KVPY SX SB Exam 2018 Question Paper

$\mathit{P}:\mathit{R}\to \mathit{R},\mathit{P}\left(0\right)=0,\mathit{P}\left(\mathit{x}\right)>{\mathit{x}}^2\forall \mathit{x}\ne 0$
P(x) will be polynomial of even degree.
$\mathit{P}\left(\mathit{x}\right)={\mathit{a}}_0{\mathit{x}}^{2\mathit{n}}+{\mathit{a}}_1{\mathit{x}}^{2\mathit{n}-1}+.......$ $+{\mathit{a}}_{2\mathit{n}-2}{\mathit{x}}^2+{\mathit{a}}_{2\mathit{n}-1}\mathit{x}$
$\mathit{P}”\left(0\right)=\frac{1}{2}$
${\mathit{a}}_{2\mathit{n}-2}=\frac{1}{4}$
$\mathit{P}\left(\mathit{x}\right)={\mathit{a}}_0{\mathit{x}}^{2\mathit{n}}+{\mathit{a}}_1{\mathit{x}}^{2\mathit{n}-1}+.......$$\frac{1}{4}{\mathit{x}}^2+{\mathit{a}}_{2\mathit{n}-1}\mathit{x}>{\mathit{x}}^2\forall \mathit{x}\ne 0$
$\mathit{P}\left(\mathit{x}\right)={\mathit{a}}_0{\mathit{x}}^{2\mathit{n}}+{\mathit{a}}_1{\mathit{x}}^{2\mathit{n}-1}+.......$$-\frac{3}{4}{\mathit{x}}^2+{\mathit{a}}_{2\mathit{n}-1}\mathit{x}>0\forall \mathit{x}\ne 0$
Which is not always true
Method II: $\mathit{P}\left(\mathit{x}\right)>{\mathit{x}}^2$
$\mathit{P}\left(\mathit{x}\right)={\mathit{x}}^2+\mathrm{f}\left(\mathit{x}\right)\mathrm{f}\left(\mathit{x}\right)>0\forall \mathit{x}\in {\mathit{R}}_0$
∵ $\mathit{P}\left(0\right)=0\Rightarrow \mathrm{f}\left(0\right)=0$
Now, P”(x) = 2 + f”(x)
f”(0) = negative
⇒ f(x) is concave down so f(x) can’t be +ve always
⇒ which is contradiction