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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
