If $$f(x) = \begin{cases} \frac{1}{|x|} & ; |x| \geq 1 \\ ax^2 + b & ; |x| < 1 \end{cases}$$ is differentiable at every point of the domain, then the values of $$a$$ and $$b$$ are respectively:

For $$f(x)$$ to be differentiable everywhere, it must first be continuous. At $$x = 1$$: from the right, $$f(1) = \frac{1}{|1|} = 1$$, and from the left, $$f(1) = a(1)^2 + b = a + b$$. Continuity requires $$a + b = 1$$.

By symmetry (since $$|x|$$ and $$ax^2 + b$$ are both even functions), the condition at $$x = -1$$ gives the same equation $$a + b = 1$$.

For differentiability at $$x = 1$$: the derivative from the right is $$\frac{d}{dx}\left(\frac{1}{x}\right)\bigg|_{x=1} = -\frac{1}{x^2}\bigg|_{x=1} = -1$$ (for $$x > 1$$, $$|x| = x$$). The derivative from the left is $$2ax\big|_{x=1} = 2a$$. Setting these equal: $$2a = -1$$, so $$a = -\frac{1}{2}$$.

From $$a + b = 1$$: $$b = 1 - (-\frac{1}{2}) = \frac{3}{2}$$.

Therefore $$a = -\frac{1}{2}$$ and $$b = \frac{3}{2}$$.

The answer is Option D: $$-\frac{1}{2}, \frac{3}{2}$$.