Answer : D
Let's analyze each language one by one.
1. $max(L) :$ Informally, and in simple words, $max(L)$ is just a subset of strings of $L$ such that those strings are maximal i.e. those strings cannot be appended with anything. i.e. If $w$ is a string in $L$ then it will be in $max(L)$ if and only if $wx$ is Not in $L$ for any(all) $x$ such that $|x| \geq 1$
2. $L1 = $ $\left \{ a^ib^jc^k| k \leq i \,\, or\,\,k\leq j; i,j,k \geq 0 \right \}$
$L1$ consists of the Union of the following languages : $a^*,b^*,a^*b^*,$ $a^{\geq m} b^nc^m$ and $a^n b^{\geq m}c^m \,\,such \,\,that \,\,m\geq 1$
Now, If you analyze each of the sublanguages of $L1, $ You can always append something to $a^*$ (append one more $a$), You can always append something to $b^*$ (append one more $b$), You can always append something to $a^*b^*$ (append one more $b$).
Coming to last two sublanguages i.e. $a^{\geq m} b^nc^m$ and $a^n b^{\geq m}c^m \,\,such \,\,that \,\,m\geq 1,$ You can only append $c's$ to the strings of these languages, that too only when $\#(c)$ is strictly less than $max(\#(a), \#(b)). $ (Try, check and see)
So, we can say that the maximal strings of $L1$ are of the form $a^m b^nc^m, such\,\,\, that\,\,\, n\leq m$ Or $a^n b^m c^m, such\,\,\, that\,\,\, n\leq m \,\,;Where \,\,m\geq 1$
Hence, $max(L1) = $ $\left \{ a^mb^nc^m| n\leq m, m \geq 1 \right \}$ $\cup$ $\left \{ a^nb^mc^m| n\leq m, m \geq 1 \right \}$
Now, It is easy to see that $max(L1) $ is Non-CFL ( Because Two comparison conditions).
3. $L2 = $ $\left \{ a^ib^jc^k| i,j,k \geq 0 \right \}$
$Max(L2)$ is empty language because You can always append something at end of every string in $L2$ (Hint : Just append the last symbol of $w$ at the end of $w$ )
Hence, $max(L2)$ is Empty, Hence, Regular, hence, CFL.
One thing to Note from $max(L1)$ is That "set of $CFL's$ is Not closed under $max$ Operation."
