1.當(dāng)a=0時,f(x)=|2x+1|+|2x+3|>=6
(a)當(dāng)x屬于(-無窮,-3/2]時,f(x)=-2x-1+(-2x-3)=-4x-4>=6,得x屬于(-無窮,-5/2]
(b)當(dāng)x屬于(-3/2,-1/2]時,f(x)=-2x-1+2x+3=2,無解
(c)當(dāng)x屬于(-1/2,+無窮)時,f(x)=4x+4>=6,x屬于[1/2,+無窮)
綜上:x屬于(-無窮,-5/2]U[1/2,+無窮)
2.原式化為:|2x+1|+|2x+3|>=a^2-a
設(shè)左邊的新函數(shù)為g(x)=|2x+1|+|2x+3|,右邊為h(a)=a^2-a
(a)當(dāng)x屬于(-無窮,-3/2]時,g(x)=-2x-1+(-2x-3)=-4x-4,g(x)的范圍為[2,+無窮)
(b)當(dāng)x屬于(-3/2,-1/2]時,g(x)=-2x-1+2x+3=2
(c)當(dāng)x屬于(-1/2,+無窮)時,g(x)=4x+4,范圍為[2,+無窮)
所以g(x)min=2
因此,a^2-a