Question

The number of polynomials $p:R\to R$ satisfying $p\left(0\right)=0,p\left(x\right)>x^2$ for all $x\ne 0$, and $p^{\prime \prime }\left(0\right)=\frac{1}{2}$ is

• A. 0.0
• B. 1
• C. more than 1, but finite
• D. infinite

Answer 1

The correct option is A 0.0

Assuming a polynomial $g\left(x\right)=p\left(x\right)-x^2$
from the given conditions,
$p\left(x\right)>x^2\Rightarrow p\left(x\right)-x^2>0\Rightarrow g\left(x\right)>0\forall x\ne 0$
We know that $p\left(0\right)=0$
$\therefore g\left(0\right)=p\left(0\right)-0^2=0$
So $x=0$ should be minima
$\therefore g^{\prime \prime }\left(0\right)>0\Rightarrow p^{\prime \prime }\left(0\right)-2>0\Rightarrow p^{\prime \prime }\left(0\right)>2$
which contradict the given condition $p^{\prime \prime }\left(0\right)=\frac{1}{2}$
Hence no such polynomial exists.