= {a,b}, L = { s : s starts and ends with b, #a(s) = n, 0}. For example, s = babaabab L because s starts and ends with b, and #a(s) = 4 = 2; b L; s = baabab L because #a(s) = 3, and 3 n for any n 0; and s = babaaba L because s does not end with b.

Use the CF Pumping Theorem to show that L CFLs. If you wish, you can also use the closure property of the CFLs under intersection with an RL.