JEE Main 2024 — 27 January, Shift 1 — Question 13

$f\left(3^{-}\right)=\frac{a}{b} \frac{\left(7 x-12-x^{2}\right)}{\left|x^{2}-7 x+12\right|}$ (for $f(x)$ to be cont.)

$\Rightarrow \mathrm{f}\left(3^{-}\right)=\frac{-\mathrm{a}}{\mathrm{b}} \frac{(\mathrm{x}-3)(\mathrm{x}-4)}{(\mathrm{x}-3)(\mathrm{x}-4)} ; \mathrm{x}<3 \Rightarrow \frac{-\mathrm{a}}{\mathrm{b}}$

Hence $\mathrm{f}\left(3^{-}\right)=\frac{-\mathrm{a}}{\mathrm{b}}$

Then $\mathrm{f}\left(3^{+}\right)=2^{\lim _{x \rightarrow 3^{+}}\left(\frac{\sin (x-3)}{x-3}\right)}=2$ and $f(3)=b$.

Hence $f(3)=f\left(3^{+}\right)=f\left(3^{-}\right)$

$\Rightarrow \mathrm{b}=2=-\frac{\mathrm{a}}{\mathrm{b}}$

$b=2, a=-4$

Hence only 1 ordered pair $(-4,2)$.