因为：2^(x^2)<=2^x
所以x^2<=x
即0<=x<=1.
f(x)=-(2^x)^2+(1/2)x^2-m
=-(2^x-1/4)^2+(1/16)-m
设2^x=t,则有1<=t<=2.
f(x)=-(t-1/4)^2+(1/16)-m
开口向下,对称轴t=1/4,区间[1,2]在对称轴的右方,要使f(x)>=0,则有：
f(2)>=0
代入即得到：
m<=-3.