Differentiate both sides with respect to x, we get  f(x) = x – x² f (x)

$\therefore \mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{1+{\mathrm{x}}^2}$

$\text{ Also, } {\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)=\frac{1-{\mathrm{x}}^2}{1+{\mathrm{x}}^2}=0\Rightarrow \mathrm{x}\pm 1$

Clearly, distance between the local maximum and local minimum of the curve

$\mathrm{y}=\mathrm{f}\left(\mathrm{x}\right)\text{ is }\left|\frac{1}{2}-\left(\frac{-1}{2}\right)\right|=1$

Now, verify alternatives.