Consider the function. f(x)= cc a (7 x-12-x^2 ) b |x^2-7 x+12 | &… - Vidyadip

MCQ

Consider the function. $f(x)=\left\{\begin{array}{cc} \frac{a\left(7 x-12-x^2\right)}{b\left|x^2-7 x+12\right|} & , x<3 \\ 2^{\frac{\sin (x-3)}{x-[x]}} & , x>3 \\ b & , x=3 \end{array}\right.$ 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 $\mathrm{S}$ is :

A
$2$
B
Infinitely many
C
$4$
✓
$1$

✓

Answer

Correct option: D.

$1$

d
$f\left(3^{-}\right)=\frac{a}{b} \frac{\left(7 x-12-x^2\right)}{\left|x^2-7 x+12\right|}$ (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}}$ $\mathrm{b}=2, \mathrm{a}=-4$

Hence only 1 ordered pair $(-4,2)$.

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