Question

已知定義域?yàn)镽的函數(shù)f(x)滿足f(-x)=-f(x+4),且當(dāng)x＞2時(shí),f(x)單調(diào)遞增.如果x₁+x₂＜4且(x₁-2)(x₂-2)＜0,則f(x₁)+f(x₂)的值(    )

A.恒小于0               B.恒大于0             C.可能為0           D.可正可負(fù)

Answer 1

A

解析：由(x₁-2)(x₂-2)＜0,

不妨設(shè)x₁＜2,x₂＞2.

由4-x₁＞2,

又x₁+x₂＜4,即4-x₁＞x₂.

由題意f(4-x₁)＞f(x₂).

又f(x₁)=f［-(-x₁)］=-f(4-x₁),

∴-f(x₁)＞f(x₂),即f(x₁)+f(x₂)＜0.