分析:根據(jù)偶函數(shù)的定義,函數(shù)的定義域關于原點對稱,且f(-x)=f(x),可知A,函數(shù)的定義域不關于原點對稱;B,D,函數(shù)滿足f(-x)=-f(x);C,函數(shù)滿足f(-x)=f(x),故可判斷.
解答:解:根據(jù)偶函數(shù)的定義,函數(shù)的定義域關于原點對稱,且f(-x)=f(x),可知
A,函數(shù)的定義域不關于原點對稱,故函數(shù)非奇非偶
B,D,函數(shù)滿足f(-x)=-f(x),故函數(shù)是奇函數(shù)
C,函數(shù)滿足f(-x)=f(x),故函數(shù)是偶函數(shù)
故選C.
點評:本題的考點是函數(shù)奇偶性的判斷,考查偶函數(shù)的定義,解題的關鍵是利用偶函數(shù)的定義,函數(shù)的定義域關于原點對稱,且f(-x)=f(x).