Consider the function $f\left(x\right)=\left\{\begin{array}{l}\frac{P\left(x\right)}{x-2},x<2\\ 2x+3,x\ge 2\end{array}\right.$ where P(x) is a polynomial such that $P\text{'}\text{'}\text{'}$(x) is identically equal to 0 for all x and p(3) =9. If $\underset{x\to 2}{\lim }f\left(x\right)$exists then P(x) is

$p^{\prime \prime \mathrm{\prime }}\left(\mathrm{x}\right)=0\text{ means }\mathrm{p}\left(\mathrm{x}\right)$ is a two – degree polynomial

Since  $\underset{x\to 2}{lim} f\left(x\right)\text{ exists, }f\left(x\right)$ will be divisible by x =2

$\Rightarrow \mathrm{p}\left(\mathrm{x}\right)=\mathrm{a}(\mathrm{x}-2)(\mathrm{x}-\mathrm{b})$

$\text{ so, }a(2-b)=7$

$\mathrm{p}\left(3\right)=9,\mathrm{a}(3-\mathrm{b})=9$

Since $$\begin{gathered} \text{ dividing them we will get }\mathrm{b}=-\frac{3}{2}\text{ and }\mathrm{x}=2 \\ \text{ so }p\left(x\right)=2x^2-x-6 \end{gathered}$$