每日一题_191128

已知\(\forall x>0,x\mathrm{e}^{2x}-kx-{\ln}x-1\geqslant 0\),求实数\(k\)的取值范围.
解析:
法一 原不等式等价于\[\forall x>0, k\leqslant \dfrac{1}{x}\cdot\left(x\mathrm{e}^{2x}-{\ln}x-1\right)= \dfrac{1}{x}\cdot\left(\mathrm{e}^{2x+{\ln}x}-{\ln}x-1\right).\]而我们孰知\(\forall x\in\mathbb{R},\mathrm{e}^x\geqslant x+1.\)所以\[RHS\geqslant \dfrac{1}{x}\cdot\left(2x+{\ln}x+1-{\ln}x-1\right)=2.\]当且仅当\(2x+{\ln}x=0\)时上述不等式取得等号,显然存在\(x_0\in\left(\dfrac{1}{\mathrm{e}},1\right)\)满足前式, 所以\(k\)的取值范围为\((-\infty,2]\).
法二 分离参数则原不等式等价于\[\forall x>0,g(x)=\mathrm{e}^{2x}-\dfrac{1}{x}-\dfrac{{\ln}x}{x}\geqslant k.\]求导可得\[g'(x)=2\mathrm{e}^{2x}+\dfrac{{\ln}x}{x^2},x>0.\]再次求导可得\[g''(x)=4\mathrm{e}^{2x}+\dfrac{1-2{\ln}x}{x^3},x>0.\]于是\[\forall x>0,g''(x)\geqslant 4+0-\dfrac 23 \cdot\dfrac{1}{\mathrm{e}}>0.\]所以\(g'(x)\)在区间\((0,+\infty)\)单调递增,又因为\[g'(1)>0>g'\left(\dfrac{1}{\mathrm{e}^3}\right).\]所以\(g'(x)\)必存在唯一的零点\(x_0\in\left(\dfrac{1}{\mathrm{e}^3},1\right)\), 即有\[2x_0^2\mathrm{e}^{2x_0}=-{\ln}x_0.\]设\(t=x_0^2\mathrm{e}^{2x_0}\),则\(2t=-{\ln}x_0\),而\[{\ln}t={\ln}(x_0^2\mathrm{e}^{2x_0})=2{\ln}x_0+2x_0.\]所以\({\ln}t+2t={\ln}x_0+2x_0,\)所以\(x_0=t\),于是我们有\[\forall x>0,g(x)\geqslant g(x_0)=\dfrac{t}{x_0^2}-\dfrac{1}{x_0}+\dfrac{2t}{x_0}=\dfrac{1}{x_0}-\dfrac{1}{x_0}+\dfrac{2x_0}{x_0}=2.\]
因此\(k\)的取值范围为\((-\infty,2]\).