Twice continuously differentiable function such that $f’’-f<0, \forall x\in(0,1)$.

Let $$f:[0,1]\to\Bbb R$$ be twice continuously differentiable function such that $$f’’(x)-f(x)<0, \forall x\in(0,1)$$ and $$f(0)=f(1)=0$$, then which of the following statements is/are true about $$f$$?

$$1.$$ $$f$$ has at least one zero in $$(0,1).$$

$$2.$$ $$f$$ has at least two zeros in $$(0,1).$$

$$3.$$ $$f(x)>0, \forall x\in (0,1)$$.

$$4.$$ $$f(x)<0, \forall x\in (0,1)$$.

If i consider the example $$x(1-x)$$ on $$[0,1]$$ , then only option $$3$$ is correct one , but i want to solve the problem theoretically without using example or counter examples. It seems that last option is false because that $$f$$ can’t be both negative and concave . Please suggest how to discard rest options. Thanks .

Assume that $$f$$ attains its minimum at some point $$c \in (0, 1)$$. Then $$f''(c) < f(c) \le f(0) = 0 \, ,$$ contradicting the fact that the second derivative is $$\ge 0$$ at a minimum in the interior of the interval.
It follows that $$f$$ attains its minimum on $$[0, 1]$$ only at the boundary points $$x=0$$ and $$x=1$$.
• but second derivative is not given $\geq 0$ then of what contradiction? Dec 5, 2020 at 13:40
• @neelkanth: It is given that $f'' < f$, and also $f(c) \le 0$ at the minimum. It follows that $f''(c) < 0$. On the other hand, $f''(c) \ge 0$ holds generally at a minimum, this is a contradiction. Dec 5, 2020 at 13:57