\(g(x) = \dfrac{1}{2}ax^2+(1-a)x-lnx有两个零点，求a的取值范围.\)
\(g(x) = 0,即a(\dfrac{1}{2}x^2-x) = lnx-x\)
\(①a = 0时，即lnx-x=0,没有零点\)
\(②a\neq0时，即\dfrac{1}{a} = \dfrac{\dfrac{1}{2}x^2-x}{lnx-x}\)
\(令h(x) = \dfrac{\dfrac{1}{2}x^2-x}{lnx-x},h'(x) = \dfrac{(x-1)(lnx-\dfrac{1}{2}x-1)}{(lnx-x)^2}\)
\(令m(x) = lnx-\dfrac{1}{2}x-1,m'(x) = \dfrac{1}{x}-\dfrac{1}{2}\)
\(x\in(0,2)时，m'(x)>0,m(x)递增，当x>2时，m'(x)<0,m(x)递减\)
\(所以m(x)\le m(2) = ln2-2<0\)
\(因此x\in(0,1)时，h'(x)>0,h(x)递增，x\in(1,+\infty)时，h'(x)<0,h(x)递减\)
\(所以h(x)_{max} = h(1) = \dfrac{1}{2}\)
\(x\to 0,h(x)\to 0,x\to +\infty,h(x)\to -\infty\)
\(所以\dfrac{1}{a}\in(0,\dfrac{1}{2})\Rightarrow a\in(2,+\infty)\)
标签：infty,lnx,dfrac,递增,零点,2.27 From： https://www.cnblogs.com/yuanhongyi/p/17160418.html