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
$(1986,2 M)$
(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. $(a, c)$
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 $(-ve)$ to (+ve) at $x=0$.
$\therefore \quad P(x)$ attains minimum at $x=0$.
Hence, it has only one minimum at $x=0$.
