Question

已知-1＜a+b＜3,且2＜a-b＜4,则2a+3b的范围是(    )

A.(-,)         B.(-,)          C.(-,)           D.(-,)

Answer 1

D

解析：令2a+3b=k₁(a+b)+k₂(a-b),则k₁+k₂=2,k₁-k₂=3.

解得k₁=,k₂=-,结合不等式性质可知D选项正确.