每日导数52

非常好的一道界值分析，数学分析浓郁的综合性题

已知函数\(f(x)=e^{1-x}-ax\ln x,a>\dfrac{1}{4}\)

(1)若\(f(x)\leq 3-2x\)，求\(a\)

(2)证明：\(f(x)\leq 1-(a+1)\ln x\)

(3)证明：\(f(m)+f\left(\dfrac{1}{n}\right)+f\left(\dfrac{n}{m}\right)\leq 3\)

解

(1) \(e^{1-x}-ax\ln x\leq 3-2x\)

即\(e^{1-x}-ax\ln x+2x-3\leq 0\)

即\(x\left(\dfrac{e^{1-x}}{x}-a\ln x-\dfrac{3}{x}+2\right)\leq 0\)

记\(h(x)=\dfrac{e^{1-x}}{x}-a\ln x-\dfrac{3}{x}+2\)

因\(h(1)=0\)

\(h^{\prime}(x)=\dfrac{-xe^{1-x}-e^{1-x}}{x^2}-\dfrac{a}{x}+\dfrac{3}{x^2}=\dfrac{-e^{1-x}(x+1)-ax+3}{x^2}\)

记\(\varphi(x)=-e^{1-x}(x+1)-ax+3\)

则\(\varphi(1)=1-a\)

现说明\(a>1\)不合题，\(a<1\)也不合题,只有\(a=1\)合题

Case1当\(a>1\)时，\(\varphi(1)<0\)

\(\varphi^{\prime}(x)=e^{1-x}(x+1)-e^{1-x}-a=xe^{1-x}-a\)

因\(\left(xe^{1-x}\right)^{\prime}=e^{1-x}(1-x)\)，则\(xe^{1-x}<1\)

并且\(xe^{1-x}\)在\((0,1)\)上单调增

从而\(\varphi^{\prime}(x)<0\)，则\(\varphi(x)\)单调递减

因\(\varphi(1)=1-a<0\)，则一定存在区间\((x_0,1)\)上\(\varphi(x)<0\)(因为\(\varphi(x)\)是光滑连续的)

从而\(h(x)\)在\(x\in(x_0,1)\)上单调递减

从而\(h(x)>h(1)=0\)，不合题

Case2 当\(a<1\)时，\(\varphi(1)>0\)

\(\varphi(x)=-e^{1-x}(x+1)-ax+3\)，\(\varphi^{\prime}(x)=e^{1-x}(x+1)-e^{1-x}-a=xe^{1-x}-a\)

\(\varphi^{\prime}(1)=1-a>0\)

而因\(xe^{1-x}<1\)，并且在\(x>1\)上单调递减

从而一定存在\((1,x_1)\)，使得\(\varphi^{\prime}(x)>0\)

\(\varphi(x)\)在\((1,x_1)\)上单调增，则\(\varphi(x)>\varphi(1)=1-a>0\)

从而\(h(x)>h(1)=0\)，不合题

Case3 当\(a=1\)时，\(\varphi^{\prime}(x)\leq 0\)

则\(\varphi(x)\)单调递减，因\(\varphi(1)=0\)

则\(h(x)\)在\((0,1)\)增，在\((1,+\infty)\)上减

从而\(h(x)\leq h(1)=0\)

综上，\(a=1\)

(2)令\(g(x)=e^{1-x}-ax\ln x- 1+(a+1)\ln x,g(1)=0\)

\(g^{\prime}(x)=-e^{1-x}-a\ln x-a+\dfrac{a+1}{x},g^{\prime}(1)=0\)

\(g^{\prime\prime}(x)=e^{1-x}-\dfrac{a}{x}-\dfrac{a+1}{x^2}=\dfrac{x^2e^{1-x}-ax-a-1}{x^2}\)

记\(\mu(x)=x^2e^{1-x}-ax-a-1\)

因\((x^2e^{1-x})^{\prime}=xe^{1-x}(-x+2)\)

则\(x^2e^{1-x}-1\leq \dfrac{4}{e}-1,x>0\)

而在\(x\in(0,1),x^2e^{1-x}-1<0\)，从而\(\mu(x)<0\)

则\(g^{\prime}(x)>g^{\prime}(1)=0\)

则\(g(x)\)在\((0,1)\)上增

当\(x>1\)时，\(x^2e^{1-x}-ax-a-1<\dfrac{4}{e}-\dfrac{1}{4}-\dfrac{1}{4}-1<0\)

从而\(x>1,g^{\prime}(x)\)单调递减，即\(g^{\prime}(x)<g^{\prime}(1)=0\)

从而\(g(x)\)在\((1,+\infty)\)上单调递减

综上\(g(x)\leq g(1)=0\)

得证!

(3)由(2) \(f(x)\leq 1-(a+1)\ln x\)

则\(f(m)+f\left(\dfrac{1}{n}\right)+f\left(\dfrac{n}{m}\right)\leq 3-(a+1)\ln \left(m\cdot\dfrac{1}{n}\cdot\dfrac{n}{m}\right)=3\)