看著嚇人,式子變形
設\(f(x)=e^x,g(x)=\ln x\)
(1)已知\(e^x\geq kx\geq \ln x\)恆成立,求\(k\)取值範圍
(2)已知直線\(l\)與曲線\(f(x),g(x)\)分別切於點\((x_1,f(x_1)),(x_2,g(x_2))\),其中\(x_1>0\)
(I)求證:\(e^{-2}<x_2<e^{-1}\)
(II)已知\((\lambda x_2-x+1)e^x+x\leq 0\)對任意 \(x\in[x_1,+\infty)\)恆成立,求\(\lambda\)取值範圍.
解
(1)因\(e^x,\ln x\)互為反函式,並且\(y=ex\)與\(e^x\)相切,\(y=\dfrac{1}{e}x\)與\(\ln x\)相切,則\(k\in\left[\dfrac{1}{e},e\right]\)
(2)
(I)由題設\(l\)的斜率為\(k\),則有
則有$$\dfrac{\ln x_2-\dfrac{1}{x_2}}{x_2-(-\ln x_2)}=\dfrac{1}{x_2}$$
整理有$$x_2\ln x_2-\ln x_2-x_2-1=0$$
記\(\varphi(t)=t\ln t-\ln t-t-1,\varphi^{\prime}(t)=\ln t+1-\dfrac{1}{t}-1=\ln t-\dfrac{1}{t}\)單調遞增
而\(\varphi^{\prime}(e^{-1})=-1-e<0\)
則\(\varphi(t)\)在\((0,e^{-1})\)上單調遞減
而
從而由零點存在定理,則\(t\in(e^{-1},e^{-2})\),即\(x_2\in(e^{-1},e^{-2})\)
(II)由(I),得\(x_1\in(1,2)\)
原不等式轉化為\(\lambda x_2\leq -xe^{-x}+x-1\)
記\(\gamma(x)=-xe^{-x}+x-1,\gamma^{\prime}(x)=e^{-x}(x-1)+1=\dfrac{x-1+e^x}{e^x}>\dfrac{x-1+x+1}{e^x}=\dfrac{2x}{e^x}>0\)
則\(\lambda x_2\leq \gamma(x)\leq \gamma(x_1)=-x_1e^{-x_1}+x_1-1\)
即\(\lambda\leq -x_1+(x_1-1)e^{x_1}\)
記\(F(t)=-t+(t-1)e^t,F^{\prime}(t)=te^t-1>0\)
則\(\lambda\leq F(t)<F(0)=-1\)
注:下面答案的第三問寫法,比我的答案範圍大,目前看不出我錯在哪裡.
由(I)得\(e^{x_1}(x_1-1)=x_1+1\)
前面都一樣,則有\(\lambda x_1\leq\gamma(x) \leq \gamma(x_1)=\dfrac{e^{x_1}(x_1-1)-x_1}{e^{x_1}}=\dfrac{1}{e^{x_1}}\)
則\(\lambda\leq 1\)