UGC NET CSE | January 2017 | Part 3 | Question: 20

Question

Consider the languages $L_{1}= \phi$ and $L_{2}=\{1\}$. Which one of the following represents $L_{1}^{\ast}\cup L_{2}^{\ast} L_{1}^{\ast}$?

• A. $\{\in \}$
• B. $\{\in,1\}$
• C. $\phi$
• D. $1^{\ast}$

Comments

L ₁* ∪ L ₂* L ₁*

 φ* ∪ (1)*.φ*

* > . >  ∪ (priority)

ε ∪ (1)*.ε

ε ∪ (1)*

1*

Answer 1

Correct option is (4) 1*

Given $L_1 = \phi $, so, $L_1^* = \{ \epsilon \}$. Remember, Kleene's star operator means to repeat the strings any number of times (including 0 times).

$L_2 = \{ 1 \}$, therefore, $L_2^* = \{ \epsilon, 1, 11, 111, ... \}$, i.e $ = 1^*$.

Now, $L_1^* \cup L_2^* . L_1^*$

$= \{ \epsilon \} \cup \{ 1^* \} . \{ \epsilon \}$

$= \{ \epsilon \} \cup \{ 1^* \}$

$= \{ 1^* \}$

Answer 2

L1=ϕ,

$L1^{*}$= {$\epsilon$,$\phi$,$\phi$ $\phi$,$\phi$$\phi$$\phi$.......}={$\epsilon$}

 $L2={1}$

$L2^{*}$={$\epsilon$,1,11,111,1111............}=$1^{*}$

$L1^{*}\cup L2^{*} L1^{*}$

$\epsilon^{*}\cup1^{*}{\epsilon}$

$1^{*}$

Answer 3

option D) is correct 1*,

As, L1* ∪  L2*.L1*

 ={ϵ} ∪ {1∗}.{ϵ}

 ={ϵ} ∪ {1∗}

 ={1∗}