KVPY 2019 SB SX Exam Previous Papers

It is given that $\mathit{P}$ is a non-zero
polynomial such that, $\mathit{P}\left(1+\mathit{x}\right)=\mathit{P}\left(1-\mathit{x}\right)$
On differentiating both sides w.r.t. $\mathit{x}$,
we get
${\mathit{P}}^{\prime }\left(1+\mathit{x}\right)=-{\mathit{P}}^{\prime }\left(1-\mathit{x}\right)$
Now on putting $\mathit{x}=0$, we get
${\mathit{P}}^{\prime }\left(1\right)=-{\mathit{P}}^{\prime }\left(1\right)$
$\Rightarrow {\mathit{P}}^{\prime }\left(1\right)=0$
i.e the polynomial $\mathit{P}$ touches the $\mathit{X}$-axis at
$\mathit{x}=1$ and $\mathit{P}\left(1\right)=0($ given $)$
$\therefore \mathit{P}\left(\mathit{x}\right)={\left(\mathit{x}-1\right)}^2\mathit{Q}\left(\mathit{x}\right)$
$\therefore$ Largest integral value of ' $\mathit{m}$ ' such that
${\left(\mathit{x}-1\right)}^{\mathit{m}}$ divides $\mathit{P}\left(\mathit{x}\right)$ is $2.$