题目内容
(12分) 已知函数
=loga
(a>0且a≠1)是奇函数
(1)求
,(
(2)讨论
在(1,+∞)上的单调性,并予以证明
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(1)求
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(2)讨论
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(1)
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(2)当a>1时,f(x)=loga
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解:
(1)
(2)设u=
,任取x2>x1>1,则
u2-u1=
=
=
.
∵x1>1,x2>1,∴x1-1>0,x2-1>0.
又∵x1<x2,∴x1-x2<0.
∴
<0,即u2<u1.
当a>1时,y=logax是增函数,∴logau2<logau1,
即f(x2)<f(x1);
当0<a<1时,y=logax是减函数,∴logau2>logau1,
即f(x2)>f(x1).
综上可知,当a>1时,f(x)=loga
在(1,+∞)上为减函数;当0<a<1时,f(x)=loga
在(1,+∞)上为增函数.
(1)
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(2)设u=
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u2-u1=
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=
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=
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∵x1>1,x2>1,∴x1-1>0,x2-1>0.
又∵x1<x2,∴x1-x2<0.
∴
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当a>1时,y=logax是增函数,∴logau2<logau1,
即f(x2)<f(x1);
当0<a<1时,y=logax是减函数,∴logau2>logau1,
即f(x2)>f(x1).
综上可知,当a>1时,f(x)=loga

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