Step 1: Understanding the Concept:
For a function to be continuous at \( x = c \), the left-hand limit (LHL), right-hand limit (RHL), and the function value \( f(c) \) must be equal.
Step 2: Detailed Explanation:
1. LHL: \( \lim_{x \to 2^-} (\frac{x-2}{-(x-2)} + a) = -1 + a \).
2. RHL: \( \lim_{x \to 2^+} (\frac{x-2}{x-2} + b) = 1 + b \).
3. Value: \( f(2) = a + b \).
Equating LHL = RHL = \( f(2) \):
\( -1 + a = 1 + b = a + b \).
From \( -1 + a = a + b \), we get \( b = -1 \).
From \( 1 + b = a + b \), we get \( a = 1 \).
Wait, let's re-check: \( -1 + a = a + b \implies b = -1 \). \( -1 + a = 1 + b \implies a - b = 2 \). If \( b = -1 \), then \( a = 1 \). Checking \( f(2) = a + b = 1 - 1 = 0 \).
Setting \( LHL = RHL = f(2) \implies -1+a = 1+b = a+b \).
\( -1+a = a+b \implies b = -1 \).
\( 1+b = a+b \implies a = 1 \).
Result: \( a=1, b=-1 \).
Step 3: Final Answer:
The correct values are \( a = 1, b = -1 \) (Option B).