= Solution
{c}
First take $a\ne0$, put $A=|a|$, and suppose $k>0$. The <force> is the negative <derivative> of the <potential energy>:
$$
U(x)=-\frac{kx}{A^2+x^2},\qquad
\boxed{F(x)=-U'(x)=\frac{k(A^2-x^2)}{(A^2+x^2)^2}.}
$$
The <potential energy> is an <odd function>, passes through $U(0)=0$, and approaches zero from above as $x\to-\infty$ and from below as $x\to+\infty$. Its <stationary points> are $x=\pm A$:
$$
U(-A)=\frac{k}{2A},\qquad U(A)=-\frac{k}{2A},\qquad
U''(-A)=-\frac{k}{2A^3},\qquad U''(A)=\frac{k}{2A^3}.
$$
Thus the sole <stable equilibrium> is the <local minimum> at $x=A$; the <local maximum> at $x=-A$ is an <unstable equilibrium>. The requested values at the <stable equilibrium> are
$$
\boxed{x_{\rm stable}=A,\qquad U_{\rm stable}=-\frac{k}{2A},\qquad F_{\rm stable}=0.}
$$
The <force> is an <even function>, positive for $|x|<A$, negative for $|x|>A$, and zero at $x=\pm A$. Its central maximum is $F(0)=k/A^2$. Since
$$
F'(x)=\frac{2kx(x^2-3A^2)}{(A^2+x^2)^3},
$$
its two minima occur at $x=\pm\sqrt3A$, with $F=-k/(8A^2)$, and $F\to0^-$ at both ends. These values fix the shapes of both requested sketches.
\Image[/past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2021-potential-and-force.png]
{title=Dimensionless potential energy and force for positive k and nonzero a, with the stable and unstable equilibria marked}
In one dimension, \b[attraction towards the origin] means $xF(x)<0$, while \b[repulsion from the origin] means $xF(x)>0$. With that explicitly stated convention,
$$
\boxed{\begin{aligned}
\text{attraction:}&\quad -A<x<0\quad\text{or}\quad x>A,\\
\text{repulsion:}&\quad x<-A\quad\text{or}\quad 0<x<A.
\end{aligned}}
$$
A positive <force> alone means rightward, not necessarily outward. At the origin $F(0)\ne0$, so it is not an <equilibrium>; the inward/outward distinction is undefined exactly there.
The sign assumptions matter because the paper calls $a,k$ merely constants. For $k<0$ both curves reverse sign, the <stable equilibrium> is $x=-A$, its <potential energy> is $-|k|/(2A)$, and the attraction/repulsion intervals interchange. For $k=0$ the <force> vanishes everywhere on the nonsingular domain and there is no isolated <potential energy> minimum. If $a=0$ and $k\ne0$, then $U=-k/x$ for $x\ne0$ and $F=-k/x^2$, with no finite <equilibrium>.
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