f(x) is a non negative function 
Let $f(x)$be a non-negative function on $[0,\infty)$ such that $f(0)=0$ and $f'(x)\le 2f(x)$ for all $x>0$. Then, on $[0,∞)$,
A. $f(x)$ is always a constant function
B. $f(x)$ is strictly increasing
C. $f(x)$ is strictly decreasing
C. $f'(x)$ changes sign


My attempt

*

*multiply by $e^{-2x}$
so by solving the diff we get $0≤f(x)≤ce^{2x}$
where $c$ is a positive constant


*therefore none of the options above match


*but the answer given is $f(x)$ is a constant function


*the given answer could have been correct only if it was given $f(x)$ is negative


*but f(x) is given to be non negative
please help

As an alternative approach: division by $ f $ shows $(\log f (x))'\leq 2$ and thus $ f (x)\leq f (0) \exp 2x$.
Since $ f (x)=0$ you are not allowed to divide by $ f $. To fix this apply the argument to $ f_{\epsilon}=f+\epsilon $
