JEE MAIN - Mathematics (2024 - 27th January Morning Shift - No. 9)
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 :
$$ f(x)=\left\{\begin{array}{cc} \frac{\mathrm{a}\left(7 x-12-x^2\right)}{\mathrm{b}\left|x^2-7 x+12\right|} & , x<3 \\\\ 2^{\frac{\sin (x-3)}{x-[x]}} & , x>3 \\\\ \mathrm{~b} & , x=3, \end{array}\right. $$
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 :
Infinitely many
4
2
1
Explanation
$$f\left(3^{-}\right)=\frac{a}{b} \frac{\left(7 x-12-x^2\right)}{\left|x^2-7 x+12\right|} \quad$$ (for $$f(x)$$ to be cont.)
$$\Rightarrow \mathrm{f}\left(3^{-}\right)=\frac{-\mathrm{a}}{\mathrm{b}} \frac{(\mathrm{x}-3)(\mathrm{x}-4)}{(\mathrm{x}-3)(\mathrm{x}-4)} ; \mathrm{x}<3 \Rightarrow \frac{-\mathrm{a}}{\mathrm{b}}$$
Hence $$f\left(3^{-}\right)=\frac{-a}{b}$$
Then $$f\left(3^{+}\right)=2^{\lim _\limits{x \rightarrow 3^{+}}\left(\frac{\sin (x-3)}{x-3}\right)}=2$$ and $$f(3)=b$$.
Hence $$\mathrm{f}(3)=\mathrm{f}\left(3^{+}\right)=\mathrm{f}\left(3^{-}\right)$$
$$\begin{aligned} & \Rightarrow \mathrm{b}=2=-\frac{\mathrm{a}}{\mathrm{b}} \\ & \mathrm{b}=2, \mathrm{a}=-4 \end{aligned}$$
Hence only 1 ordered pair $$(-4,2)$$.
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