(1) f(1)=0f(1)=0f(1)=0, f(x)≤x2−1f(x)\leq x^2-1f(x)≤x2−1 解集;
(2) f(x)f(x)f(x) 在 (0,+∞)(0,+\infty)(0,+∞) 存在极大值, m\mathbf{m}m 取值范围.
(1) m=−1m=-1m=−1, f(x)=x2−x−lnxf(x)=x^2-x-\ln xf(x)=x2−x−lnx. f(x)≤x2−1⇔x+lnx≥1f(x)\leq x^2-1 \Leftrightarrow x+\ln x\geq 1f(x)≤x2−1⇔x+lnx≥1. s(x)=x+lnxs(x)=x+\ln xs(x)=x+lnx, s′(x)=1+1/x>0s'(x)=1+1/x>0s′(x)=1+1/x>0, s(x)≥s(1)=1⇔x≥1s(x)\geq s(1)=1 \Leftrightarrow x\geq 1s(x)≥s(1)=1⇔x≥1. 解集 [1,+∞)[1,+\infty)[1,+∞).
(2) f′(x)=(2x−m)(x−1)/xf'(x)=(2x-m)(x-1)/xf′(x)=(2x−m)(x−1)/x. m≤0m\leq0m≤0: x=1x=1x=1 极小, 无极大. 0<m<20<m<20<m<2: x=m2x=\frac{m}{2}x=2m 极大. m=2m=2m=2: 无极值. m>2m>2m>2: x=1x=1x=1 极大. 答案: m>0m>0m>0 and\text{and}and m≠2m\neq 2m=2.