Proving of regular language

Question

Is this regular or not

L = {w1w^R | w ∈ {0,1}* (where for any word w ∈ {0,1})*, w^R denotes the reverse of w)

Answer 1

L is not a regular language.

We can contradict by pumping lemma for regular languages.

Let assume L is regular, so there exists an integer $ k $ of pummping lemma.

$ w = a^kb $

$ w^R = ba^k $

$ ww^R = a^kbba^k \in L $

$ w' = a^{k-r}bba^k \notin L , r>0 $

Contradiction to $ w' \in L $.