= Solution
{c}
Use the <dimensionless variables>
$$
A=\frac{a^2}{4R^2},\qquad x=\frac{r^2}{4R^2},\qquad\beta=\frac{v}{\sqrt{2gR}}.
$$
The stated $R>a/2$ gives $0<A<1$. Since $f(r)=4Rx(1-x)$, the inequality from <5/b> factorizes as
$$
\boxed{D(r)=4R^3(x-A)\left[4x^2-4(1-A)x+\beta^2\right]\geq0.}
$$
The two additional possible <turning points> are
$$
x_\pm=\frac{1-A\pm\sqrt{(1-A)^2-\beta^2}}2.
$$
They are real exactly when $\beta+A\leq1$. This exposes the intended critical value
$$
\boxed{\left(\frac{v}{\sqrt{2gR}}+\frac{a^2}{4R^2}\right)_{\rm crit}=1,}
$$
but the ordering of the initial radius $A$ relative to these roots must also be checked.
First suppose $0<A<1/3$. Then $A<(1-A)/2$. If $\beta<1-A$, the allowed branch starting at $A$ is trapped either between $x_-$ and $A$ or between $A$ and $x_-$, according to its initial direction. At the coincident-root value $\beta^2=4A(1-2A)$ it is exactly a <circular orbit>. For $\beta=0$, the inner radial bound is the axis rather than a nonzero <turning point>.
At $\beta=1-A$, the additional roots coalesce at $x_*=(1-A)/2>A$. The particle starts outward and approaches $x_*$, or
$$
r_*=R\sqrt{2(1-A)},
$$
without reaching it in finite time. Indeed $D$ is proportional to $(x-A)(x-x_*)^2$ and the radial <energy> equation gives $dt/dx$ proportional to $1/|x-x_*|$ near $x_*$. Its integral diverges. This is a bounded <separatrix> approaching an unstable <circular orbit>. For $\beta>1-A$, there is no outer root and the particle escapes. Thus \b[for $0<A<1/3$, boundedness holds exactly when $\beta+A\leq1$], with the equality case included. At $A=1/3$, the same inequality characterizes boundedness, but equality is the exact <circular orbit> at the starting radius.
\b[The source's weaker assumption $A<1$ is insufficient to make this single threshold a general criterion.] For $1/3<A<1/2$, the initial radius lies beyond the coalescing-root position. The transition is instead the unstable <circular orbit> at
$$
\beta_{\rm circ}=2\sqrt{A(1-2A)}<1-A.
$$
Below this <speed> the initial radial <acceleration> is inward and the particle remains in the inner allowed interval. At equality it stays exactly circular; above it the initial <acceleration> is outward and the accessible branch extends to infinity, even if $\beta+A<1$. For $A>1/2$, the <surface> slopes downward outward and every such launch escapes. At $A=1/2$, only $\beta=0$ remains at rest on the summit circle.
Consequently the complete classification for the intended azimuthal launch, including zero launch <speed> and exact equilibria, is
$$
\boxed{\begin{array}{ll}
0<A\leq1/3:&\text{bounded iff }0\leq\beta\leq1-A,\\
1/3<A\leq1/2:&\text{bounded iff }0\leq\beta\leq2\sqrt{A(1-2A)},\\
1/2<A<1:&\text{unbounded for every }\beta\geq0.
\end{array}}
$$
For a concrete counterexample to the unqualified proposed criterion, take $A=3/4$ and $\beta=1/10$. Then $\beta+A=17/20<1$, but $f'(a)<0$ and the outward motion has no accessible outer barrier. This satisfies the printed radius assumption and escapes. A sufficient additional radius restriction for the intended threshold is $a\leq2R/\sqrt3$, with the equality and <circular orbit> qualifications above.
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