Question

Consider the function $f(x)=\begin{cases}\frac{P(x)}{\sin (x-2)}, & x \neq 2 \\ 7, & x=2\end{cases}$ Where $P(x)$ is a polynomial such that $P''(x)$ is always a constant and $P(3)=9$. If $f(x)$ is continuous at $x=2$, then $P(5)$ is equal to________.

Answer 1

$f(x)=\begin{cases}\frac{P(x)}{\sin (x-2)}, & x \neq 2 \\ 7, & x=2\end{cases}$
$P''(x)=$ const. $\Rightarrow P(x)$ is a 2 degree polynomial
$f(x)$ is cont. at $x=2$
$f\left(2^{+}\right)=f\left(2^{-}\right)$
$\displaystyle\lim _{x \rightarrow 2^{+}} \frac{(x-2)(a x+b)}{\sin (x-2)}=7$
$\Rightarrow 2 a +b=7$
$P(x)=(x-2)(a x+ b)$
$P(3)=(3-2)(3 a +b)=9$
$\Rightarrow 3 a +b=9$
$a=2, b=3$
$P(5)=(5-2)(2.5+3)=3.13=39$