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!
 A: 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.
A: 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 $\textbf{D}$.   
A: Here's a direct proof of D.
Let $F(x) =\displaystyle \int_{0}^{x}f(t)dt$ then by the Fundamental Theorem of Calculus, $F'(x)$ exists $\forall \, x \in [0,1].$ and $F'(x) =f(x.)$ Define $g:[0,1] \to \mathbb{R}$ by $g(x)= F(x) -f(x).$
Claim: $\exists \, a, b \in [0,1]$ with $a \neq b$ such that $g(a) = g(b).$
Proof of claim: Suppose not. Then $g(a)=g(b)$ implies $a=b$ so $g$ is injective and since it is clearly continuous too, $g$ is monotone. WLOG, let $g$ be monotone increasing. Then since $g$ is differentiable, $g'(x) \geq 0 \, \forall \, x \in [0,1].$ If there exists at least one $x$ for which $g'(x) =0$ we are done so assume $g'(x)>0.$ Since $g(0) =0,$ we have $g(x)>0$ for all $x \in [0,1].$ Now onsider:
$\begin{align} \displaystyle \int_{0}^{1}g(x)dx&= \int_{0}^{1}F(x)dx -\int_{0}^{1}f(x)dx\\&= \displaystyle \int_{0}^{1}\int_{0}^{x}f(t)dtdx -\int_{0}^{1}f(x)dx\\&= \displaystyle\int_{0}^{1}\int_{t}^{1}f(t)dxdt-\int_{0}^{1}f(x)dx\\&= \displaystyle \int_{0}^{1}f(t)(1-t)dt -\int_{0}^{1}f(x)dx\\&= \displaystyle -\int_{0}^{1}tf(t)dt\end{align}$
$g(x)>0 \,\forall \, x \in [0,1] \implies \displaystyle \int_{0}^{1}g(x)dx >0$
$\Rightarrow \displaystyle -\int_{0}^{1}tf(t)dt>0$ 
$\Rightarrow \, \exists \, c \in (0, 1)$ such that $cf(c) <0$ and therefore $f(c)<0.$
Since $f$ is a continuous function on a closed and bounded interval, it attains its bounds. In particular $\exists \, d \in [0, 1]$ such that $f(d)\leq f(x) \, \forall \, x \in [0,1].$ Clearly $d\neq 0, 1$ or else $f(x) \geq 0 \, \forall x \in [0,1]$ contradicting the fact that $f(c) <0.$ Therefore $d \in (0,1)$ and since it is a point of minimisation, $f'(d) =0.$ Then $f(d)= f(d) -f'(d) =g'(d)>0> f(c)$ contradicting the fact that $d$ is a point of minimisation of $f.$ Therefore our hypothesis that $g$ is injective is false and hence $g$ is not injective and there exists $a, b \in [0, 1]$ with $a \neq b$ such that $g(a) =g(b).$
Then since $g$ restricted to $[a, b]$ satisfies the conditions for Rolle's Theorem, there exists at least one $x_0 \in [0,1]$ such that $g'(x_0)=0$ which implies $f(x_0)=f'(x_0).$
Note that the proof follows almost identically if we assume $g$ to be monotone decreasing in the proof of the claim.
