Find non-regular $L$ such that $L \cup L^R$ is regular?
Hint: try to come up with a language $L$, such that $L \cup L^R = \Sigma^*$.

Hint2:


   Try to think on complements of known non-regular languages. 


Solution:


   Let $C = \{a^nb^n \mid n&gt;0\}$, and define $L$ to be all the strings EXCEPT for those in $C$. That is, $L = \Sigma^*\setminus C$. It is easy to see that $L$ is not Regular (closure under complement). Now let's think about $L^R$. It contains all the strings in $\Sigma^*$, EXCEPT for strings of the form $b^na^n$.
 Finally, $L\cup L^R = \Sigma^*$. Strings of the form $a^nb^n$ are not in $L$ but they are in $L^R$. Strings of the form $b^na^n$ are not in $L^R$ but they are in $L$.. al the rest of the strings appear in both languages. 
 Since $\Sigma^*$ is regular, we are done.