数学参考答案与评分标准第1页，共6页高中2023级第一次诊断性考试数学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．1．C2．D3．B4．D5．A6．B7．A8．C二、选择题：本大题共3小题，每小题6分，共18分．全部选对的得6分，选对但不全的得部分分，有选错的得0分．9．BCD10．ACD11．AC三、填空题：本题共3个小题，每小题5分，共15分．12．1；13．112；14．(10)(01)，，四、解答题：本题共5小题，第15题13分，第16、17小题15分，第18、19小题17分，共77分．解答应写出文字说明、证明过程或演算步骤．15．解：（1） ()sin()fxx的最小正周期为，∴2，·······················································································2分又3()22f，即3sin()2，则3sin2，···························4分又()2，，则23，······························································5分∴2()sin(2)3fxx；····································································6分（2）由题知：2()()sin[2()]663gxfxxsin(2)3x，···········8分由[0]2，x，则42333x≤≤，··················································10分∴3sin(2)123x≤≤，································································12分故()gx的值域为3[1]2，．·····························································13分数学参考答案与评分标准第2页，共6页16．解：（1）当2m时，(2)2()|2|(2)2xxxfxxxxxx，≥，，，·····················3分故()fx的单调递增区间为：(2]，，[1)，，单调递减区间为：[−2，−1]；·····································································································6分（2）由题知x∈[1，2]，()2fx≤，即对任意x∈[1，2]，2xxm≤恒成立，············································7分∴2xmx≤，即22xmxx≤≤，则22xmxxx≤≤，∴22xmxxx≤≤，·····································································9分令2()gxxx，易知()gx在[12]，x单调递增，故max()(2)1gxg，∴1m≥，则1m≤，····································································11分令2()hxxx，易知()gx在[12]，x单调递减，在[22]，x单调递增，故min()(2)22hxh，·······························································13分∴22m≤，即22m≥，···························································14分综上：221m≤≤．··································································15分17．解：（1）因为：当2n≥时，122nnaan，所以：)2)](1([21nnanann，······...