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3 Answers

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Original RegExp :: $0(0+1)^*0$ and minimal string generated by this RegExp is $00$ and it generates all strings that start and end with $0$ except string $0$.

Now RegExp's in option (B) and (C) generate $000$ and $0000$ as their minimal string respectively and generate all strings that start and end with $0$ afterwards. (none of these two accepts $00$)
So, Language accepted by $0(0+1)^*00$ and $00(0+1)^*00$ is a proper subset of language generated by $0(0+1)^*0$

Regular Expression in option (A) $0(0+1)^*(0+00)$ is equivalent to $0(0+1)^*0 + 0(0+1)^*00$. Our original RegExp and and additional RegExp that is a proper subset of $0(0+1)^*0$.
Hence, (A) is correct answer.

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Options (a) , (b) , (c) are accpeted by same FA but

Regular expression 0(0|1)* 0 can generate minimal string as 00.

So Minimal String generated by option (a) is 00 or 000 , option (b) is 000 and by option (c) is 0000 .

So most appropriate answer  would be option (a) .

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