$\mathrm{Let} \mathrm{f}:\mathrm{R}\to \mathrm{R}$ be a continuous onto function satisfying, $\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(-\mathrm{x}\right)=0\forall \in \mathrm{R}. \mathrm{If} \mathrm{f}\left(-3\right)=2 \mathrm{and} \mathrm{f}\left(5\right)=4 \mathrm{in} \left[-5,5\right],$ then the minimum number of roots of the equation f(x) = 0 is

$\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(-\mathrm{x}\right)=0$ 

Therefore, f(x) is an odd function. 

Since points (-3, 2) and (5,4) lie on the curve, 

For minimum number of roots, graph of continuous function f(x) is as follows:

From the above graph of(x), it is clear that equation f(x) = 0  has at least three real roots.