Question

If the equation $a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots a_{1}x=0,(a_1\ne 0,n\ge 2)$ has a positive root $x=\alpha ,$ then the equation $n{a}_n{x}^{n-1}+(n-1)a_{n-1}{x}^{n-2}+\cdots +a_1=0$ has a positive root, which is

• A. Equal to $\frac{\alpha }{2}$
• B. Equal to $\alpha$
• C. less than $\alpha .$
• D. Greater than $\alpha$

Answer 1

The correct option is C less than $\alpha .$

Let $f\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots a_{1}x$
Clearly $,f\left(0\right)=0$ and $f\left(\alpha \right)=0$
Also $f$ is continuous in $[0,\alpha ]$ and differentiable in $(0,\alpha )$
$\therefore$ From Rolle's theorem , $f^{\prime }\left(x\right)=0$ for atleast one value in $(0,\alpha )$
$\Rightarrow n{a}_n{x}^{n-1}+(n-1)a_{n-1}{x}^{n-2}+\cdots +a_1=0$ has a positive real root less than $\alpha .$