在△AFC与△GFA中，因为?FAG??FCA，?AFG??CFA，
所以△AFG∽△CFA， ······································································ 7分 所以
FAFG，即FA2?FG?FC．………………………………………………………9分 ?FCFA113AB，FG?GC，FC?GC， 222因为FA?所以
13AB2?GC2，即AB?3GC， 44又GC?1，所以AB?3． ········································································ 10分 解法二：（Ⅰ）同解法一． ······································································· 5分 (Ⅱ) 由(Ⅰ) 知，?BAD??ACG，
因为D,C,E,G四点共圆，所以?ADB??CEG， ·········································· 6分 所以△ABD∽△CGE，所以
ABAD， ……………………………………………7分 ?CGCE由割线定理，AG?AD?AE?AC， ······························································ 9分 又因为AD,BE是△ABC的中线，所以G是△ABC的重心， 所以AG?所以
2AD,又AC=2AE=2EC， 32ADAD2=2EC2，所以?3， 3CEAB···································· 10分 ?3，因为CG?1，所以AB?3． ·
CG所以
（23）选修4?4；坐标系与参数方程
本小题考查直线的极坐标方程和参数方程、椭圆的参数方程等基础知识，考查运算求解能力，考查数形结合思想、化归与转化思想等． 满分10分．
?x?3cos?,x2?y2?1， 解法一：（Ⅰ）由?消去参数?，得9?y?sin?x2?y2?1． ·即C的普通方程为····························································· 2分 9由?sin???????··················· 3分 ??2，得?sin???cos??2，………（＊） ·
4?将??x??cos?,代入（＊），化简得y?x?2， ·········································· 4分
?y??sin??． ····································································· 5分 4所以直线l的倾斜角为
??x?tcos,??4 （Ⅱ）由（Ⅰ）知，点P?0,2?在直线l上， 可设直线l的参数方程为?（t为参数），
??y?2?tsin??4?2t,?x??2即?（t为参数）， ····························································· 7分 ?y?2?2t??2x2?y2?1并化简，得5t2?182t?27?0． ·代入································· 8分 9??182??2?4?5?27?108?0．
设A,B两点对应的参数分别为t1,t2， 则t1?t2??1827·························· 9分 2?0,t1t2??0，所以t1?0,t2?0, ·
5518··································· 10分 2． ·
5所以PA?PB?t1?t2???t1?t2??解法二：（Ⅰ）同解法一. ············································································ 5分 （Ⅱ）直线l的普通方程为y?x?2.
由??y?x?2,22?x?9y?9消去y得10x2?36x?27?0， ······································ 7分
于是??362?4?10?27?216?0. 设A(x1,y1),B(x2,y2)，则x1?x2??1827?0,x1x2??0，所以x1?0,x2?0， 510 ····································································································· 8分 故PA?PB?1?1|x1?0|?1?1|x2?0|?
（24）选修4?5：不等式选讲
本小题考查绝对值不等式的解法与性质、不等式的证明等基础知识，考查运算求解能力、推理论证能力，考查分类与整合思想、化归与转化思想等． 满分10分．
解法一：（Ⅰ）（ⅰ） 当x≤?1时，原不等式可化为?x?1??2x?2，解得x??1，
此时原不等式的解是x??1； ······························································· 2分 （ⅱ）当?1?x??222|x1?x2|?182. ······ 10分 51时，原不等式可化为x?1??2x?2，解得x??1， 2此时原不等式无解； ··········································································· 3分 （ⅲ）当x≥?1时，原不等式可化为x?1?2x，解得x?1， 2此时原不等式的解是x?1； ································································· 4分 综上，M?xx??1或x?1． ··························································· 5分 （Ⅱ）因为f?ab??ab?1??ab?b???1?b? ······································ 6分
≥??········································· 7分 ab?b?1?b ·
····································· 8分 ?ba?1?1?b． ·
因为a,b?M，所以b?1，a?1?0， ··············································· 9分 所以f?ab??a?1?1?b，即f?ab??f?a??f??b?． ······················ 10分 解法二：（Ⅰ）同解法一．
（Ⅱ）因为f?a??f??b??a?1??b?1≤a?1???b?1??a?b， ······· 7分 所以，要证f?ab??f?a??f??b?，只需证ab?1?a?b，
即证ab?1?a?b， ······································································ 8分 即证a2b2?2ab?1?a2?2ab?b2，
即证a2b2?a2?b2?1?0，即证a2?1b2?1?0． ····························· 9分 因为a,b?M，所以a?1,b?1，所以a2?1b2?1?0成立，
所以原不等式成立． ········································································· 10分
2222????????高考模拟数学试卷
时间：120分钟 满分：100分
一、选择题（本大题10小题，每小题4分，共40分，在每小题给出的四个选项中，只有一项符合题目要求）.
1．一个年级有12个班，每个班有学生50名，并从1至50排学号，为了交流学习经验，要求每班学号为14的同学留下进行交流，这里运用的是
A．分层抽样
B．抽签抽样
（ ）
D．系统抽样
（ ）
C．随机抽样
2．设全集U?R，集合A?xx?2,B?x0?x?5，则集合AIB= A．x0?x
??????B. x0?x?2 C．x0?x?2 D．x2?x?5
（ ）
??????３．已知a、b是两条异面直线，直线c∥a，那么c与b的位置关系
A．一定是异面 4．已知函数f(x)??
B．一定是相交
C．不可能平行 D．不可能相交
（ ）
??log3x,x?0，则x??2, x?0
B．4
?f???1??f???= ?9??
C．2
D．
A．
1 4
2 2（ ）
5．已知倾斜角为θ的直线，与直线x-3y?1?0垂直，则tan?=
A．
1 B．3 3 C．?3 D．?
1 3
（ ）
6．设M=2a?a?2?,N??a?1??a?3?,则有
A．M>N
B． M≥N
C．M （ ） 7．在△ABC中，A︰B︰C＝1︰2︰3，则a︰b︰c等于 A．1︰2︰3 B．3︰2︰1 C．1︰3︰2 D．2︰3︰1 （ ） ?x??1?2, x?0 ,8．已知函数f(x)??x，则该函数是 ??2?1 , x?0 ,A．偶函数，且单调递增 B．奇函数，且单调递增 C．偶函数，且单调递减 D．奇函数，且单调递减 uuuruuuruuur9．如图，在△ABC中，已知BD?2DC，则AD= （ ） r3uuur1uuuA．?AB?AC 22 r2uuur1uuuB．AB?AC 33r2uuur1uuuD．AB?AC 33r3uuur1uuuC．AB?AC 22 （第９题图） 10．已知函数f(x)?2sin??x??????0,0????? 的图象上相邻两个最高点的距离为?．若将函数 f(x)的图象向左平移 ?个单位长度后，所得图象关于y轴对称．则函数f(x)的解析式为 6