摘要:(1)当a≥1时.f(x)=a?x+ln(1?x), f′(x)=?1+<0, f(x)是减函数.无极值,----------------------------------
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设a≥0,f (x)=x-1-ln2 x+2a ln x(x>0).
(Ⅰ)令F(x)=xf'(x),讨论F(x)在(0.+
)内的单调性并求极值;
(Ⅱ)求证:当x>1时,恒有x>ln2x-2a ln x+1.
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