Question

高一数学题。在线等！！要有详细过程！！

Answer 1

1
a=1,f(x)=1+(1/2)^x+(1/4)^x=1+(1/2)^x+[(1/2)^x]^2=[(1/2)^x+1/2]^2+3/4=[2^(-x)+1/2]^2+3/4
当x∈(-∞,0)，f(x)是减函数，值域为(∞,3)，所以f(x)是有界函数，其上界为3
2
f(x)=[(1/2)^x+a/2]^2+1-a^2/4>=3,1/2^x+a/2>=√(2+a^2/4)或1/2^x+a/2<=-√(2+a^2/4),
x>=0,所以1/2^x>0>=√(2+a^2/4)-a/2或a/2+√(2+a^2/4)<=-1<=-1/2^x
所以得(2+a^2/4)-(-1-a/2)^2<=0,得a>=1
3
m>-1,g(x)=(1-mx^2)/(1+mx^2)
0>=m>-1,+∞>g(1)=(1-m)/(1+m)>=g(x)=(1-mx^2)/(1+mx^2)>=g(0)=1;
m>=0,g(0)=1>=g(x)=(1-mx^2)/(1+mx^2)>=g(1)=(1-m)/(1+m)>-∞
所以T(m)=(1-m)/(1+m),m>-1
0>=m>-1,T(m)>=1;0<=m<=1,1>=T(m)>=0;m>=1,T(m)<=0