Question

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

• A. 0
• B. 1
• C. more than 1, but finite
• D. infinite

Answer 1

Answer: (A)

We have,
$p(x)>x^2,p(0)=0,p^{\prime \prime }(0)=\frac{1}{2}$
Let $g(x)=p(x)-x^2$
$g(x)>0,\forall x\ne 0$
and $g(0)=P(0)-0=0$
$\Rightarrow x=0$ should be minima
$\because g^{\prime \prime }(x)$ should be $\ge 0$ at $x=0$
Now, $g^{\prime }(x)=p^{\prime }(x)-2x$
$g^{\prime \prime }(x{)}^2=p^{\prime \prime }(x)-2$
$g^{\prime \prime }(0)=P^{\prime \prime }(0)-2=\frac{1}{2}-2=-\frac{3}{2}$
But $g^{\prime \prime }(0)\ne 0$
$\because$ No polynomial exists