Woodin's HOD Dichotomy Theorem says that if an extendible cardinal exists, then either $V$ and $HOD$ are rather close or rather far apart. My question is whether the "far" case can be strengthened in analogy to Jensen's Covering Lemma to say something more about large singulars. Suppose $\delta$ is extendible and every regular $\kappa \geq \delta$ is measurable in $HOD$. Does there exist a singular $\lambda > \delta$ which is singular in $HOD$?

Woodin's HOD Dichotomy Theorem says that if an extendible cardinal exists, then either $V$ and $HOD$ are rather close or rather far apart. My question is whether the "far" case can be strengthened in analogy to Jensen's Covering Lemma to say something more about large singulars. Suppose $\delta$ is extendible and every regular $\kappa \geq \delta$ is measurable in $HOD$. Does there exist a singular $\lambda > \delta$ which is singular in $HOD$?

EDIT: The comment by Gabe Goldberg gives an easy "yes" answer. Here's a harder version of the question. Suppose $\delta$ is extendible and $\lambda$ is the least cardinal above $\delta$ such that $V_\lambda \models ZFC$. Are all singular cardinals in the interval $(\delta,\lambda]$ singular in $HOD$?