A) neither a maximum nor a minimum
B) only one maximum
C) only one minimum
D) only one maximum and only one minimum
Correct Answer: C
Solution :
\[P(x)={{a}_{0}}+{{a}_{1}}{{x}^{2}}+{{a}_{2}}{{x}^{4}}+....+{{a}_{n}}{{x}^{2n}}\] where, \[{{a}_{n}}>{{a}_{n-1}}>{{a}_{n-2}}....>{{a}_{2}}>{{a}_{1}}>{{a}_{0}}>0\] \[\Rightarrow \] \[P(x)=2{{a}_{1}}x+4{{a}_{2}}{{x}^{3}}+....+2n{{a}_{n}}{{x}^{2n-1}}\] \[=2x\{{{a}_{1}}+2{{a}_{2}}{{x}^{2}}+.....+n{{a}_{n}}{{x}^{2n-2}}\}\] ...(i) where, \[({{a}_{1}}+2{{a}_{2}}{{x}^{2}}+3{{a}_{3}}{{x}^{4}}+....+n{{a}_{n}}{{x}^{2n-2}})>0\] For all\[x\in R\] Thus, \[\left\{ \begin{matrix} P(x)>0,when\,x>0 \\ P(x)<0,\,when\,x<0 \\ \end{matrix} \right.\] ie,\[P(x)\]changes sign from\[(-ve)\]to\[(+ve)\]at \[x=0\] Hence,\[P(x)\]attains minimum at\[x=0\]. \[\Rightarrow \]Only one minimum at\[x=0\].
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