分析:先根据判断出(x2-x1)(f(x2)-f(x1))>0,进而可推断f(x)在x1,x2∈[0,+∞)(x1≠x2)上单调递增,又由于f(x)是偶函数,可知在x1,x2∈(-∞,0](x1≠x2)单调递增.进而可判断出f(3),f(-2)和f(1)的大小. 解:∵(x2-x1)(f(x2)-f(x1))>0, ∴>则f(x)在x1,x2∈[0,+∞)(x1≠x2)上单调递增, 又f(x)是偶函数,故f(x)在x1,x2∈(-∞,0](x1≠x2)单调递减. 且满足n∈N*时,f(-2)=f(2),3>2>1>0, 得f(1)<f(-2)<f(3), 故选B |