x^2-mx+2＞=0在[1,2]上恒成立,求m的范围

Question

x^2-mx+2＞=0在[1,2]上恒成立,求m的范围

Answer 1

设f(x)=x^2-mx+2,函数过定点（0,2）.
当m<=2时,f(x)=x^2-mx+2在[1,2]上单调递增,要使条件成立,有f(1)=3-m>=0,即m<=3.
当m>=2且m<=4,f(x)=x^2-mx+2在[1,2]上先递减后递增,要使条件成立,则需Δ=m^2-8<=0,即
-2根号2<=m<=2根号2
当m>=4时,f(x)=x^2-mx+2在[1,2]上单调递减,要使条件成立,有f(2)=6-2m>=0,即m<=3,矛盾.
综上所述,m的取值范围为m<=2根号2