Consider the function $f\left(x\right)=\frac{P\left(x\right)}{\sin \left(x-2\right)}, x\ne 2$, and $f\left(x\right)=7, x=2$where $P\left(x\right)$ is a polynomial such that $P^{"}\left(x\right)$ is always a constant and $P\left(3\right)=9.$ If $f\left(x\right)$ is continuous at $x=2,$ then $P\left(5\right)$ is equal to __________.

We have, $f\left(x\right)=\left\{\begin{array}{l}\frac{P\left(x\right)}{\sin \left(x-2\right)},\\ 7,\end{array}\right.\begin{array}{c}x\ne 2\\ x=2\end{array}$

$P^{"}\left(x\right)=$ Constant

$\Rightarrow P\left(x\right)$ is a $2$ degree polynomial

Let $P\left(x\right)=\left(x-2\right)\left(ax+b\right)$

$f\left(x\right)$ is continuous at $x=2$

$f\left(2^{+}\right)=f\left(2^{-}\right)$

Now, $\underset{x\to 2^{+}}{\lim }\frac{P\left(x\right)}{\sin \left(x-2\right)}=7$

$\Rightarrow \underset{x\to 2^{+}}{\lim }\frac{\left(x-2\right)\left(ax+b\right)}{\sin \left(x-2\right)}=7$

$\Rightarrow 2a+b=7$

$\Rightarrow P\left(3\right)=\left(3-2\right)\left(3a+b\right)=9$

$\Rightarrow 3a+b=9$

$\Rightarrow a=2,b=3$

Hence, $P\left(5\right)=\left(5-2\right)\left(2.5+3\right)$

$=3.13$

$=39.$