= Solution
{c}
The <function> $F$ is an <odd function>, passes through $(0,0)$, and has the horizontal <asymptote> $F=0$ at both infinities. Its <derivative> is
$$
F'(x)=\frac{1-x^2}{(1+x^2)^2}.
$$
It decreases for $|x|>1$ and increases for $|x|<1$, with its lowest value $F(-1)=-1/2$ and highest value $F(1)=1/2$. These features determine the <graph of a function> shown below.
For $0<\epsilon<1/2$, the equation is the <quadratic equation> $\epsilon x^2-x+\epsilon=0$. The <quadratic formula> gives exactly two real <equation roots>:
$$
x_\pm=\frac{1\pm\sqrt{1-4\epsilon^2}}{2\epsilon}.
$$
The <binomial series> gives $\sqrt{1-4\epsilon^2}=1-2\epsilon^2-2\epsilon^4-4\epsilon^6+O(\epsilon^8)$, hence the two requested <asymptotic expansions> are
$$
\boxed{x_-=\epsilon+\epsilon^3+O(\epsilon^5),\qquad
x_+=\epsilon^{-1}-\epsilon+O(\epsilon^3).}
$$
Their product is exactly one. Both are positive, with $x_-<1<x_+$; there are no negative <equation roots> because $F(x)<0$ for $x<0$. At $\epsilon=1/2$ the two real <equation roots> coincide at one, and for $\epsilon>1/2$ there are none. In the stated limit $\epsilon\to0^+$, these two branches are therefore exhaustive.
\Image[/past-exam-of-the-differential-equations-2-course-of-the-university-of-oxford/2021-roots.png]
{title=Two positive intersections of the rational function with a small positive level}
{description=Original sketch of $F(x)=x/(1+x^2)$ with the two real <equation roots> marked for $\epsilon=0.15$. The curve also shows its negative lowest point and positive highest point.}
Back to article page
