每日导数63

高考题还是质量高，观察式子进行选值放缩

已知函数\(f(x)=xe^{ax}-e^x\)

(1)当\(x>0\)时，\(f(x)<-1\)，求\(a\)取值范围

(2)证明:\(\dfrac{1}{\sqrt{1^2+1}}+\dfrac{1}{\sqrt{2^2+2}}+\cdots+\dfrac{1}{\sqrt{n^2+n}}>\ln(n+1)\)

解

(1) \(f^{\prime}(x)=e^{ax}(1+ax)-e^x,f(0)=-1,f^{\prime}(0)=0\)

\(f^{\prime\prime}(x)=e^{ax}(2a+a^2x)-e^x,f^{\prime\prime}(0)=2a-1\)

现在说明\(a\leq \dfrac{1}{2}\)合题，\(a>\dfrac{1}{2}\)不合题

当\(a>\dfrac{1}{2}\)时，\(f^{\prime\prime}(0)=2a-1>0\)

则由保号性，一定存在\((0,x_0)\)上,\(f^{\prime\prime}(x)>0\)

从而在\(x\in\left(0,x_0\right)\)上，\(f^{\prime}(x)\)单调递增，则\(f^{\prime}(x)>f^{\prime}(0)=0\)

则\(f(x)>f(0)=-1\)不合题

当\(a\leq\dfrac{1}{2}\)时，考虑换主元

\(\varphi(a)=xe^{ax}-e^x\leq xe^{\frac{x}{2}}-e^x\)

记\(h(x)=xe^{\frac{x}{2}}-e^x,h^{\prime}(x)=e^{\frac{x}{2}}\left(1+\dfrac{x}{2}-e^{\frac{x}{2}}\right)\leq 0\)

则\(h(x)< h(0)=-1\)

从而\(f(x)=\varphi(a)\leq h(x)< h(0)=-1\)

综上\(a\leq \dfrac{1}{2}\)

(2)取\(a=\dfrac{1}{2}\),则\(xe^{\frac{x}{2}}-e^x<-1\)

则取\(e^{\frac{x}{2}}=m\)，即有

\(\ln m<\dfrac{1}{2}\left(m-\dfrac{1}{m}\right),m>1\)(常见的放缩，没有题给也应该想到)

再观察式子:\(\dfrac{1}{\sqrt{n^2+n}}=\dfrac{\sqrt{n^2+n}}{n^2+n}=\dfrac{\sqrt{n^2+n}}{n(n+1)}=\dfrac{\sqrt{n^2+n}}{n}-\dfrac{\sqrt{n^2+n}}{n+1}=\sqrt{\dfrac{n+1}{n}}-\sqrt{\dfrac{n}{n+1}}\)

从而带入\(\sqrt{\dfrac{k+1}{k}}\)进不等式有

\(\ln\sqrt{\dfrac{k+1}{k}}<\dfrac{1}{2}\left(\sqrt{\dfrac{k+1}{k}}-\sqrt{\dfrac{k}{k+1}}\right)\)

即\(\ln\dfrac{k+1}{k}<\sqrt{\dfrac{k+1}{k}}-\sqrt{\dfrac{k}{k+1}}\)

即\(\ln(k+1)-\ln k<\sqrt{\dfrac{k+1}{k}}-\sqrt{\dfrac{k}{k+1}}=\dfrac{1}{\sqrt{k^2+k}}\)

取\(k=1,2,3\cdots ,n\)累加有

\(\ln(n+1)<\dfrac{1}{\sqrt{1^2+1}}+\dfrac{1}{\sqrt{2^2+2}}+\cdots+\dfrac{1}{\sqrt{n^2+n}}\)