1.方程f(x)=2^|x-m|=2^|m|在x≥4恒有一解,
即|x-m|=|m|在x≥4恒有一解,
平方得x^2-2mx=0在x≥4恒有一解x=2m>=4,
∴m>=2.
2.x=x>=m;①
{-x^2+mx+2m-8=-(x-m/2)^2+m^2/4+2m-8,x=4时f(x)(x>=4)的值域是[1,+∞),命题不成立;
m
1.方程f(x)=2^|x-m|=2^|m|在x≥4恒有一解,
即|x-m|=|m|在x≥4恒有一解,
平方得x^2-2mx=0在x≥4恒有一解x=2m>=4,
∴m>=2.
2.x=x>=m;①
{-x^2+mx+2m-8=-(x-m/2)^2+m^2/4+2m-8,x=4时f(x)(x>=4)的值域是[1,+∞),命题不成立;
m