JEE Main 2024 (Online) 27th January Morning ShiftQuestion 1
Consider the function $f(x)=\begin{cases}\frac{a\left(7x-12-x^{2}\right)}{b\left|x^{2}-7x+12\right|}, & x<3\\ 2^{\frac{\sin(x-3)}{x-[x]}}, & x>3\\ b, & x=3\end{cases}$ where $[x]$ denotes the greatest integer less than or equal to $x$. If $S$ denotes the set of all ordered pairs $(a,b)$ such that $f(x)$ is continuous at $x=3$, then the number of elements in $S$ is :
Limits & Continuity
Solution
Concept. Find the left and right limits at $3$ and equate to $b$. Formulas used. (1) $x^2-7x+12=(x-3)(x-4)$(2) $\lim_{u\to0}\frac{\sin u}u=1$Step 1 - Left of 3. For $x<3$ near $3$, $(x-3)(x-4)>0$, so $|x^2-7x+12|=x^2-7x+12$. Also $7x-12-x^2=-(x^2-7x+12)$. So $f(x)=-\frac ab$ (needs $b\ne0$). Step 2 - Right of 3. For $3<x<4$, $[x]=3$, so $x-[x]=x-3$ and by (2): $$f=2^{\frac{\sin(x-3)}{x-3}}\to2^1=2$$ Step 3 - Continuity. $-\frac ab=2=b\Rightarrow b=2$, $a=-4$. Step 4. Exactly one pair $(-4,2)$. Answer: C