Let x and y be as in part $($a$),$ and let l be the largest index such that xl $!=$ yl$.$

Without loss of generality, assume that xl $=0$ and yl $=1.$

Consider the strings x$\text{'}=$ x$0^($l$-1)$ and y$\text{'}=$ y$0^($l$-1),$ that is$,$ we append l$-1$ zeros to the end of x and y$,$ respectively.

Then x$\text{'}$ and y$\text{'}$ are both in L$\_$r$,$ since their r$-$th symbols from the right end are $0.$ Moreover, x$\text{'}$ and y$\text{'}$ differ in their $($r$-$l$)-$th symbols from the right end, which is a $1$ in x$\text{'}$ and a $0$ in y$\text{'}.$

Therefore, x$\text{'}$ and y$\text{'}$ are not equivalent with respect to L$\_$r$.$

On the other hand, since xl $!=$ yl$,$ the strings x$\text{'}$ and y$\text{'}$ must differ in their l$-$th symbols from the right end, which is a $0$ in x$\text{'}$ and a $1$ in y$\text{'}.$

Therefore, any DFA that recognizes L$\_$r must distinguish between x$\text{'}$ and y$\text{'},$ and hence must have at least $2^($r$-$l$)$ states. But this contradicts the assumption that D has less than $2^$r states. Therefore, such a DFA for L$\_$r with less than $2^$r states cannot exist.