Let $p$ denote a positive real number. Can the function

$$
f(x)=-\frac{1}{2 p} x^{2}+x
$$

defined on the closed interval $\left[0 ; \frac{4}{p}\right]$ take a value greater than 1?
Consider the curve given by the equation $y=-\frac{1}{2 p} x^{2}+x$. This is a parabola that opens downwards $(p>0)$, with roots at $x_{1}=0, x_{2}=2 p$, and its vertex at the point $T\left(p ; \frac{p}{2}\right)$. When examining the graph of the function $f(x)$, we obtain an arc of the parabola in question. One endpoint of this arc is the origin of the coordinate system, and the other endpoint is the point $P\left(\frac{4}{p} ; \frac{4 p^{2}-8}{p^{3}}\right)$.

a) If the arc in question contains the vertex, then the maximum value of $f(x)$ is clearly $\frac{p}{2}$. This holds if $0<\frac{4}{p}<2 p$, which simplifies to $0<p<2$. For $p>2$, it follows. Let's check if $\frac{4 p^{2}-8}{p^{3}}>1$ can hold in this case, i.e.,

$$
p^{3}-4 p^{2}+8>0
$$

This can be rewritten as:

$$
(p-2)\left(p^{2}-2 p-4\right)>0
$$

Since $p>2$, we have $p^{2}-2 p-4<0$, or $(p-1)^{2}<5$. The solution to this is $1-\sqrt{5}<p<1+\sqrt{5}$.

Combining all of this, the function $f(x)$ can take values greater than 1 if and only if $2<p<1+\sqrt{5}$.

For example, when $p=3$, the maximum value of $f(x)$ is $\frac{28}{27}$, which is attained at $x=\frac{4}{3}$, but $f(x)$ is also greater than 1 at any point between $3-\sqrt{3}$ and $\frac{4}{3}$.

Based on the work of Gergely Bábosik (Vác, Industrial Secondary School III. grade)