Consider the following languages. $$\begin{aligned} L_1 &= \{\langle M \rangle \mid M \text{ takes at least 2016 steps on some input}\}, L_2 &= \{\langle M \rangle \mid M \text{ takes at least 2016 steps on all inputs}\} \text{ and} L_3 &= \{\langle M \rangle \mid M \text{ accepts } \varepsilon\}, \end{aligned}$$ where for each Turing machine $M$, $\langle M \rangle$ denotes a specific encoding of $M$. Which one of the following is TRUE?

To determine the decidability (recursiveness) of each language:

1. Language $L_1$: In $2016$ steps, a Turing Machine $M$ can scan at most the first $2016$ cells of its tape. Therefore, its behavior within the first $2016$ steps depends exclusively on the prefix of the input string of length at most $2016$. Over a finite input alphabet $\Sigma$, the set of all possible input strings of length at most $2016$ is finite. A decider can simulate $M$ on each of these finitely many input strings for up to $2016$ steps. If $M$ takes at least $2016$ steps on at least one such input, accept; otherwise, reject. Since the simulation is guaranteed to terminate in finite time, $L_1$ is recursive.

2. Language $L_2$: Using the exact same reasoning, if $M$ takes at least $2016$ steps on all inputs of length at most $2016$, it will take at least $2016$ steps on every possible input (since longer inputs cannot affect the first $2016$ steps). A decider can simulate $M$ on all inputs of length $\le 2016$ up to $2016$ steps. If $M$ executes at least $2016$ steps on every one of them, accept; otherwise, reject. Thus, $L_2$ is recursive.

3. Language $L_3$: $L_3$ is the empty string (or blank tape) acceptance problem for Turing machines. By Rice's Theorem (or via standard reduction from the Halting Problem $H_{\text{TM}}$), deciding whether a Turing machine accepts the empty string $\varepsilon$ is undecidable. Thus, $L_3$ is not recursive.

Therefore, $L_1, L_2$ are recursive and $L_3$ is not recursive, matching option (C).