Consider the function $$f(x) = \begin{cases} \frac{a(7x - 12 - x^2)}{b|x^2 - 7x + 12|}, & x < 3 \\ 2^{\frac{\sin(x-3)}{x - [x]}}, & x > 3 \\ b, & x = 3 \end{cases}$$, where $$[x]$$ denotes the greatest integer less than or equal to $$x$$. If $$S$$ denotes the set of all ordered pairs $$(a, b)$$ such that $$f(x)$$ is continuous at $$x = 3$$, then the number of elements in $$S$$ is :

Left Hand Limit (LHL).

$$f(x) = \frac{a(7x-12-x^2)}{b|x^2-7x+12|} = \frac{-a(x^2-7x+12)}{b|x^2-7x+12|}$$.

For $$x < 3$$, $$x^2-7x+12$$ is positive. So $$|x^2-7x+12| = (x^2-7x+12)$$.

$$LHL = \frac{-a}{b}$$.

 Right Hand Limit (RHL).

$$2^{\frac{\sin(x-3)}{x-\lfloor x \rfloor}}$$. For $$x \to 3^+$$, $$\lfloor x \rfloor = 3$$.

$$RHL = 2^{\lim_{x \to 3} \frac{\sin(x-3)}{x-3}} = 2^1 = 2$$.

For continuity, $$LHL = RHL = f(3) \implies \frac{-a}{b} = 2$$ and $$b = 2$$.

If $$b = 2$$, then $$\frac{-a}{2} = 2 \implies a = -4$$.

There is only 1 such pair: $$(-4, 2)$$.