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Step-by-Step Solution
Step 1: Examine the Given Piecewise Function
The function is defined as:
$f(x) =
\begin{cases}
\frac{a\bigl(7x - 12 - x^2\bigr)}{\,b\,\bigl|\,x^2 - 7x + 12\,\bigr|}, & x < 3,\\[6pt]
2^{\frac{\sin(x-3)}{\,x - [x]\,}}, & x > 3,\\[6pt]
b, & x = 3,
\end{cases}$
where $[x]$ denotes the greatest integer less than or equal to $x.$ We want the function to be continuous at $x = 3.$
Step 2: Condition for Continuity
For $f(x)$ to be continuous at $x = 3,$ the left-hand limit, the function value at $3,$ and the right-hand limit at $3$ must be equal:
$\displaystyle \lim_{x \to 3^-} f(x) \;=\; f(3) \;=\; \lim_{x \to 3^+} f(x).$
Step 3: Compute the Left-Hand Limit ($x \to 3^-$)
For $x < 3,$
$f(x) \;=\; \dfrac{a\bigl(7x - 12 - x^2\bigr)}{\,b\,\bigl|\,x^2 - 7x + 12\,\bigr|}\,.$
Notice that
$x^2 - 7x + 12 = (x-3)(x-4)\text{,}$
and also
$7x - 12 - x^2 = -(x^2 - 7x + 12) = -(x-3)(x-4).$
Hence
$
f(x) \;=\; \dfrac{-\,a\,(x-3)(x-4)}{\,b\,\bigl|\,(x-3)(x-4)\bigr|}.
$
For $x$ just to the left of $3,$ $(x-3)$ is negative and $(x-4)$ is also negative. So their product $(x-3)(x-4)$ is positive. That means
$
\bigl|\,(x-3)(x-4)\bigr| = (x-3)(x-4).
$
Thus, for $x$ approaching $3$ from the left:
$
\lim_{x \to 3^-} f(x)
= \lim_{x \to 3^-} \dfrac{-\,a\,(x-3)(x-4)}{\,b\,\bigl[(x-3)(x-4)\bigr]} = -\,\dfrac{a}{b}.
$
So the left-hand limit is $-\tfrac{a}{b}.$
Step 4: Compute the Right-Hand Limit ($x \to 3^+$)
For $x > 3,$
$
f(x) = 2^{\,\frac{\sin(x-3)}{\,x-[\,x\,]}},
$
where $[\,x\,]$ is the greatest integer $\leq x.$ When $3 < x < 4,$ we have $[\,x\,] = 3,$ so the exponent simplifies to
$
\dfrac{\sin(x-3)}{\,x-3}.
$
As $x$ approaches $3$ from the right,
$
(x-3) \to 0^{+}.
$
We use the standard limit
$
\lim_{t \to 0} \dfrac{\sin t}{t} = 1.
$
Therefore,
$
\lim_{x \to 3^+} 2^{\,\frac{\sin(x-3)}{\,x-3}}
= 2^1 = 2.
$
So the right-hand limit is $2.$
Step 5: Value of the Function at $x=3$
By definition, $f(3) = b.$
Step 6: Impose Continuity at $x=3$
Continuity requires:
$
-\,\dfrac{a}{b} = f(3^-) = f(3) = b = f(3^+) = 2.
$
Hence we get two conditions from this equality:
1) $b = 2,$
2) $-\tfrac{a}{b} = b.$
Substituting $b = 2$ into $-\tfrac{a}{b} = b$ gives:
$
-\,\dfrac{a}{2} = 2
\; \Longrightarrow \;
a = -4.
$
Step 7: Number of Possible $(a,b)$ Pairs
We find a single unique solution: $(a,b) = (-4, 2).$ Since there is exactly one pair satisfying continuity at $x=3,$ the number of elements in the set $S$ is 1.
Final Answer
The correct choice is 1, corresponding to the unique pair $(-4,\,2).$
