= Solution
In a smooth scalar <Landau free energy>, a line of coexistence between two distinct minima can end by those minima and the intervening maximum coalescing. At a generic endpoint with no competing third phase,
$$
\boxed{V'(M_c)=V''(M_c)=V^{(3)}(M_c)=0,\qquad V^{(4)}(M_c)>0.}
$$
The first condition is stationarity, the second removes the inverse susceptibility, and the third is needed because nonzero cubic curvature at a zero-quadratic point would prevent it being a stable minimum. After shifting $M=M_c+\psi$ and mixing two control parameters, the local potential is $V_c+R\psi^2/2+U\psi^4/4-H\psi+\cdots$, $U>0$. Its two-phase coexistence line $H=0,R<0$ reaches $R=H=0$, with $\Delta M=2\sqrt{-R/U}\to0$ and diverging <magnetic susceptibility>. This is the familiar <critical endpoint of a quartic Landau free energy>.
\b[The word “must” needs the two-phase endpoint assumptions.] A first-order line can instead end at a triple point or a boundary of the available control parameters. There is an explicit scalar counterexample in the sextic potential: fix $u<0<v$ and vary $(r,h)$. The $+M/-M$ coexistence line at $h=0$, $r<3u^2/(16v)$ ends when the $M=0$ phase also becomes globally stable. At that triple point the nonzero minima remain $\pm\sqrt{-3u/(4v)}$; their jump does not vanish. Thus the literal unrestricted assertion is false. The continuous-endpoint argument above establishes the intended case and distinguishes it from this counterexample.
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