JEE Main 2024 — Mathematics Calculus

Given: f(x)={\begin{matrix}\frac{a(7x-12-{x}^{2})}{b|{x}^{2}-7x+12|} & ,x<3 \\ {2}^{\frac{\mathrm{sin}(x-3)}{x-[x]}} & ,x>3 \\ b & ,x=3\end{matrix}

$$\Rightarrow f({3}^{-})=\frac{a(7x-12-{x}^{2})}{b|{x}^{2}-7x+12|}$$⇒f(3−)=b∣x2−7x+12∣a(7x−12−x2)​

$$\Rightarrow f({3}^{-})=\frac{-a(x-3)(x-4)}{b(x-3)(x-4)}$$⇒f(3−)=b(x−3)(x−4)−a(x−3)(x−4)​

$$\Rightarrow f({3}^{-})=\frac{-a}{b}...(i)$$⇒f(3−)=b−a​...(i)

Now, $f({3}^{+})={2}^{\underset{x\rightarrow {3}^{+}}{\mathrm{lim}}(\frac{\mathrm{sin}(x-3)}{x-3})}$f(3+)=2x→3+lim​(x−3sin(x−3)​)

$$\Rightarrow f({3}^{+})=2...(\mathrm{ii})$$⇒f(3+)=2...(ii)

Also, $f(3)=b...(\mathrm{iii})$f(3)=b...(iii)

It is given that $f(x)$f(x) is continuous at $x=3$x=3.

Hence, $f(3)=f({3}^{+})=f({3}^{-})$f(3)=f(3+)=f(3−)

$$\Rightarrow b=2=\frac{-a}{b}$$⇒b=2=b−a​

$$\Rightarrow b=2,a=-4$$⇒b=2,a=−4

Hence, there is only $1$1 elements in $S$S, which is ordered pair $(-4,2)$(−4,2).