4小题，每小题5分，共20分）13．1【解析】当散点图的所有点都在一条斜率为非0实数的直线上时，它的残差为0，残差的平方和为0，所以它的相关系数为1．故答案为：1．14．4【解

析】由题意，得𝑆𝑘=−3[1−(−2)𝑘]1−(−2)=15，解得𝑘=4．故答案为：4．15．75【解析】设𝑏𝑛={1𝑎𝑛+1}，则𝑏3=1𝑎3+1=13，𝑏7=1𝑎7+1=12．又因为数列{1𝑎𝑛+1

}为等差数列，所以2𝑏5=𝑏3+𝑏7=13+12=56，即𝑏5=1𝑎5+1=512，解得𝑎5=75．故答案为：75．16．0.130.46【解析】设事件A为购买的零件是甲厂产品，事件B为购买的零件是乙厂产品，事件C为购买的零件是次品，则�

�(𝐴)=0.7，𝑃(𝐵)=0.3，𝑃(𝐶|𝐴)=0.1，𝑃(𝐶|𝐵)=0.2，所以𝑃(𝐶)=𝑃(𝐴)𝑃(𝐶|𝐴)+𝑃(𝐵)𝑃(𝐶|𝐵)=0.7×0.1+0.3×0.2=0.13．因为𝑃(𝐵𝐶)=0.3×0.2=0.06，所以𝑃(𝐵|𝐶)

=𝑃(𝐵𝐶)𝑃(𝐶)=0.060.13≈0.46．故答案为：0.13；0.46．四、解答题（本大题共6小题，共70分）规范解答评分细则17．（1）证明：因为𝑎𝑛+1=2𝑎𝑛+2𝑛+1，所以

𝑎𝑛+1−2𝑎𝑛=2𝑛+1，····················································2分所以𝑎𝑛+12𝑛+1−𝑎𝑛2𝑛=1，·····

······················································4分即𝑏𝑛+1−𝑏𝑛=1．···································

··························5分又因为𝑏1=𝑎12=2，·························································6分所以数列{𝑏𝑛}是首项为2、公差为1的等差数列．··············

···········7分未求出𝑏1扣1分；未说明首项、公差扣1分；第3页共5页（2）解：由（1）知𝑏𝑛=2+(𝑛−1)×1=𝑛+1，·····················8分所以𝑐𝑛=1(𝑛+1)(𝑛+2)=1𝑛+1−1𝑛+2，················

·························9分所以𝑆𝑛=12−13+13−14+⋯+1𝑛+1−1𝑛+2=12−1𝑛+2．···················10分18．解：（1）记事件𝐴1为“甲在第一轮比赛中胜出”，

𝐴2为“甲在第二轮比赛中胜出”，𝐵1为“乙在第一轮比赛中胜出”，𝐵2为“乙在第二轮比赛中胜出”，则𝐴1，𝐴2，𝐵1，𝐵2相互独立，且𝑃(𝐴1)=45，𝑃(𝐴2)=23，𝑃(𝐵1)=35，𝑃(𝐵2)=34．···················3分因为

在两轮比赛中均胜出视为赢得比赛，则𝐴1𝐴2为“甲赢得比赛”，𝐵1𝐵2为“乙赢得比赛”，所以𝑃(𝐴1𝐴2)=𝑃(𝐴1)𝑃(𝐴2)=45×23=815，···································4分𝑃(𝐵1𝐵2)=𝑃(𝐵1)

𝑃(𝐵2)=35×34=920．····································5分因为815>920，所以派甲参赛赢得比赛的概率更大．·······················6分（2）记事件𝐷为“甲

赢得比赛”，𝐸为“乙赢得比赛”，则“两人中至少有一人赢得比赛”=𝐷∪𝐸．由（2）知，𝑃(𝐷)=𝑃(𝐴1𝐴2)=815，𝑃(𝐸)=𝑃(𝐵1𝐵2)=920，所以𝑃(𝐷)=1−𝑃(𝐷)=1−815=715，····················

·················8分𝑃(𝐸)=1−𝑃(𝐸)=1−920=1120，·········································10分所以𝑃(𝐷∪𝐸)=1−𝑃(𝐷𝐸)=1−𝑃(𝐷)

𝑃(𝐸)=1−715×1120=223300．故两人中至少有一人赢得比赛的概率为223300．····························12分19．解：（1）由题意，可知𝑞≠1．···········································

·····1分因为𝑆6𝑆3=2627，所以𝑎1(1−𝑞6)1−𝑞÷𝑎1(1−𝑞3)1−𝑞=2627，················································3分即1−𝑞61−𝑞3=1+𝑞

3=2627，解得𝑞=−13，····································5分所以𝑎𝑛=−1×(−13)𝑛−1=−1(−3)𝑛−1．························

···········7分（2）由（1）知𝑎𝑛=−1(−3)𝑛−1，则𝑎𝑛2=19𝑛−1，······························8分所以𝑎𝑛2𝑎𝑛−12=19，且𝑎12=1，········

··············································10分所以数列{𝑎𝑛2}是以1为首项、公比为19的等比数列，·····················11分所以𝑎12+𝑎

