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#bb9eff6e-e5e6-4eb2-8a3f-0bab4e1a8b97中等解答题恒成立与端点效应导数

30.(2023春•天津期末)已知函数f(x)=(a1)lnx(a1)x+1f(x)=(a-1)lnx-(a-1)x+1
(1)证明:当a=2a=2时,f(x)0f(x)\leqslant 0恒成立;
(2)若g(x)=f(x)+x22x1+a(a>2)g(x)=f(x)+\frac{{x}^{2}}{2}-x-1+a(a>2)g(m)=12(m1)g(m)=\frac{1}{2}(m\ne 1),证明:x(1\forall x\in (1m]m]m<2a3m<2a-3