In the same place you're injecting form elements, inject a JavaScript variable that your $(document).ready()
can pick up
In PHP
if ($form_failed)
echo '<script>var showAdditionalQuestion = 1;</script>';
Then in JS
$(document).ready(function(){
if (showAdditionalQuestion)
$('#showAdditionalQuestion').show();
});
(Or just inject the code so show the section directly)
This is probably more flexible and robust than just checking if the checkbox is checked on load (what if a form input is set but you don't want to show that section on load?), but that's an alternative