Question

设f(x)=x+x/4-6(x>0)和g(x)=-x^2+ax+m(a,m均为实数）,且对于任意实数x,都有g(x)=g

Answer 1

∵g(x)=g(4-x)∴g(x-4)=-（x-4）^2+a（x-4）+mg(x-4)=-x^2+8x-16+ax-4a+m-x^2+8x-16+ax-4a+m=-x^2+ax+m8x-16-4a=0 a=2x-4∴g(x)=-x^2+(2x-4)x+mg(x)=x^2-4x+mF(x)=f(x)-g(x)=x+x/4-6-(x^2-4x+m)=-x^2+21/4x-6-mF(x)=-x^2+21/4x-(6+m)（1）函数F(x)在（0,+∞)内有一个零点对称轴为x=21/8所以-(6+m)>0 m0 m