Question

已知函数f x=x2+a1对任意x1 x2属于R比较1/2[f(x1)+f(x2)与f(x1+x2/2)的大小2若x属于

Answer 1

(1)

f(x1)=x1^2+a f(x2)=x2^2+a

f(x1)+f(x2)=x1^2+x2^2+2a

f((x1+x2)/2)=((x1+x2)/2)^2+a

1/2[f(x1)+f(x2)]-f((x1+x2)/2)=1/2(x1^2+x2^2)+a-a-((x1+x2)/2)^2

=1/2(x1^2+x2^2)-((x1+x2)/2)^2

=1/2(x1^2+x2^2-1/2(x^2+x^2+x1x2))

=1/4(x1^2+x2^2-2x1x2)

=1/4(x1-x2)^2>=0

所以1/2[f(x1)+f(x2)]>=f((x1+x2)/2)

(2)

lf(x)l=-1 且1+a