设偶函数f(x)=loga|x+b|在(0,+∞)上单调递增,则f(b-2)与f(a+1)的

设偶函数f(x)=loga|x+b|在(0,+∞)上单调递增,则f(b-2)与f(a+1)的大小关系为

A.f(b-2)=f(a+1)                 B.f(b-2)>f(a+1) 

C.f(b-2)<f(a+1)                  D.不能确

答案

C


解析:

∵函数f(x)是偶函数,∴b=0,此时f(x)=loga|x|.

当a>1时,函数f(x)=loga|x|在(0,+∞)上是增函数,∴f(a+1)>f(2)=f(b-2);

当0<a<1时,函数f(x)=loga|x|在(0,+∞)上是减函数,∴f(a+1)>f(2)=f(b-2).

综上,可知f(b-2)<f(a+1).

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