\section*{Problem 7 - V01107 = V601215}

For which values of \(a\) does the curve

\[
y=\frac{1}{4}\left(a x-x^{3}\right)
\]

intersect the \(x\)-axis at an angle of \(45^{\circ}\) ?
}

The x-axis is intersected at an angle of \(45^{\circ}\) if the slope \(m\) of the tangent at the zeros of (1) is equal to \(\tan 45^{\circ}=1\) or \(\tan 135^{\circ}=-1\). This means that the value of the first derivative of (1) must be \(\pm 1\) at the zeros.

1. First derivative function: \(\quad y^{\prime}=f^{\prime}(x)=\frac{1}{4}\left(a-3 x^{2}\right)\)

Zeros of \(f(x)\) : \(\quad x_{1 ; 2}= \pm \sqrt{a} ; \quad x_{3}=0\)

Function value of \(f^{\prime}(x)\) at the zeros:

\[
f^{\prime}\left(x_{1}\right)=\frac{1}{4}\left(a-3 \sqrt{-a}^{2}\right)=-\frac{a}{2} ; \quad f^{\prime}\left(x_{2}\right)=\frac{1}{4}\left(a-3 \sqrt{a}^{2}\right)=-\frac{a}{2} ; \quad f^{\prime}\left(x_{3}\right)=\frac{1}{4}\left(a-3 \cdot 0^{2}\right)=\frac{a}{4}
\]

From \(-\frac{a}{2}= \pm 1\) we get the values \(a=-2\) and \(a=2\). However, the zeros \(x_{1}\) and \(x_{2}\) do not exist for \(a<0\), so only \(a=2\) remains as a solution. From \(\frac{a}{4}= \pm 1\) we get the values \(a=-4\) and \(a=4\) (the zero \(x_{3}\) exists for all \(a\)), with the solution set of the problem being

\[
a \in\{-4,2,4\}
\]