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Given the following two languages :

$L_{1}=\left\{a^{n}b^{n}|n\geq 1\right\} \cup \left\{a\right\}$

$L_{2}=\left\{w C w^{R}|w \in \left\{a, b\right\}^{*}\right\}$

Which statement is correct ?

1. Both $L_{1}$ and $L_{2}$ are not deterministic.
2. $L_{1}$ is not deterministic and $L_{2}$ is deterministic.
3. $L_{1}$ is deterministic and $L_{2}$ is not deterministic.
4. Both $L_{1}$ and $L_{2}$ are deterministic.

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I think A is the answer.

L1 can be accepted by NPDA. // It is CFG

L2 can be accepeted by DPDA or NPDA // It is CFG

Correct me if I am wrong

Both are deterministic CFL

L2... Insert symbol into stack until C doesn't appear... If C the skip it and compare top of stack with input symbol.. Because of C it becomes deterministic...

L1..

If "a " comes then we can't decide ...but after "a"  , "a" or "b" comes then first part of given language is considered and if epsilon comes then second part need to consider... We can easily recognize it with the help of DPDA...
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@papesh

What about union {a}  in L1 ?

I think answer should be (B).

Because, We know that deterministic CFL is not closed under Union.

Please Correct me if i m wrong.

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I think you are correct. For L1 we will not able to give DCFL.So,option B is correct option.