L1={anbmcmdn|m,n≥1} : It is clearly DCFL (store a, then store b, then remove b if c comes, then remove a if d comes)
L2={0p1q |p>q≥0} ∪ {0p1q |q>p≥0}: It is as good as {0p1q | either p is greater than q or q is greater than p, i,e. p!=q}, It is also DCFL.
L3=L1∪L2 = DCFL ∪ DCFL = CFL (bcz DCFL is not closed under union, but CFL is, we have upgraded DCFL to CFL)
L4=L1L2 = DCFL.DCFL = CFL (bcz DCFL is not closed under concatenation, but CFL is, we have upgraded DCFL to CFL)
A. L3∩L4 : CFL ∩ CFL = CSL (bcz CFL is not closed under intersection, but CSL is, we have upgraded CFL to CSL)
B. comp(L3∩L4) : CFL ∩ CFL = CSL, comp(CSL) = CSL
C. comp(L3)∩L4 : comp(CFL) ∩ CFL = CSL ∩ CFL = CSL
D. L3.L4 : CFL.CFL = CFL (bcz CFL is closed under concatenation)
Answer should be D
follow this link for closure property: http://gatecse.in/closure-property-of-language-families/