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 :
Given, \[p(x)={{a}_{0}}+{{a}_{1}}{{x}^{2}}+{{a}_{2}}{{x}^{4}}+....+{{a}_{n}}{{x}^{2n}}\]?(i) where\[{{a}_{n}}>{{a}_{n-1}}>{{a}_{n-2}}>...>{{a}_{2}}>{{a}_{1}}>{{a}_{0}}>0\] On differentiating Eq. (i) w.r.t. x, we get \[p'(x)=2{{a}_{1}}x+4{{a}_{2}}{{x}^{3}}+...+2n{{a}_{n}}x{{2}^{n-1}}\] \[=2x({{a}_{1}}+2{{a}_{2}}{{x}^{2}}+...+n{{a}_{n}}x{{2}^{n-2}})\]?.(ii) where\[({{a}_{1}}+2{{a}^{2}}+...+n{{a}_{n}}{{x}^{2n-2}})>0,\]\[\forall x\in R\] Thus \[\forall x\in R\]when\[x<0\]when\[x>0\] ie, ff (x) changes sign from (-ve) to (+ve) at\[x=0.\] Therefore, p (x) attains minimum at x = 0. Hence, it has only one minimum at x = 0.
You need to login to perform this action.
You will be redirected in
3 sec
