The candidate key is $AB$.
Since $A$ is a proper subset of $AB$, first consider $A^+$.
We have $A\to I$.
$\therefore A^+=\{A,I\}$.
Now the FD $AI\to IJ$ can be used because both $A$ and $I$ are available.
Thus $J$ is also obtained:
$A^+=\{A,I,J\}$.
Therefore, both $A\to I$ and $A\to J$ hold in $F^+$.
Since $I$ and $J$ are non-prime attributes and $A$ is a proper subset of candidate key $AB$, both represent partial dependency information that must be removed from the main relation.
Hence create $R_1(A,I,J)$.
The remaining relation is $R_2(A,B,C,D,E,F,G,H)$.
Now check the options.
In B, $J$ remains in $R_2$. But $A\to J$ is implied by $F$, so $R_2$ would still contain a partial dependency.
In C, $I$ remains in $R_2$, and the explicit partial dependency $A\to I$ remains.
Thus neither is in $\text{2NF}$.
The important trap is that a $\text{2NF}$ violation does not have to appear explicitly in the original FD list.
Dependencies in $F^+$ matter as well.
Therefore, the correct answer is A.
