由图设f(x)=a(x+1)(x-2)且a>0,求出f(x1)和f(x2),代入(x1+1)﹒f(x2)-(x2+1)﹒f(x1)化简,再由已知的范围判断出各个因式的符号即可.
【解析】
由图设f(x)=a(x+1)(x-2),(a>0)
则f(x1)=a(x1+1)(x1-2),f(x2)=a(x2+1)(x2-2),
∴(x1+1)﹒f(x2)-(x2+1)﹒f(x1)
=(x1+1)﹒a(x2+1)(x2-2)-(x2+1)﹒a(x1+1)(x1-2)
=a(x1+1)(x2+1)[(x2-2)-(x1-2)]
=a(x1+1)(x2+1)(x2-x1)
∵-1<x1<x2<2,∴x1+1>0,x2+1>0,x2-x1>0,
又a>0,则a(x1+1)(x2+1)(x2-x1)>0,
∴(x1+1)﹒f(x2)-(x2+1)﹒f(x1)>0,
故选A.