Define $\mathrm{f}(x)=\left\{\begin{array}{cl}\mathrm{b}-\mathrm{a} x & , \text { if } x 2\end{array}\right…

The limit $\lim_{x \to 2} f(x)$ exists if and only if the left-hand and right-hand limits at $x = 2$ are equal.

For $x < 2$, $f(x) = b - a x$, so the left-hand limit is $\lim_{x \to 2^-} f(x) = b - 2a$.

For $x > 2$, $f(x) = a + 2b x$, giving the right-hand limit $\lim_{x \to 2^+} f(x) = a + 4b$.

Setting these equal: $b - 2a = a + 4b$.

This simplifies to $-3b = 3a$, and dividing both sides by $3b$ (since $b \neq 0$ for the ratio to be defined) yields $\frac{a}{b} = -1$.

Final answer: $\boxed{\text{B}}$