22+⋯+𝑎𝑛2=1×(1−19𝑛)1−19=98−18×9𝑛−1．··························12分直接给出𝑏𝑛=𝑛+1的扣0.5分；或𝑆𝑛=𝑛2(𝑛+2)．未说明“

相互独立”扣0.5分．或𝑎𝑛=−(−13)𝑛−1；𝑎𝑛的通项公式未化简的扣0.5分；未说明首项、公比扣1分；或9𝑛−88×9𝑛−1．第4页共5页20．解：（1）正··············

·····················································2分（2）令𝑡=𝑥2，则552411979iiiitx，··············

······················3分552112805iiiiiiyytx，·····················································4分552111

11155iiiittx，······················································5分𝑦=34．··························································

··················6分所以5152215ˆ5iiiiitybyttt=2805−5×11×34979−5×112=935374=2.5，························7分𝑎̂=𝑦−𝑏̂𝑡=34−

2.5×11=6.5，············································8分所以𝑦关于𝑡的回归方程为𝑦̂=6.5+2.5𝑡，故𝑦关于𝑥的回归方程为𝑦̂=6.5+2.5𝑥2．··················

···················9分（3）由（2）可得𝑦̂=6.5+2.5𝑥2．令𝑥=6，则𝑦̂=6.5+2.5×62=96.5．·····································11分故预测2023年该公司

新能源汽车的年销售量为96.5万辆．·············12分21．解：（1）恰好取到3种颜色的球的概率𝑃=C21×C21×C21C63=25．···············3分（2）随机变量𝑋的可能

取值为4，5，6，7，8．·························4分当𝑋＝4时，取出2个红球和1个白球，则𝑃(𝑋=4)=C22×C21C63=110；·····················································5分当𝑋＝5时

，取出2个红球和1个黑球或1个红球和2个白球，则𝑃(𝑋=5)=C22×C21+C21×C22C63=15；···············································6分当𝑋＝6时，取出1个红球和1个白球和1个黑球，则𝑃(𝑋=6)

=C21×C21×C21C63=25；····················································7分当𝑋＝7时，取出1个红球和2个黑球或2个白球和1个黑球，则𝑃(𝑋=7)=C21×C22+C22×C21C63=15；

················································8分当𝑋＝8时，取出2个黑球和1个白球，则𝑃(𝑋=8)=C22×C21C63=110，···································

····················9分所以随机变量𝑋的分布列如下：·······································10分𝑋45678P110152515110所以𝐸(𝑋)=4×110+5×15+6×25+7×15+8×110=6．···

·········12分未给出随机变量X的所有取值的扣1分；未写出“当X＝4时，2个红球和1个白球”等文字情况的不扣分；未写出𝐸(𝑋)的计算过程的扣0.5分．第5页共5页22．（1）证明：因为𝑇𝑛为数列{𝑆𝑛}的前n项和，当𝑛=1时，𝑆1

+𝑇1=𝑆1+𝑆1=2𝑆1=2，则𝑆1=1；···················1分当𝑛≥2时，𝑇𝑛−𝑇𝑛−1=𝑆𝑛．𝑆𝑛+𝑇𝑛=2①，𝑆𝑛−1+𝑇𝑛−1=2②，···························

···········2分①－②，得2𝑆𝑛=𝑆𝑛−1(𝑛≥2)，则𝑆𝑛𝑆𝑛−1=12(𝑛≥2)，·····················3分所以数列{𝑆𝑛}是首项为1、公比为12的等比数列．·······················

····4分（2）解：由（1）可得，𝑆𝑛=(12)𝑛−1．当𝑛=1时，𝑎1=𝑆1=1；······················································5分当𝑛≥2时，𝑎𝑛=𝑆𝑛−𝑆𝑛−1=(12)𝑛−1−(12

)𝑛−2=−(12)𝑛−1，·········6分显然对于𝑛=1不成立，所以𝑎𝑛={1，𝑛=1，−(12)𝑛−1，𝑛≥2．············································7分当𝑛=1时，𝐴1=𝑎1=1；··········

··········································8分当𝑛≥2时，𝐴𝑛=1−[2×12+3×(12)2+⋯+𝑛(12)𝑛−1]③，则12𝐴𝑛=12−[2×(12)2+3×(12)3+⋯+𝑛⋅(12)𝑛]

④．·················9分由③−④，得12𝐴𝑛=12−[1+(12)2+(12)3+⋯+(12)𝑛−1−𝑛⋅(12)𝑛]=12−{1+14[1−(12)𝑛−2]1−12−𝑛⋅(12)𝑛}=(𝑛+2)⋅(

12)𝑛−1，·································10分所以𝐴𝑛=(𝑛+2)⋅(12)𝑛−1−2．···········································11分又因为𝑛=1时，𝐴1=3×1−2=1

．所以𝐴𝑛=(𝑛+2)⋅(12)𝑛−1−2．·············································12分未讨论𝑛=1的情况的扣1分；未说明首项、公比扣1分；未讨论𝑛=1的情况的扣1分；𝑎𝑛的

通项公式未归纳成分段形式的，扣0.5分；③−④的最后结果对，计算过程错误的扣0.5分；最后结果未化简的扣0.5分；未给出𝑛=1时𝐴1的值的扣1分．获得更多资源请扫码加入享学资源网微信公众号www.xi

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