= Solution
{c}
In the extremal case, the squared norm of the <Killing vector field> is
$$
g(K,K)=-\left(1-\frac Mr\right)^2.
$$
For $M=|Q|>0$, it is negative for every regular $r\ne M$ and vanishes precisely at $r=M$ in the extended <spacetime>. Thus
$$
\boxed{K\text{ is null exactly on }r=M.}
$$
The static chart cannot itself test the geometry at its horizon. In the regular <Eddington-Finkelstein coordinates>, the normal to $h=r-M=0$ is $n=dr$, and
$$
g^{-1}(n,n)=g^{rr}=F,\qquad n^\sharp=\partial_v+F\partial_r.
$$
At $r=M$ this gives $n^\sharp=\partial_v=K$, a nonzero <null vector> tangent to the surface. Therefore \b[$r=M$ is a <null hypersurface>], even though $F$ has the same sign on its two sides.
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