= Solution
{c}
The oscillatory numerator $\cos(\nu\theta)+\nu\sin(\nu\theta)$ has <amplitude> $\sqrt{1+\nu^2}$. Therefore the reciprocal <radius> stays bounded away from zero exactly when
$$
\nu^2-1>\sqrt{1+\nu^2}
\quad\Longleftrightarrow\quad\nu^2>3
\quad\Longleftrightarrow\quad L^2<\frac12.
$$
In this case the radial bounds are
$$
\frac{a\nu^2}{\nu^2-1+\sqrt{1+\nu^2}}
\leq r\leq
\frac{a\nu^2}{\nu^2-1-\sqrt{1+\nu^2}}.
$$
For completeness, the <potential energy> per unit <mass> is $\Phi=-1/(ar)+1/(2r^2)$, and <conservation of mechanical energy> gives
$$
\frac12\dot r^2+U(r)=\mathcal E,\qquad
U(r)=\frac{1+L^2}{2r^2}-\frac1{ar},\qquad
\mathcal E=\frac{L^2-1/2}{a^2}.
$$
This <effective potential> diverges positively at zero, has one minimum at $r=a(1+L^2)$, and tends to zero from below at infinity. Since the initial radial <velocity> is negative, the particle first reaches an inner <turning point>. If $L^2\geq1/2$, the subsequent outward motion has no finite outer <turning point> and escapes. At equality the escape <speed> tends to zero, but the <radius> is still unbounded; infinity is approached only as time tends to infinity. Hence
$$
\boxed{\text{The path is bounded exactly for }0<L<1/\sqrt2.}
$$
The minimum of $U$ permits a <circular orbit> for other initial data, but the nonzero initial radial <velocity> here excludes that exceptional solution.
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