Gujarati
5. Continuity and Differentiation
normal

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

A

$0$

B

$1$

C

more than $1,$ but finite

D

infinite

(KVPY-2018)

Solution

(a)

We have,

$p(x) > x^2, p(0)=0, p^{\prime \prime}(0)=\frac{1}{2}$

Let $g(x)=p(x)-x^2$

$g(x) > 0, \forall x \neq 0$

$\text { and } \quad g(0)=p(0)-0=0$

$\Rightarrow x=0 \text { should be minima. }$

$\because g^{\prime \prime}(x) \text { should be } \geq 0 \text { at } x=0$

$\text { Now, } \quad g^{\prime}(x)=p^{\prime}(x)-2 x$

$\quad g^{\prime \prime}(x)^2=p^{\prime \prime}(x)-2$

$g^{\prime \prime}(0)=p^{\prime \prime}(0)-2=\frac{1}{2}-2=-\frac{3}{2}$

But $g^{\prime \prime}(0) \geq 0$

$\because$ No polynomial exists.

Standard 12
Mathematics

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