 Solution

The stationary Killing vector field is $k={\partial }_t={\partial }_v$, with $k^2=-f$. The normal to $r=\text{constant}$ has squared norm $g^{rr}=f$, so $r=r_{+}$ is a null hypersurface; there $k$ is both tangent and normal. It is therefore a Killing horizon.

For a static metric of this form, the surface gravity is $\kappa =f^{\prime }(r_{+})/2$. Since $r_{+}^2{r}_{-}^2=Q^2$ and $r_{+}^2+r_{-}^2=2M$,For the extremal black hole $M=Q$, the horizons coincide and $\kappa =0$.