= Solution
Write $F(x)=\mu^2-(x^2-a)^2$. Its <fixed points> satisfy $x^2=a\pm|\mu|$, retaining only nonnegative right sides. At a simple <fixed point>,
$$
F'(x)=-4x(x^2-a).
$$
Thus on the outer branches $x^2=a+|\mu|$, the positive point is stable and the negative point unstable. On the inner branches $x^2=a-|\mu|$, the positive point is unstable and the negative point stable.
For $a>0$, there are four simple points for $0<|\mu|<a$, two for $|\mu|>a$, and a saddle-node at each $(\mu,x)=(\pm a,0)$, where the two inner points coalesce. At $\mu=0$, there are also bifurcations at $x=\pm\sqrt a$. The smooth branches $x=\pm\sqrt{a+\mu}$ and $x=\pm\sqrt{a-\mu}$ cross and exchange stability. Locally $F=\mu^2-4a\xi^2+\text{higher terms}$, where $\xi=x-x_0$. After a parameter-dependent shift this is a <transcritical bifurcation>. It is not an ordinary saddle-node: two nearby branches exist on both sides of zero. At zero itself each crossing point is semistable, attracting from the right and repelling from the left.
For $a<0$, the only possible branch is $x=\pm\sqrt{a+|\mu|}$, existing for $|\mu|\geq|a|$. Each point $(\pm|a|,0)$ is a <saddle-node bifurcation>. There are no <fixed points> for $|\mu|<|a|$; outside that interval the positive point is stable and the negative point unstable. At either threshold $F\sim2ax^2<0$, so the zero point is semistable. The first row of the diagram displays these branches; solid curves are stable and dashed curves unstable.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/ii/paper-2-bifurcations.png]
{title=Fixed-point bifurcation diagrams for both scalar systems with positive and negative a; stable branches are solid and unstable branches dashed}
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