對于0<a<1,給出下列四個不等式:
①
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②

③

④

.其中成立的是( )
A.①③
B.①④
C.②③
D.②④
【答案】
分析:根據(jù)題意,∵0<a<1∴

>1∴

又∵y=log
ax此時在定義域上是減函數(shù),∴①log
a(1+a)<log
a(1+

)錯誤;②log
a(1+a)>log
a(1+

)正確;又∵y=a
x此時在定義域上是減函數(shù),∴③a
1+a<a
1
錯誤;④a
1+a>a

正確.
解答:解:∵0<a<1,∴a<
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,從而1+a<1+

.
∴l(xiāng)og
a(1+a)>log
a(1+

).
又∵0<a<1,∴a
1+a>a

.
故②與④成立.
點(diǎn)評:此題充分考查了不等式的性質(zhì),同時結(jié)合函數(shù)單調(diào)性對不等關(guān)系進(jìn)行了綜合判斷.
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