A question concerning linear algebra (Multiple options could be correct!)

Q. Let $A$ be a $4\times 4$ matrix over $\mathbb R$ such that
$\rho(A)=2$, and $A^3=A^2\ne O$. Suppose that $A$ is not diagonalizable.
Then
(a) One of the Jordan blocks of the Jordan Canonical form of $A$ is
$\pmatrix{ 0 & 1 \\ 0 & 0}.$
(b) $A^2=A\ne O.$
(c) $\exists$ a vector $v$ such that $A^2v=O.$
(d) The characteristic polynomial of $A$ is $x^4-x^3.$

Here, $\rho(A)$ means the rank of $A,$ and $O$ denotes the zero matrix of order $4$. My attempt:
If $f(x)=x^3-x^2=x^2(x-1),$ then $f(A)=O.$ So, the minimal polynomial of $A, m(x)$ (say), must divide $f(x).$ Since $A$ is not diagonalizable, the only possible choice for $m(x)$ is $x^2(x-1).$ This imples that there is a Jordan block of size $2$ in the Jordan Canonical form of $A$ corresponding to the eigen value $0$ of $A$. Therefore, option (a) is correct.
If $A^2=A\ne O$ were true, that would mean that the matrix $A$ is idempotent, hence diagonalizable. However, it is not so. Thus, we see that option (b) is false. (right?)
$A^2v=O$ represents a homogeneous system of linear equations. As such, it is always consistent. So, option (c) is correct! (right?)
Since $m(x)\mid c(x),$ the characteristic polynomial of $A.$ There are two possible choices for $c(x)$, viz., $x^3(x-1)$ and    $x^2(x-1)^2.$ In addition, the geometric multiplicity of the eigen value $0, \gamma(0)$ (say), equals the nullity of $A-0I_4,$ that is: $\gamma(0)=\eta(A-0I_4)=\eta(A)=4-\rho(A)=4-2=2.$ But, $A$ is not diagonalizable. Therefore, the algebraic multiplicity of the eigen value $0$, $\alpha(0)$ (say), must not equal $\gamma(0).$ Now, we find that among the two feasible choices for $c(x),$ only the former one, i.e, $c(x)=x^3(x-1)=x^4-x^3,$ satisfies this condition. Because $\alpha(0)=3\ne 2=\gamma(0)$ in this case. Hence, option (d) is correct as well. (right?)
 A: Option (a): I know you didn’t ask, but correct! The minimal polynomial must divide $x^2(x - 1)$, and cannot be square-free (otherwise it’s diagonalisable). So yes, the minimal polynomial must be $x^2(x - 1)$. The power of $x$ represents the largest Jordan block corresponding to eigenvalue $0$, so a $\pmatrix{0&1\\0&0}$ block must occur in the JNF.
Option (b): Correct! Alternatively, if $A^2 = A$, then the minimal polynomial must divide $x^2 - x = x(x - 1)$, which would contradict $x^3 - x^2$ being the minimal polynomial.
Option (c): Technically correct! Your answer perfectly suits the question as written, but I’m sure the intention behind the question was to specify $v \neq 0$. A homogeneous system may only have the $0$ solution, so eliminating $v = 0$ would make your answer invalid. In this particular case, however, there will be a non-trivial solution to $A^2v = 0$, simply because $0$ is an eigenvalue of $A$. We can even choose $v \neq 0$ and $Av \neq 0$ (though the first condition is redundant), by choosing a generalised eigenvector of index $2$. So, your answer is $100\%$ correct, but I think whomever wrote the question was hoping for more, and just wrote the question badly.
Option (d): Incorrect! Yes, there are two possibilities of characteristic polynomials, and both are possible. For example, consider the following JNF matrices:
$$J_1 = \pmatrix{0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1}, \qquad J_2 = \pmatrix{0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1}.$$
The characteristic polynomials are, respectively, $x^3(x - 1)$ and $x^2(x - 1)^2$. The geometric multiplicities of $\lambda = 0$ of these matrices are $2$ and $1$ respectively, while their algebraic multiplicities are $3$ and $2$ respectively. Neither are diagonalisable (which can be seen by the $2 \times 2$ Jordan block in each of them).
EDIT: As pointed out in the comments, the rank is assumed to be $2$. This eliminates $J_2$, and anything similar to it, from being $A$. Thus, the answer given in the question is indeed correct.
