theory of computation

Question

Let L = {aⁿb^la^k: n = l or l ≠ k}. The language is 

A) regular 

B) DCFL but not regular.

C) NDCFL but not DCFL.

D) context sensitive but not CFL.

The Option is C) right ...??? Please verify ...

Answer 1

$L1=\left \{a^{n}b^{n}a^{*} \right \}$

$L2=\left \{ a^{*}b^{m}a^{n}|m\neq n\right \}$

both L1 and L2 individually are DCFL but $L1\bigcup L2$ are not we can't  discriminate working of both on seeing a,

Comments

so it is a NDCFL right ?

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@vignesh yes..

Answer 2

n=l or l!=k means it can be n=l, l!=k, n=l and l!=k..in last case 2 comparisons are required so it cannot be cfl..it should be csl..plz correct me if I am wrong.