Application of Derivatives 4 Question 26
####27. If $P(x)=a_{0}+a_{1} x^{2}+a_{2} x^{4}+\ldots+a_{n} x^{2 n}$ is a polynomial in a real variable $x$ with $0<a_{0}<a_{1}<a_{2}<\ldots<a_{n}$. Then, the function $P(x)$ has
$(1986,2 \mathrm{M})$
(a) neither a maximum nor a minimum
(b) only one maximum
(c) only one minimum
(d) only one maximum and only one minimum
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Answer:
Correct Answer: 27. $(\mathrm{a}, \mathrm{c})$
Solution:
- Given, $P(x)=a_{0}+a_{1} x^{2}+a_{2} x^{4}+\ldots+a_{n} x^{2 n}$ where, $a_{n}>a_{n-1}>a_{n-2}>\ldots>a_{2}>a_{1}>a_{0}>0$
$\Rightarrow P^{\prime}(x)=2 a_{1} x+4 a_{2} x^{3}+\ldots+2 n a_{n} x^{2 n-1}$
$=2 x\left(a_{1}+2 a_{2} x^{2}+\ldots+n a_{n} x^{2 n-2}\right)$
where, $\left(a_{1}+2 a_{2} x^{2}+3 a_{3} x^{4}+\ldots+n a_{n} x^{2 n-2}\right)>0, \forall x \in R$.
Thus,
$ P^{\prime}(x)>0 \text {, when } x>0 $
$ P^{\prime}(x)<0 \text {, when } x<0 $
i.e. $P^{\prime}(x)$ changes sign from $(-\mathrm{ve})$ to (+ve) at $x=0$.
$\therefore \quad P(x)$ attains minimum at $x=0$.
Hence, it has only one minimum at $x=0$.
