高2024级数学试题答案一、单选题CDABCCBA二、多选题9.ABD10.ABD11.ACD三、填空题12.1,213.1,2,3,614.453四、解答题15.解：（1）由已知得1020xx，解得1x且2x.....................................................5分函数fx的定义域为1,22,....................................................6分（2）由已知0x，令1tx，则21xt，1t221112gxgtttt22,1,gxxxx....................................................13分注：没写x的范围扣1分16.解：（1）2|1030|520|52Axxxxxxxx.........................................3分若1m，则|21Bxx..........................................................................................................4分|21RCBxxx或..................................................................................................................5分|5212RACBxxx或...................................................................................7分（2）若ABA，则BA...................................................................................................8分当B时，有121mm，即2m...........................................................................................10分当B时，215212mmm，解得122m.....................................................................................13分实数m的取值集合是1,2............................................................................................................15分17.解：（1）设直线0xtt与x轴交于点,0Ct，与线段OB或BA交于点D由已知有1,3B，2,0A，3OABS当01t时，21133222OCDftSOCCDttt...........................................................2分{#{QQABYYQQgggAAABAAAgCQQ2ACgGQkgCAAagOhBAIIAABCAFABAA=}#}当12t时，211333232233222OABACDftSSACCDtttt........................................................................4分当2t时，3OABftS..................................................................................................6分223,0123233,1223,2ttyfttttt..............................................................................7分（2）由（1）得3,0123223,1223,2ttftgttttttt...............................................................8分当01t时，302gt............................................................................................................11分当12t时，22222tttt，当且仅当2tt，即2t时取等号此时max322362gtg..........................................................................................13分当2t时，332gtt..............................................................................................................14分当2t时，gt有最大值为236.....................................................................................15分18.解：（1）函数Gx的定义域为R，则不等式221110axax的解集为R当210a时，1a若1a，20010xx的解集为R若1a，210x的解集不是R..........................................................................................