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For the function $f(x)$ to be continuous at $x=2$, the left-hand limit (LHL) and the right-hand limit (RHL) at $x=2$ must be equal, and also equal to the value of the function at $x=2$.
The left-hand limit is given by: $$LHL = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{x-2}{x-2} = 1$$
The right-hand limit is given by: $$RHL = \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (ax^2 - bx + 3)$$ Since the function is a polynomial for $x > 2$, we can directly substitute $x=2$: $$RHL = a(2)^2 - b(2) + 3 = 4a - 2b + 3$$
The value of the function at $x=2$ is given as $f(2) = 1 + a$.
For continuity, we must have $LHL = RHL = f(2)$. Therefore, we have two equations: 1. $1 = 4a - 2b + 3$ 2. $1 = 1 + a$
From equation (2), we get: $1 = 1 + a \implies a = 0$ Substitute $a = 0$ into equation (1): $1 = 4(0) - 2b + 3 \implies 1 = -2b + 3 \implies 2b = 2 \implies b = 1$
Therefore, $a = 0$ and $b = 1$.
Final Answer: a = 0, b = 1
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