Continuity And Differentiability Question 3
Question 3 - 2024 (27 Jan Shift 1)
Consider the function.
$f(x)=\left{\begin{array}{cc}\frac{a\left(7 x-12-x^{2}\right)}{b\left|x^{2}-7 x+12\right|} & , x<3 \ 2^{\frac{\sin (x-3)}{x-[x]}} & , \quad x>3 \ b & , \quad x=3\end{array}\right.$
Where $[\mathrm{x}]$ 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 :
(1) 2
(2) Infinitely many
(3) 4
(4) 1
Show Answer
Answer (4)
Solution
$f\left(3^{-}\right)=\frac{a}{b} \frac{\left(7 x-12-x^{2}\right)}{\left|x^{2}-7 x+12\right|} \quad$ (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 $f\left(3^{+}\right)=2^{\lim _{x \rightarrow 3^{+}}\left(\frac{\sin (x-3)}{x-3}\right)}=2$ 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 $(-4,2)$.
