# Let $y:[0,1] \rightarrow \mathbb R$ be a twice continuously differentiable function such that $y''(x)-y(x)<0$

I came across the following problem that says:

Let $y:[0,1] \rightarrow \mathbb R$ be a twice continuously differentiable function such that $y''(x)-y(x)<0$ for all $x \in (0,1)$ and $y(0)=y(1)=0.$ Then which of the following statement(s) is/are true?

(a) $y$ has at least two zeros in $(0,1)$.
(b) $y$ has at least one zero in $(0,1)$.
(c) $y(x)>0$ for all $x \in (0,1)$.
(d) $y(x)<0$ for all $x \in (0,1)$.

Can someone point me in the right direction? Thanks in advance for your time.

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This example $y=1-(2x-1)^2$ eliminates all but one. – P.. Jan 8 at 12:55
@Pambos I have one question.Solving $y(x)=0,$ we see that $x=0,1$ which does not lie in $(0,1)$. So how can option $(a)$ be correct? – learner Jan 8 at 13:06
Hint: (a) is not the correct answer. – P.. Jan 8 at 13:07
@Pambos I have got it.It is not $(a),(b)$ or $(d)$. So it must be $(c)$. – learner Jan 8 at 13:11