Continuity And Differentiability Question 3
Question 3 - 2024 (27 Jan Shift 1)
Consider the function.
$f(x)=\begin{cases}\frac{a\left(7 x-12-x^{2}\right)}{b\left|x^{2}-7 x+12\right|} & , x<3 \\ 2^{\frac{\sin (x-3)}{x-[x]}} & , \quad x>3 \\ b & , \quad x=3\end{cases}$
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 :
(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.)
$\Rightarrow f\left(3^{-}\right)=\frac{-a}{b} \frac{(x-3)(x-4)}{(x-3)(x-4)} ; x<3 \Rightarrow \frac{-a}{b}$
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 $f(3)=f\left(3^{+}\right)=f\left(3^{-}\right)$
$\Rightarrow b=2=-\frac{a}{b}$
$b=2, a=-4$
Hence only 1 ordered pair $(-4,2)$.
