Consider the function $f(x)=\frac{P(x)}{\sin (x-2)}, \quad x \neq 2$
$=7 \quad, \quad x=2$
Where $\mathrm{P}(\mathrm{x})$ is a polynomial such that $\mathrm{P}^{\prime \prime}(\mathrm{x})$ is always a constant and $\mathrm{P}(3)=9$. If $\mathrm{f}(\mathrm{x})$ is continuous at $x=2$, then $P(5)$ is equal to________.
$f(x)= \begin{cases}\frac{P(x)}{\sin (x-2)}, & x \neq 2 \\ 7, & x=2\end{cases}$
$P^{\prime \prime}(x)=$ const. $\Rightarrow P(x)$ is a 2 degree polynomial
$f(x)$ is cont. at $x=2$
$f\left(2^{+}\right)=f\left(2^{-}\right)$
$\lim _{x \rightarrow 2^{+}} \frac{P(x)}{\sin (x-2)}=7$
$\lim _{x \rightarrow 2^{+}} \frac{(x-2)(a x+b)}{\sin (x-2)}=7 \Rightarrow 2 a+b=7$
$P(x)=(x-2)(a x+b)$
$\mathrm{P}(3)=(3-2)(3 \mathrm{a}+\mathrm{b})=9 \Rightarrow 3 \mathrm{a}+\mathrm{b}=9$
$a=2, b=3$
$P(5)=(5-2)(2.5+3)=3.13=39$