Question

关于x的不等式x^2+px+2＞=3x+p对任意p属于【0,4】恒成立,则x的取值范围

Answer 1

(x-1)p+x²-3x+2≥0

x=1时,不等式成立.

x≠1时,设f(p)=(x-1)p+x²-3x+2,只需f(0)≥0,且f(1)≥0

所以x²-3x+2≥0且x²-3x+2+x-1≥0

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