解:函數(shù)中存在“倍值區(qū)間”,則:①f(x)在[a,b]內是單調函數(shù);②
f(a)="2a," f(b)=2b或f(a)="2b," f(b)=2a
①f(x)=x
2(x≥0),若存在“倍值區(qū)間”[a,b],則
A=0,b=2
∴f(x)=x
2(x≥0),若存在“倍值區(qū)間”[0,2];
②f(x)=e
x(x∈R),若存在“倍值區(qū)間”[a,b],則f(a)="2a," f(b)=2b
構建函數(shù)g(x)=e
x-x,∴g′(x)=e
x-1,∴函數(shù)在(-∞,0)上單調減,在(0,+∞)上單調增,∴函數(shù)在x=0處取得極小值,且為最小值.∵g(0)=1,∴,g(x)>0,∴e
x-x=0無解,故函數(shù)不存在“倍值區(qū)間”;
③f(x)=
若存在“倍值區(qū)間”[a,b]⊆[0,1],則f(a)="2a," f(b)=2b
∴a=0,b=1,若存在“倍值區(qū)間”[0,1];
④f(x)=log
a(a
x-
),log
a(a
m-)=2m,log
a(a
n-
)="2n" (a>0,a≠1).不妨設a>1,則函數(shù)在定義域內為單調增函數(shù)
若存在“倍值區(qū)間”[m,n],則log
a(a
n-
)=2n,log
a(a
m-)=2m
∴2m,2n是方程log
a(a
x-
)=2x的兩個根,∴2m,2n是方程a
2x-a
x+
=0的兩個根,由于該方程有兩個不等的正根,故存在“倍值區(qū)間”[m,n];綜上知,所給函數(shù)中存在“倍值區(qū)間”的有①③④
故選C.