[tex]\displaystyle f(x)=2x-1\\\\f(m^2)-2f(m)+f(2m)=\frac{m}{2}\\\\2m^2-1-2(2m-1)+2\cdot 2m-1=\frac{m}{2}\\\\2m^2-2-4m+2+4m=\frac{m}{2}\\\\2m^2=\frac{m}{2}\\\\4m^2=m\\\\4m^2-m=0\\\\\\x=\frac{-b\pm\sqrt{\Delta}}{2a}\\\\x=\frac{-(-1)\pm\sqrt{b^2-4ac}}{2\cdot 4}\\\\x=\frac{1\pm\sqrt{(-1)^2-4\cdot 4 \cdot 0}}{8}\\\\x=\frac{1\pm\sqrt{1}}{8}\\\\x=\frac{1\pm 1}{8}\\\\\\x_1=\frac{1+1}{8}=\frac{2}{8}=\frac{1}{4}\\\\x_2=\frac{1-1}{8}=\frac{0}{8}=0\\\\S=\left\{0, \frac{1}{4}\right\}[/tex]
Com isso, a igualdade é válida para m = 0 e m = 1/4.
