面ACE的距离），d?分
221221?，所以直线A1B与平面EAC之间的距离为
7 ……12720、（本小题满分12分）
?22?【解析】（1）设B22,y0，由题意得：???22??????222?y0?12 ……2分
?2py0解之得：??y0?2，所以抛物线的方程为x2?4y． ……4分
?p?2（2）设点P由题意知PP2?x2,y2?，P4?x4,y4?，P2，P4P3?x3,y3?，1?x1,y1?，1，P3在圆上，在抛物线上．因为直线l过点F且斜率为1，所以直线的方程为y?x?1． ……5分
?y?x?1112x?x??1,xx??2x?2x?11?0联立?2，得，所以 131322x?y?12??P1P3?1?12??x1?x3?2?4x1x3?2?11???1?2－4???????2?46 ……7分
?y?x?12同理：由?2，得x－4x?4?0，所以x2?x4?4,x2x4??4
?x?4y?P2P4?1?12??x2?x4?2?4x2x4?2?42－4???4??8 ……9分
由题意易知：P1P2?P1P3?P2P3……①，P3P4?P2P4?P2P3……②
①—②得：P1P2?P3P4?P1P3?P2P4 ……11分
46?8 ……12分 ?P1P2?P3P4?21、（本小题满分12分）
【解析】（1）根据题意可得，f?e??2， ……1分 ef??x??1?lnx?lne1???fe???k??，所以，即， ……3分 2222xeee21??2?x?e?，即x?e2y?3e?0． ……4ee所以在点?e,f?e??处的切线方程为y?数学试题（文科） 第 11页，共 13页
分
1ax2?1lnx?ax2?1??0在x?1恒成立， （2）根据题意可得，f?x???xxx令g?x??lnx?ax2?1，?x?1?，所以g??x????????1?2ax， ……5分 x当a?0时，g??x??0，所以函数y?g?x?在?1,???上是单调递增，所以g?x??g?1??0，
1ax2?1所以不等式f?x???成立，即a?0符合题意； ……7分
xx当a?0时，令
??1?2ax?0，解得x?x111，令?1，解得a?，
22a2a① 当0?a??11?1?上g??x??0，在上g??x??0，所1,时，?1，所以g??x?在??22a?2a????1?1????，???以函数y?g?x?在1,上单调递增，在????上单调递减， 2a2a??????1?2?111?1??g???ln?a???1???lna??a，令h?a???lna??a， ??a??aaa?a???111a2?a?10?a?h??a????2?1??0恒成立，又， 22aaa所以h?a??h????ln所以0?a??1??2?113?1??2??ln2??0，所以存在g???0， 222?a?1不符合题意； ……10分 2②当a?11时，?1g??x??0在?1,???上恒成立，所以函数y?g?x?在?1,???上是22a1不符合题意； 2单调递减，所以g?x??g?1??0，显然a?综上所述，a的取值范围为aa?0． ……12分 22．（本小题满分10分）
数学试题（文科） 第 12页，共 13页
??【解析】（1）曲线C1的普通方程为4x?3y?2?0; ···························································· 2分
曲线C2的直角坐标方程为:y?x2. ················································································· 5分 3?x?2?t,??5(t为参数)代入y?x2得 （2）C1的参数方程??y??2?4t.?5?9t2?80t?150?0, ············································································································ 6分
设t1,t2是A、B对应的参数,则t1?t2??8050············································· 7分 ，t1t2??0. ·
9311|PA|?|PB||t1?t2|8????. ······························································· 10分 |PA||PB||PA|?|PB||t1t2|1523．（本小题满分10分）
1?3x,x?,?2?1?【解析】（1）f(x)??2?x,?1?x?, ················································································· 2分
2???3x,x?1.??11???x?,??1?x?,?x??1,f(x)?9等价于? ························································ 3分 2或?2或??3x?0???3x?9??2?x?9综上,原不等式的解集为{x|x?3或x??3}. ··································································· 5分 （2）|x?a|?|x?a|?2|a|. ························································································ 7分
13由(Ⅰ)知f(x)?f()?.
22所以2|a|?3， ················································································································ 9分 233实数a的取值范围是[?,].························································································ 10分
44
数学试题（文科） 第 13页，共 13页