Consider the function f(x) = \ cases a(7x - 12 - x ^ 2 ) b| x ^ 2 -...

Given: f\left(x\right) = \left\{\begin{cases}\dfrac{a\left(7x - 12 - {x}^{2}\right)}{b\left|{x}^{2} - 7x + 12\right|} & , x < 3 \\ 2\dfrac{\sin ( x - 3 )}{x - [ x ]} & , x > 3 \\ b & , x = 3\end{cases}\right)</p><p>⇒ f\left({3}^{-}\right) = \dfrac{a\left(7x - 12 - {x}^{2}\right)}{b | {x}^{2} - 7x + 12 |}$</p><p>$⇒ f\left({3}^{-}\right) = \dfrac{- a ( x - 3 ) ( x - 4 )}{b ( x - 3 ) ( x - 4 )}$</p><p>$⇒ f\left({3}^{-}\right) = \dfrac{- a}{b} . . . \left(i\right)$</p><p>Now,$f\left({3}^{+}\right) = {2}^{\lim_{x \to {3}^{+}}\left(\dfrac{\sin ( x - 3 )}{x - 3}\right)}$</p><p>$⇒ f\left({3}^{+}\right) = 2 . . . \left(ii\right)$</p><p>Also,$f ( 3 ) = b . . . \left(iii\right)$</p><p>It is given that$f\left(x\right)$is continuous at$x = 3$.</p><p>Hence,$f ( 3 ) = f ( {3}^{+} ) = f ( {3}^{-} )$</p><p>$⇒ b = 2 = \dfrac{- a}{b}$</p><p>$⇒ b = 2, a = - 4$</p><p>Hence, there is only$1$elements in$S$, which is ordered pair$( - 4, 2 )$.