ISRO CS 2025 Q.78: solution and answer

Official answer: CSource: ISRO's final answer key of 15 May 2026

1. Each part of $R$ is a star of a block. $(a^{\ast }b{)}^{\ast }$ is a concatenation of blocks of the form $a^{\ast }b$, so every block ends in $b$. Hence this part generates $\epsilon$ and only strings ending in $b$.

2. Every nonempty string ending in $b$ is in $(a^{\ast }b{)}^{\ast }$. Cut the string just after each $b$. Each piece is some $a$'s followed by one $b$, which is a block $a^{\ast }b$.

3. By symmetry, $(b^{\ast }a{)}^{\ast }$ generates exactly $\epsilon$ together with all strings that end in $a$.

4. Every string over $\{a,b\}$ is empty, ends in $b$, or ends in $a$. So $L(R)=\{\epsilon \}\cup \{\text{ends in }b\}\cup \{\text{ends in }a\}=\{a,b{\}}^{\ast }$.

5. Hence $R$ is equivalent to $(a+b{)}^{\ast }$, which is option (C).

6. Option (A), $a^{\ast }{b}^{\ast }+b^{\ast }{a}^{\ast }$: the string $aba$ is in neither part. It is not of the form $a^{\ast }{b}^{\ast }$ because an $a$ follows a $b$. It is not of the form $b^{\ast }{a}^{\ast }$ because a $b$ follows an $a$. So (A) misses a string of $L(R)$.

7. Option (B), $a^{\ast }+b^{\ast }$: the string $ab$ contains both letters, so it is in neither part, yet it ends in $b$ and is in $L(R)$.

8. Option (D), $a^{\ast }b+b^{\ast }a$: it does not generate $\epsilon$. It also misses $aa$ and $abab$. Each is in $L(R)$.