Question:

Step 1: State the condition for continuity at a point. For a function $f(x)$ to be continuous at a point $x = c$, the following three conditions must be satisfied: 1. $f(c)$ must be defined. 2. The limit of $f(x)$ as $x$ approaches $c$ must exist, i.e., $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)$. 3. The limit must be equal to the function value, i.e., $\lim_{x \to c} f(x) = f(c)$. Step 2: Identify the value of the function at $x=1$. From the definition of $f(x)$, when $x=1$, the function value is given as: $$f(1) = 3$$ Step 3: Calculate the left-hand limit of $f(x)$ at $x=1$. For $x < 1$, the function is defined as $f(x) = ax + 1$. We calculate the left-hand limit as $x$ approaches $1$: $$ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (ax + 1) $$ Substitute $x=1$ into the expression: $$ \lim_{x \to 1^-} f(x) = a(1) + 1 = a + 1 $$ Step 4: Calculate the right-hand limit of $f(x)$ at $x=1$. For $x > 1$, the function is defined as $f(x) = bx^2 + 1$. We calculate the right-hand limit as $x$ approaches $1$: $$ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (bx^2 + 1) $$ Substitute $x=1$ into the expression: $$ \lim_{x \to 1^+} f(x) = b(1)^2 + 1 = b + 1 $$ Step 5: Equate the function value and the limits to find the values of $a$ and $b$. For $f(x)$ to be continuous at $x=1$, the left-hand limit, the right-hand limit, and the function value at $x=1$ must all be equal: $$ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) $$ Substituting the values calculated in the previous steps: $$ a + 1 = b + 1 = 3 $$ From $a + 1 = 3$: $$ a = 3 - 1 $$ $$ a = 2 $$ From $b + 1 = 3$: $$ b = 3 - 1 $$ $$ b = 2 $$ Step 6: Calculate the required expression $(a - b)$. Now we substitute the values of $a$ and $b$ into the expression $(a - b)$: $$ a - b = 2 - 2 $$ $$ a - b = 0 $$ The final answer is $\boxed{0}$.