解:(1)當(dāng)a=1時(shí),f(x)=2·4x-2x-1.
令f(x)=0,即2·(2x)2-2x-1=0,解得2x=1或2x=-
(舍去).
∴x=0,∴函數(shù)f(x)的零點(diǎn)為x=0. --------------------------4
(2)解法一:若f(x)有零點(diǎn),則方程2a·4x-2x-1=0有解----------------6
于是2a=
=(
)x+(
)x ----------------------------------------------------------10
∵(
)x>0,∴2a>
-
=0,即 a>0.------------------------------12
解法二:令t=2x,∵x∈R,∴t>0,
則方程2at2-t-1=0在(0,+∞)上有解. ------------------------6
①當(dāng)a=0時(shí),方程為t+1=0,即t=-1<0,
此時(shí)方程在(0,+∞)無(wú)解.-----------------------------------------8
②當(dāng)a≠0時(shí),令g(t)=2at2-t-1,
若方程g(t)=0在(0,+∞)上有一解,則ag(0)<0,即-a<0,解得a>0.
若方程g(t)=0在(0,+∞)上有兩解,則
無(wú)解
-------------------------------------------10
綜上所述,所求實(shí)數(shù)a的范圍是(0,+∞). --------------------------12