每日导数90

函数\(f(x)=\dfrac{e^x}{x}-a\)的图像与\(x\)轴的两交点为\(A(x_1,0),B(x_2,0)(x_2>x_1)\)

(1)令\(h(x)=f(x)-\ln x+x\)，若\(h(x)\)有两个零点，求\(a\)的取值范围

(2)证明:\(x_1x_2<1\)

(3)证明:当\(a\geq 5\)时，以\(AB\)为直径的圆与直线\(y=\dfrac{\sqrt{3}}{4}(x+1)\)恒有公共点

(参考数据:$e^{0.25}\approx 1.3,e^{2.5}\approx 12.2 $)

解

(1)\(h(x)=\dfrac{e^x}{x}-a-\ln x+\ln e^x=\dfrac{e^x}{x}+\ln\dfrac{e^x}{x}-a\xlongequal[]{\frac{e^x}{x}=t}t+\ln t-a,t\in [e,+\infty)\)

因\(t+\ln t-a\)单调递增，则要使得\(h(x)\)有两个零点

则要有\(t+\ln t-a\)有一个零点即可

则要有\(e+\ln e-a<0\)即\(a>e+1\)

(2)由题\(\begin{cases} \dfrac{e^{x_1}}{x_1}=a\\ \dfrac{e^{x_2}}{x_2}=a \end{cases}\)即$ \dfrac{e^{x_1}}{x_1}= \dfrac{e^{x_2}}{x_2}$

即\(x_1-\ln x_1=x_2-\ln x_2\)

即\(\dfrac{x_1-x_2}{\ln x_1-\ln x_2}=1\)

由对数均值不等式:

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

有\(1=\dfrac{x_1-x_2}{\ln x_1-\ln x_2}>\sqrt{x_1x_2}\)

即\(x_1x_2<1\)

如果不想使用对数均值不等式，可如下操作:

设\(\dfrac{x_2}{x_1}=t,t>1\)

则有\(x_2=tx_1\)

\(\begin{cases} \dfrac{e^{x_1}}{x_1}=a\\ \dfrac{e^{x_2}}{x_2}=a \end{cases}\)，两式做比\(\dfrac{x_2}{x_1}e^{x_1-x_2}=1\)

即\(te^{x_1(1-t)}=1\)，即\(\ln t+x_1(1-t)=0\)得\(x_1=\dfrac{\ln t}{t-1}\)，则\(x_2=\dfrac{t\ln t}{t-1}\)

从而\(x_1x_2=\dfrac{t\ln^2t}{(t-1)^2}\)，记\(\varphi(t)=\dfrac{t\ln^2t}{(t-1)^2}\)

\[\varphi^{\prime}(t)=\dfrac{(t-1)\ln t\left[-(t+1)\ln t+2t-2\right]}{(t-1)^4} \]

\[=\dfrac{(t-1)(t+1)\ln t\left[-\ln t+\dfrac{2(t-1)}{t+1}\right]}{(t-1)^4} \]

\[=\dfrac{(t-1)(t+1)\ln t\left[-\ln t+\dfrac{2(t+1-2)}{t+1}\right]}{(t-1)^4} \]

\[=\dfrac{(t-1)(t+1)\ln t\left[-\ln t+2-\dfrac{4}{x+1}\right]}{(t-1)^4} \]

记\(\gamma(t)=-\ln t+2-\dfrac{4}{x+1},\gamma^{\prime}(t)=\dfrac{-x^2+2x-1}{x(x+1)^2}<0\)，则\(\gamma(t)<\gamma(1)=0\)

从而\(\varphi^{\prime}(t)<0\)，则\(\varphi(t)<\varphi(1)=0\)

得证

(3)根据平面几何知识，则要使得其有两个交点，则有

\[\dfrac{\left|\dfrac{\sqrt{3}}{4}\cdot\dfrac{x_1+x_2}{2}+\dfrac{\sqrt{3}}{4}\right|}{\dfrac{\sqrt{19}}{4}}<\left|\dfrac{x_1-x_2}{2}\right| \]

整理得

\[16(x_1+x_2)^2-12(x_1+x_2)76x_1x_2>16(x_1+x_2)^2-12(x_1+x_2)-88>0 \]

再整理得

\[4\left[4(x_1+x_2)-11\right](x_1+x_2+2)>0 \]

即证\((x_1+x_2)>\dfrac{11}{4}\)

由(2)的第二个操作，有\(x_1+x_2=\dfrac{\ln t}{t-1},x_2=\dfrac{t\ln t}{t-1}\)

则\(x_1+x_2=\dfrac{\ln t}{t-1}+\dfrac{t\ln t}{t-1}>\dfrac{11}{4}\)

即证:\((1+t)\dfrac{\ln t}{t-1}>\dfrac{11}{4}\)

记\(\zeta(t)=(1+t)\dfrac{\ln t}{t-1},\zeta^{\prime}(t)=\dfrac{-\left(2\ln t-t+\dfrac{1}{t}\right)}{(t-1)^2}\)

因\(\ln x<\dfrac{1}{2}\left(x-\dfrac{1}{x}\right)\)

则\(\zeta^{\prime}(t)>0\)

从而\(\zeta(t)\)单调递增

现在开始估算:

因\(a\)越大，\(x_1+x_2\)的值也越大

当\(a=5\),\(f\left(\dfrac{1}{4}\right)=4e^{0.25}-5>0,f(2.5)=-4e^{2.5}-5<0\)

则\(1>x_1>0.25,x_2>2.5\)

从而$x_1+x_2>\dfrac{11}{4} $

因而\(a\geq 5\)时,\(x_1+x_2>\dfrac{11}{4}\)