东北三省精准教学2024年12月高三联考 数学-简版答案及考点细目表.docx

东北三省精准教学2024年12月高三联考数学41AKABAC所以93，·································································································11分参考答案439ADABACADAK又77，所以7·················································································12分，1234567891011所以AD过点K，即AD，BE和CF三线交于一点K.································································13分（备注：按考点给分，漏步骤相应扣分；可使用建系方法解题，过程酌情给分）CABDBBCCABCBCDACD22216．【答案】（1）证明见解析（6分）（2）（9分）1112.21(5分）【解析】（1）由勾股定理，AD2＝AC2﹣CD2=4，··········································································1分13.5/6/7（写出一个即可得满分5分）满足PA2+PD2＝AD2，所以PA⊥PD．························································································2分14.（0,1）（5分）因为平面PAD⊥平面ABCD，平面PAD∩平面ABCD＝AD，CD⊂平面ABCD，CD⊥AD，所以CD⊥平面PAD，············································································································4分15．【答案】（1）a7，b5，c6（6分）（2）证明见解析（7分）又PA⊂平面PAD，所以PA⊥CD．····························································································5分【解析】（1）因为△ABC的周长为18，所以abc18，···········································1分因为PD，CD⊂平面PCD，且PD∩CD＝D，所以PA⊥平面PCD.···················································6分由于b，c，a是递增的等差数列，故ab2c，·························································2分（备注：证明PA⊥PD得2分，证明PA⊥CD得3分，证明PA⊥平面PCD得1分）所以c6，ab12(ab)①，········································································3分222（）方法一：取的中点，作交于，连接，则⊥，1bca2ADOOM//CDBCMOPOPAD又cosA②，··········································································4分52bc因为平面PAD⊥平面ABCD，平面PAD∩平面ABCD＝AD，OP⊂平面PAD，所以OP⊥平面ABCD，以O为原点，由①②，解得a7，b5，c6··············································································6分.OA，OM，OP所在直线分别为x轴，y轴，z轴建立空间直角坐标系Oxyz，······································7分（备注：推导出①得分，否则按考点给分；推导出②得分；联立①②得出得分）31a,b,c2151所以A1,0,0,P0,0,1,C1,5,0,E,,，设Ba,b,0，a＞0，b＞0，·················8分222（2）由题意可得BDAE3,CDAF4,BFCE2，23151所以，，分AFABAEAC······························································································7则EBa,b,，ABa1,b,0，CBa1,b5,0，CP1,5,1.······9分35222设和交于点，由，，三点共线，得3，分BECFKBKEAKt1AB1t1AEt1AB1t1AC··········85易知平面PAD的一个法向量为e0,1,0，因为BE//平面PAD，2由C，K，F三点共线，得·······································9分AKt2AF1t2ACt2AB1t2AC53，所以EBe0，解得b.·························································································10分224tt，t，132193355所以解得···················································································10分因为AB⊥BC，所以，解得a或a（舍），即CB,,0.···········11分32ABCB01t1t，t，222251223设平面PBC的一个法向量为mx,y,z，第1页共4页{#{QQABbYSEggCgABBAAAgCUwEQCgEQkgACCagOwBAIoAAAyRFABAA=}#}x5yz0,5m·CP0,因为⊥，所以⊥，且＝＝．分则令，得，可得，分ADCDBTCDCTDT···································································1255x1y5,z4m1,5,4··························122m·CB0,xy0,2230由CT2+BT2＝BC2，可知BC，2易知平面ABC的一个法向量n0,0,1，··············································································13分QNCT6由△BTC∽△BNQ得，所以QN，····································································13分mn222BQBC3则cosm,n，····························································································14分mn1111EQ6EQPO，因此tanENQ，22222QN4因为二面角PBCA为锐二面角，所以二面角PBCA的余弦值为.······························15分11222（备注：建系给分，写出点的坐标给分，计算出平面的法向量给分，求出二面角的余弦值给分，cosENQ··········································································································14分11PBC4311，过程酌情给分）222所以二面角P﹣BC﹣A的余弦值为．·············································································15分方法二：取AD的中点O，连接OP和OC，再取OC的中点Q，连接QE，11（备注：分析出∠ENQ为二面角给3分，分析出TB∥AD给2分，计算出∠ENQ的余弦值给3分，过程酌情给在平面ABCD内过点Q作BC的垂线，垂足为点N，连接EN，分）因为＝，且是的中点，所以⊥．PAPDOADPOAD5ln217．【答案】（1）m，n（6分）（2）14小时（9分）又平面PAD⊥平面ABCD，平面PAD∩平面ABCD＝AD，PO⊂平面PAD，26所以PO⊥平面ABCD．···········································································································7分【解析】（1）由正午12点的城市活力度为20，知M(12)20，···················································1分因为EQ是△POC的中位线，则EQ∥PO，5代入数据得12m56m20，解得m，··········································································3分所以EQ⊥平面ABCD．224点到次日早上6点期间的城市活力度均为工作日内活力度的最低值，因为BC⊂平面ABCD，所以BC⊥EQ．·······················································································8分ln2因为⊥，，⊂平面，且＝，故M(24)M(6)5，代入数据得520e12n，解得n.······················································6分BCQNQNEQENQQN∩EQQ6所以BC⊥平面ENQ．又EN⊂平面ENQ，所以BC⊥EN，5t10,6„t„12,2由二面角的平面角的定义可知，∠即为二面角﹣﹣的平面角．分（2）由（1）知MtENQPBCA·······································9ln2(t12)20e6,12t24.连接BQ，并延长BQ交CD于点T．„由EQ∥PO，PO⊂平面PAD，EQ⊄平面PAD，所以EQ∥平面PAD．···········································10分当6t12时，令M(t)10，解得t[6,8]，···········································································8分ln2当EB∥平面PAD时，EQ，EB⊂平面EBQ，且EQ∩EB＝E，(t12)6当12t