参考答案
一、选择题（每小题5分,共60分） BCCB ACDC CBBA
二、填空题（每小题4分,共16分）
1x??13. 14. ??1,1??(1,4]; 15. y?54.8?1??;
5100??16. (??,0] [???.0?也给满分] 三、解答题:
131?loga2?loga2?0 2lg25?lg2?1?lg5?2 （2）lg20?log10025?lg2?1?2解：（1）loga2?loga1????3?3?632633?2?3?12????2?3?2?3?6 （3）23?12?22??61216132111118．证明：任取x1,x2?(0,1],且x1?x2， 则f?x1??f?x2????x1???1??1??x1?x2??x1x2?1??? ?x??2x???x1?xx2?12???0?x1?x2?1,?x1?x2?0,x1x2?1?0,x1x2?0 ?f?x1??f?x2??0,即f?x1??f?x2?
所以函数?f?x??x?1在区间(0,1]上是减函数。 x?60t,0?t?2.5,?19．解：y??150,2.5?t?3.5,
?150?50?t?3.5?,3.5?t?6.5??60t,0?t?2.5,?则y??150,2.5?t?3.5,
??50t?325,3.5?t?6.5?函数的图象如右
y 150 100 50 o 1 2 3 4 5 6 t
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20．；解：（1）?1?xx?1?0,??0,即?x?1??x?1??0. 1?xx?1??1，??1?x?1,?f?x?的定义域为1?
（2）证明：
1?x1?x?1?x??f?x??loga,?f??x??loga?loga??1?x1?x?1?x?奇函数.
（3）解:当a>1时, f?x?>0,则
?1??loga1?x??f?x??f?x?中为1?x1?x1?x2x?1，则?1?0,?0 1?xx?1x?1?2x?x?1??0,?0?x?1
因此当a>1时,使f?x??0的x的取值范围为（0,1）.
当0?a?1时, f?x??0,则0?1?x?1 1?x1?x
?1?0,1?x则 1?x
?0,1?x
解得?1?x?0
因此当0?a?1时, 使f?x??0的x的取值范围为（-1,0）. 21.证明:设函数使f?x??2x?3x?6.
?f?1???1?0,f?2??4?0
又?f?x?是增函数,所以函数f?x??2x?3x?6在区间[1,2]有唯一的零点, 则方程6?3x?2在区间[1,2]有唯一一个实数解.设该解为x0,则x0?[1,2] 取x1?1.5,f?1.5??0.33?0,f?1?f?1.5??0.?x0??1,1.5? 取x2?1.25,f?1.25??0.128?0,f?1?f?1.25??0.?x0??1,1.25?
取x3?1.125,f?1.125???0.44?0,f?1.125?f?1.25??0.?x0??1.125,1.25? 取x4?1.1875???0.16?0,f?1.1875?f?1.25??0.?x0??1.1875,f?1.1875,1.25?
x ~7~
?1.25?1.1875?0.0625?0.1
?可取x0?1.2则方程的实数解为x0?1.2
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