解:因?yàn)楹瘮?shù)f(x)是定義在R上的奇函數(shù),當(dāng)x>0時(shí),f(x)=e
-x(x-1),
設(shè)x<0,則-x>0,所以-f(x)=f(-x)=e
x(-x-1),即f(x)=e
x(x+1),故①正確;
對(duì)x<0時(shí)的解析式求導(dǎo)數(shù)可得,f′(x)=e
x(x+2),令其等于0,解得x=-2,
且當(dāng)x∈(-∞,-2)上導(dǎo)數(shù)小于0,函數(shù)單調(diào)遞減;當(dāng)x∈(-2,+∞)上導(dǎo)數(shù)大于0,函數(shù)單調(diào)遞增,
x=-2處為極小值點(diǎn),且f(-2)>-1,且在x=1處函數(shù)值為0,且當(dāng)x<-1是函數(shù)值為負(fù).
又因?yàn)槠婧瘮?shù)的圖象關(guān)于原點(diǎn)中心對(duì)稱,故函數(shù)f(x)的圖象應(yīng)如圖所示:
由圖象可知:函數(shù)f(x)有3個(gè)零點(diǎn),故 ②錯(cuò)誤;
若關(guān)于x的方程f(x)=m有解,則實(shí)數(shù)m的取值范圍是-1<m<1,故③錯(cuò)誤;
由于函數(shù)-1<f(x)<1,故有對(duì)?x
1,x
2∈R,|f(x
2)-f(x
1)|<2恒成立,即④正確.
故正確的命題為①④.
分析:設(shè)x<0,則-x>0,由函數(shù)得性質(zhì)可得解析式,可判①的真假,再由性質(zhì)作出圖象可對(duì)其他命題作出判斷.
點(diǎn)評(píng):本題考查奇函數(shù)的性質(zhì),由圖象作出函數(shù)的圖象是解決問(wèn)題的關(guān)鍵,屬基礎(chǔ)題.