74th Romanian Mathematical Olympiad

Consider the set $A=\{x\in I\mid f^{\prime \prime }(x)\ne 0\}$. If, by way of contradiction, $A\ne \varnothing$, as $f^{\prime \prime }=(f^{\prime }{)}^{\prime }$ has the intermediate point value on $I$, the set $A$ cannot be a singleton.

So, let $a,b\in A$, $a<b$. We get $f(a)=f(b)=0$. The function $g:I\to \mathbb{R}$, $g(x)=f(x)f^{\prime }(x)$, for all $x\in I$, is increasing on $I$ as $g^{\prime }(x)=(f^{\prime }(x){)}^2+f(x)f^{\prime \prime }(x)=(f^{\prime }(x){)}^2\ge 0$, for any $x\in I$. As a consequence, by $g(a)=g(b)=0$, we obtain $g(x)=0$, for all $x\in [a,b]$, so $g^{\prime }=0$. Then $f^{\prime \prime }(a)={\lim }_{x\searrow a}\frac{f^{\prime }(x)-f^{\prime }(a)}{x-a}=0$, a contradiction. This shows that $A=\varnothing$.