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贵州新高考协作体2023届上学期入学质量监测理科数学参考答案1—5：DDCDB6—10：BBCBA11—12：CD13、2;14、12;15、3216、6(1,]217.解析：（1）每名同学得分低于70分的概率：21(0.040.02)105

，不低于80分的概率：10.02105，·····························2分122145525C······························5分（2）60,70的人数：2人，70,90的人数：6人···

·························7分X可取1，2，32126386156CCPXC12263830256CCPXC363820356CPXC··························10分

X123P65630562056495660606)(XE···························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11111a···

·······················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P

AD，PA面PADEOPA，又//EOBDPABD······································································

····3分且PAPD，BDPDDPA面PBD······································································5分（2）PO底面ABCDPB在底面ABCD内

的射影是OB，PBO是线面角，连接OB，在RtPBO中，1tan,22POPBOPOADOB即525OB，25OB·······························7分,//EOADBDEO，BDAD在RtOBD中，22(25)24BD··

·····························8分分别以,DADB为,xy轴，过点D作OP的平行线为z轴，建立空间直角坐标系：则：(2,0,2),(4,0,0)(0,0,0)(4,4,0)(2,2,0)PADCE平面DAP的法向量(0

,1,0)m平面PCE的法向量(1,3,3)n···················10分319cos,19||||mnmnmn所以锐二面角的余弦值为31919···········································

·12分20.解析：（1）12||||2||||QFQAaQFQA2||||4QAQF······················2分225||||||||2QAQFFA，255||||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,(,),(,)lykxMxyNx

y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······························4分2ffe

f，ln12111ln21ln2224e，cba······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：10

xe·····························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：222d···················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13111xxx

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）23,43,123,1x

axafxxaaxxax·····································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