Verified Solution • 2025–26 Session moderate

QJEEM24Ma1859: JEE Main 2024 Mathematics - Class 12 JEEM24Ma1859 Solution: Consider the function. where denotes the greatest integer less…

Consider the function.

f(x)={a(7x−12−x2)b∣x2−7x+12∣amp;,xlt;32sin⁡(x−3)x−[x]amp;,xgt;3 bamp;,x=3,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][x] denotes the greatest integer less than or equal to xx . If S\mathrm{S} denotes the set of all ordered pairs (a, b) such that f(x)f(x) is continuous at x=3x=3 , then the number of elements in S\mathrm{S} is :

Multiple Choice Options

  • A Infinitely many
  • B 4
  • C 2
  • D 1

Step-by-Step Verified Solution

f(3−)=ab(7x−12−x2)∣x2−7x+12∣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)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(3−)=−abf\left(3^{-}\right)=\frac{-a}{b}

Then f(3+)=2limx→3+(sin⁡(x−3)x−3)=2 and f(3)=bf(3)=b .

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

amp;⇒b=2=−abamp;b=2,a=−4\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)(-4,2) .

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