Answer is B
LR(0) ⊂ LR(1) = LR(2) ........LR(k) = LR(k+1) ; for k >= 1 : LR(k)=LR(k+1)
so, Languages of grammars parsed by LR(2) parsers is not a strict super set of the languages of grammars parsed by LR(1) parsers
although they both are same ;
if you compare their grammars :
LR(0) ⊂ LR(1) ⊂ LR(2) ........LR(k) ⊂ LR(k+1)