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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 .

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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 $x=1$.

So (3) is true, which implies that the other statements are wrong.

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    $\begingroup$ but second derivative is not given $\geq 0$ then of what contradiction? $\endgroup$
    – neelkanth
    Dec 5, 2020 at 13:40
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    $\begingroup$ @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. $\endgroup$
    – Martin R
    Dec 5, 2020 at 13:57
  • $\begingroup$ yes sir now cleared . $\endgroup$
    – neelkanth
    Dec 5, 2020 at 14:06
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    $\begingroup$ thank you sir ... $\endgroup$
    – neelkanth
    Dec 5, 2020 at 14:08

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