Question

已知-1＜a+b＜3,且2＜a-b＜4,則2a+3b的范圍是(    )

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

Answer 1

D

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

解得k₁=,k₂=-,結合不等式性質可知D選項正確.