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[函数] 一道方程实根个数类问题

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aishuxue Posted 2015-5-14 21:35 |Read mode
Last edited by hbghlyj 2025-4-7 01:35若函数 $f(x)=-\ln x+a x^2+b x-a-2 b$ 有两个极值点 $x_1, x_2$,其中 $-\frac{1}{2}<a<0, b>0$,且 $f(x_2)=x_2>x_1$,则方程 $2 a[f(x)]^2+b f(x)-1=0$ 的实根个数为

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zqr1815 Posted 2015-6-1 00:09
回复 1# aishuxue

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