函式\(f(x)=\dfrac{e^x}{x}-a\)的影像與\(x\)軸的兩交點為\(A(x_1,0),B(x_2,0)(x_2>x_1)\)
(1)令\(h(x)=f(x)-\ln x+x\)，若\(h(x)\)有兩個零點，求\(a\)的取值範圍
(2)證明:\(x_1x_2<1\)
(3)證明:當\(a\geq 5\)時，以\(AB\)為直徑的圓與直線\(y=\dfrac{\sqrt{3}}{4}(x+1)\)恆有公共點
(參考資料:$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\(t+\ln t-a\)單調遞增，則要使得\(h(x)\)有兩個零點
則要有\(t+\ln t-a\)有一個零點即可
則要有\(e+\ln e-a<0\)即\(a>e+1\)
(2)由題\(\begin{cases} \dfrac{e^{x_1}}{x_1}=a\\ \dfrac{e^{x_2}}{x_2}=a \end{cases}\)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\(\dfrac{x_1-x_2}{\ln x_1-\ln x_2}=1\)
由對數均值不等式:
有\(1=\dfrac{x_1-x_2}{\ln x_1-\ln x_2}>\sqrt{x_1x_2}\)
即\(x_1x_2<1\)
如果不想使用對數均值不等式，可如下操作:
設\(\dfrac{x_2}{x_1}=t,t>1\)
則有\(x_2=tx_1\)
\(\begin{cases} \dfrac{e^{x_1}}{x_1}=a\\ \dfrac{e^{x_2}}{x_2}=a \end{cases}\)，兩式做比\(\dfrac{x_2}{x_1}e^{x_1-x_2}=1\)
即\(te^{x_1(1-t)}=1\)，即\(\ln t+x_1(1-t)=0\)得\(x_1=\dfrac{\ln t}{t-1}\)，則\(x_2=\dfrac{t\ln t}{t-1}\)
從而\(x_1x_2=\dfrac{t\ln^2t}{(t-1)^2}\)，記\(\varphi(t)=\dfrac{t\ln^2t}{(t-1)^2}\)
記\(\gamma(t)=-\ln t+2-\dfrac{4}{x+1},\gamma^{\prime}(t)=\dfrac{-x^2+2x-1}{x(x+1)^2}<0\)，則\(\gamma(t)<\gamma(1)=0\)
從而\(\varphi^{\prime}(t)<0\)，則\(\varphi(t)<\varphi(1)=0\)
得證
(3)根據平面幾何知識，則要使得其有兩個交點，則有
整理得
再整理得
即證\((x_1+x_2)>\dfrac{11}{4}\)
由(2)的第二個操作，有\(x_1+x_2=\dfrac{\ln t}{t-1},x_2=\dfrac{t\ln t}{t-1}\)
則\(x_1+x_2=\dfrac{\ln t}{t-1}+\dfrac{t\ln t}{t-1}>\dfrac{11}{4}\)
即證:\((1+t)\dfrac{\ln t}{t-1}>\dfrac{11}{4}\)
記\(\zeta(t)=(1+t)\dfrac{\ln t}{t-1},\zeta^{\prime}(t)=\dfrac{-\left(2\ln t-t+\dfrac{1}{t}\right)}{(t-1)^2}\)
因\(\ln x<\dfrac{1}{2}\left(x-\dfrac{1}{x}\right)\)
則\(\zeta^{\prime}(t)>0\)
從而\(\zeta(t)\)單調遞增
現在開始估算:
因\(a\)越大，\(x_1+x_2\)的值也越大
當\(a=5\),\(f\left(\dfrac{1}{4}\right)=4e^{0.25}-5>0,f(2.5)=-4e^{2.5}-5<0\)
則\(1>x_1>0.25,x_2>2.5\)
從而$x_1+x_2>\dfrac{11}{4} $
因而\(a\geq 5\)時,\(x_1+x_2>\dfrac{11}{4}\)