$\begin{array}{ll} L_1 = \{ \langle M \rangle \mid M \text{ takes more than } 2021 \text{ steps on all inputs} \} \\ L_2 = \{ \langle M \rangle \mid M\text{ takes more than } 2021 \text{ steps on some input} \} \end{array}$

Here, both $L_1$ and $L_2$ are decidable as we can have a systematic procedure in deciding them (correctly saying if an input is in $L$ or not)

For both $L_1$ and $L_2$ we have to monitor the $\textsf{TM}$ for $2021+1$ steps for all possible inputs of size $2021$ (if the input set is having $k$ alphabets we will have $k^{2020}$ possible strings which is still a finite number.)

If for all the inputs $M$ is taking more than $2021$ steps, then it means for all larger strings also it must take more than $2021$ steps and we can answer “yes” for $L_1$ or else “no”.

If for none of the inputs $M$ is taking more than $2021$ steps then it means even for any larger string $M$ won’t be taking more than $2021$ steps. So, we can answer “no” for $L_2$ or else “yes”.

Thus we correctly decided both $L_1$ and $L_2.$

NOTE: There is intersection between

- “yes” case of $L_1$ and “yes” case of $L_2$
- “no” case of $L_1$ and “yes” case of $L_2$
- “no” case of $L_2$ and “no” case of $L_1$

but not between the “yes” case of $L_1$ and “no” case of $L_2.$

Correct Option: A.