题意,y=2x2+bx+3,在[-1,1]上是减函数,Р故-b4≥12+b+3≥0→-5≤b≤-4;Р(Ⅱ)由题意f(x)=2x+1,①x∈(-12,0),则f(x)∈(0,1),Р由函数y=x+1x的图象及性质可知f(x)+ 1f(x)∈(-2, +∞),Р即h(x)的值域是(2, +∞);Р②h(x)表达式变形可得h(x)=2(x+12)+ 12x+12,可知h(x)是由双沟函数y=2x+12x向左平移12个单位可得,Р即h(x)的图象关于点(-12,0)对称,(也可严格证明h(-1-x)+h(x)=0),Р由g(-1-x)+g(x)=0可得g(x)的图象关于点(-12,0)对称,Р故y1+y2+⋯+ym=0, x1+ x2+⋯+xm =-m2 Р则所求(x1+y1) + (x2+y2)+⋯+ (xm+ym)=-m2.Р22、解:(1)由题意得A(2,log32),B(4,log34),C(4, logm4).Р又AC与x轴平行,∴logm4=log32,Р解得m=9.Р(2)由题意得A(a, logca),B(b, logcb),C(b, logmb).Р∵AC与x轴平行,∴logmb=logca.Р∵b=a2,∴m=c2,Р∴mb-2ca=c2a2-2ca=(ca-1)2-1Р∴ca=1时,mb-2ca取得最小值-1.Р(3)h(f(x2))=alogcx2,φ(x1)= blogcx1,Р∵a<x1<x2<b,且c>1,∴logca<logcx1<logcx2<logcb.Р又∵a>1,b>1,∴alogcx2<alogcb, blogca<blogcx1.Р又∵logcb∙logca=logca∙logcb,Р∴logcalogcb=logc blogca.Р∴alogcb=blogca,∴alogcx2<blogcx1.Р即h(f(x2))<φ(f(x1)).