隐零点与类偏移，第三问太难想不出来，不知道考哪里

\(f(x)=x^2+\dfrac{1-\ln x}{x}-a\)有两个零点\(x_1,x_2\)且\(x_1<x_2\)

(1)求\(a\)的取值范围

(2)证\(x_1x_2<1\)

(3)证\(x_2-x_1<\sqrt{a^2-4}<x_2^2-x_1^2\)

解

(1)\(x^2+\dfrac{1-\ln x}{x}-a=0\)

即\(F(x)=x^3+1-\ln x-ax=0\)

\(F^{\prime}(x)=3x^2-a-\dfrac{1}{x}\)是单调递增

因\(x\to 0,F^{\prime}(x)\to -\infty,x\to -\infty,F^{\prime}(x)\to +\infty\)

则存在唯一的\(x_0\)使\(F^{\prime}(x_0)=0\)

并且\(x=x_0\)是\(F(x)\)$的极小值点

从而要使得其有两个零点，则要有\(F(x_0)<0\)

因\(F^{\prime}(x_0)=3x_0^2-a-\dfrac{1}{x_0}=0\)

得\(a=3x_0^2-\dfrac{1}{x_0}\)

则\(F(x_0)=x_0^3-x_0\left(3x_0^2-\dfrac{1}{x_0}\right)-\ln x_0+1=-2x_0^3+2-\ln x_0<0\)

因\(-2x_0^3+2-\ln x_0\)单调递减

从而\(F(x_0)<0\)解得\(x_0>1\)

因\(a=3x_0^2-\dfrac{1}{x^2_0}\)

从而\(a>3-1=2\)

(2) 由(1)得\(0<x_1<1<x_0<x_2\)

则要证\(x_1x_2<1\)

即证\(x_1<\dfrac{1}{x_2}\)

即证\(0=F(x_1)<F\left(\dfrac{1}{x_2}\right)\)

令\(\dfrac{1}{x_2}=t\in(0,1),F(t)=t^3-at-\ln t+1<t^3-2t-\ln t+1\)

记\(H(t)=t^3-2t-\ln t+1,H^{\prime}(t)=3t^2-2-\dfrac{1}{t}<3-2-1=0\)

从而\(H(t)<H(1)=1-2+0+1=0\)

得证

(3) 想不出来，今后有水平了再回来订正