1030

设函数 ${\displaystyle f(x)=x\ln x}$.

1. 求 ${\displaystyle f(x)}$ 图像上点 ${\displaystyle (1,f(1))}$ 处点切线方程；
2. 若 ${\displaystyle f(x)\ge a(x-\sqrt{x})}$ 在 ${\displaystyle x\in (0,+\mathrm{\infty })}$ 时恒成立，求 ${\displaystyle a}$ 的取值范围;
3. 若 ${\displaystyle x_1,x_2\in (0,1)}$, 证明: ${\displaystyle {\displaystyle |f(x_1)-f(x_2)|\le |x_1-x_2{|}^{\frac{1}{2}}}}$.

1. 解 容易求得答案 ${\displaystyle y=x-1}$.

2. 解 不等式两边消去 ${\displaystyle \sqrt{x}}$，再令 ${\displaystyle t=\sqrt{x}}$, 有 $$\begin{array}{r}g(t)=2t\ln t-a(t-1)\ge 0,\mathrm{\forall }t\in (0,+\mathrm{\infty }).\end{array}$$

   • 当 ${\displaystyle a=2}$, 则$$\begin{array}{r}g(t)=2(t\ln t-(t-1))\ge 2(t(1-\frac{1}{t})-(t-1))=0,\end{array}$$
     符合题意；
   • 当 ${\displaystyle a>2}$, 记 ${\displaystyle a^{\prime }={\mathrm{e}}^{\frac{a}{2}-1}>1}$, 则 ${\displaystyle g^{\prime }(t)=2(1+\ln t)-a}$ 在$ (1,a')$上小于 ${\displaystyle 0}$, 从而 ${\displaystyle g(a^{\prime })<g(1)=0}$, 不成立;
   • 当 ${\displaystyle a<2}$, 则 ${\displaystyle 0<a^{\prime }<1}$, ${\displaystyle g^{\prime }(t)}$ 在 ${\displaystyle (a^{\prime },1)}$ 上大于 ${\displaystyle 0}$, 从而 ${\displaystyle g(a^{\prime })<g(1)=0}$, 不成立.

   综上，${\displaystyle a=2}$.

3. 证明 不妨设 ${\displaystyle x_1\ge x_2}$.

   • 若 ${\displaystyle f(x_2)\ge f(x_1)}$, 依据不等式 ${\displaystyle \sqrt{x_1-x_2}\ge \sqrt{x_1}-\sqrt{x_2}}$ (将 ${\displaystyle \sqrt{x_2}}$ 左移后两边平方即证), 只需要证 $$\begin{array}{r}f(x_2)-f(x_1)\le \sqrt{x_1}-\sqrt{x_2}⟺x_2\ln x_2+\sqrt{x_2}\le x_1\ln x_1+\sqrt{x_1},\end{array}$$
     那只需要说明 ${\displaystyle u(x)=x\ln x+\sqrt{x}}$ 在 ${\displaystyle (0,1)}$ 上单增. 事实上 $$\begin{array}{r}{u}^{\prime }(x)=1+\ln x+\frac{1}{2\sqrt{x}},u^{″}(x)=\frac{4\sqrt{x}-1}{4x^{\frac{3}{2}}},\end{array}$$
     因此 ${\displaystyle u^{\prime }(x)}$ 有唯一极小值点 ${\displaystyle x={\displaystyle \frac{1}{16}}}$, 故 $$\begin{array}{r}{u}^{\prime }(x)\ge u^{\prime }\left(\frac{1}{16}\right)=3-4\ln 2>3-4\cdot \frac{1}{2}(2-\frac{1}{2})=0.\end{array}$$
   • 若 ${\displaystyle f(x_2)<f(x_1)}$, 依据不等式 ${\displaystyle \sqrt{x_1-x_2}\ge x_1-x_2}$, 只需要证 $$\begin{array}{r}f(x_1)-f(x_2)\le x_1-x_2⟺x_1\ln x_1-x_1\le x_2\ln x_2-x_2.\end{array}$$
     而 ${\displaystyle v(x)=x\ln x-x}$ 满足 ${\displaystyle v^{\prime }(x)=\ln x<0,\mathrm{\forall }0<x<1}$, 因此上式得证.