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Step-by-Step Solution
Step 1: State the condition for continuity at x = 3
A function $f(x)$ is continuous at $x = 3$ if:
(1) $\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x)$
(2) The common limit equals $f(3)$.
Here, $f(3) = b$. So we need:
$
\lim_{x \to 3^-} f(x) = b
\quad\text{and}\quad
\lim_{x \to 3^+} f(x) = b.
$
Step 2: Compute the left-hand limit, $\lim_{x \to 3^-} f(x)$
For $x < 3$, the function is given by:
$
f(x) = \frac{a \bigl(7x - 12 - x^2\bigr)}{b \,\bigl|\,x^2 - 7x + 12\,\bigr|}.
$
First, factor the denominator:
$
x^2 - 7x + 12 = (x - 3)(x - 4).
$
For $x$ just less than 3, both $(x - 3)$ and $(x - 4)$ are negative, so their product is positive. Hence,
$
|x^2 - 7x + 12| = (x - 3)(x - 4)
\quad \text{for } x < 3.
$
Next, notice:
$
7x - 12 - x^2 = -(x^2 - 7x + 12) = -(x - 3)(x - 4).
$
Substituting these into $f(x)$ gives:
$
f(x)
= \frac{a\,\bigl[-(x - 3)(x - 4)\bigr]}{b\,(x - 3)(x - 4)}
= \frac{-a \,(x - 3)(x - 4)}{b\,(x - 3)(x - 4)}
= -\frac{a}{b}, \quad x \neq 3,4.
$
Thus,
$
\lim_{x \to 3^-} f(x) = -\frac{a}{b}.
$
Step 3: Compute the right-hand limit, $\lim_{x \to 3^+} f(x)$
For $x > 3$, the function is given by:
$
f(x) = 2^{\frac{\sin(x - 3)}{x - [x]}}.
$
As $x \to 3^+$ and $3 < x < 4$, the greatest integer function $[x] = 3$. So,
$
x - [x] = x - 3.
$
The exponent becomes:
$
\frac{\sin(x - 3)}{x - 3}.
$
Using the standard limit $\lim_{u \to 0} \frac{\sin u}{u} = 1,$ we get
$
\lim_{x \to 3^+} \frac{\sin(x - 3)}{x - 3} = 1.
$
Hence,
$
\lim_{x \to 3^+} f(x)
= \lim_{x \to 3^+} 2^{\frac{\sin(x - 3)}{x - 3}}
= 2^1
= 2.
$
Step 4: Apply the continuity conditions
Continuity at $x = 3$ requires:
$
\lim_{x \to 3^-} f(x) = b
\quad \text{and} \quad
\lim_{x \to 3^+} f(x) = b.
$
(a) From the left-hand limit:
$
-\frac{a}{b} = b
\quad\Longrightarrow\quad
-a = b^2
\quad\Longrightarrow\quad
a = -b^2.
$
(b) From the right-hand limit:
$
2 = b
\quad\Longrightarrow\quad
b = 2.
$
Substitute $b = 2$ into $a = -b^2$:
$
a = - (2^2) = -4.
$
Step 5: Conclude the ordered pair(s)
There is exactly one pair $(a, b)$ that makes $f(x)$ continuous at $x = 3$:
$
(a, b) = (-4, 2).
$
Therefore, the set $S$ has only one element, and the correct answer is 1.
