山东省青岛市黄岛区高三数学上学期期中试卷答案 山东省青岛市黄岛区高三数学上学期期中试卷(PDF) 山东省青岛市黄岛区高三数学上学期期中试卷(PDF)VIP免费

高三数学答案第1页共4页（数学是有生命的，题目是有经典的）2019-2020学年度第一学期期中学业水平检测高三数学参考答案一、单项选择题：本大题共10小题，每小题4分，共40分。110：CBACCADADB二、多项选择题：本大题共3小题，每小题4分，共12分。11.BCD;12.AD;13.BCD;三、填空题：本大题共4个小题，每小题4分，共16分。14.（1）60；（2）120,0；15.0；16.2c；17.3(31)2；四、解答题：共82分。解答应写出文字说明，证明过程或演算步骤。18.（12分）解：（1）在ABC中：ABEF//，所以32AFE························2分在AFE中由正弦定理知：21sinsinsinEAFEAFEFAFEAE··················5分又因为32AFE为钝角，所以6EAF··················································6分（2）因为32AFE，6EAF，所以6AEF，2EFAF·············8分又因为ABEF//，3AB，2EF，所以2AFCF，即6AC··························9分在ABC中由余弦定理知：2222cos27BCABACABACBAC······································11分33BC·····························································································12分19.解：（1）取AB中点F，连接DF，//DCAB且12DCAB//DCBF且DCBF所以四边形BCDF为平行四边形，又ABBC,1BCCD所以四边形BCDF为正方形···········································································2分在RtΔAFD中，因为1DFAF，所以2AD在RtΔBCD中，因为1BCCD，所以2BD因为2AB，所以222ADBDAB，BDAD···········································4分因为BD面ABCD，面PAD面ABCDAD，面PAD面ABCD所以BD面PAD·······················································································6分因为PA面PAD所以PABD·····························································································7分CBFADEPG高三数学答案第2页共4页（数学是有生命的，题目是有经典的）（2）线段AB上存在一点G，满足14AGAB即G为AF中点时，BC∥面PEG·································································9分证明如下：连结EG，E为AD的中点,G为AF中点,//GEDF又//DFBC，所以//GEBC，································································12分GE面PEG，BC面PEG，BC∥面PEG·······································14分20.（14分）解：（1）因为211lglg2lg2lglglgnnnnnnnnbaaabaaa·····························2分又因为11lg1ba，···················································································3分所以nb是首项为1公比2的等比数列······························································4分（2）由（1）得：12lgnnnab········...