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

You should have gone one step further and determined the constant $c$ as $f(0)$ which is $0$ so that indeed you find the constant zero function as the only possible solution.
\begin{align}
&(e^{-2x}f(x))'=e^{-2x}(f'(x)-2f(x))\le0
\\\implies& 
\int_0^x(e^{-2s}f(s))'\,ds=e^{-2x}f(x)- e^{-2·0}f(0)\le0
\\\implies& 
f(x)\le f(0)\,e^{2x}=0\qquad\text{(per $f(0)=0$)}
\end{align}
See Gronwall lemma, but your idea using the integrating factor is equivalent to it in this case.
