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!

Let us take $f(x) = \sin(2k\pi x)$, where $k$ is an integer constant. Clearly this function is differentiable on $[0,1]$ and $f(0) = f(1) = 0$. The derivative is given by: $f'(x) = (2k\pi) \cdot \cos(2k\pi x)$.
If we plot $f(x)$ and $f'(x)$ together as a function of $x$, we see that in every interval $[j/k, (j+1)/k]$, with $j= 0, 1, 2, \ldots , (k-1)$, there are exactly two intersections. Hence in total there are $2k$ values of $x$ where $f'(x) = f(x)$. Since $k$ can be chosen arbitrarily, the number of solutions is unbounded.
Therefore answers $\textbf{A}$ and $\textbf{C}$ are incorrect.
On the other hand, it is clear that the sign of $f'(x)$ must change at least once on the interval $[0,1]$ (unless it is a equal to zero everywhere). In doing so it will intersect $f(x)$. It follows that there is at least one value of $x$ for which $f'(x) = f(x)$. An example of a function with one intersection is $f(x) = \sin(\pi x)$. It follows that answer $\textbf{B}$ is incorrect.
The right answer is $\textbf{D}$.   
