每日导数40

出得比较臭的题

若函数\(f(x)\)在定义域内存在两个不同的数\(x_1,x_2\)满足\(f(x_1)=f(x_2)\)且\(f(x)\)在点\(\left(x_1,f(x_1)\right),(x_2,f(x_2))\)处的切线相同，则称\(f(x)\)是切合函数

\((1)\) 证明:\(f(x)=2x^3-6x\)为切合函数

\((2)\) 若\(g(x)=x\ln x-\dfrac{1}{e}x^2+ax\)为切合函数，并设满足条件的两个数为\(x_1,x_2\)

①求证：\(x_1x_2<\dfrac{e^2}{4}\)
②求证 ：\((a+1)^2x_1x_2-\sqrt{x_1x_2}<\dfrac{3}{4}\)

解

\((1)\) 由题\(\begin{cases}2x_1^3-6x_1=2x_2^3-6x_2\\ \\6x_1^2-6=6x_2^2-6\end{cases}\)解得\(\begin{cases}x_1=-\sqrt3\\ \\x_2=\sqrt3\end{cases}\)
\((2)\)

① 已知\(g(x)\)是切合函数，\(g^{\prime}(x)=\ln x+1-\dfrac{2}{e}x+a\)，有

\(\begin{cases}x_1\ln x_1-\dfrac{1}{e}x_1^2+ax_1=x_2\ln x_2-\dfrac{1}{e}x_2^2+ax_2 \\ \\ \ln x_1+1-\dfrac{2}{e}x_1+a=\ln x_2+1-\dfrac{2}{e}x_2+a \end{cases}\Rightarrow\begin{cases}x_1\ln x_1-\dfrac{1}{e}x_1^2+ax_1=x_2\ln x_2-\dfrac{1}{e}x_2^2+ax_2 \\ \\ \ln x_1-\dfrac{2}{e}x_1=\ln x_2-\dfrac{2}{e}x_2 \end{cases}\)

从而\(\dfrac{x_1-x_2}{\ln x_1-\ln x_2}=\dfrac{e}{2}\)，从而有对数均值不等式:

\[\dfrac{a+b}{2}\geq \dfrac{a-b}{\ln a-\ln b}\geq \sqrt{ab} \]

则\(\sqrt{x_1x_2}<\dfrac{x_1-x_2}{\ln x_1-\ln x_2}=\dfrac{e}{2}\)，即\(x_1x_2<\dfrac{e^2}{4}\)

② 由式(1)得到\(a=\dfrac{x_1\ln x_1-x_2\ln x_2}{x_2-x_1}+\dfrac{1}{e}(x_2+x_1)\)

由式(2)得到\(\dfrac{1}{e}=\dfrac{\ln x_1-\ln x_2}{2(x_1-x_2)}\)

联立得到\(a=\dfrac{x_1\ln x_1-x_2\ln x_2}{x_2-x_1}+\dfrac{\ln x_1-\ln x_2}{2(x_1-x_2)}(x_2+x_1)=\dfrac{2x_1\ln x_1-2x_2\ln x_2-x_2\ln _1-x_1\ln x_1+x_2\ln x_2+x_1\ln x_2}{2(x_2-x_1)}\)
\(=\dfrac{x_1\ln x_1-x_2\ln x_2-x_2\ln x_1+x_1\ln x_2}{(2x_2-x_1)}=\dfrac{x_1\ln x_1x_2-x_2\ln x_1x_2}{2(x_1-x_2)}=-\dfrac{\ln x_1x_2}{2}\)

则\(x_1x_2=e^{-2a}<\dfrac{e}{2}\)即\(a>\ln\dfrac{2}{e}\)，代回要证不等式，整理得
\(\dfrac{3}{4}e^{2a}+e^a-(a+1)^2>0\)

记\(h(a)=\dfrac{3}{4}e^{2a}+e^a-(a+1)^2,h^{\prime}(a)=\dfrac{3}{2}e^{2a}+e^a-2(a+1),h^{\prime\prime}(a)=3e^{2a}+e^a-2=(3e^a-2)(e^a+1)\)

则\(h^{\prime}(a)>h^{\prime}\left(\ln\dfrac{2}{3}\right)=2\left(\dfrac{2}{3}-\ln\dfrac{2}{3}-1\right)>0\) \(\left(\ln\dfrac{2}{e}>\ln\dfrac{2}{3}\right)\)

从而\(h(a)>h\left(\ln\dfrac{2}{e}\right)>h\left(\ln\dfrac{2}{3}\right)=\ln\dfrac{2}{3}\left(-\ln\dfrac{2}{3}-2\right)>0\)