Context free and regular languages

Question

Let L be a given context-free language over the alphabet {a, b}. Construct L1, L2 as follows.
Let L1 = L − {xyx | x, y ∈ {a, b}^∗},
and L2 = L·L. Then,

	(A) Both L1 and L2 are regular.
	(B) Both L1 and L2 are context free but not necessarily regular.
	(C) L1 is regular and L2 is context free.
	(D) L1 and L2 both may not be context free

Comments

what is L1

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answered

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L1 = phi, since L1 = L -(a+b)* //no matter what is L

Answer 1

let $L_3=\left \{ xyx| x,y\,\, \epsilon (a+b)^{*}  \right \}$ 

$L_3=(a+b)^{*}$ is regular.

$L$ is CFL.

$L_1=L-L_3=L\bigcap (L_3)^{'}$

$(L_3)^{'}$ is regular

hence $L_1$ is CFL.

[edit]

thanks to  Anu007

But if we see carefully,

$L_1=L-(a+b)^{*}=\phi$i.e Regular

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CFL are closed under concatenation hence $L_2$ is CFl.

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Most appropriate answer is option $C$

Comments

I am not getting at all what you are saying .

Please express mathematically what you are trying to say or write clear.

what is this ?

Any language(on a,b) -(a+b)* = phi so regular why only cfl

What do you want to say ?

who is regular?$L_1$, $L_3$?

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nothing :)

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sorry , i don't know how and why i did not recognize the point ,you want to say.Thanks.Updated the answer.