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If the equation  $a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots .+a_{1} x=0, a_{1} \neq 0, n \geq 2$ has a positive root $x=\alpha$, then the equation $n a_{n} x^{n-1}+(n-1) a_{n-1} x^{n-2}+\ldots .+a_{1}=0$ has a positive roots, which is

1) greater than $\alpha$
2) smaller than $\alpha$
3) greater than or equal to $\alpha$
4) equal to $\alpha$
in Mathematics by Bronze Status (9,486 points) | 521 views

1 Answer

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Answer

(2)

Explanation

$f(0)=0, f(\alpha)=0$ and $f(0)$ is continous on $[0, \alpha]$ and differentiable on $(0, \alpha)$


$\Rightarrow f^{\prime}(k)=0$ for some $k \in(0, \alpha) .$

by Bronze Status (9,486 points)

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