Question

设A>1,函数F(X)=logaX在区间[A,2A]上的最大值与最小值之差为0.5,则A=

A.根号2 B.2 C.2倍根号2 D.4

Answer 1

(1)当0<a<1时,
f(x)=loga x在(0,+∞)单调递减
f(x)max=loga A
f(x)min=loga 2A
loga 2A-loga A=0.5
loga 2=0.5
a=4
与0<a<1矛盾,舍
(2)当a>1时,
f(x)=loga x在(0,+∞)单调递增
f(x)max=loga 2A
f(x)min=loga A
loga 2A-loga A=0.5
loga 2=0.5
a=4