Question

已知函數(shù)y＝f(x－1)的圖象關(guān)于直線x＝1對(duì)稱，且當(dāng)x∈(－∞，0)，f(x)＋xf′(x)<0成立，若a＝(2^0.2)·f(2^0.2)，b＝(ln 2)·f(ln 2)，c＝·f，則a，b，c的大小關(guān)系是(　　)．

A．a>b>c	B．b>a>c
C．c>a>b	D．a>c>b

Answer 1

B

因?yàn)楹瘮?shù)y＝f(x－1)的圖象關(guān)于直線x＝1對(duì)稱，則y＝f(x)關(guān)于y軸對(duì)稱，所以函數(shù)y＝xf(x)為奇函數(shù)．又因?yàn)閇xf(x)]′＝f(x)＋xf′(x)，所以當(dāng)x∈(－∞，0)時(shí)，[xf(x)]′＝f(x)＋xf′(x)<0，函數(shù)y＝xf(x)單調(diào)遞減；則當(dāng)x∈(0，＋∞)時(shí)，函數(shù)y＝xf(x)單調(diào)遞減．因?yàn)?<2^0.2<2,0<ln 2<1，log ＝2，所以0<ln 2<2^0.2<log ，所以b>a>c.