\section*{Problem 2 - 151242}

Determine the set of all positive real numbers \(r\) for which the following statement is true:

For every positive real number \(a\), the function \(f\) defined for all real \(x\) by \(f(x)=4-x^{2}-a x^{3}\) has a root between the numbers 2 - ar and 2.
}
Given \(f(0)=4>0\) and since \(a>0, f(2)=-8 a<0\), we need to consider \(f(x)=-2 x-3 a x^{2}<0\).

In the interval \(2-a r<2\), it holds. Now,

\[
\begin{aligned}
f(2-a r) & =4-(2-a r)^{2}-a(2-a r)^{3} \\
& =a\left[a^{3} r^{3}+a r(12-r)-6 a^{2} r^{2}-4(2-r)\right]
\end{aligned}
\]
is greater than zero for all positive \(a\) if and only if for all positive \(a\),

\[
g(a)=a^{3} r^{3}+a r(12-r)-6 a^{2} r^{2}-4(2-r)>0
\]

holds. If \(r>0\) is a real number that satisfies the conditions of the problem, then:

\[
\lim _{a \rightarrow 0} g(a)=4 r-8 \geq 0
\]

This means \(r \geq 2\).

Conversely, any \(r \geq 2\) has the required property. For \(r=2\), we have:

\[
\begin{array}{r}
\left(a-\frac{3}{2}\right)^{2}+\frac{1}{4}>0 \\
a^{2}-3 a+\frac{5}{2}>0 \quad \text { so } \\
8 a^{3}+20 a-24 a^{2}=f(2-2 a)>0
\end{array}
\]

for all positive real \(a\).

If \(r>2\), then due to the monotonic decrease of \(f\) and the inequality \(f(2-2 a)>0\) just proven, it follows from \(0<2-a r<2\) that \(f(2-a r)>0\). Therefore, exactly all \(r \geq 2\) have the required property.

Adapted from [3]