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贵州新高考协作体2023届上学期入学质量监测文科数学参考答案1—5：DDCDC6—10：BBCBA11—12：CD13、2;14、12;15、3216、6(1,]217.解析：（1）前三组的频率之和：1(0.10.20.40.2)0.1，前三组的频

率分别为0.025,0.025,0.05···························3分250.025350.025450.05550.1650.2750.4850.269.25x······6分（2）60,70的人数：2人，70,

80的人数：4人····························8分设事件A：至少一名学生的得分在60,70从6人中抽取2人，基本事件总数：15设事件A包含的基本事件个数：9所以93155PA························

····12分18.解析：（1）由已知条件：12nnSnan当2n时：12(1)1nnSnan两式相减得：1211nnnanana，即：1(1)1nnnana······

·························2分左右同除1nn得：111(1)nnaannnn11111nannn即：11111nnaannnn，且11111

a···························5分所以数列1nann是首项为1，公差为0的等差数列，即常数列11nann，1nan···························6

分（2）左边1111112334(1)212nnn························12分19.解析：（1）证明：取AD中点O，连接,OPOEEAED且O是AD中点，EOAD

面PAD面ABCD，交线是AD，EO面ABCDEO面PAD，PA面PADEOPA，又//EOBDPABD·····································

·······································3分且PAPD，BDPDDPA面PBD········································································5分（2）,//EOAD

BDEO，BDAD在RtABD中，42,4ABAD，4,22BDDE在RtPOE中，2POOE，22PE在PDE中，三边都等于22，所以其面积为23，·······································7分设

点C到平面PDE的距离为h在三棱锥PCDE中，42CD,//DEABABCDDECD，22DECDE的面积1422282S···················································9分PO底面ABCD，锥体

的高2PO················································10分由等体积法：11238233h，解得：833h所以点C到平面PDE的距离为833····································

·················12分20.解析：（1）12||||2||||QFQAaQFQA2||||4QAQF···························2分225||||||||2QAQFFA，2

55||||22QAQF，当2,,AFQ三点共线时，“”成立，1554||||422QFQA，所以1||||QFQA的最大值是542···················5分（2）由题意：直线l的斜率存

在，设1122:1,(,),(,)lykxMxyNxy由22(1)143ykxxy得：2222(43)84120kxkxk2122843kxxk，212241243kxxk························

···················6分1111xPMMFx···········································7分2121xPNNFx·······················

···················8分22122212122282243228412114343kxxkkkxxxxkk28233···············

··12分21解析：（1）fx的定义域为0,，2ln'xfxx令'0fx，则1x································2分当01x，'0fx，fx单调递增当

1x，'0fx，fx单调递减································5分（2）由题意：lngxmxx有两个零点12,xx即：lnxxm有两个实数根12,xx令lnhxxx，则'1lnhxx由'

0hx得：1xe；由'0hx得：10xe·······························6分不妨设：21,xx，则21110xxehx在点22,2ee处的切线方程为2yxe设直线2yxe与直线ym的交点横坐标为3x，且31(

0,)xe，23xme，·······················8分以下证明：31xx因为hx在1(0,)e上递减，只需证310hxhxhx在点1,0处的切线方程为1yx设直线1yx与直线ym的交点横坐标为4x，且41(,1)xe，41xm

，·······················10分以下证明：42xx综上，221431xxxxmem，即22112xxem··············12分22解析：（1）曲线1:cos2C········

···············2分曲线2:2sinC·······················4分由题意：点P为23,CC的交点，联立2sin4得：(2,)4P，直角坐标(1,1)·················

································································5分直线43:4CR点M为41,CC的交点，联立cos234得：3(22,)4M···················

6分点N为42,CC的交点，联立2sin34得：3(2,)4N···················7分所以，||2222MN···················8分点(1,1)P到直线4:0Cxy的距离：2

22d···················9分PMN的面积12212S，所以PMN的面积为1···················10分23．解析：（1）求解不等式3|1||1|1fxxx①13111x

xx解得：1x····································1分②113111xxx解得：114x····································2分③13111xxx

解得：52x····································3分综上，15()(,)42x·······························

·················5分（2）23,43,123,1xaxafxxaaxxax·····································7分当1ax时，令0fx得：34ax

当1x时，令0fx得：32ax当1x，11fafx与x轴围成的三角形的面积2311333|||1|22482aaaSa解得：1a或3a（舍）综上所述：1a·····································

10分