数学试题答案第1页（共4页）高三年级9月学习质量综合评估数学试题参考答案及评分标准一、选择题：本题共8小题，每小题5分，共40分．在每小题给出的四个选项中，只有一项是符合题目要求的。题号12345678答案BBDCBAAD二、选择题：本题共3小题，每小题6分，共18分。在每小题给出的四个选项中，有多项符合题目要求。全部选对的得6分，部分选对的得部分分，有选错的得0分。题号91011答案ABDBCDBCD三、填空题：本题共3小题，每小题5分，共15分。12．25；13．2；14．8．四、解答题：共77分.解答应写出文字说明、证明过程或演算步骤。15．【解析】（1）()fx为偶函数．···············································································2分证明：()fx的定义域为()，，关于原点对称；()cos()()sin()cossin()fxxxxxxxfx，所以()fx为偶函数．················································································5分（2）因为()-sin+sin+coscosfxxxxxxx，··············································7分因为2x，，所以cos0x≤，································································8分所以()fx在2，上单调递减，所以'()cos0fxxx＜．·············································································9分因为()0()=1022ff＞，＜，······························································11分由零点存在定理可得()fx在2，有且仅有一个零点．·······························13分16．【解析】（1）取AC中点为M，连接11BMAMAB，，，由ABC△和1AAC△为正三角形可知MBAC，1MAAC，···························2分数学试题答案第2页（共4页）又1MBMA，平面1MBA，且1MBMAM.故1ACMBA平面.···················································································4分又1AB平面1MBA，故1ACBA．····························································6分（2）由题意可知1AABC为正四面体，所以点1A在平面ABC上的射影ABC的中心，记该点为H，·······························································································8分所以1AMH为二面角1AACB的平面角，记为.······································9分在ABC△中，解得33AH，所以221163AHAAAH．·······························11分又因为1326HMHA，则1tan22AHHM．所以平面1AAC与平面ABC夹角的正切值为22.········································15分17．【解析】（1）由题意：24b，所以2b．因为53cea，所以35ac，，所以E的方程为194xy22．·····································································4分（2）证明：由（1）可知(30)A，，(02)B，．直线PA的方程为00(3)3yyxx，令0x，得00303yMx...