Consider the function.

Question

$f(x)=\begin{array}{|c|c|} \frac{a(7x-12-{x}^{2})}{b|{x}^{2}-7x+12|} & ,x3 \  \ b & ,x=3, \end{array}$

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 :

Accepted Answer

$f({3}^{-})=\frac{a}{b}\frac{(7x-12-{x}^{2})}{|{x}^{2}-7x+12|}\,$ (for $f(x)$ to be cont.)

$\Rightarrow f({3}^{-})=\frac{-a}{b}\frac{(x-3)(x-4)}{(x-3)(x-4)};xHence $f({3}\^{-})=\frac{-a}{b}$

Then $f({3}\^{+})={2}\^{\underset{x\to {3}\^{+}}{lim}(\frac{\sin (x-3)}{x-3})}=2$ and $f(3)=b$.

Hence $f(3)=f({3}\^{+})=f({3}\^{-})$

$$$\begin{gathered} \Rightarrow b=2=-\frac{a}{b} \ b=2,a=-4 \end{gathered}$$$

Hence only 1 ordered pair $

Solutions · 1