 Solution

At coexistence a nonzero $Q_t$ must be stationary and have the same free energy as $Q=0$. Dividing $f(Q_t)=0$ by $6Q_t^2$ and $f^{\prime }(Q_t)=0$ by $6Q_t$ givesEliminating $a$ yields $Q_t=-b/(6c)$ and then $b^2=12ac$. The allowed nonpositive root therefore givesAs $b$ is decreased through $b_t$, the equilibrium scalar order jumps from $0$ for $b>b_t$ to $Q_t$ just on the $b<b_t$ side. At $b=b_t$ both values coexist; the stable aligned branch below it is $Q_{+}$ from part (e). As a global-minimum check, at coexistenceThere are two equal minima and a barrier at $Q_t/2$, of height $a^2/(8c)$. Comparing these minima, rather than finding only the stationary-point discriminant, determines the transition.

This is the unrestricted Landau answer, compatible with the physical range of a uniaxial nematic order parameter when $a\le 3c$ so that $Q_t\le 1$. If one instead enforces $Q\in [-1/2,1]$ for arbitrary positive coefficients, negative $Q$ cannot win for $b\le 0$, and bounded quartic nematic coexistence minimizes $a/Q+3cQ$ over $0<Q\le 1$. For $a>3c$ its minimum is at the boundary, giving $b_t=-(a+3c)$ and a jump $0\to 1$. No coefficients were supplied to choose this boundary case; the near-transition Landau approximation normally uses the interior result.