(1)
f(1)＝2＋(1－m)|1－m| ≥ 4
當(dāng)m＞1時,(1－m)(m－1) ≥ 2,無解；
當(dāng)m ≤ 1時,(1－m)(1－m) ≥ 2,解得：m ≤ 1－√2
∴m的取值范圍是：m ≤ 1－√2
(2)
∵m＞0,x ≥ m
∴h(x)＝f(x)/x
＝[2x²＋(x－m)(x－m)]/x
＝[2x²＋x²＋m²－2mx]/x
＝(3x²＋m²－2mx)/x
＝3x＋(m²/x)－2m
任取m ≤ x1 ≤ x2,
則h(x2)－h(huán)(x1)
＝[3x2＋(m²/x2)－2m]－[3x1＋(m²/x1)－2m]
＝(3x2－3x1)＋[ m²(x1－x2) /(x1x2) ]
＝[3x1x2(x2－x1)＋m²(x1－x2)]/(x1x2)
＝[(x2－x1)(3x1x2－m²)]/(x1x2)
＝(x2－x1)[(3x1x2－m²)/(x1x2)]
∵x2－x1＞0,3x1x2－m²＞3m²－m²＞0,x1x2＞0
∴h(x2)－h(huán)(x1)＞0
即h(x1)＜h(x2)
即h(x)在[m,＋∞)為單調(diào)遞增函數(shù)