L _{1}* ∪ L _{2}* L _{1}*

φ* ∪ (1)*.φ*

*** > . > ∪ (priority)**

**ε **∪ (1)*.**ε**

**ε **∪ (1)*

1*

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

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

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

$= \{ 1^* \}$

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

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

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

$= \{ 1^* \}$