Question

Consider the function $f\left(x\right)=\{\begin{array}{cc}\frac{P\left(x\right)}{\sin (x-2)},& x\ne 2\\ 7,& x=2\end{array},$ where $P\left(x\right)$ is polynomial such that $P^{\prime \prime }\left(x\right)$ 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

Answer 1

$P\left(x\right)=k(x-2)(x-b)$
$\underset{x\to 2}{lim}f\left(x\right)=7$
$\Rightarrow \underset{x\to 2}{lim}\frac{k(x-2)(x-b)}{\sin (x-2)}=7$
$\Rightarrow \underset{x\to 2}{lim}k(x-b)=7$
$(\because \underset{x\to 0}{lim}\frac{\sin x}{x}=1)$
$\Rightarrow k(2-b)=7\cdots \left(1\right)$
and $P\left(3\right)=k(3-2)(3-b)=9$
$\Rightarrow k(3-b)=9\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$
$\frac{2-b}{3-b}=\frac{7}{9}$
$\Rightarrow 18-9b=21-7b$
$\Rightarrow -2b=3$
$\Rightarrow b=-\frac{3}{2}$
and
$k(2-b)=7$
$\Rightarrow k(2+\frac{3}{2})=7$
$\Rightarrow k=2$
$\therefore P\left(x\right)=2(x-2)(x+\frac{3}{2})$
$\Rightarrow P\left(5\right)=3\times 13$
$\therefore P\left(5\right)=39$