Question

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

Answer 1

Answer: 39

$f(x)=\{\begin{array}{ll}\frac{P(x)}{\sin (x-2)},& x\ne 2\\ 7,& x=2\end{array}$
$P^{\prime \prime }(x)=$ const. $\Rightarrow P(x)$ is a 2 degree polynomial
$f(x)$ is cont. at $x=2$
$f\left(2^{+}\right)=f\left(2^{-}\right)$
$\underset{x\to 2^{+}}{\lim }\frac{(x-2)(ax+b)}{\sin (x-2)}=7$
$\Rightarrow 2a+b=7$
$P(x)=(x-2)(ax+b)$
$P(3)=(3-2)(3a+b)=9$
$\Rightarrow 3a+b=9$
$a=2,b=3$
$P(5)=(5-2)(2.5+3)=3.13=39$