山东省青岛市高三数学第三次模拟(自主检测)试卷参考答案(PDF) 山东省青岛市高三数学第三次模拟试卷(PDF) 山东省青岛市高三数学第三次模拟试卷(PDF)VIP专享VIP免费

数学答案第1页（共6页）青岛市2020年高三自主检测数学参考答案及评分标准一、单项选择题：本题共8小题，每小题5分，共40分。BABCACBD二、多项选择题：本题共4小题，每小题5分，共20分。9．ABC10．BD11．AC12．ABD三、填空题：本题共4个小题，每小题5分，共20分。13．102m；14．01x；15．（1）1721；（2）1617(12)；16．233；四、解答题：本题共6小题，共70分。解答应写出文字说明，证明过程或演算步骤。17.（本小题满分10分）解：（1）由题知：在直角梯形12AOOC中，2221212()20ACAOCOOO······1分所以在三棱锥OABC中，222ACAOOC所以AOOC····························································································2分又因为AOOB，COOBO，所以AO平面BOC···················································································4分又因为AO平面AOB所以，平面AOB平面BOC········································································5分（2）由（1）知：AOOC，AOOB，又BOOC；以O为坐标原点，以OC，OB，OA的方向分别作为x轴，y轴，z轴的正方向，建立如图空间直角坐标系Oxyz，································································6分所以(0,0,4),(0,2,0),(2,0,0)ABC，(2,0,0)OC··········································7分设(,,)nxyz为平面ABC的法向量，(0,2,4)AB，(2,2,0)BC由00nABnBC可得：240220yzxy，令2x得：(2,2,1)n················································································9分设直线OC与平面ABC所成角为，所以||2sin=3||||OCnOCn，所以直线OC与平面ABC所成角的正弦值为23················································10分zyxOABC数学答案第2页（共6页）18．（本小题满分12分）解：（1）若选择条件①12a，则放在第一行的任何一列，满足条件的等差数列{}na都不存在······························1分若选择条件②11a，则放在第一行第二列，结合条件可知12314,7aaa，，································2分则*32,Nnann····················································································4分若选择条件③13a，则放在第一行的任何一列，满足条件的等差数列{}na都不存在综上可知：32nan··················································································5分（2）由（1）知，12(1)(32)nnbn所以当n为偶数时，123nnTbbbb22222212341nnaaaaaa1212343411()()()())()nnnnaaaaaaaaaaaa（1233()naaaa(132)32nn29322nn···························9分当n为奇数时，1nnnTTb2293(1)(1)(32)22nnn293222nn·····················11分2293,22932,22nnnnTnnn...