Closure against the operator $A(L)=\{ww^Rw \mid w \in L \wedge |w| \lt 2007\}$ I would like your help with the following question:


  Let $L$ be a language, and operator $A(L)=\{\,ww^Rw \mid w \in L\ \wedge\ |w| \lt 2007\,\}$ where $x^R$ is the  reversed string of $x$. Which of the
  following statements are correct?
  
  
  If $L$ is regular so $A(L)$ is regular.
  If $L$ is a CFL which is not regular then $A(L)$ is CFL which is not regular.
  If $L$ is a CFL which is not regular, then $A(L)$ is a CFL which may or may not be regular.
  If $L$ is not a CFL then $A(L)$ is not CFL.
  


What does the fact that $|w|&lt; 2007$ help me with the decision? 
For (2) I can choose $O^n1^n$ and I get that $0^n1^{2n}0^{2n}1^n$, which is not regular, but for (3),(4) I can't find an examples to refute it. The answer is 3, but I can't understand why, since $A(L)= ww^R \circ  w$  but $ww^R$ is not regular.
Since $|w| &lt; 2007$, the number of strings like $ww^Rw$ is finite. So $A(L)$ is finite for all $L$ and is hence regular.