The solution of the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x}=2xy^2$ Question from pg 32 of Barron's AP Calculus

The solution of the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x}=2xy^2$ for which $y = -1$ when $x = 1$ is
(A) $y = -\frac{1}{x^2}$ for $x \neq 0$
(B) $y = -\frac{1}{x^2}$ for $x > 0$

I think A is a possible solution; however the answer given is

This function is discontinuous at $x=0$. Since the particular solution much be differentiable in an interval containing the inital value $x = 1$, the domain is $x > 0$.

Giving (B) as the correct answer.
Even from the answer given in the book, should not (A) be an acceptiable answer as well because its domain contains the domain of (B)?
 A: If we want to use a domain that is as large as possible, the general solution of the DE is $y=-\frac{1}{x^2}+C$ when $x\gt 0$, $y=-\frac{1}{x^2}+D$ when $x\lt 0$, where $C$ and $D$ are arbitrary constants.
So if we want to use this maximal domain, there is no such thing as the solution of the DE for which $y=-1$ when $x=1$. We would have to talk about the solutions,, and write $y=-\frac{1}{x^2}$ when $x\gt 0$, $y=-\frac{1}{x^2}+D$ for $x\lt 0$.
On the basis of the information given, including the initial condition, we can make no definite claim about $y(a)$ for any negative $a$.  Thus, to claim that the solution is $-\frac{1}{x^2}$ for all $x\ne 0$ is incorrect.  
A: I'd agree with mrf's comment ... if the book somewhere explicitly requires solutions to have connected domains always, otherwise I think your objection is justified and $\,A\,$ should be the correct answer.
But for the almost trivial requirement that the solution to a diff. eq. with a given condition at some point ( say, $\,(1\,,\,-1)\,$ in your case ) must be defined in some open interval around that point, I can't remeber anything about compulsory connectedness of the solution's domain, but check this in Barron's book.
