Question

已知函数f(x)=2^x-.

(1)若f(x)=2,求x的值;

(2)若2^tf(2t)+mf(t)≥0对于t∈［1,2］恒成立,求实数m的取值范围.

Answer 1

解:(1)当x＜0时,f(x)=0;

当x≥0时,f(x)=2^x-.                                                       

由条件可知2^x-=2,即2^2x-2·2^x-1=0,

解得2^x=1±.                                                               ∵2^x＞0,∴x=log₂(1+).                                                     

(2)当t∈［1,2］时,2^t(2^2t-)+m(2^t-)≥0,                                     

即m(2^2t-1)≥-(2^4t-1).

∵2^2t-1＞0,∴m≥-(2^2t+1).                                                      

∵t∈［1,2］,∴-(1+2^2t)∈［-17,-5］,

故m的取值范围是［-5,+∞).