= Solution
{c}
With $\tau$ the <proper time> of the massive <test particle>, the <four-velocity> has squared norm $-1$:
$$
-1=-F\dot t^2+F^{-1}\dot r^2+r^2\dot\phi^2.
$$
The <conserved quantities> in part (i) give
$$
F\dot t=\frac1m\left(E-\frac{eQ}{r}\right),\qquad \dot\phi=\frac l{r^2}.
$$
Multiplying the normalization by $F$ and substituting gives the <charged-particle radial equation in Reissner-Nordstrom spacetime>:
$$
\boxed{\dot r^2=\frac1{m^2}\left(E-\frac{eQ}{r}\right)^2-F(r)\left(1+\frac{l^2}{r^2}\right).}
$$
At a radial <turning point> $r_*>0$, the left side is zero, so
$$
F(r_*)\left(1+\frac{l^2}{r_*^2}\right)
=\frac1{m^2}\left(E-\frac{eQ}{r_*}\right)^2\ge0.
$$
The parenthesis on the left is strictly positive. Hence $F(r_*)\ge0$. The factorization in part (b) now yields
$$
\boxed{r_*\ge r_+\quad\text{or}\quad 0<r_*\le r_-.}
$$
\b[There is no radial <turning point> between the horizons.] Indeed, for $r_-<r<r_+$, the right-hand side of the radial equation is strictly positive, independently of the <electric charge> $e$ of the <test particle>. This is the <charged-particle turning-point exclusion in Reissner-Nordstrom spacetime>.
The inequalities are necessary, not a claim that a radial <turning point> exists for every $E,l,e$. At a regular horizon crossing a <timelike vector> cannot be tangent to a <null hypersurface>, so a massive <test particle> actually has $\dot r\ne0$ there. Algebraic equality at $r_\pm$ would require $E=eQ/r_\pm$ and does not describe a regular finite-<proper time> radial <turning point> on the horizon. In the <Schwarzschild spacetime> limit $Q=0$, the positive-radius inner alternative disappears.
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