= Solution
Using $K^2=2q\mu_0\rho\Omega^2/B_z^2$, the <finite-thickness magnetorotational instability criterion> is precisely
$$
\boxed{KH>\frac\pi2}.
$$
For the midplane-symmetric equilibrium with $B_+\ne0$, the absolute radial <magnetic field> is
$$
|B_x(z)|=\frac{|B_+|}{|\sin(KH)|}|\sin(Kz)|,\qquad 0\leq z\leq H.
$$
Its first maximum occurs at $z_* =\pi/(2K)$. The instability condition puts $z_*$ strictly inside the disc: $z_*<H$. The magnitude rises from zero to this maximum, then decreases on a nonempty interval before the surface. Thus the unstable equilibrium exhibits <nonmonotonic magnetic bending>, as illustrated below.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-321-magnetic-bending.png]
{title=Monotonic and nonmonotonic magnetic bending}
{description=The symmetric equilibrium profiles have $B_x(H)=B_+$. The weak-field example has an interior maximum and admits a growing magnetic mode.}
The link between bending and instability concerns a nonzero imposed surface inclination. If $B_+=0$, the unbent equilibrium $B_x=0$ can still have <magnetorotational instability>; the literal nonmonotonic-bending conclusion does not apply to that degenerate case. Resonances with $\sin(KH)=0$ also require the separate equilibrium analysis in part (b).
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