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**Option A.**

$L_1 = \{ \varepsilon, a, aa, aaa, aaaa, \ldots \}$

$L_2 = \{ \varepsilon, b, bb, bbb, bbbb, \ldots \}$

$\begin{align}

L_1 \cdot L_2 &= \left \{ \begin{array}{c} \varepsilon , \\ a, &b,\\ aa, &ab, &bb\\ aaa, &aab, &abb, &bbb,\\ aaaa, &aaab, &aabb, &abbb, &bbbb, & \ldots \end{array}\right \}\\[1em]

L_1 \cdot L_2 &= a^*b^*

\end{align}$

Thus, $L_1 \cdot L_2$ is Regular.

(Also, since both $L_1$ and $L_2$ are Regular, their concatenation has to be Regular since Regular languages are closed under concatenation)

However, $L_1 \cdot L_2 \neq a^nb^n$. This is because in $a^*b^*$, the number of $a$'s and $b$'s can be different whereas in $a^nb^n$ they have to be the same.

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