How many $f(x)$ are possible satisfying $f(x)=f'(x)$ and $f(0)=f(1)=0$. 
Let $f:[0,1]\to\Bbb{R}$ be a fixed continuous function such that $f$ is differentiable on $(0,1)$ and $f(0)=f(1)=0$. Then the equation $f(x)=f'(x)$ admits how many solutions?

The only solution that I am getting is $y=0$. This is because $y=y'$ implies $y=Ae^x$, and $A=0$ when accounting for boundary conditions. However, I am not sure if this is the right answer. Most other examinees were saying that multiple such functions are possible. An explanation of this would be great.
EDIT: The options were

A. No solution $x\in (0,1)$
B. More than one solution $x\in (0,1)$
C. Exactly one solution $x\in (0,1)$
D. At least one solution $x\in (0,1)$

I feel that if C is right, then so is D!

Shorter proof of D): 
Consider $g(x)=e^{-x}f(x)$. Then also $g(0)=g(1)=0$ and by Rolle's theorem/lemma there exists at least one $x\in(0,1)$ with $g'(x)=0$. But as
$$
g'(x)=e^{-x}(f'(x)-f(x))
$$
this also means $f'(x)=f(x)$ there. Thus D) is correct. Use the counter-examples to exclude the other cases.
