Consider the function. f(x) = { a(7x-12-x^2)/b|x^2 - 7x + 12|, x<3 {2^{sin(x-3)/x-[x]}, x > 3} {b, x = 3}

Question

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 [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

Answer 1

Correct option is (4) 1

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

Then \(f\left(3^{+}\right) = 2^{\lim\limits_{x \rightarrow 3^{+}}\left(\frac{\sin (x-3)}{x-3}\right)}=2\) 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 \((-4,2)\).