Let $p(x) = a_{0} + a_{1}x^{2} + a_{2}x^{4} + \ldots + a_{n}{x^{2}}^{n}$ be a polynomial in a real variable $x$ with $0

Question

Let $p(x) = a_{0} + a_{1}x^{2} + a_{2}x^{4} + \ldots + a_{n}{x^{2}}^{n}$ be a polynomial in a real variable $x$ with $0 < a_{0} < a_{1} < a_{2}\ldots < a_{n}.\$The function $p(x)$ has

Dr. Abubacker · Accepted Answer

(c)
$P(x) = a_{0} + a_{1}x^{2} + a_{2}x^{4} + \ldots + a_{n}x^{2n}$
Where, $a_{n} > a_{n - 1} > a_{n - 2}\ldots$>$a_{2} > a_{1} > a_{0} > 0$
$\Rightarrow \ \ \ P^{'}(x) = 2a_{1}x + 4a_{2}x^{3} + \ldots + 2na_{n}x^{2n - 1}$
$= 2x\{ a_{1} + 2a_{2}x^{2} + \ldots + na_{n}x^{2n - 2}\}$ ….. (i)
Where,
$\left( a_{1} + 2a_{2}x^{2} + 3a_{3}x^{4} + \ldots + na_{n}x^{2n - 2} ight) > 0$
For all $x\epsilon R$
Thus, $\left\{ \begin{array}{r}
P^{'}(x) > 0,\ whenx > 0 \
P^{'}(x) < 0,\ whenx > 0
\end{array} ight.\$
$ie$ $P'$ ($x$) changes sign from $( -$ve) to (+ve) at $x = 0$
Hence, $P^{\ }(x)$ attains minimum at $x = 0$
$\Rightarrow \ \ \$Only one minimum at$x = 0$.

Q. Let $p(x) = a_{0} + a_{1}x^{2} + a_{2}x^{4} + \ldots + a_{n}{x^{2}}^{n}$ be a polynomial in a real variable $x$ with $0 < a_{0} < a_{1} < a_{2}\ldots < a_{n}.\$The function $p(x)$ has

Explanation

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