= Solution
{c}
Use an <affine parameter> $s$ on the <geodesic>, so its <tangent vector> $U$ obeys $\nabla_UU=0$. The <product rule>, <metric compatibility>, and the <Conformal Killing equation> give
$$
\begin{aligned}
\frac{d}{ds}(X_\mu U^\mu)
&=U^\nu U^\mu\nabla_\nu X_\mu+X_\mu(\nabla_UU)^\mu\\
&=U^\nu U^\mu\nabla_{(\nu}X_{\mu)}
=\lambda g_{\mu\nu}U^\mu U^\nu.
\end{aligned}
$$
Hence
$$
\boxed{\frac{d}{ds}(X\cdot U)=\lambda\,g(U,U).}
$$
\b[It is conserved on every <null geodesic> with an <affine parameter>, but generally not on other <geodesics>.] On a <timelike geodesic> parametrized by <proper time>, $g(U,U)=-1$, so the derivative is $-\lambda$. A <Killing vector field>, or a <geodesic> on which $\lambda$ vanishes, supplies the exceptional conservation. This is the <conformal Killing geodesic identity>.
For an arbitrary parameter with $\nabla_UU=fU$, the right-hand side acquires $f(X\cdot U)$. Thus even for a <null geodesic>, the simple conserved expression presupposes an <affine parameter>.
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