每日导数58

常见的处理技巧与端点效应

已知函数\(f(x)=(x-1)\ln(x-2)-a(x-3)\)，若当\(x>3\)，\(f(x)>0\)恒成立，求\(a\)范围

解
\((x-1)\ln(x-2)-a(x-3)>0\)

即\(\ln(x-2)-a\dfrac{x-3}{x-1}>0\)

记\(g(x)=\ln(x-2)-a\dfrac{x-1-2}{x-1}=\ln(x-2)+\dfrac{2a}{x-1}-a,g^{\prime}(3)=0\)

\(g^{\prime}(x)=\dfrac{1}{x-2}-\dfrac{2a}{(x-1)^2}\)单调递增，\(g^{\prime}(3)=1-\dfrac{a}{2}\)

Case1 当\(a\leq 2\)时，\(g^{\prime}(x)\geq 0\)，则\(g(x)\)单调增

\(g(x)\geq g(3)=0\)

Case2 当\(a>2\)时，\(g^{\prime}(3)<0\)，因\(g^{\prime}(x)\)单调增

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

则一定存在\(x_0>3\)，使得\(g^{\prime}(x_0)=0\)

从而在\((3,x_0)\)上，\(g^{\prime}(x)<0\)

则\(g(x)<g(3)=0\)，不合题

综上\(a\leq 2\)