Question

已知函数f(x)=Iog4(4^x+1)+kx(k∈R)是偶函数

Answer 1

(1) 因为f(x)是偶函数,则：f(-1)= Iog4(1/4+1)-k=f(1)= Iog4(4+1)+k,

Iog4(5/4)-k= Iog4(5)+k,

2k+ Iog4(5)- Iog4(5/4)=0,

2k+ Iog4(5*4/5)=0,

2k+ Iog4(4)=0,

2k+1=0,

k=-1/2

(2) f(x)-m= Iog4(4^x+1)-1/2*x-m=0,

Iog4(4^x+1)=1/2*x+m,

4^x+1=4^(1/2*x+m),

4^x+1-4^(1/2*x+m)=0,

2^(2x)-2^x*4^m+1=0,

令2^x=y,则方程化为y^2-4^m*y+1=0,

判别式＝(4^m)^2-4=4^(2m)-4≥0,

2m≥1,

m≥1/2