每日导数74

稀疏平常的一道比值代换

已知函数\(f(x)=e^x-\dfrac{1}{2}ax^2\)

(1)若\(f(x)\)在\((0,+\infty)\)上单调递增，求\(a\)范围

(2)若\(f(x)\)有两个极值点分别为\(x_1,x_2(x_1<x_2)\)，当\(\lambda>1\)证明:\(x_1+\lambda x_2>\lambda+1\)

解
(1)\(f^{\prime}(x)=e^x-ax\geq 0\)，不难得到\(a\leq e\)

(2)由题\(\begin{cases} e^{x_1}=ax_1\\ e^{x_2}=ax_2 \end{cases}\)，设\(\dfrac{x_1}{x_2}=t<1\)，即\(x_1=tx_2\)

则有\(e^{x_1-x_2}=t\)，即\(e^{x_2(t-1)}=t\)

即\(x_2(t-1)=\ln t\)，则得到\(x_2=\dfrac{\ln t}{t-1}\)

进而\(x_1=\dfrac{t\ln t}{t-1}\)

则\(x_1+\lambda x_2-\lambda-1=\dfrac{t\ln t}{t-1}+\dfrac{\lambda\ln t}{t-1}-\lambda-1>0\)

整理有\(t\ln t+\lambda\ln t-(\lambda+1)(t-1)<0\)

记\(H(\lambda)=t\ln t+\lambda\ln t-(\lambda+1)(t-1)\)，\(H^{\prime}(\lambda)=\ln t-(t-1)<t-1-(t-1)=0\)

则\(H(\lambda)<H(1)=t\ln t+\ln t-2(t-1)\)

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

因\(\ln t>1-\dfrac{1}{t}\)，则\(\varphi^{\prime}(t)>1-\dfrac{1}{t}+\dfrac{1}{t}-1=0\)

从而\(\varphi(t)\)单调递增

则\(\varphi(t)<\varphi(1)=0\)

得证.