([b]4[/b]) Let $ a$, $ b$ be constants such that $ \lim_{x\rightarrow1}\frac {(\ln(2 \minus{} x))^2}{x^2 \plus{} ax \plus{} b} \equal{} 1$. Determine the pair $ (a,b)$.
1. We start with the given limit:
   \[
   \lim_{x \rightarrow 1} \frac{(\ln(2 - x))^2}{x^2 + ax + b} = 1
   \]
   As \( x \to 1 \), the numerator \((\ln(2 - x))^2\) approaches \((\ln(1))^2 = 0\). For the limit to be finite and equal to 1, the denominator must also approach 0. Therefore, we have:
   \[
   1^2 + a \cdot 1 + b = 0 \implies 1 + a + b = 0 \implies b = -1 - a
   \]

2. Next, we apply L'Hopital's Rule because both the numerator and the denominator approach 0 as \( x \to 1 \). L'Hopital's Rule states that if \(\lim_{x \to c} \frac{f(x)}{g(x)}\) results in an indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then:
   \[
   \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}
   \]
   provided the limit on the right exists.

3. We differentiate the numerator and the denominator:
   \[
   f(x) = (\ln(2 - x))^2 \implies f'(x) = 2 \ln(2 - x) \cdot \frac{-1}{2 - x} = -\frac{2 \ln(2 - x)}{2 - x}
   \]
   \[
   g(x) = x^2 + ax + b \implies g'(x) = 2x + a
   \]

4. Applying L'Hopital's Rule, we get:
   \[
   \lim_{x \rightarrow 1} \frac{(\ln(2 - x))^2}{x^2 + ax + b} = \lim_{x \rightarrow 1} \frac{-\frac{2 \ln(2 - x)}{2 - x}}{2x + a}
   \]

5. Simplifying the limit:
   \[
   \lim_{x \rightarrow 1} \frac{-2 \ln(2 - x)}{(2 - x)(2x + a)}
   \]
   As \( x \to 1 \), \(\ln(2 - x) \to \ln(1) = 0\). For the limit to be equal to 1, the denominator must also approach 0. Therefore:
   \[
   2 \cdot 1 + a = 0 \implies 2 + a = 0 \implies a = -2
   \]
   Substituting \( a = -2 \) into \( b = -1 - a \):
   \[
   b = -1 - (-2) = -1 + 2 = 1
   \]

6. Therefore, the pair \((a, b)\) is:
   \[
   (a, b) = (-2, 1)
   \]

The final answer is \( \boxed{ (a, b) = (-2, 1) } \)