Consider the function.

$$ f(x)=\left\{\begin{array}{cc} \frac{\mathrm{a}\left(7 x-12-x^2\right)}{\mathrm{b}\left|x^2-7 x+12\right|} & , x<3 \\\\ 2^{\frac{\sin (x-3)}{x-[x]}} & , x>3 \\\\ \mathrm{~b} & , x=3, \end{array}\right. $$

where $[x]$ denotes the greatest integer less than or equal to $x$. If $\mathrm{S}$ denotes the set of all ordered pairs (a, b) such that $f(x)$ is continuous at $x=3$, then the number of elements in $\mathrm{S}$ is :

<p>$$f\left(3^{-}\right)=\frac{a}{b} \frac{\left(7 x-12-x^2\right)}{\left|x^2-7 x+12\right|} \quad$$ (for $f(x)$ to be cont.)</p> <p>$$\Rightarrow \mathrm{f}\left(3^{-}\right)=\frac{-\mathrm{a}}{\mathrm{b}} \frac{(\mathrm{x}-3)(\mathrm{x}-4)}{(\mathrm{x}-3)(\mathrm{x}-4)} ; \mathrm{x}<3 \Rightarrow \frac{-\mathrm{a}}{\mathrm{b}}$$</p> <p>Hence $f\left(3^{-}\right)=\frac{-a}{b}$</p> <p>Then $$f\left(3^{+}\right)=2^{\lim _\limits{x \rightarrow 3^{+}}\left(\frac{\sin (x-3)}{x-3}\right)}=2$$ and $f(3)=b$.</p> <p>Hence $\mathrm{f}(3)=\mathrm{f}\left(3^{+}\right)=\mathrm{f}\left(3^{-}\right)$</p> <p>$$\begin{aligned} & \Rightarrow \mathrm{b}=2=-\frac{\mathrm{a}}{\mathrm{b}} \\ & \mathrm{b}=2, \mathrm{a}=-4 \end{aligned}$$</p> <p>Hence only 1 ordered pair $(-4,2)$.</p>